Outline — click chapter to expand
2.1 Noise and BER
- 2.1.1 Classification of noise sources
- 2.1.2 Thermal noise voltage and power
- Practice 1 — Thermal noise power
- 2.1.3 White Gaussian noise & PSD
- Practice 2 — PSD & noise power
- 2.1.3b Parseval's theorem
- Practice 2b — PSD ↔ time-domain
- 2.1.4 SSB / DSB-SC SNR
- Practice 3 — SSB vs DSB-SC SNR
- 2.1.5 Bit error rate (BER)
- Practice 4 — BER calculation
- 2.1.6 Integrate-and-dump
- Practice 5 — Integrator BER
- 2.1.7 Modulation comparison
- Lecture — QPSK min SNR (IS-54)
- Practice 6 — 16-QAM min SNR
- 2.1.8 16-QAM implementation
2.2 Noise Temperature & Figure
- 2.2.1 Equivalent noise temperature
- Practice 7 — $T_e$ of amplifier
- Practice 7b — $T_e$, source ≠ 0 K
- 2.2.2 Y-factor measurement
- Lecture — Y-factor
- Practice 8 — Y-factor
- 2.2.3 Noise figure $F$
- Eq. (1) — $F = 1+T_e/T_0$
- Eq. (2) — full $F$ derivation
- Practice 9 — $F \leftrightarrow T_e$
- 2.2.4 Lossy line noise figure
- Practice 10 — Lossy line
- 2.2.5 Friis cascade
- Lecture — Cascaded receiver
- Practice 11 — Loss before LNA
- 2.2.6 Mismatch / Wilkinson
- Practice 12 — Wilkinson noise
- 2.2.7 Antenna $T_a$ & Friis transmission
- Practice 13 — Friis & antenna noise
- 2.2.8 G/T ratio
- Lecture — DBS satellite
- Practice 14 — G/T & CNR
- Example — min-detectable QPSK
- Example — wireless receiver
- 2.2.9 Full receiver ($T_a \neq T_0$)
- Lecture — full receiver analysis
- HW H3(a) — BPSK / 64-QAM
- HW H3(b) — Pozar 14.6 receiver
2.3 Distortion (IP3)
- 2.3.1 Taylor-series model
- 2.3.2 Gain compression & $P_{1\text{dB}}$
- Practice 16 — 1-dB compression
- 2.3.3 Intermodulation
- Practice 17 — IM products
- HW H4 — ACPR vs IMR
- 2.3.4 IP3
- Practice 18 — IP3 / IM3 level
- 2.3.5 $P_{1\text{dB}}$ ↔ IP3
- Lecture — OP1dB & OPsf
- Practice 19 — $P_{1\text{dB}}$ from IP3
- 2.3.6 Dynamic range
- Lecture — dynamic ranges
- Practice 20 — SFDR
- 2.3.7 Intercept point of cascaded components
- Lecture — cascaded IP3
- Practice 21 — 3-stage cascade IP3
- 2.3.8 Passive intermodulation (PIM)
Problems & Exercises
2.1 Noise and BER in Linear Receivers
2.1.1 Classification of Noise Sources
| Noise Type | Origin | Key Characteristics |
|---|---|---|
| Thermal (Johnson–Nyquist) | Thermal vibration of bound charges | White spectrum; present in all resistors, lossy lines, atmosphere |
| Shot | Discrete charge carriers crossing a junction | $\propto$ DC current; dominant in diodes & transistors |
| Flicker (1/f) | Surface defects in semiconductors | Power $\propto 1/f$; sets oscillator phase noise floor |
| Plasma | Random motion in ionized gas | Ionospheric propagation, sparking contacts |
| Quantum | Quantized photons | Usually negligible at microwave freq ($hf \ll kT$) |
2.1.2 Thermal Noise Voltage and Power
Planck Expression
where each variable is:
| Symbol | Meaning | Unit |
|---|---|---|
| $T$ | Temperature of the resistor | kelvin (K) |
| $B$ | Bandwidth of the system | Hz |
| $f$ | Center frequency of the bandwidth | Hz |
| $R$ | Resistance | $\Omega$ |
| $h$ | Planck's constant $= 6.626\times10^{-34}$ | J·s |
| $k$ | Boltzmann's constant $= 1.380\times10^{-23}$ | J/K |
Rayleigh–Jeans Approximation ($hf \ll kT$)
Taylor expand: $e^{hf/kT}-1 \approx hf/kT$, yielding:
Available Noise Power
Matched load receives half the open-circuit voltage → power delivered:
Thévenin Model & Maximum Power Transfer
A noisy resistor $R$ is modeled as a Thévenin equivalent: an ideal voltage source $v_n^s$ (carrying all the noise) in series with a noise-free internal resistance $R_s = R$. This is identical to any other Thévenin source — the internal resistance is simply the impedance looking back into the source.
When the load $R_L$ is connected, a voltage divider forms. The voltage across the load is:
The power delivered to the load in general:
This can also be written in terms of the ratio $r = R_L/R_s$:
By the maximum power transfer theorem, $P_{load}$ is maximized when $R_L = R_s$ (i.e., $r = 1$). Taking the derivative $\frac{d}{dr}\!\left[\frac{r}{(1+r)^2}\right] = \frac{1-r}{(1+r)^3} = 0$ confirms the peak at $r=1$.
Where Does the $1/4$ Come From?
Set $R_L = R_s = R$ (matched) in the general power expression:
The 4 comes from $(R_s + R_L)^2 = (R + R)^2 = (2R)^2 = 4R^2$. Physically: the voltage divider gives the load only half the open-circuit voltage ($v_n = v_n^s/2$), and squaring that half gives $1/4$ of the open-circuit voltage squared.
Now substitute $v_n^s = \sqrt{4kTBR}$ (from the Rayleigh–Jeans result):
The factor of 4 in the numerator (from Planck/Rayleigh–Jeans) and the factor of 4 in the denominator (from the matched voltage divider) cancel exactly — which is why the maximum available noise power $kTB$ is independent of resistance $R$.
The chart below shows this relationship. Hover over the curve to read exact values. The red dot marks the matched condition $R_L = R_s$ where maximum power ($= kTB$) is delivered:
Maximum Power Transfer Theorem — click to show chart
The maximum power transfer theorem is why impedance matching ($Z_L = Z_0 = 50\;\Omega$) is always assumed throughout this chapter. Any mismatch means less noise power is delivered — but also less signal power. When we write $P_n = kTB$, we are implicitly assuming perfect matching. The same principle applies to amplifier inputs, antenna feedlines, and every junction in the receiver chain.
Problem: A 75 Ω resistor at $T = 350$ K is connected to a matched 75 Ω load through an ideal bandpass filter with bandwidth $B = 2$ MHz. Find (a) the rms open-circuit noise voltage, (b) the noise power delivered to the load, and (c) the two-sided power spectral density.
Click to reveal solution
(a) $v_n^s = \sqrt{4kTBR} = \sqrt{4 \times 1.38\times10^{-23} \times 350 \times 2\times10^6 \times 75}$
$= \sqrt{4 \times 1.38\times10^{-23} \times 350 \times 2\times10^6 \times 75} = \sqrt{2.898\times10^{-12}} = 1.70\;\mu\text{V (rms)}$
(b) $P_n = kTB = 1.38\times10^{-23} \times 350 \times 2\times10^6 = 9.66\times10^{-15}$ W $= -110.1$ dBm
(Note: independent of the 75 Ω resistance!)
(c) $S_n = kT/2 = 1.38\times10^{-23}\times350/2 = 2.415\times10^{-21}$ W/Hz (two-sided)
2.1.3 White Gaussian Noise and PSD
Background: What is a PDF, and why is thermal noise Gaussian? (click to expand)
What is a PDF (Probability Density Function)?
PDF here stands for Probability Density Function — it has nothing to do with PDF files! It describes how likely the noise voltage is to take any particular value at any instant.
The noise voltage $v_n$ is random — at any moment it could be positive or negative, large or small. The PDF $f(v_n)$ tells you the probability of finding $v_n$ in any range: specifically, $f(v_n)\,dv_n$ gives the probability that the noise voltage falls between $v_n$ and $v_n + dv_n$.
Why Is Thermal Noise Gaussian?
A physical resistor can be thought of as a series connection of many tiny resistors, each contributing a small independent random voltage. By the Central Limit Theorem, the sum of many independent random variables always converges to a Gaussian (bell-curve) distribution — regardless of the distribution of each individual source. Therefore, thermal noise has the Gaussian PDF:
with zero mean (the bell curve is centered at $v_n = 0$).
The full normal distribution for a random variable $X$ with mean $\mu$ and standard deviation $\sigma$ is written $X \sim \mathcal{N}(\mu,\,\sigma^2)$ with PDF:
$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-\dfrac{(x-\mu)^2}{2\sigma^2}}$$For thermal noise, $\mu = 0$ (no DC offset), so the formula simplifies to the equation above. The standard normal is the special case $\mu=0,\,\sigma=1$:
$$f(x) = \frac{1}{\sqrt{2\pi}}\,e^{-x^2/2} \qquad Z \sim \mathcal{N}(0,\,1)$$| Property | Value |
|---|---|
| Mean | $\mu$ (center of bell) |
| Standard deviation | $\sigma$ (width of bell) |
| Variance | $\sigma^2$ |
| Total area | $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx = 1$ |
| Symmetry | $f(\mu+x) = f(\mu-x)$ — bell is symmetric about $\mu$ |
| Peak value | $\dfrac{1}{\sigma\sqrt{2\pi}}$ at $x = \mu$ |
Connection to BER: BPSK noise $n \sim \mathcal{N}(0,\sigma^2)$, so the error probability is the tail integral beyond the threshold $v_0$:
$$P_e = \int_{v_0}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}\,e^{-n^2/2\sigma^2}\,dn = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{v_0}{\sigma\sqrt{2}}\right)$$The erfc is simply the Gaussian tail integral — which is why it appears in every BER formula.
Key Parameters of the Gaussian PDF
| Parameter | Symbol | Physical Meaning |
|---|---|---|
| Variance | $\sigma^2$ | Noise power for band-limited noise: $\sigma^2 = n_0 B$ |
| Standard deviation | $\sigma$ | RMS noise voltage (when $R = 1\;\Omega$); wider $\sigma$ = noisier |
| Mean | $0$ | No DC offset — noise averages to zero over time |
For a Gaussian distribution: 68% of the time $v_n$ falls within $\pm\sigma$, 95% within $\pm 2\sigma$, 99.7% within $\pm 3\sigma$. Noise exceeds $3\sigma$ only 0.3% of the time — but in a communication system running at millions of bits per second, even rare events cause bit errors.
Power Spectral Density (PSD)
The two-sided PSD of thermal noise is:
This PSD is flat — it has the same value at every frequency (within the Rayleigh–Jeans approximation). A signal with flat PSD is called white noise, by analogy with white light which contains all colors equally.
The total noise power is recovered by integrating the PSD over bandwidth:
Autocorrelation and the Wiener–Khinchin Theorem
The autocorrelation function $R(\tau)$ measures how similar a noise signal is to a time-shifted copy of itself:
If the noise at time $t$ is completely unrelated to the noise at time $t+\tau$, then $R(\tau) = 0$. White noise has exactly this property — samples at any two different times are uncorrelated.
The Wiener–Khinchin theorem states that the PSD and autocorrelation are a Fourier transform pair:
Step-by-Step Derivation: $R(\tau) = \frac{n_0}{2}\,\delta(\tau)$
We start with ideal white noise: $S(\omega) = n_0/2$ (constant for all $\omega$).
Step 1 — Substitute into the Wiener–Khinchin integral:
Step 2 — Pull out the constant ($n_0/2$ does not depend on $\omega$):
Step 3 — Recognize the standard Fourier identity:
Why Does This Integral Equal $\delta(\tau)$? (click to expand)
Using Euler's formula, expand the integrand $e^{j\omega\tau} = \cos(\omega\tau) + j\sin(\omega\tau)$:
The $\sin$ part: always zero by symmetry
$\sin(\omega\tau)$ is an odd function of $\omega$ (since $\sin(-x) = -\sin(x)$). When you integrate any odd function over the symmetric interval $(-\infty, +\infty)$, the negative-$\omega$ half exactly cancels the positive-$\omega$ half:
The $\cos$ part when $\tau \neq 0$: perfect cancellation
$\cos(\omega\tau)$ oscillates between $+1$ and $-1$ as $\omega$ varies. Think of it as a wave along the $\omega$-axis:
- From $\omega = 0$ to $\omega = \pi/\tau$: one positive half-cycle (area $= +$)
- From $\omega = \pi/\tau$ to $\omega = 2\pi/\tau$: one negative half-cycle (area $= -$)
- From $\omega = 2\pi/\tau$ to $\omega = 3\pi/\tau$: positive again...
Every positive half-cycle is immediately followed by a negative half-cycle of exactly equal area. Over $(-\infty,+\infty)$, they all pair up and cancel:
The $\cos$ part when $\tau = 0$: no oscillation
$\cos(\omega\cdot 0) = \cos(0) = 1$ for every $\omega$. The integrand is just the constant $1$ — there is no oscillation at all, so nothing cancels:
Summary:
| Integrand | Oscillates? | Cancellation? | Result | |
|---|---|---|---|---|
| $\tau \neq 0$ | $\cos(\omega\tau)$ | Yes — every $+$ half-cycle has a matching $-$ half-cycle | Perfect cancellation | $0$ |
| $\tau = 0$ | $\cos(0) = 1$ | No — flat constant | Nothing to cancel | $\infty$ |
Zero everywhere except infinite at one point — that is exactly the Dirac delta function $\delta(\tau)$. At $\tau = 0$ everything adds constructively (infinite result); at any $\tau \neq 0$ the oscillations destroy themselves completely (zero result).
The Sifting Property of $\delta(\tau)$ (click to expand)
The Dirac delta function is defined by two properties: $\delta(\tau) = 0$ for all $\tau \neq 0$, and $\int_{-\infty}^{\infty}\delta(\tau)\,d\tau = 1$.
From these, the sifting property follows directly. If $\delta(\tau)$ is zero everywhere except at $\tau = 0$, then multiplying it by any function $g(\tau)$ only "survives" at $\tau = 0$:
$$\int_{-\infty}^{\infty} \delta(\tau)\cdot g(\tau)\,d\tau$$Since $\delta(\tau) = 0$ for $\tau \neq 0$, the only point contributing to this integral is $\tau = 0$. At that point, $g(\tau)$ has the value $g(0)$, so we can pull it out as a constant:
The delta function "sifts out" or "samples" the value of $g$ at the origin — hence the name.
Sanity Check — Verify via Reverse Fourier Transform (click to expand)
Apply the sifting property with $g(\tau) = e^{-j\omega\tau}$:
Step by step: (1) $\delta(\tau) = 0$ for $\tau \neq 0$, so only $\tau = 0$ contributes. (2) Evaluate $e^{-j\omega\tau}$ at $\tau = 0$: $e^{-j\omega\cdot 0} = e^0 = 1$. (3) Result is $1$, regardless of $\omega$.
This confirms $\delta(\tau) \leftrightarrow 1$ is a Fourier pair: a delta in time ↔ a constant in frequency.
Step 4 — Combine:
The entire derivation reduces to recognizing the delta function identity — everything else is just pulling out a constant. Summarizing the Fourier pair:
Frequency Domain
$S(\omega) = \dfrac{n_0}{2}$
Constant (flat PSD)
Time Domain
$R(\tau) = \dfrac{n_0}{2}\,\delta(\tau)$
Delta (uncorrelated)
This is a Dirac delta function — infinite at $\tau = 0$ and zero everywhere else:
| Time shift | $R(\tau)$ | Physical meaning |
|---|---|---|
| $\tau = 0$ | $R(0) = \frac{n_0}{2}\delta(0) \to \infty$ | Total noise power (signal compared with itself) |
| $\tau \neq 0$ | $R(\tau) = 0$ | Perfectly uncorrelated at any time shift |
The Problem: $R(0) \to \infty$ (click to expand)
Setting $\tau = 0$ gives infinite total power — which is physically impossible! The issue is that $S(\omega) = n_0/2$ being constant at all frequencies (out to infinity) is only a mathematical idealization. In reality, the Rayleigh–Jeans approximation breaks down at very high frequencies, and the PSD must roll off. The flat-PSD model is valid only over a finite, though very wide, frequency band.
Real (Band-Limited) Noise
For noise limited to a finite bandwidth $B$, the autocorrelation becomes a sinc function (not a delta), and the total power at $\tau = 0$ is finite:
The two-sided PSD $S_n(f) = n_0/2$ is defined over both positive and negative frequencies. For a system with one-sided bandwidth $B$ (the physical bandwidth you'd see on a spectrum analyzer), the two-sided frequency range extends from $-B$ to $+B$:
The $2$ in the numerator of $2B$ cancels the $2$ in the denominator of $n_0/2$, giving $n_0 B = kTB$.
Alternatively, using the one-sided PSD $S_n(f) = n_0$ (defined only for $f \geq 0$), you integrate over just $B$ and get the same result directly:
$\displaystyle R(0) = \int_0^{B} n_0\,df = n_0\cdot B = kTB$
The factor of 2 is purely a bookkeeping difference between the two PSD conventions — the physical noise power is always $kTB$.
Key Takeaway (click to expand)
We use the "white" (flat PSD) model because it's an excellent approximation within the system bandwidth. But we must always integrate over the finite bandwidth $B$ to get physically meaningful power. The delta function $\delta(\tau)$ and infinite power are mathematical artifacts of extending the flat-PSD model to infinite bandwidth — they never occur in real systems.
AWGN — The Standard Channel Model
Combining "white" (flat PSD) + "Gaussian" (amplitude PDF) gives AWGN — Additive White Gaussian Noise:
Additive
Noise adds to the signal: $r(t) = s(t) + v_n(t)$
White
Flat PSD → all frequencies contribute equally
Gaussian
Amplitude follows the bell-curve PDF above
Why it matters
AWGN is the standard noise model used in all BER calculations in this chapter
By definition, AWGN has a flat two-sided PSD at every frequency:
$$S(f) = \frac{n_0}{2} \quad \text{for all } f \in (-\infty,\, +\infty)$$The $\frac{1}{2}$ is always present because this is the two-sided convention — the total noise density $n_0$ is split equally between positive and negative frequencies. The value of $n_0$ depends on the physical source:
$$n_0 = kT \quad \Rightarrow \quad \frac{n_0}{2} = \frac{kT}{2} \approx 2\times10^{-21}\;\text{W/Hz at }T_0=290\;\text{K}$$| Property | Expression | Note |
|---|---|---|
| Two-sided PSD | $S(f) = n_0/2$ | Defined for all $f$, positive and negative |
| One-sided PSD | $S(f) = n_0$ | Defined for $f \geq 0$ only — double the two-sided value |
| Noise power in BW $B$ | $N = \frac{n_0}{2}\times 2B = n_0 B = kTB$ | Two-sided integral over $[-B, +B]$ |
| Noise power in BW $B$ | $N = n_0 \times B = kTB$ | One-sided integral over $[0, B]$ — same result |
The two-sided PSD $n_0/2$ integrates over a bandwidth of $2B$ (from $-B$ to $+B$): $\frac{n_0}{2}\times 2B = n_0 B$. The one-sided PSD $n_0$ integrates over just $B$ (from $0$ to $B$): $n_0 \times B = n_0 B$. The factor of 2 in the PSD and the factor of 2 in the bandwidth always cancel — physical noise power is always $n_0 B = kTB$ regardless of which convention you use.
Problem: A receiver has a system noise temperature of 600 K and a bandwidth of 5 MHz. Find the total noise power and the one-sided PSD $n_0$.
Click to reveal solution
$n_0 = kT = 1.38\times10^{-23}\times 600 = 8.28\times10^{-21}$ W/Hz
$N = n_0 B = 8.28\times10^{-21}\times 5\times10^6 = 4.14\times10^{-14}$ W $= -103.8$ dBm
2.1.3b Why Integrating PSD Gives Power — Parseval's Theorem
Power is defined as a time-domain quantity — $P = \lim_{T\to\infty}\frac{1}{T}\int v^2(t)\,dt$. Yet we routinely compute it by integrating the PSD over frequency: $P = \int S(f)\,df$. Why are these two expressions equal? The answer is Parseval's theorem.
The Three Equivalent Definitions of Average Power
For a noise signal $v(t)$, average power (across a $1\;\Omega$ load) can be computed three equivalent ways:
| Domain | Expression | What It Uses |
|---|---|---|
| Time | $\displaystyle P = \lim_{T\to\infty}\frac{1}{T}\int_0^T v^2(t)\,dt$ | Waveform directly |
| Frequency (PSD) | $\displaystyle P = \int_{-\infty}^{\infty} S(f)\,df$ | Power per Hz — measurable with spectrum analyzer |
| Statistical | $\displaystyle P = E[v^2] = \sigma^2$ (for zero-mean noise) | Variance of the amplitude distribution |
For an ergodic stationary random process — which thermal noise is — all three give the same numerical result.
Why $E[v_n^2] = \sigma^2$ for noise (click to expand)
Variance is defined as the average squared deviation from the mean:
$$\text{Var}(v_n) = E[(v_n - \mu)^2]$$Expanding the square and using linearity of expectation:
$$= E[v_n^2] - 2\mu\,E[v_n] + \mu^2 = E[v_n^2] - \mu^2$$For noise, $\mu = E[v_n] = 0$ (zero mean — noise averages to zero over time), so:
$$\sigma^2 = \text{Var}(v_n) = E[v_n^2] - 0 = E[v_n^2]$$And $E[v_n^2]$ is the mean square value, which equals average power across $1\,\Omega$ by the time-domain definition. So all three are the same:
$$\underbrace{\sigma^2}_{\text{variance}} = \underbrace{E[v_n^2]}_{\text{mean square}} = \underbrace{P}_{\text{noise power}}$$Parseval's Theorem — Energy Signals
For a deterministic signal $v(t)$ with Fourier transform $V(f)$, total energy is preserved across the transform:
The left side is energy computed in time. The right side is the same energy computed from the spectrum. $|V(f)|^2$ is the Energy Spectral Density (J/Hz).
Proof Sketch
Write $|v(t)|^2 = v(t)\cdot v^*(t)$, then substitute the inverse Fourier transform for one factor of $v(t)$:
The swap of integration order (Fubini's theorem) is justified for square-integrable signals. The inner bracket in the last step is exactly $V^*(f)$ by the definition of the Fourier transform.
Extension to Power Signals (Noise)
Thermal noise has infinite total energy, so Parseval in its energy form diverges. Instead, we truncate the signal to a window of duration $T$ and define:
where $V_T(f)$ is the Fourier transform of the windowed signal $v_T(t)$. Dividing Parseval's energy identity by $T$ and taking the limit:
A prism splits white light into a spectrum of colors. Parseval's theorem says: the total brightness of the white beam equals the sum of brightnesses of every color band. Whether you measure the combined beam directly (time domain) or sum all the spectral slices (frequency domain), you get the same total power. The Fourier transform is the mathematical prism.
The Wiener–Khinchin Connection
The PSD is also the Fourier transform of the autocorrelation function $R_v(\tau) = E[v(t)v(t+\tau)]$:
Setting $\tau = 0$: $R_v(0) = E[v^2] = P_{avg}$. And by the inverse Fourier transform at $\tau=0$: $R_v(0) = \int S(f)\,df$. These are consistent — confirming again that $\int S(f)\,df = P_{avg}$, this time via the autocorrelation route.
Integrating PSD over frequency is computing time-averaged power. It is not a new definition — it is a direct consequence of Parseval's theorem applied to power signals. The time-domain average and the frequency-domain PSD integral are two views of the same quantity, linked by the Fourier transform.
Problem: Thermal noise from a $50\;\Omega$ resistor at $T=290$ K passes through an ideal bandpass filter of bandwidth $B = 1$ MHz centered at $f_0 = 2.4$ GHz. (a) Write the two-sided PSD $S(f)$. (b) Use Parseval's theorem to find the total average noise power. (c) Verify using $P_n = kTB$.
Click to reveal solution
(a) Within the filter passband $[f_0 - B/2,\; f_0 + B/2]$ (and its negative-frequency image), $S(f) = kT/2 = n_0/2$. Outside the passband $S(f) = 0$.
(b) By Parseval / definition of PSD:
$P = \displaystyle\int_{-\infty}^{\infty} S(f)\,df = \frac{n_0}{2}\times 2B = n_0 B = kTB$
(Factor of 2 because the two-sided PSD integrates over both positive and negative frequency bands of width $B$.)
(c) Direct check: $P_n = kTB = 1.38\times10^{-23}\times290\times10^6 = 4.0\times10^{-15}$ W $= -113.9$ dBm. ✓
The Parseval integral and the $kTB$ formula give identical results — confirming the equivalence.
2.1.4 SNR Through SSB and DSB-SC Demodulators
What Does "SC" Mean in DSB-SC?
SC = Suppressed Carrier. To understand why, compare standard AM with DSB-SC:
In conventional AM (double-sideband with carrier), the transmitted signal is:
The $A_c\cos(2\pi f_c t)$ term is the carrier — it carries zero information yet consumes a large fraction of the total transmit power. DSB-SC suppresses this term entirely, transmitting only the sidebands:
| DSB (with carrier) | DSB-SC | |
|---|---|---|
| Carrier transmitted | Yes | No — suppressed |
| Power efficiency | Low — carrier wastes power | High — all power in sidebands |
| Demodulation | Simple envelope detector | Coherent local oscillator required |
| Bandwidth | $2f_m$ | $2f_m$ (same) |
Removing the carrier saves significant transmit power, but the receiver loses its phase reference. A coherent local oscillator — phase-locked to the original carrier frequency — is now required at the receiver to demodulate. SSB takes this one step further: it also removes one of the two sidebands, halving the bandwidth as well, but requiring an even more precise receiver.
SSB System Block Diagram
The complete SSB communication system has three stages: modulator, channel, and demodulator. Understanding what each block does to the signal and noise is essential for the SNR analysis.
What Each Block Does
| Block | Function | Signal At Output |
|---|---|---|
| Mixer × (modulator) | Multiplies $m(t)$ by carrier $\cos(2\pi f_{IF}t)$ | DSB signal: both upper and lower sidebands around $f_{IF}$ |
| Sideband filter | Removes one sideband | SSB signal: half the bandwidth of DSB |
| + $n_0/2$ (channel) | AWGN added by the channel | SSB signal + white noise |
| IF filter | Bandpass filter: selects desired signal, limits noise BW | $v_i(t) = A\cos(\omega_{IF}+\omega_m)t + v_n(t)$ with power $S_i, N_i$ |
| Mixer × (demod) | Multiplies $v_i(t)$ by $\cos(2\pi f_{IF}t)$ → translates to baseband | Baseband signal + noise + double-frequency terms |
| Baseband filter | Low-pass filter removes $2f_{IF}$ terms | $v_o(t) = \frac{A}{2}\cos\omega_m t + \frac{1}{2}x(t)$ |
Narrowband Noise Representation
The noise at the IF filter output is represented using its in-phase and quadrature components:
where $x(t)$ and $y(t)$ are band-limited, independent, zero-mean random processes with equal power: $E\{x^2\} = E\{y^2\} = N_i$.
SSB Demodulation — SNR Analysis
The received signal plus noise enters the demodulator:
After mixing with $\cos\omega_{IF}t$ and low-pass filtering (removing all $2\omega_{IF}$ terms):
Both the signal and noise are scaled by the same factor of $1/2$:
| Quantity | Expression | How it changes |
|---|---|---|
| Input signal power $S_i$ | $A^2/2$ | |
| Output signal power $S_o$ | $(A/2)^2/2 = A^2/8$ | $= S_i/4$ |
| Input noise power $N_i$ | $E\{x^2\} = E\{y^2\}$ | |
| Output noise power $N_o$ | $E\{[x(t)/2]^2\} = N_i/4$ | $= N_i/4$ |
DSB-SC Demodulation — SNR Analysis
The DSB-SC system is similar to SSB but without the sideband filter — both sidebands are transmitted. The IF filter bandwidth is $2f_m$ (twice that of SSB). Note there is no sideband filter in the modulator:
Key differences from SSB: no sideband filter in the modulator (both sidebands transmitted), and the IF filter has bandwidth $2f_m$ instead of $f_m$. The input signal has two components at $\omega_{IF}\pm\omega_m$, each with amplitude $A/\sqrt{2}$. After mixing and filtering:
| Quantity | Expression | How it changes |
|---|---|---|
| Input signal power $S_i$ | $\frac{1}{2}(A/\sqrt{2})^2 + \frac{1}{2}(A/\sqrt{2})^2 = A^2/2$ | |
| Output signal power $S_o$ | $\frac{1}{2}(A/\sqrt{2})^2 = A^2/4$ | $= S_i/2$ |
| Output noise power $N_o$ | $N_i/4$ | $= N_i/4$ (same as SSB) |
In DSB-SC, both sidebands carry the same signal information. At the demodulator, these two copies add coherently (voltages add → power doubles). But the noise from each sideband is independent, so noise adds only incoherently (powers add, not voltages). The result: signal power is doubled relative to noise, giving a 3 dB advantage. The trade-off is that DSB-SC requires twice the bandwidth of SSB.
Mirror Image Noise Folding at the Mixer (click to expand)
To truly understand why SSB preserves SNR while DSB-SC improves it by 3 dB, you need to see what happens to noise at the mixer — the mirror image folding effect.
Before the mixer: the signal sits at IF frequency $f_{IF}$. Noise with two-sided PSD $n_0/2$ surrounds it symmetrically — both the upper sideband ($f_{IF}+f_m$) and lower sideband ($f_{IF}-f_m$) contain independent noise:
After the mixer: multiplying by $\cos(2\pi f_{IF}t)$ shifts the entire spectrum both down and up by $f_{IF}$. The upper sideband translates down to baseband at $+f_m$. The lower sideband translates to $-f_m$, which mirrors (folds) onto $+f_m$. Both land on the same baseband frequencies:
Since the two sideband noise sources are independent (uncorrelated), their powers add:
This folding is why the two-sided PSD ($n_0/2$) matters — after mixing, both sides fold together to give $n_0$ per unit bandwidth at baseband.
SSB vs DSB-SC: Signal Folding Comparison
The noise folds identically in both cases. The difference is what happens to the signal:
| Signal at upper sideband | Signal at lower sideband | After mixing (signal) | After mixing (noise) | SNR change | |
|---|---|---|---|---|---|
| SSB | Signal present | Empty (filtered out) | Signal $\times 1$ only | Noise from both sides folds: $\times 2$ | No improvement |
| DSB-SC | Signal copy 1 | Signal copy 2 (same info) | Two copies add coherently: voltage $\times 2$ → power $\times 4$ | Noise from both sides folds: $\times 2$ | $\times 2$ → 3 dB better |
The mixer always folds noise from both sidebands into baseband — this is unavoidable. In SSB, only one sideband has signal, so the signal doesn't benefit from the folding. In DSB-SC, the signal is in both sidebands and adds coherently (voltages, not powers), beating the noise folding by a factor of 2.
This is also why the two-sided PSD ($n_0/2$) is essential for demodulation analysis — you must account for noise from both sidebands folding onto the same baseband frequencies.
Problem: A communication link has an input SNR of 25 dB. Compare the output SNR for (a) SSB and (b) DSB-SC demodulation.
Click to reveal solution
(a) SSB: $\text{SNR}_{out} = \text{SNR}_{in} = 25$ dB (no change)
(b) DSB-SC: $\text{SNR}_{out} = \text{SNR}_{in} + 3 = 28$ dB
The DSB-SC gives 3 dB better output SNR, at the cost of requiring twice the bandwidth.
DSB (with Carrier) — Block Diagram and SNR
DSB adds the carrier term $A_c\cos(\omega_{IF}t)$ back into the transmitted signal. The modulator no longer suppresses it — the carrier goes through alongside both sidebands:
The key difference: the demodulator is a simple envelope detector (diode + RC) — no coherent local oscillator needed. The carrier provides a built-in phase reference, so the envelope of $v_i(t)$ directly tracks $m(t)$. The DC block afterwards removes the DC offset $A_c/2$ left by the carrier.
How the Envelope Detector Works
The envelope detector has three components — diode, capacitor, and resistor — each playing a distinct role:
| Component | Role | What happens |
|---|---|---|
| Diode D | Half-wave rectifier | Blocks negative half-cycles — only positive peaks pass through |
| Capacitor C | Peak holder | Charges rapidly to each positive peak (through diode's low resistance); holds voltage between peaks |
| Resistor R | Discharge control | C discharges slowly through R — time constant $\tau = RC$ controls how fast it can follow envelope changes |
| DC block | Offset removal | Series capacitor removes the $A_c$ DC offset, leaving only $m(t)$ |
The RC Time Constant Condition
The time constant $\tau = RC$ must satisfy:
- $RC \gg 1/f_c$ — C doesn't discharge significantly between carrier cycles. The carrier ripple is smoothed out, leaving only the envelope.
- $RC \ll 1/f_m$ — C can still follow the slow changes of the message envelope $m(t)$. If RC is too large, the detector can't track a falling envelope (diagonal clipping distortion).
In DSB-SC, $s(t) = m(t)\cos(\omega_c t)$ — no carrier. When $m(t)$ goes negative, the phase of the carrier flips 180°. The envelope of the signal is $|m(t)|$, not $m(t)$. An envelope detector would output $|m(t)|$ — a full-wave rectified version of the message, which is completely wrong. This is why DSB-SC and SSB require a coherent local oscillator phase-locked to the carrier frequency.
DSB Signal Power and SNR
The transmitted signal is $[A_c + m(t)]\cos(\omega_{IF}t)$. Total input power at the IF filter output:
Only the sideband power reaches the output as signal. The modulation efficiency is:
where $k_a = A/A_c$ is the modulation index ($\leq 1$ to avoid over-modulation). Even at $k_a = 1$, at most 50% of transmit power carries information.
Noise analysis is identical to DSB-SC: IF bandwidth is $2f_m$, so $N_i = 2n_0 f_m$. But SNR is degraded because part of $S_i$ is the useless carrier:
DSB and DSB-SC have the same noise floor ($N_i = 2n_0 f_m$, same IF bandwidth). The only difference is that DSB wastes at least half its transmit power on the carrier. To achieve the same SNR as DSB-SC, a DSB transmitter must use at least twice the power. This is why DSB-SC and SSB replaced DSB in all modern systems where power efficiency matters.
Full Three-Way Comparison
| DSB (AM) | DSB-SC | SSB | |
|---|---|---|---|
| Carrier transmitted | Yes | No | No |
| Sideband filter | No | No | Yes |
| IF bandwidth | $2f_m$ | $2f_m$ | $f_m$ |
| $N_i$ | $2n_0 f_m$ | $2n_0 f_m$ | $\frac{n_0 f_m}{2}$ |
| Wasted power | $A_c^2/2$ (carrier) | 0 | 0 |
| Power efficiency $\eta$ | $\leq 50\%$ | 100% | 100% |
| SNR$_{out}$ / SNR$_{in}$ | $< 1$ (worse) | 2 (+3 dB) | 1 (0 dB) |
| Demodulation | Envelope detector | Coherent LO | Coherent LO |
| Receiver complexity | Simplest | Medium | Most complex |
| Historical use | AM broadcast radio | Data/audio links | HF voice, satellite |
2.1.5 Bit Error Rate (BER)
What is Binary PSK (BPSK)?
PSK = Phase Shift Keying — information is encoded in the phase of the carrier, not its amplitude or frequency. Binary means only two possible phases, one for bit 0 and one for bit 1:
| Bit | Signal | Phase |
|---|---|---|
| 1 | $+v_0\cos(\omega_c t)$ | 0° |
| 0 | $-v_0\cos(\omega_c t)$ | 180° |
The two signals are simply negatives of each other — same amplitude, same frequency, opposite phase. Written compactly: $s(t) = \pm v_0\cos(\omega_c t)$.
Amplitude is easily distorted by channel fading and attenuation. Phase is more robust — the receiver only needs to detect which of two phase states the signal is in, not measure an exact amplitude. BPSK achieves the best BER performance of all binary modulation schemes for a given $E_b/n_0$ — it is the most power-efficient binary modulation.
BPSK Detection
The receiver multiplies the received signal by a local carrier $\cos(\omega_c t)$ (coherent detection) and integrates over one bit period $T_b$:
Why divided by 2?
For bit "1", $r(t) = v_0\cos(\omega_c t)$. Multiplying by the local carrier and integrating:
Where does $\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}$ come from? (click to expand)
Start from the cosine double angle formula:
$$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$$Substitute $\sin^2(\theta) = 1 - \cos^2(\theta)$:
$$\cos(2\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2\cos^2(\theta) - 1$$Rearrange to isolate $\cos^2(\theta)$:
$$\boxed{\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}}$$This is a pure algebraic rearrangement — no approximation. Replace $\theta = \omega_c t$ and you get the substitution used in the integral above.
Interactive: $\cos^2(\omega_c t) = \tfrac12 + \tfrac12\cos(2\omega_c t)$ — splitting the integrand (click to expand)
The identity $\cos^2(\theta) = \frac{1}{2} + \frac{1}{2}\cos(2\theta)$ splits the integrand into two terms. The $\cos(2\omega_c t)$ term oscillates at twice the carrier frequency and averages to zero over any whole number of carrier cycles. Only the constant $\frac{1}{2}$ survives — that is where the $\div 2$ comes from. The interactive plot below shows both terms:
Increase the number of cycles — the red term always cancels to zero. The surviving ½ (green dashed) gives the Tb/2 result.
The result is compared against a threshold at zero. An error occurs when noise pushes the result across zero to the wrong side — this probability is the BER.
For binary PSK with threshold detection, the received signal is $r(t) = s(t) + v_n(t)$, where $s(t) = +v_0$ for a "1" and $s(t) = -v_0$ for a "0". The receiver compares the received voltage against a threshold (at zero) to decide which bit was sent. An error occurs when noise pushes the signal past the threshold:
The noisy received signal fluctuates around $+v_0$ during a "1" bit and around $-v_0$ during a "0" bit. If noise exceeds $v_0$ in magnitude, the signal crosses the threshold and causes a bit error.
Signal power $S_o$: after the coherent detector the signal output is a DC level $\pm v_0$. The power of a DC level across $1\,\Omega$ is simply:
$$S_o = v_0^2$$No factor of $\frac{1}{2}$ — that only appears for sinusoids. A DC level has the same instantaneous and average power.
Noise power $N_o$: the output noise is Gaussian with zero mean and variance $\sigma^2$. Since $E[v_n] = 0$:
$$N_o = E[v_n^2] = \sigma^2$$Therefore:
$$\frac{S_o}{N_o} = \frac{v_0^2}{\sigma^2} \qquad \Rightarrow \qquad \sqrt{\frac{S_o}{N_o}} = \frac{v_0}{\sigma}$$This ratio is exactly "how many noise standard deviations the signal sits away from the threshold" — the deeper this is into the Gaussian tail, the lower the BER.
What is erfc?
The complementary error function is the tail area of a Gaussian curve from $x$ to $\infty$:
It appears in BER because noise is Gaussian — the probability of an error is exactly the tail area of the noise distribution beyond the decision threshold. A larger argument $x$ means the threshold is further into the tail → smaller area → lower BER.
直覺解釋（用文科生也懂的方式）（click to expand）
想像你在射飛鏢，靶心是「0」這條判斷線。你丟出的飛鏢本來應該落在 +v₀ 的位置（代表 bit "1"），但噪音會讓飛鏢隨機偏移。
噪音是高斯分布，意思是：大部分時候偏移很小，偶爾才偏很多——就像鐘形曲線，中間高、兩邊低：
什麼時候出錯？ 噪音把 +v₀ 的訊號往左推，推過了 0 這條線 → 接收器誤判為 "0" → 出錯。那個跑到門檻左邊的面積就是「尾巴（tail）」，也就是出錯的機率：
erfc 的括號裡是 $\sqrt{S_o/N_o}$，就是訊號強度除以噪音強度。這個數字越大，代表訊號比噪音強很多，訊號中心離門檻很遠，鐘形曲線幾乎沒有尾巴跑到門檻那邊：
| 情況 | 鐘形曲線 | 尾巴大小 | 結果 |
|---|---|---|---|
| SNR 小（訊號弱） | 中心靠近門檻，分布又寬 | 大尾巴 | 常常出錯 |
| SNR 大（訊號強） | 中心遠離門檻，分布又窄 | 幾乎沒有尾巴 | 幾乎不出錯 |
一句話： erfc 就是在算「鐘形曲線有多少面積跑到了危險區」。訊號越強、噪音越小，危險區的面積就越小，錯誤就越少。
那對 $-v_0$（sent "0"）而言呢？
送 "0" 時，訊號在 $-v_0$，整個圖左右翻轉。出錯條件變成噪音把訊號往右推過 0，誤判為 "1"：
高斯分布是對稱的，所以：
$$P(\text{error}|\text{sent "1"}) = P(n < -v_0) = P(n > v_0) = P(\text{error}|\text{sent "0"})$$不管送 "0" 還是 "1"，出錯機率完全相同。這就是為什麼 BER 公式只需要寫一個——高斯的對稱性保證它們相等，不需要分兩種情況討論。
為什麼不直接把門檻移開，讓錯誤不再發生？
直覺上很合理——把門檻移到 $+v_0$ 旁邊，送 "1" 的那側就幾乎不會錯。但這有一個致命的代價：
把門檻往左移，$+v_0$ 那側的尾巴確實縮小了——但門檻同時往 $-v_0$ 靠近，$-v_0$ 那側的尾巴等量暴增。總 BER 不變：
$$P_e^{total} = \frac{1}{2}P_e(\text{"1"}) + \frac{1}{2}P_e(\text{"0"})$$對稱情況下，任何偏離 0 的門檻都只是把錯誤從一側搬到另一側，門檻在 0 是數學上最優的決策點。
真正的解法：拉開兩個訊號之間的距離
縮小尾巴的正確方法不是移門檻，而是增加 $+v_0$ 和 $-v_0$ 之間的間距，也就是提高訊號功率。這樣兩個分布同時遠離門檻，兩側的尾巴都縮小：
$E_b/n_0$ 衡量的正是「兩個訊號分布離門檻有多遠」相對於「噪音分布有多寬」。$E_b/n_0$ 越大，兩個鐘形曲線離得越遠、重疊越少、BER 越低——這是提升通訊可靠度的根本方法，而不是調整門檻位置。
In BPSK, an error occurs when Gaussian noise $v_n$ exceeds the signal level $v_0$ and flips the decision. The probability of this is:
$$P_e = P(v_n > v_0) = \int_{v_0}^{\infty} \frac{1}{\sqrt{2\pi}\,\sigma}e^{-v^2/2\sigma^2}\,dv$$Substituting $t = v/\sigma$ converts this exactly into $\frac{1}{2}\mathrm{erfc}(v_0/\sigma\sqrt{2})$. The argument $\sqrt{S_o/N_o}$ is precisely this ratio — signal level over noise standard deviation.
Key erfc Values
| $x$ | $\mathrm{erfc}(x)$ | $P_e = \frac{1}{2}\mathrm{erfc}(x)$ | Note |
|---|---|---|---|
| 0 | 1 | 0.5 | 50% error — no SNR at all |
| 1 | 0.1573 | 0.079 | |
| 2 | 0.00468 | $2.3\times10^{-3}$ | |
| 3 | $2.2\times10^{-5}$ | $1.1\times10^{-5}$ | $\approx P_e = 10^{-5}$ target |
| 3.12 | $\approx 2\times10^{-5}$ | $10^{-5}$ | Requires $E_b/n_0 = 9.6$ dB |
| 4 | $1.5\times10^{-8}$ | $7.5\times10^{-9}$ | Very reliable link |
The Two Forms — Why They Are Equal
The two expressions in the equation above are related by:
| Term | Meaning | Units |
|---|---|---|
| $S_o/N_o$ | Output SNR at the detector | dimensionless |
| $E_b/n_0$ | Energy per bit / noise density — fundamental link quality | dimensionless |
| $R_b/B$ | Spectral efficiency — bits per second per Hz of bandwidth | bps/Hz |
Higher SNR → larger erfc argument → smaller tail → lower $P_e$. The erfc function drops very steeply — a few dB improvement in $E_b/n_0$ near the $10^{-5}$ target changes $P_e$ by orders of magnitude.
$E_b = S_o/R_b$ = energy per bit (J); $n_0 = N_o/B$ = noise PSD (W/Hz)
$E_b/n_0$ = bit energy to noise density ratio — fundamental link quality metric
Spectral efficiency: $\eta = R_b/B$ (bps/Hz). Then $S_o/N_o = (E_b/n_0)\cdot(R_b/B)$.
Derivation of the BER Formula
Step 1 — The received signal. After the BPSK detector, the received value is signal plus Gaussian noise:
Step 2 — Probability of error. For "1" sent, error occurs when $r < 0$, i.e. when noise $n < -v_0$. By symmetry of the Gaussian this equals the right tail $P(n > v_0)$:
Step 3 — Substitute $t = n/\sigma$, then $u = t/\sqrt{2}$.
Step 4 — Express in terms of $S_o/N_o$. Signal power $S_o = v_0^2$ and noise power $N_o = \sigma^2$, so:
Step 5 — Express in terms of $E_b/n_0$. By definition $E_b = S_o/R_b$ and $n_0 = N_o/B$, so:
$\underbrace{P(\text{noise crosses threshold})}_{\text{Gaussian tail integral}}$ $\xrightarrow{\;t=n/\sigma,\;u=t/\sqrt{2}\;}$ $\dfrac{1}{2}\mathrm{erfc}\!\left(\dfrac{v_0}{\sigma\sqrt{2}}\right)$ $\xrightarrow{\;S_o=v_0^2,\;N_o=\sigma^2\;}$ $\dfrac{1}{2}\mathrm{erfc}\!\left(\sqrt{\dfrac{S_o}{N_o}}\right)$ $\xrightarrow{\;E_b,\,n_0\;\text{defs}\;}$ $\dfrac{1}{2}\mathrm{erfc}\!\left(\sqrt{\dfrac{E_b}{n_0}\cdot\dfrac{R_b}{B}}\right)$
Physical Meaning of "Smaller Tail" (click to expand)
The "tail" is the probability that noise is large enough to push the signal past the decision threshold — causing a wrong bit decision. Visually, it is the shaded area under the noise distribution curve beyond the threshold:
| Tail size | Physical meaning |
|---|---|
| Large tail | Noise is strong relative to signal — frequently pushes past threshold → many bit errors |
| Small tail | Noise is weak relative to signal — rarely crosses threshold → very few errors |
Two things shrink the tail — both increase the erfc argument $\sqrt{S_o/N_o}$, moving the threshold deeper into the distribution:
① More signal power — the distribution center $+v_0$ moves further from the threshold at zero. The curve shape stays the same width, but the whole bell shifts right — so less of it spills past the threshold:
② Less noise — the distribution stays centred at the same $+v_0$, but the bell becomes narrower. Even though the center-to-threshold distance is unchanged, the narrower shape puts far less probability in the tail region:
Why signal and noise scale differently (click to expand)
When you double signal power, the amplitude $v_0$ grows as $\sqrt{2}$ — the distribution shifts right by $\sqrt{2}$. But when noise power doubles, $\sigma$ also grows as $\sqrt{2}$ — the bell gets wider by $\sqrt{2}$. The key asymmetry: signal shifting the center is a linear move of the whole bell, while noise widening is a spread that directly inflates the tail.
$S_o/N_o = v_0^2/\sigma^2$ captures exactly this asymmetry. The erfc argument is $\sqrt{S_o/N_o} = v_0/\sigma$:
- Doubling signal power: $v_0 \to \sqrt{2}\,v_0$, so $v_0/\sigma \to \sqrt{2}\,(v_0/\sigma)$ — argument grows, tail shrinks, BER drops
- Doubling noise power: $\sigma \to \sqrt{2}\,\sigma$, so $v_0/\sigma \to v_0/(\sqrt{2}\,\sigma)$ — argument shrinks, tail grows, BER rises
Both effects enter through the ratio $v_0/\sigma$. The SNR is simply asking: "how many noise standard deviations away is the signal from the threshold?"
Because the Gaussian tail falls off as $e^{-x^2}$ (faster than exponential), a small increase in SNR causes a dramatic collapse in BER:
| $E_b/n_0$ | $P_e$ |
|---|---|
| 8 dB | $\sim 10^{-3}$ |
| 9.6 dB | $10^{-5}$ |
| 12 dB | $\sim 10^{-9}$ |
Just 3–4 dB of SNR improvement changes the error rate by 6 orders of magnitude. This cliff-like behaviour is why communication systems are designed to operate just above the BER target threshold — adding more power beyond that gives diminishing returns compared to the rapid gains near the knee.
Problem: A BPSK system has $E_b/n_0 = 12$ dB, $R_b = 100$ kbps, $B = 100$ kHz. Find the output SNR and estimate $P_e$.
Click to reveal solution
$S_o/N_o = E_b/n_0 \cdot R_b/B = 12\text{ dB} + 10\log(100\text{k}/100\text{k}) = 12 + 0 = 12$ dB $= 15.85$ (linear)
$P_e = \frac{1}{2}\,\mathrm{erfc}(\sqrt{15.85}) = \frac{1}{2}\,\mathrm{erfc}(3.98) \approx \frac{1}{2}\times 1.1\times10^{-8} \approx 5.5\times10^{-9}$
(Very low error rate — well below the typical $10^{-5}$ requirement.)
2.1.6 BER Improvement by Integrate-and-Dump
$T_b$ = bit period — the duration of one bit, equal to the reciprocal of the bit rate:
$$T_b = \frac{1}{R_b}$$For example, $R_b = 1$ Mbps → $T_b = 1\;\mu\text{s}$ per bit. The integrate-and-dump integrates exactly over one $T_b$, then resets. Longer $T_b$ (slower bit rate) means more integration time, better SNR, and lower BER — the fundamental tradeoff between data rate and reliability.
In PSK demodulators, the integrate-and-dump detector dramatically improves BER by accumulating the signal over one bit period before making a decision. The block diagram and waveforms:
Left: Input signal $s(t)$ — binary PSK with bits 1, 1, 0, 1. Right: Integrator output $s_o(t)$ — ramps to $+VT$ for "1" and $-VT$ for "0", then resets (dumps) each bit period.
The integrator accumulates the signal over one bit period $T$, then the sampler reads the value at $t_s = N\cdot T$ and resets. The signal accumulates to $\pm VT$, while noise only grows as $\sqrt{T}$ (standard deviation), giving a much better SNR at the sample instant:
Why signal grows as $T$ but noise only as $\sqrt{T}$
Signal is constant $\pm v_0$ every moment — every infinitesimal slice $dt$ adds in the same direction, so contributions stack linearly: $\int_0^T v_0\,dt = v_0 T$. Noise is random — each slice has a random sign and cancels partially with the next. This is a random walk: variance adds linearly, but standard deviation (amplitude) only grows as $\sqrt{T}$. The interactive plot below shows both curves side by side, and the resulting SNR improvement at the bottom:
SNR improves as √Tb — the longer the integration, the better the detection.
x-axis: t / Tb = normalized time — 0 is the start of the bit period, 1 is the end (when the sampler reads and resets). The chart shows one full bit period regardless of the actual value of Tb.
Full Derivation of the BER Formula
Step 1 — Error condition for bit "0". When bit "0" is sent, the integrator output signal is $s_o(T) = -VT$. An error occurs when the output noise $v_n^{out}$ pushes the total above zero:
Step 2 — Write as a Gaussian tail integral. The output noise is Gaussian with zero mean and variance $\sigma_{out}^2$:
Step 3 — Convert to erfc. Apply the substitution $u = v_n^{out}/(\sigma_{out}\sqrt{2})$. Three things change simultaneously:
| What changes | How |
|---|---|
| Variable | $v_n^{out} = \sigma_{out}\sqrt{2}\,u$ |
| Differential | $dv_n^{out} = \sigma_{out}\sqrt{2}\,du$ |
| Lower limit | $v_n^{out} = VT \;\Rightarrow\; u = \dfrac{VT}{\sigma_{out}\sqrt{2}}$ |
| Upper limit | $\infty \;\Rightarrow\; \infty$ (unchanged) |
In any definite integral substitution, the limits must transform the same way as the variable. If $u = g(v)$, then every limit $v = a$ becomes $u = g(a)$:
$$\int_{v=a}^{v=b} f(v)\,dv \xrightarrow{u=g(v)} \int_{u=g(a)}^{u=g(b)} \tilde{f}(u)\,du$$Here the lower limit is $v_n^{out} = VT$. Plug it into the substitution formula $u = v_n^{out}/(\sigma_{out}\sqrt{2})$:
$$u_{\min} = \frac{VT}{\sigma_{out}\sqrt{2}}$$That's the new lower limit — just the old value plugged in.
Now transform the exponent and the prefactor × differential separately:
Exponent: $\dfrac{(v_n^{out})^2}{2\sigma_{out}^2} = \dfrac{(\sigma_{out}\sqrt{2}\,u)^2}{2\sigma_{out}^2} = \dfrac{2\sigma_{out}^2 u^2}{2\sigma_{out}^2} = u^2$ ✓
Prefactor × differential:
Assembled:
Comparing with the definition $\mathrm{erfc}(x) = \dfrac{2}{\sqrt{\pi}}\displaystyle\int_x^\infty e^{-u^2}\,du$, we see that $\dfrac{1}{\sqrt{\pi}}\displaystyle\int_x^\infty e^{-u^2}\,du = \dfrac{1}{2}\,\mathrm{erfc}(x)$, so:
Step 4 — Substitute $E_b$ and $\sigma_{out}^2$. Two substitutions from the physics of the integrator:
The transmitted BPSK signal is $v(t) = V\cos(\omega_c t)$, a sinusoid of amplitude $V$. The average power of a sinusoid is $V^2/2$ (the ½ comes from $\langle\cos^2\rangle = \frac{1}{2}$). Energy per bit = power × bit duration $T$:
$$E_b = \frac{V^2}{2} \cdot T = \frac{V^2 T}{2}$$The factor of ½ is entirely due to the sinusoidal carrier — a DC level $V$ would give $E_b = V^2T$ with no ½.
The integrator computes $y = \frac{1}{T}\int_0^T n(t)\,dt$ (normalized by $T$). The variance of this is:
$$\sigma_{out}^2 = \frac{1}{T^2}E\!\left[\left(\int_0^T n(t)\,dt\right)^2\right] = \frac{1}{T^2}\int_0^T\int_0^T E[n(t_1)n(t_2)]\,dt_1\,dt_2$$For white noise, $E[n(t_1)n(t_2)] = n_0\,\delta(t_1-t_2)$ (autocorrelation is a delta function). The double integral collapses:
$$= \frac{1}{T^2}\int_0^T n_0\,dt = \frac{n_0 T}{T^2} = \frac{n_0}{T}$$Physically: the integrator averages noise over $T$ — longer integration means more cancellation of random fluctuations, so variance falls as $1/T$.
| Substitution | Origin |
|---|---|
| $E_b = V^2T/2$ | Sinusoidal carrier: average power = $V^2/2$, times bit period $T$ |
| $\sigma_{out}^2 = n_0/T$ | Normalized integrator: noise variance falls as $1/T$ due to averaging |
Plugging in:
The slide uses a normalized integrator output where the integration result is divided by $T$, giving a per-unit-time output. Under this convention the signal output is $V$ (not $VT$) and $\sigma_{out}^2 = n_0/T$, so:
$$\left(\frac{V}{\sqrt{2}\,\sigma_{out}}\right)^2 = \frac{V^2}{2n_0/T} = \frac{V^2 T}{2n_0} = \frac{E_b}{n_0}$$The $T^2$ cancels cleanly. The final result is the same regardless of normalization convention:
$$\boxed{P_e = \frac{1}{2}\,\mathrm{erfc}\!\left(\sqrt{\frac{E_b}{n_0}}\right)}$$$P(\text{noise} > VT)$ $\;\xrightarrow{\text{Gaussian tail}}\;$ $\dfrac{1}{2}\mathrm{erfc}\!\left(\dfrac{VT}{\sqrt{2}\,\sigma_{out}}\right)$ $\;\xrightarrow{E_b = V^2T/2,\;\sigma_{out}^2 = n_0/T}\;$ $\dfrac{1}{2}\mathrm{erfc}\!\left(\sqrt{\dfrac{E_b}{n_0}}\right)$
Signal accumulates to $VT$, noise variance is $n_0/T$ — their ratio gives $E_b/n_0$, the fundamental figure of merit.
Problem: A PSK system has $V = 0.1$ V, $T = 10\;\mu$s, $n_0 = 4\times10^{-18}$ W/Hz. Find $E_b/n_0$ and $P_e$ with the integrate-and-dump detector.
Click to reveal solution
$E_b = V^2T/2 = (0.1)^2 \times 10\times10^{-6}/2 = 5\times10^{-8}$ J
$E_b/n_0 = 5\times10^{-8}/(4\times10^{-18}) = 1.25\times10^{10}$ → $10\log(1.25\times10^{10}) = 101$ dB
$P_e = \frac{1}{2}\,\mathrm{erfc}(\sqrt{1.25\times10^{10}}) \approx 0$ (essentially error-free)
2.1.7 Comparison of Modulation Methods
BPSK vs QPSK
BPSK (Binary PSK) — 1 bit per symbol, 2 phase states (0°, 180°): $s(t) = \pm v_0\cos(\omega_c t)$
QPSK (Quadrature PSK) — 2 bits per symbol, 4 phase states (0°, 90°, 180°, 270°), modulating both I and Q axes independently.
| BPSK | QPSK | |
|---|---|---|
| Bits per symbol | 1 | 2 |
| Phase states | 2 (0°, 180°) | 4 (0°, 90°, 180°, 270°) |
| Bandwidth | $R_b$ | $R_b/2$ — half the bandwidth for the same bit rate |
| BER formula | $\frac{1}{2}\mathrm{erfc}\!\left(\sqrt{E_b/n_0}\right)$ | $\frac{1}{2}\mathrm{erfc}\!\left(\sqrt{E_b/n_0}\right)$ |
| $E_b/n_0$ for $P_e=10^{-5}$ | 9.6 dB | 9.6 dB |
| Complexity | Simple | Needs I/Q demodulator |
QPSK is essentially two independent BPSK channels running on orthogonal I and Q axes. Each axis carries 1 bit with the same noise margin $v_0/\sigma$ as BPSK. Since the axes are orthogonal (I and Q do not interfere with each other), the BER of each is identical to BPSK — so the overall BER formula is the same.
QPSK achieves twice the spectral efficiency with no BER penalty — this is why it dominates in practice (4G/5G, satellite, cable modem).
| Modulation | $E_b/n_0$ for $P_e=10^{-5}$ | Notes |
|---|---|---|
| Coherent PSK | 9.6 dB | Best; requires sync detection |
| Coherent FSK | 12.6 dB | 3 dB worse than PSK |
| Noncoherent FSK | 13.4 dB | No carrier recovery needed |
| Coherent ASK | 15.6 dB | 6 dB worse than PSK |
| Noncoherent ASK | 16.4 dB | Simplest implementation |
Given: QPSK, $R_b = 48.6$ kbps, $B = 30$ kHz, required BER $= 10^{-5}$.
For QPSK: $P_e = \frac{1}{2}\mathrm{erfc}(\sqrt{E_b/n_0}) = 10^{-5}$ → $E_b/n_0 = 9.6$ dB
$\frac{S_o}{N_o} = \frac{E_b}{n_0}\cdot\frac{R_b}{B} = 9.6 + 10\log\!\left(\frac{48.6}{30}\right) = 9.6 + 2.1 = 11.7$ dB
2.1 Noise and BER in Linear Receivers (continued)
Problem: A 16-QAM system has $R_b = 200$ kbps and $B = 50$ kHz. If the required $E_b/n_0$ for the target BER is 14.4 dB (from Table 14.1 in Pozar), find the minimum detectable output SNR.
Click to reveal solution
$\frac{S_o}{N_o} = \frac{E_b}{n_0}\cdot\frac{R_b}{B} = 14.4 + 10\log\!\left(\frac{200\text{k}}{50\text{k}}\right) = 14.4 + 6.02 = 20.4$ dB
16-QAM has spectral efficiency $R_b/B = 4$ bps/Hz (since each symbol carries 4 bits), which increases the required SNR by 6 dB compared to $E_b/n_0$.
2.1.8 16-QAM Implementation
16-QAM carries 4 bits per symbol using both amplitude and phase — it is not enough to just demodulate I and Q like QPSK.
16-QAM uses a 4×4 constellation — 4 amplitude levels on each of I and Q axes, giving $4\times4 = 16$ points total. Each axis independently carries 2 bits (not 1 like QPSK), using 4 amplitude levels: $\pm1, \pm3$.
Q
+3 • • • •
+1 • • • •
-1 • • • •
-3 • • • •
-3 -1 +1 +3 → I
Each • = one unique 4-bit symbol
Transmitter
- Split the bitstream into groups of 4 bits
- Map 2 bits → I amplitude level (one of $\pm1, \pm3$)
- Map 2 bits → Q amplitude level (one of $\pm1, \pm3$)
- Modulate: $s(t) = I\cos(\omega_c t) - Q\sin(\omega_c t)$
Receiver
- I/Q demodulation — multiply by $\cos(\omega_c t)$ and $\sin(\omega_c t)$, integrate over $T_s$
- Get two numbers: one I value and one Q value
- Decision — compare each against 3 thresholds (not 1 like BPSK/QPSK): $-2, 0, +2$
- Map each detected level back to 2 bits → combine for 4 bits total
Yes and no. The I/Q demodulation step is the same as QPSK — multiply by local carrier, integrate. That part is sufficient to extract the I and Q values. But the decision step is harder:
| BPSK | QPSK | 16-QAM | |
|---|---|---|---|
| Axes | 1 | 2 (I, Q) | 2 (I, Q) |
| Levels per axis | 2 | 2 | 4 |
| Thresholds per axis | 1 | 1 | 3 |
| Bits per symbol | 1 | 2 | 4 |
In QPSK each axis only needs to decide $+$ or $-$ (one threshold at 0). In 16-QAM each axis must decide among 4 levels — requiring 3 thresholds and an amplitude detector, not just a sign detector.
The constellation points are closer together relative to their power. The minimum distance between adjacent points is smaller, so noise is more likely to push a point to a wrong neighbor.
This is why 16-QAM requires about 4 dB more $E_b/n_0$ than QPSK for the same BER — you pay an SNR penalty for the spectral efficiency gain.
On top of that, converting from $E_b/n_0$ to output SNR adds another 6 dB because of the spectral efficiency $R_b/B = 4$ bps/Hz:
So 16-QAM pays twice: ~4 dB more $E_b/n_0$ for the denser constellation, plus 6 dB from packing 4 bits into each symbol period — a total of ~10 dB more output SNR than QPSK.
2.2 Noise Temperature and Noise Figure
2.2.1 Equivalent Noise Temperature
Problem: An amplifier with gain $G = 15$ dB and bandwidth $B = 500$ MHz produces an output noise power of $N_o = -65$ dBm when its input is terminated with a 0 K source. Find $T_e$.
Why 0 K source? At 0 K the thermal noise from the source is zero (since thermal noise $= kTB$, and $T=0$ gives 0). This means all the output noise $N_o = -65$ dBm comes purely from the amplifier's internal noise, not from the input. This isolates the amplifier's own noise contribution, making it possible to directly solve for $T_e$.
Click to reveal solution
$G = 10^{1.5} = 31.62$; $N_o = 10^{(-65-30)/10} = 3.16\times10^{-10}$ W
$T_e = \frac{N_o}{GkB} = \frac{3.16\times10^{-10}}{31.62\times1.38\times10^{-23}\times500\times10^6} = \frac{3.16\times10^{-10}}{2.18\times10^{-13}} = 1449$ K
Problem: An amplifier with gain $G = 10$ dB and bandwidth $B = 100$ MHz produces an output noise power of $N_o = -80$ dBm when its input is terminated with a source at $T_s = 290$ K. Find $T_e$.
Key difference from Example 7: Since the source is not at 0 K, the output noise now contains both amplified source thermal noise and the amplifier's own noise: $N_o = G \cdot k \cdot (T_s + T_e) \cdot B$. We must subtract the source contribution to isolate $T_e$.
Click to reveal solution
$G = 10^{10/10} = 10$; $N_o = 10^{(-80-30)/10} = 10^{-11}$ W
$T_e = \frac{N_o}{GkB} - T_s = \frac{10^{-11}}{10\times1.38\times10^{-23}\times100\times10^6} - 290 = \frac{10^{-11}}{1.38\times10^{-14}} - 290 = 724.6 - 290 = 435$ K
2.2.2 Y-Factor Measurement
Given: X-band amp, $G=20$ dB, $B=1$ GHz. $T_1=290$ K → $N_1=-62$ dBm; $T_2=77$ K → $N_2=-64.7$ dBm.
$Y = 10^{2.7/10} = 1.86$
$T_e = (290 - 1.86\times77)/(1.86-1) = (290-143.2)/0.86 = 170$ K
If a source with an equivalent noise temperature of $T_s = 450$ K drives the amplifier with a gain of 20 dB (=100), the source into the amplifier is $kT_sB$. The total noise power out of the amplifier can be calculated as:
$N_o = G\,k\,B(T_s+T_e) = 100\times1.38\times10^{-23}\times10^9\times620 = 8.56\times10^{-10}$ W $= -60.7$ dBm
Here $10^9$ comes from $B = 1\text{ GHz} = 1\times10^9\text{ Hz}$, converting to base SI units.
- It is close to room temperature (~17 °C), making lab measurements natural
- It gives a clean thermal noise density: $kT_0 = 1.38\times10^{-23}\times 290 = 4.00\times10^{-21}$ W/Hz exactly
So whenever you see $(F-1)\times 290$, the 290 is not an approximation — it is the defined constant $T_0$.
Understanding $F$ and $NF$ (click to expand)
What $F$ means: The noise figure $F$ is defined as the SNR degradation factor: $F = SNR_{in}/SNR_{out}$. "Noise figure" and "SNR degradation" are the same concept — $F$ measures how much the device worsens the signal-to-noise ratio.
- $F = 1$ — no degradation, ideal (noiseless) device
- $F = 2$ — SNR cut in half; device added as much noise as was already there
- $F = 10$ — SNR reduced by 10×
Naming convention: Some textbooks call the linear ratio $F$ the noise factor and reserve noise figure for the dB value. Others (including Pozar) use "noise figure" for both and let context distinguish them.
Why $NF = 10\log_{10}(F)$? Because $F$ is a power ratio ($SNR_{in}/SNR_{out}$, where each SNR is signal power / noise power). The dB scale for power ratios is always $10\log_{10}$ — same as gain ($G_{\text{dB}} = 10\log_{10}G$) or loss ($L_{\text{dB}} = 10\log_{10}L$). For voltage ratios it would be $20\log_{10}$ instead, because power $\propto V^2$.
Derivation: why $F = 1 + T_e/T_0$?
From the definition $F = \frac{S_i/N_i}{S_o/N_o}$, substitute $S_o = G\cdot S_i$ and $N_o = G\cdot kB(T_0 + T_e)$:
$$F = \frac{S_i/(kT_0B)}{G\cdot S_i/[G\cdot kB(T_0+T_e)]} = \frac{T_0+T_e}{T_0} = 1 + \frac{T_e}{T_0}$$The gain $G$ and signal $S_i$ cancel — $F$ depends only on how much extra noise $T_e$ the network adds relative to $T_0 = 290$ K.
Rearranging:
$$F - 1 = \frac{T_e}{T_0} \;\;\Longrightarrow\;\; T_e = (F-1)\cdot 290\text{ K}$$Problem: A low-noise amplifier is tested with $T_1 = 290$ K ($N_1 = -70$ dBm) and $T_2 = 77$ K ($N_2 = -71.5$ dBm). Find $T_e$ and the noise figure $F$.
Click to reveal solution
$Y = 10^{((-70)-(-71.5))/10} = 10^{0.15} = 1.413$
$T_e = \frac{290 - 1.413\times77}{1.413-1} = \frac{290 - 108.8}{0.413} = \frac{181.2}{0.413} = 438.7$ K
$F = 1 + T_e/T_0 = 1 + 438.7/290 = 2.513$
$\text{NF} = 10\log_{10}(F) = 10\log_{10}(2.513) = 4.0\text{ dB}$
2.2.3 Noise Figure
General output noise:
$$N_o = G\,(N_i + N_{added}) \;\;\Rightarrow\;\; T_e = \frac{N_{added}}{k\,B}$$General noise figure (depends on $N_i$):
$$F = \frac{S_i / N_i}{S_o / N_o} = \frac{S_i / N_i}{G\,S_i\,/\,[G(N_i + N_{added})]} = 1 + \frac{N_{added}}{N_i}$$Since in general $N_{added}$ cannot be proportional to $N_i$, the noise figure depends on $N_i$. But in real noise figure measurement, $N_i$ is defined at $kT_0B$ with $T_0 = 290$ K, such that a fixed value of noise figure can be taken:
$$F \equiv 1 + \frac{N_{added}}{N_i(T_0)} = 1 + \frac{k\,T_e\,B}{k\,T_0\,B} = 1 + \frac{T_e}{T_0} = 1 + \frac{T_e}{290}$$Output noise with $T_e$ model:
$N_i$: input noise power; $k\,T_e\,B$: internally generated noise from amplifier.
$$N_o = G\cdot(\underbrace{N_i}_{\text{input}} + \underbrace{k\,T_e\,B}_{\text{amplifier}}) = G\cdot(N_i + N_{added}) = k\,G\,B\,(T_0 + T_e)$$Noise figure from SNR ratio:
$$F = \frac{S_i}{N_i}\cdot\frac{N_o}{S_o} = \frac{S_i}{k\,T_0\,B}\cdot\frac{k\,G\,B\,(T_0+T_e)}{G\cdot S_i} = 1 + \frac{T_e}{T_0} \;\;\Longleftrightarrow\;\; T_e = (F-1)\,T_0 \tag{2}$$Problem: An LNA has a noise figure of 0.8 dB. What is its equivalent noise temperature? If a second amplifier has $T_e = 500$ K, what is its noise figure in dB?
Click to reveal solution
LNA (NF → $T_e$):
Convert dB to linear: $F = 10^{NF/10} = 10^{0.8/10} = 10^{0.08} = 1.202$
Apply Eq. (1): $T_e = (F-1)\,T_0 = (1.202-1)\times 290 = 0.202\times 290 = 58.6$ K
Second amp ($T_e$ → NF):
Rearrange Eq. (1) the other way — solve for $F$: $$F = 1 + \frac{T_e}{T_0} = 1 + \frac{500}{290} = 1 + 1.724 = 2.724$$
Convert to dB: $NF = 10\log_{10}(F) = 10\log_{10}(2.724) = 4.35$ dB
2.2.4 Noise Figure of a Lossy Line
Physical meaning of $G$: A lossy transmission line with insertion loss $L$ is treated as a two-port with available gain $G = 1/L < 1$. A matched source (resistor $R$ at temperature $T$) delivers input noise $N_i = kTB$. In thermal equilibrium the output noise must also be $N_o = kTB$, because the lossy line at temperature $T$ looks like another resistor at $T$ from the output port.
The setup. The line itself sits at temperature $T$, has matched characteristic impedance $Z_0 = R$, and is fed by a source also at temperature $T$. Here $L$ is the insertion loss and $G = 1/L$ is the available gain.
The thermodynamic argument. The second law of thermodynamics states that the entropy of an isolated system always increases over time, and heat spontaneously flows from hotter to colder regions. When every part of the system is at the same temperature $T$, the 2nd law forbids any net flow of energy between the parts. If $N_o$ were greater or less than $N_i$, you could place a load at one end and extract work from a single-temperature reservoir — a perpetual motion machine of the second kind. So the noise entering and leaving must balance.
Step-by-step noise balance at the output port:
- Input noise. The matched source at $T$ delivers Johnson–Nyquist noise $N_i = kTB$ into the line.
- Attenuated input. After traversing the loss $L$, only $N_i / L$ of that noise remains.
- Line's own thermal emission. Because the line is lossy and at temperature $T$, it radiates its own Johnson noise, contributing exactly $kTB\,(1 - 1/L)$ at the output.
- Total output.
The attenuated input plus the line's self-emitted noise sum to exactly $kTB$, independent of how lossy the line is.
Due to thermodynamic equilibrium:
$$N_o = N_i = k\,T\,B = G\,(N_i + N_{added})$$ $$\Rightarrow\; N_{added} = \frac{1-G}{G}\,k\,T\,B = \left(\frac{1}{G}-1\right)k\,T\,B = (L-1)\,k\,T\,B$$Equivalent noise temperature:
$$T_e = \frac{N_{added}}{k\,B} = (L-1)\cdot T$$Noise figure:
$$F = 1 + \frac{N_{added}}{N_i(T_0)} = 1 + \frac{(L-1)\,k\,T\,B}{k\,T_0\,B} = 1 + (L-1)\frac{T}{T_0} \;\xrightarrow{T=T_0}\; F = L$$$\therefore\; F(\text{dB}) = IL(\text{dB})$ under the condition $T = T_0$, where $IL$ stands for Insertion Loss (插入損耗).
Problem: A coaxial cable has 3 dB loss and is at a physical temperature of $T = 350$ K. Find $T_e$ and $F$.
Click to reveal solution
$L = 10^{0.3} = 2.0$
$T_e = (2-1)\times350 = 350$ K
$F = 1 + (2-1)\times350/290 = 1 + 1.207 = 2.207 = 3.44$ dB
(Note: if $T=290$ K, $F$ would be exactly $L = 2 = 3$ dB. Since the cable is hotter, $F$ is slightly higher.)
2.2.5 Cascaded Noise Figure (Friis Formula)
Derivation: Friis formula for two stages
Output noise of stage 1:
$$N_1 = G_1(k\,T_0\,B + k\,T_{e1}\,B)$$Output noise of the cascade (stage 2 amplifies $N_1$ and adds its own noise):
$$N_o = G_2(N_1 + k\,T_{e2}\,B) = G_1 G_2\,k\,B\left(T_0 + T_{e1} + \frac{T_{e2}}{G_1}\right)$$ $$= G_1 G_2(k\,T_0\,B + k\,T_e\,B)$$Equivalent noise temperature of cascaded system:
$$\Rightarrow\; T_e = T_{e1} + \frac{T_{e2}}{G_1}$$Noise figure:
$$F = \frac{S_i}{N_i}\cdot\frac{N_o}{S_o} = \frac{S_i}{k\,T_0\,B}\cdot\frac{\colorbox{#f8d0d0}{$G_1 G_2$}\,k\,B\,(T_0 + T_{e1} + T_{e2}/G_1)}{\colorbox{#f8d0d0}{$G_1 G_2\,S_i$}}$$ $$= 1 + \frac{T_{e1}}{T_0} + \frac{T_{e2}}{G_1 T_0} = F_1 + \frac{1}{G_1}\left[\left(1+\frac{T_{e2}}{T_0}\right)-1\right] = F_1 + \frac{F_2-1}{G_1}$$General form for arbitrary number of cascaded stages:
$$T_e = T_{e1} + \frac{T_{e2}}{G_1} + \frac{T_{e3}}{G_1 G_2} + \cdots$$ $$F = F_1 + \frac{F_2-1}{G_1} + \frac{F_3-1}{G_1 G_2} + \cdots$$Why First Stage Dominates (click to expand)
The noise characteristics of a cascaded system are dominant by the first few stages, since the effect of later stages is reduced by the product of the gains of the preceding stages. For best overall system noise performance, the first stage of a receiver (LNA, low noise amplifier) should have a low noise figure and at least moderate gain.Adjust Parameters
Friis Formula Breakdown
System Performance
LNA: $F_1=2$ dB $=1.58$, $G_1=10$ dB $=10$. BPF: $F_2=1$ dB $=1.26$, $G_2=-1$ dB $=0.79$. Mixer: $F_3=4$ dB $=2.51$, $G_3=-3$ dB $=0.5$.
$F = 1.58 + \frac{0.26}{10} + \frac{1.51}{10\times0.79} = 1.58 + 0.026 + 0.191 = 1.80 = 2.55$ dB
From Eq. (1): $T_e = (F-1)\,T_0 = (1.80-1)\times290 = 232$ K. $G_{total} = 10\times0.79\times0.5 = 3.95 = 5.97$ dB
Problem: Consider two configurations for a receiver with an LNA ($G = 20$ dB, $F = 1.5$ dB) and a cable ($L = 2$ dB):
Config A: LNA → Cable Config B: Cable → LNA
Find the cascade noise figure for each. (Assume cable at $T_0 = 290$ K.)
Click to reveal solution
LNA: $F_1 = 10^{0.15} = 1.413$, $G_1 = 100$. Cable: $F = L = 10^{0.2} = 1.585$, $G = 1/L = 0.631$.
Config A (LNA first): $F = 1.413 + \frac{1.585-1}{100} = 1.413 + 0.006 = 1.419 = 1.52$ dB
Config B (Cable first): $F = 1.585 + \frac{1.413-1}{0.631} = 1.585 + 0.654 = 2.240 = 3.50$ dB
Conclusion: Placing the cable before the LNA adds ~2 dB to the noise figure! Always put the LNA first.
2.2.6 Noise Figure of Mismatched Networks & Wilkinson Divider
Interactive: Why does $\Gamma_L=0$ mean "matched"? (click to expand)
A transmission line has characteristic impedance $Z_0$. When the load $Z_L = Z_0$, all power is absorbed — no reflection. Drag the sliders to see what happens when $Z_L \neq Z_0$.
$$\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0}$$
S-parameters describe how an RF signal scatters at a two-port network. $L$ is the loss factor: how much power the line absorbs as heat.
Power Flow
| Parameter | Meaning | Value |
|---|---|---|
| $S_{21}$ | Forward transmission (signal passes through) | 0.707 |
| $S_{11}$ | Input reflection (bounces back at port 1) | 0.000 |
| $S_{22}$ | Output reflection (bounces back at port 2) | = 0 (matched load) |
| $S_{12}$ | Reverse transmission | 0.707 |
| $L$ | Loss factor = $1/|S_{21}|^2$ | 2.000 |
Mismatched Passive Two-Port Network
Passive networks are circuits built only from passive components (resistors, capacitors, inductors, transmission lines, cables, attenuators, filters, etc.) with no gain ($G \leq 1$). The only noise they produce is thermal noise (Johnson-Nyquist noise) from random thermal motion of electrons, characterized by $N = kTB$.
Active devices (diodes, transistors) generate additional non-thermal noises: shot noise (discrete carrier crossings at junctions), flicker/1/$f$ noise (carrier trapping at surface defects), and generation-recombination noise.
Because a lossy line is purely passive, the thermodynamic equilibrium argument ($N_o = N_i = kTB$) holds perfectly — no extra non-thermal noise sources exist. This is why $F = L$ (at $T = T_0$) works for cables and attenuators but not for amplifiers or mixers containing active devices.
Derivation: $T_e$ and $F$ for a mismatched passive network
Output noise using available power gain:
$$N_2 = G_A(\Gamma_s)\cdot(N_1 + N_{added}) = G_A(\Gamma_s)\cdot(k\,T\,B + N_{added})$$where $G_A(\Gamma_s)$ is the available power gain, which depends on the source reflection coefficient $\Gamma_s$.
Thermodynamic equilibrium requires $N_2 = k\,T\,B$, therefore:
$$\Rightarrow\; N_{added} = \frac{1 - G_A(\Gamma_s)}{G_A(\Gamma_s)}\,k\,T\,B$$Equivalent noise temperature:
$$\Rightarrow\; T_e = \frac{N_{added}}{k\,B} = \frac{1 - G_A(\Gamma_s)}{G_A(\Gamma_s)}\cdot T$$Noise figure:
$$\Rightarrow\; F = 1 + \frac{N_{added}}{N_i(T_0)} = 1 + \frac{1 - G_A(\Gamma_s)}{G_A(\Gamma_s)}\,\frac{k\,T\,B}{k\,T_0\,B} = 1 + \frac{1 - G_A(\Gamma_s)}{G_A(\Gamma_s)}\cdot\frac{T}{T_0}$$Propagation Constant $\gamma = \alpha + j\beta$
The propagation constant $\gamma$ describes how a wave changes as it travels along a transmission line. It has two parts:
| Symbol | Name | Units | What it controls |
|---|---|---|---|
| $\alpha$ | Attenuation constant | Np/m | How fast the amplitude decays — caused by conductor loss (skin effect) and dielectric loss |
| $\beta$ | Phase constant | rad/m | How fast the phase rotates — related to wavelength by $\beta = 2\pi/\lambda$ |
A voltage wave of initial amplitude $V^+$, after traveling distance $l$:
- Lossless line: $\alpha = 0 \;\Rightarrow\; \gamma = j\beta$ — only phase changes, no amplitude loss.
- Lossy line: $\alpha > 0$ — signal gets weaker as it propagates.
The $j$ in $j\beta$ is the imaginary unit ($j^2 = -1$), so $e^{-j\beta l}$ produces phase rotation (a circle on the complex plane) rather than amplitude change.
S-Matrix (Scattering Matrix) of a Two-Port Network
Why S-parameters?
At low frequencies, circuits are described by voltages and currents (Z, Y, ABCD matrices). At microwave frequencies this breaks down because:
- Total voltage/current are hard to define uniquely along a transmission line — they depend on where you measure.
- Open/short circuit conditions (needed to measure Z or Y parameters) can cause oscillation or damage active devices.
- Measuring total V and I at GHz frequencies requires knowing the standing wave pattern — impractical.
S-parameters solve this by working with traveling waves instead: the incident wave $a$ going into a port and the scattered (outgoing) wave $b$ coming out. These are easy to measure with a network analyzer and well-defined at any reference plane.
Definition
Normalized wave amplitudes at each port:
$V_i^+$ = incident voltage wave at port $i$, $\;V_i^-$ = reflected voltage wave, $\;Z_0$ = reference impedance
For a two-port network:
| Parameter | Definition | Physical meaning | Common dB name |
|---|---|---|---|
| $S_{11}$ | $b_1/a_1\big|_{a_2=0}$ | Input reflection coefficient — how much bounces back at port 1 | Return loss = $-20\log|S_{11}|$ |
| $S_{21}$ | $b_2/a_1\big|_{a_2=0}$ | Forward transmission — signal from port 1 → port 2 | Gain (or insertion loss) = $20\log|S_{21}|$ |
| $S_{12}$ | $b_1/a_2\big|_{a_1=0}$ | Reverse transmission — signal from port 2 → port 1 | Isolation = $-20\log|S_{12}|$ |
| $S_{22}$ | $b_2/a_2\big|_{a_1=0}$ | Output reflection coefficient — how much bounces back at port 2 | Output return loss = $-20\log|S_{22}|$ |
The condition "$a_2 = 0$" means port 2 is terminated with a matched load ($Z_0$) so no wave enters from that side. In practice, a network analyzer does exactly this: it sends a signal into one port and terminates the other with $50\;\Omega$.
- $|S_{ij}|^2$ = power ratio. For example, $|S_{21}|^2 = P_{\text{out}}/P_{\text{in}}$ is the power gain.
- $|S_{11}|^2$ = fraction of incident power reflected back.
- $1 - |S_{11}|^2 - |S_{21}|^2$ = power dissipated inside the network (for a two-port with $a_2=0$).
Key properties
| Property | Condition | Meaning |
|---|---|---|
| Reciprocal | $[S]^T = [S]$, i.e. $S_{ij} = S_{ji}$ | Network looks the same from both sides (passive networks without ferrites) |
| Lossless | $[S]^H[S] = [I]$ (unitary) | All power entering leaves — no dissipation. $|S_{11}|^2 + |S_{21}|^2 = 1$ |
| Matched | $S_{ii} = 0$ | No reflection at port $i$ |
S-matrices of common devices
| Device | $[S]$ | Notes |
|---|---|---|
| Matched lossy line | $\begin{bmatrix} 0 & e^{-\gamma l} \\ e^{-\gamma l} & 0 \end{bmatrix}$ | Reciprocal, matched. $|S_{21}|^2 = e^{-2\alpha l} = 1/L$ |
| Lossless line | $\begin{bmatrix} 0 & e^{-j\beta l} \\ e^{-j\beta l} & 0 \end{bmatrix}$ | $\alpha=0$, so $|S_{21}|=1$. Unitary. |
| Ideal amplifier (gain $G$) | $\begin{bmatrix} 0 & 0 \\ \sqrt{G} & 0 \end{bmatrix}$ | Unilateral ($S_{12}=0$), matched, not reciprocal |
| Ideal isolator | $\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$ | Passes forward, blocks reverse. Not reciprocal (uses ferrite). |
Diagonal = 0: Both ports see impedance $Z_0$, so $\Gamma = (Z_0 - Z_0)/(Z_0 + Z_0) = 0$ — no reflection.
Off-diagonal = $e^{-\gamma l}$: The signal propagates through with attenuation $e^{-\alpha l}$ and phase shift $e^{-j\beta l}$.
Symmetric ($S_{21} = S_{12}$): The line is reciprocal — it looks the same from both ends. This holds for all passive networks without magnetically biased materials (ferrites).
Application to a Mismatched Lossy Line
Matched means $Z_g = Z_0$ (source impedance equals the line's characteristic impedance), so $\Gamma_s = 0$ — a wave arriving at the source is fully absorbed, no bounce back.
Source mismatch means $Z_g \neq Z_0$, so $\Gamma_s \neq 0$ — the source acts like a partial mirror. When a wave hits it, part reflects back into the line.
A lossy line generates thermal noise and radiates it out of both ports. The noise going out port 2 reaches the load — that's expected. But noise also goes out port 1 toward the source:
- If matched ($\Gamma_s = 0$): the source absorbs all that noise. No extra noise at the output.
- If mismatched ($\Gamma_s \neq 0$): the source reflects some noise back into the line. That reflected noise travels through the line and arrives at port 2 as extra noise on top of the signal.
Analogy: Shouting in a hallway. If the far wall is padded (matched), your echo dies. If the far wall is concrete (mismatched), the echo bounces back and adds to the noise in the room.
The noise figure formula shows this: $F$ increases with $|\Gamma_s|^2$ because more mismatch = more reflected noise re-entering the system.
Interactive: Source Mismatch Effect on Noise (click to expand)
Derivation: Mismatched lossy line via S-parameters
Source reflection coefficient (mismatch at port 1):
$$\Gamma_s = \frac{Z_g - Z_0}{Z_g + Z_0} \neq 0$$S-matrix of the lossy line (length $l$, propagation constant $\gamma = \alpha + j\beta$):
$$[S] = \begin{bmatrix} 0 & e^{-\gamma l} \\ e^{-\gamma l} & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} e^{-(\alpha+j\beta)l} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \frac{e^{-j\beta l}}{e^{\alpha l}} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \frac{e^{-j\beta l}}{\sqrt{L(l)}}$$The power loss factor $L$ is defined as the ratio of input to output power. Since power $\propto |\text{amplitude}|^2$:
$$L(l) = \frac{P_{\text{in}}}{P_{\text{out}}} = \frac{|V^+|^2}{|V^+ e^{-\alpha l}|^2} = e^{2\alpha l}$$Taking the square root:
$$e^{\alpha l} = \sqrt{e^{2\alpha l}} = \sqrt{L(l)}$$Therefore the amplitude attenuation factor is:
$$e^{-\alpha l} = \frac{1}{\sqrt{L(l)}}$$Output Reflection Coefficient $\Gamma_{out}$
General formula (any two-port):
Imagine looking into port 2. What reflection do you see?
- Direct reflection: $S_{22}$ — the intrinsic reflection at port 2 itself.
- Indirect path (multiple bounces): A signal exits port 1 ($S_{12}$), hits the mismatched source ($\Gamma_s$), bounces back into port 1, passes through to port 2 ($S_{21}$), and this repeats. Each round trip picks up a factor $S_{11}\Gamma_s$ (reflect off the network's own port 1, then off the source). Summing the infinite geometric series:
Total: $\;\Gamma_{out} = S_{22} + \dfrac{S_{12}\,S_{21}\,\Gamma_s}{1 - S_{11}\,\Gamma_s}$
Substituting the matched lossy line ($S_{11}=0,\;S_{22}=0,\;S_{12}=S_{21}=e^{-j\beta l}/\sqrt{L}$):
- $\Gamma_s$ — the source mismatch that causes reflections in the first place.
- $e^{-2j\beta l}$ — round-trip phase shift. The signal travels port 2 → source → port 2, total path $2l$.
- $1/L$ — round-trip power loss. The signal is attenuated twice traversing the lossy line ($1/\sqrt{L}$ each way, squared = $1/L$).
A lossier line (larger $L$) makes $|\Gamma_{out}|$ smaller — the source mismatch is hidden behind the loss.
Interactive: $\Gamma_{out}$ of a Mismatched Lossy Line (click to expand)
Available power gain:
$$G_A(\Gamma_s) = \frac{|S_{21}|^2(1-|\Gamma_s|^2)}{|1-S_{11}\Gamma_s|^2(1-|\Gamma_{out}(\Gamma_s)|^2)} = \bbox[#ffcccc,3px]{\frac{(1/L)(1-|\Gamma_s|^2)}{1-|\Gamma_s/L|^2}}$$Equivalent noise temperature and noise figure:
$$T_e = \frac{1-G_A(\Gamma_s)}{G_A(\Gamma_s)}\,T = \frac{(L-1)(L+|\Gamma_s|^2)}{L(1-|\Gamma_s|^2)}\,T$$ $$F = 1 + \frac{T_e}{T_0} = 1 + \frac{(L-1)(L+|\Gamma_s|^2)}{L(1-|\Gamma_s|^2)}\cdot\frac{T}{T_0}$$If $\Gamma_s = 0$: $\;\;T_e = (L-1)\cdot T$ and $F = 1 + (L-1)\dfrac{T}{T_0}$, recovering the matched case.
When $\Gamma_s = 0$, this reduces to the expressions for a matched lossy line. Compare to see that the noise figure or equivalent noise temperature increases for a mismatched lossy line. The reason for this increase is that the lossy line actually delivers noise power out of both its ports, but when the input port is mismatched, some of the available noise power at port 1 is reflected from the source back into port 1, and appears at port 2. On the contrary, when the source is matched to port 1, none of the available power from port 1 is reflected back into the line, so the noise power available at port 2 is a minimum.
"when the source is matched to port 1" — $Z_g = Z_0$, so $\Gamma_s = 0$. The source impedance equals the line's characteristic impedance.
"none of the available power from port 1 is reflected back into the line" — The lossy line radiates thermal noise out of both ports. The noise going out port 1 hits the source. Since the source is matched, it absorbs 100% of that noise — nothing bounces back.
"so the noise power available at port 2 is a minimum" — Port 2 only receives two noise contributions:
- Attenuated input signal noise: $kTB/L$
- Line's own thermal emission toward port 2: $kTB(1-1/L)$
No extra term exists. This is the best case — the lowest possible noise at the output.
If the source were mismatched ($\Gamma_s \neq 0$), a third contribution would appear: the reflected noise from port 1 bouncing off the source, traveling back through the line, and arriving at port 2. That extra term makes the output noise larger than the minimum.
Therefore: matched source = lowest achievable noise figure for this lossy line. Any mismatch can only make it worse.
Wilkinson Power Divider (lossless, port 3 terminated)
S-matrix of the Wilkinson divider (with loss $L$ in each arm):
$$[S] = \frac{-j}{\sqrt{2L}} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$Derivation: Noise figure of the Wilkinson divider
Port roles in the Wilkinson divider:
Port 1 = input (signal enters)
Port 2 = output (half power, −3 dB)
Port 3 = output (half power, −3 dB)
It's a power splitter: 1 in → 2 out. For noise figure analysis, we use only port 2 as output and terminate port 3 with $Z_0$ to ground, dumping its power into the resistor.
Port 2 vs Port 3 — why different?
Port 3: Z₀ resistor to ground — a dummy load that absorbs power (not used, signal is thrown away).
Port 2: connects to the next stage (amplifier, instrument, etc.) which also presents $Z_0$ impedance — but it's the active output where you take the signal out, not a termination to ground.
The $Z_0$ label on port 2's line = characteristic impedance of the output transmission line going to the next stage, not a grounded resistor.
Why does port 3 need a dummy load but port 2 doesn't?
S-parameter gain/noise analysis needs a 2×2 matrix — but the Wilkinson has 3 ports. One port must be eliminated.
| If port 3 is... | $\Gamma_3$ | Result |
|---|---|---|
| Open (nothing connected) | $+1$ | Total reflection — waves bounce back, corrupt S-matrix |
| Shorted to ground | $-1$ | Total reflection (inverted) — same problem |
| Terminated with $Z_0$ | $0$ | No reflection — port cleanly disappears |
Port 2 doesn't need a dummy load because it's the port you're actually using — the signal comes out there to your real circuit (amplifier, analyzer, etc.).
Analogy: a Y-splitter for a garden hose. You want water from one output only. You don't cap the other (pressure builds up = reflection). You attach a drain so water flows freely without back-pressure.
How does port 2 physically connect to the next stage?
RF circuits use transmission lines (not discrete wires). Every signal path has 2 conductors: a signal trace and a ground plane.
Port 3 is identical physically — but instead of going to a next stage, its signal trace ends at a 50Ω SMD resistor soldered to the ground pad. Dead end.
To evaluate the noise figure, we first terminate port 3 with a matched load; this converts the 3-port device to a 2-port device.
What is the "termination" at port 3?
Zigzag = resistor, value $Z_0$ (e.g. 50 Ω)
Horizontal lines = ground symbol
In a real circuit: a 50 Ω resistor soldered between the port 3 trace and the ground plane.
This absorbs all incoming power → $\Gamma_3 = \frac{Z_0 - Z_0}{Z_0 + Z_0} = 0$ → no reflection back into port 3.
The full 3-port S-matrix of the Wilkinson divider (with lossy arms) is:
$$[S]_{3\times3} = \frac{-j}{\sqrt{2L}} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$How does 3×3 reduce to 2×2?
The S-parameter equations for 3 ports are:
$$b_1 = S_{11}a_1 + S_{12}a_2 + S_{13}a_3$$ $$b_2 = S_{21}a_1 + S_{22}a_2 + S_{23}a_3$$ $$b_3 = S_{31}a_1 + S_{32}a_2 + S_{33}a_3$$When port 3 is terminated with a matched load ($Z_L = Z_0$), the load reflection coefficient $\Gamma_3 = 0$, so $a_3 = \Gamma_3 \cdot b_3 = 0$ — no wave reflects back into port 3.
With $a_3 = 0$, the third column vanishes from the first two equations:
$$b_1 = S_{11}a_1 + S_{12}a_2$$ $$b_2 = S_{21}a_1 + S_{22}a_2$$This is a 2×2 system — just the upper-left submatrix:
$$[S]_{2\times2} = \frac{-j}{\sqrt{2L}} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$What if $\Gamma_3 \neq 0$? Then $a_3 = \Gamma_3 \cdot b_3 \neq 0$, and the signal bounces back from port 3 into ports 1 and 2. The effective 2×2 parameters become:
$$S'_{ij} = S_{ij} + \frac{S_{i3}\,S_{3j}\,\Gamma_3}{1 - S_{33}\,\Gamma_3}, \quad i,j \in \{1,2\}$$With matched termination ($\Gamma_3=0$), all correction terms vanish and $S'_{ij} = S_{ij}$.
What if we terminate port 2 instead? Set $a_2=0$ and take rows/columns $\{1,3\}$:
$$[S']_{2\times2} = \frac{-j}{\sqrt{2L}} \begin{bmatrix} S_{11} & S_{13} \\ S_{31} & S_{33} \end{bmatrix} = \frac{-j}{\sqrt{2L}} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$Exactly the same result. Ports 2 and 3 are symmetric in the Wilkinson divider (identical arm impedance $\sqrt{2}\,Z_0$ and length $\lambda/4$), so it doesn't matter which output port you use — the 2-port S-matrix, gain, and noise figure are identical.
How $\sqrt{2}\,Z_0$ and $\lambda/4$ produce the S-matrix — step by step:
Step 1: $\lambda/4$ transformer formula
| Symbol | Name | Meaning |
|---|---|---|
| $Z_{in}$ | Input impedance | What you "see" looking into one end of the line |
| $Z_T$ | Characteristic impedance | Property of the line itself (geometry/material) |
| $Z_L$ | Load impedance | What's connected at the far end |
| $\lambda/4$ | Quarter wavelength | Length of the line = one quarter of signal wavelength |
The general input impedance of a transmission line of length $l$:
$$Z_{in} = Z_T \frac{Z_L + jZ_T\tan(\beta l)}{Z_T + jZ_L\tan(\beta l)}$$For $\lambda/4$: $\;\beta l = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{4} = \frac{\pi}{2}$, so $\tan(\beta l) \to \infty$. Both numerator and denominator are dominated by the $\tan$ terms:
$$Z_{in} = Z_T \frac{\cancel{Z_L} + jZ_T\tan(\beta l)}{\cancel{Z_T} + jZ_L\tan(\beta l)} \;\approx\; Z_T \cdot\frac{jZ_T\cancel{\tan(\beta l)}}{jZ_L\cancel{\tan(\beta l)}} = \frac{Z_T^2}{Z_L}$$A $\lambda/4$ line "flips" the impedance: big load → small input, small load → big input. $Z_T$ controls the scaling.
Step 2: $\sqrt{2}\,Z_0$ gives $S_{11}=0$ (port 1 matched)
Each arm has $Z_T = \sqrt{2}\,Z_0$ and is terminated by $Z_L = Z_0$ (at port 2 or 3):
$$Z_{in,\text{arm}} = \frac{(\sqrt{2}\,Z_0)^2}{Z_0} = \frac{2Z_0^2}{Z_0} = 2Z_0$$Port 1 sees two arms in parallel:
$$Z_{in,\text{total}} = \frac{2Z_0 \;\|\; 2Z_0}{} = \frac{2Z_0}{2} = Z_0 \;\;\checkmark$$ $$\Rightarrow\; \bbox[#ffcccc,3px]{\Gamma_{in} = \frac{Z_{in}-Z_0}{Z_{in}+Z_0} = \frac{Z_0 - Z_0}{Z_0 + Z_0} = 0 \;\;\Rightarrow\;\; \boxed{S_{11}=0}}$$Step 3: $\lambda/4$ gives the $-j$ phase in $S_{21}$
A lossless $\lambda/4$ line has transmission matrix phase $e^{-j\beta l} = e^{-j\cdot(2\pi/\lambda)\cdot(\lambda/4)} = e^{-j\pi/2} = -j$.
Step 4: Combine all factors for $S_{21}$
| Factor | Origin | Value |
|---|---|---|
| Phase | $\lambda/4$ arm length | $-j$ |
| Power split | 2 equal arms → half power each → voltage $\times\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ |
| Loss | Each arm has insertion loss $L$ → voltage $\times\frac{1}{\sqrt{L}}$ | $\frac{1}{\sqrt{L}}$ |
Step 5: By symmetry, $S_{22}=0$ (port 2 also matched due to the $2Z_0$ isolation resistor), $S_{12}=S_{21}$ (reciprocal network).
So $\sqrt{2}\,Z_0$ and $\lambda/4$ are fully consumed into the S-matrix — they don't appear in later formulas because their job is done: they produced $S_{11}=S_{22}=0$ and $S_{21}=-j/\sqrt{2L}$.
Interactive: see how $Z_T$ controls the match
$\sqrt{2}\,Z_0 = $ 70.7 Ω
What does $\Gamma_s = 0$ mean?
$$\Gamma_s = \frac{Z_s - Z_0}{Z_s + Z_0} = 0 \;\;\Longleftrightarrow\;\; Z_s = Z_0$$The source impedance equals the characteristic impedance — no mismatch, no reflection at the input port. All available power from the source enters the network.
Matched source: $Z_s = Z_0$ means all power flows into the network ($a_1$), with zero reflected wave ($b_1 = \Gamma_s \cdot a_1 = 0$).
How to achieve $Z_s = Z_0$ in practice? (click to expand)
| Method | How it works |
|---|---|
| Already 50Ω | Most RF signal generators & amplifier outputs are designed with $Z_{out}=50\,\Omega$ |
| λ/4 transformer | Insert a quarter-wave line with $Z_T = \sqrt{Z_s \cdot Z_0}$ between source and network |
| L / π / T network | Lumped LC elements to transform impedance at a specific frequency |
| Stub matching | Shunt/series transmission line stubs designed on a Smith chart |
| Attenuator pad | Resistive network — forces both ports toward $Z_0$ at the cost of signal loss |
Interactive: impedance matching explorer
Why $G_A(\Gamma_s=0) = |S_{21}|^2$?
The general available gain formula is:
$$G_A = \frac{|S_{21}|^2\,(1-|\Gamma_s|^2)}{|1 - S_{11}\Gamma_s|^2\,(1 - |\Gamma_{out}|^2)}$$Set $\Gamma_s = 0$ (source matched to $Z_0$), each factor simplifies:
| Term | General | $\Gamma_s=0$ |
|---|---|---|
| Numerator factor | $1-|\Gamma_s|^2$ | $1$ |
| Input mismatch | $|1-S_{11}\Gamma_s|^2$ | $1$ |
| $\Gamma_{out}$ | $S_{22}+\frac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s}$ | $S_{22}$ |
So $G_A = \dfrac{|S_{21}|^2}{1-|S_{22}|^2}$. This equals $|S_{21}|^2$ only when $S_{22}=0$ (output port matched).
Here, the terminated Wilkinson divider has $S_{11}=S_{22}=0$ (matched at both ports), so all reflection terms vanish and the gain is purely the forward transmission $|S_{21}|^2 = \frac{1}{2L}$.
$\therefore$ When $T = T_0$, $F = 2L$ $\;\Leftrightarrow\;$ $F(\text{dB}) = 3\text{ dB} + \text{Loss of the line in dB from port 1 to 2} = \text{Insertion Loss in dB from port 1 to 2}.$
Using the concept of thermodynamic equilibrium:
$$N_o = k\,T\,B = G(N_i + N_{added}) = \frac{1}{2L}(k\,T\,B + N_{added})$$ $$\Rightarrow\; N_{added} = k\,T\,B\,(2L-1)$$ $$\Rightarrow\; T_e = \frac{N_{added}}{k\,B} = (2L-1)\,T$$ $$\Rightarrow\; F = 1 + \frac{N_{added}}{N_i(T_0)} = 1 + \frac{k\,T\,B\,(2L-1)}{k\,T_0\,B} = 1 + (2L-1)\frac{T}{T_0}$$Problem: A Wilkinson divider has quarter-wave arms with 0.5 dB loss each, at $T = 290$ K. Find the noise figure from port 1 to port 2.
Click to reveal solution
$L = 10^{0.05} = 1.122$
$F = 2L = 2\times1.122 = 2.244 = 3.51$ dB
Equivalently: $F\text{(dB)} = 3 + 0.5 = 3.5$ dB (the 3 dB is from the inherent power split).
2.2.7 Antenna Noise Temperature and Friis Transmission
$P_t$: transmit power, $P_r$: received power
$G_t$: transmit antenna gain, $G_r$: receive antenna gain
$R$: distance between the transmit and receive antenna
The power density radiated by an isotropic transmit antenna ($D = 1 = 0$ dB) at a distance $R$ is given by:
$$S_{avg} = \frac{P_t}{4\pi R^2} \;\;(\text{W/m}^2)$$The power density radiated by an arbitrary transmit antenna with an antenna gain of $G_t$ at a distance $R$ is:
$$S_{avg} = \frac{G_t P_t}{4\pi R^2} = \frac{EIRP}{4\pi R^2} \;\;(\text{W/m}^2)$$where $EIRP = G_t P_t$, representing "effective isotropic radiated power".
The power received by a receive antenna with an effective aperture area is:
$$P_r = S_{avg}\cdot A_e = \frac{G_t P_t}{4\pi R^2}\,A_e \;\;(\text{W})$$For a receive antenna gain:
$$G_r = \frac{4\pi}{\lambda^2}\,A_e$$The Friis equation can be derived from the above two expressions as:
$$P_r = \frac{G_t\,G_r\,\lambda^2}{(4\pi R)^2}\,P_t \;\;(\text{W}) \;\;\Rightarrow\;\; P_r \propto f^{-2},\;\; P_r \propto R^{-2}$$
The antenna noise temperature $T_a$ is the equivalent temperature of a matched resistor that would deliver the same noise power $N_i = kT_aB$ to the receiver as the real antenna does. It is not the physical temperature of the antenna metal — it is a weighted blend of two distinct noise sources.（中文：天線雜訊溫度 $T_a$ 是一個等效溫度，代表一個匹配電阻在此溫度下會輸出與實際天線相同的雜訊功率 $N_i = kT_aB$ 給接收機。它不是天線金屬的物理溫度，而是兩個不同雜訊來源的加權混合：一是天線指向方向的背景亮度溫度 $T_b$（透過效率 $\eta_{rad}$ 進入），二是天線本體因損耗而自發輻射的物理溫度 $T_p$（以 $1-\eta_{rad}$ 的比例貢獻）。）
More details — symbols, derivation, limiting cases, pitfalls, rigorous definition
Symbol meanings:
- $T_b$ — brightness (background) temperature of the scene the antenna is pointing at. This is the noise picked up through the main beam: cold sky ($\sim$3–50 K), ground ($\sim$290 K), the Sun ($\sim$10$^4$–10$^6$ K), atmospheric absorption, galactic background, etc.
- $T_p$ — physical temperature of the lossy parts of the antenna structure (radome, conductors, ohmic dissipation). For a ground-based antenna this is typically $\approx 290$ K.
- $\eta_{rad}$ — radiation efficiency, $0 \le \eta_{rad} \le 1$. It is the fraction of input power that is actually radiated; the remainder $(1-\eta_{rad})$ is dissipated as heat in the antenna.
Physical reasoning (reciprocity): A lossy antenna acts like a two-port mixer of noise sources. By reciprocity, only the fraction $\eta_{rad}$ of the incoming sky brightness reaches the terminals — the rest is absorbed by ohmic loss. Simultaneously, the lossy material itself emits thermal noise at its physical temperature $T_p$, and the fraction it contributes to the terminals is exactly $(1-\eta_{rad})$ (the same factor by which it absorbs). Adding these two independent contributions:
$$\underbrace{\eta_{rad}\,T_b}_{\text{sky seen through losses}} \;+\; \underbrace{(1-\eta_{rad})\,T_p}_{\text{self-emission of lossy parts}} \;=\; T_a$$Limiting cases:
- $\eta_{rad} = 1$ (lossless antenna): $T_a = T_b$. The terminals see the sky directly.
- $\eta_{rad} = 0$ (totally lossy): $T_a = T_p$. The antenna behaves like a matched resistor at room temperature — no sky information reaches the receiver.
- Cold sky ($T_b \ll T_p$): even small ohmic losses raise $T_a$ dramatically. Example: $T_b = 30$ K, $T_p = 290$ K, $\eta_{rad} = 0.95 \Rightarrow T_a = 28.5 + 14.5 = 43$ K. A 5% loss has nearly doubled the floor a hypothetical $T_b = 0$ K sky would have set.
Why it matters for receivers: In Friis-equation analysis the input noise is $N_i = kT_aB$, and $T_a$ is plugged directly into the system noise budget $T_{SYS} = T_a + T_e$. For satellite/radio-astronomy links pointed at cold sky, minimizing $(1-\eta_{rad})T_p$ — by using low-loss feed lines, cooled feeds, or higher $\eta_{rad}$ — is often more impactful than reducing the receiver $T_e$ itself.
Common pitfalls:
- Do not confuse $T_a$ with the antenna's metal temperature; they coincide only when $\eta_{rad} \to 0$.
- Do not use the standard noise-figure cascade with $T_0 = 290$ K when $T_a \neq T_0$ — the correct form is $N_o = Gk(T_a + T_e)B$.
- $T_b$ is frequency- and pointing-direction-dependent; the same antenna has very different $T_a$ when pointed at the zenith vs. the horizon.
Reference (rigorous definition of $T_b$): the brightness temperature seen by the antenna is the directivity-weighted average of the surrounding sky brightness $T_B(\theta,\phi)$ over the upper hemisphere:
$T_b = \dfrac{\displaystyle\int_{\phi=0}^{2\pi}\!\!\int_{\theta=0}^{\pi/2} T_B(\theta,\phi)\,D(\theta,\phi)\,\sin\theta\,d\theta\,d\phi}{\displaystyle\int_{\phi=0}^{2\pi}\!\!\int_{\theta=0}^{\pi/2} D(\theta,\phi)\,\sin\theta\,d\theta\,d\phi}$
where $T_B(\theta,\phi)$ is the angular distribution of the background brightness temperature and $D(\theta,\phi)$ is the antenna's directivity pattern. The simple scalar $T_b$ used in $T_a = \eta_{rad}T_b + (1-\eta_{rad})T_p$ is exactly this beam-weighted integral.
Problem: A satellite link at 4 GHz has $P_t = 10$ W, $G_t = 30$ dB, $G_r = 40$ dB, $R = 36{,}000$ km. The receive antenna has $\eta_{rad} = 0.95$, $T_b = 30$ K, $T_p = 290$ K. Find (a) EIRP, (b) received power, (c) antenna noise temperature.
Click to reveal solution
(a) EIRP $= G_t P_t = 10^3 \times 10 = 10{,}000$ W $= 40$ dBW $= 70$ dBm
(b) $\lambda = 3\times10^8/(4\times10^9) = 0.075$ m
$P_r = 10^3\times10^4\times10\times\left(\frac{0.075}{4\pi\times3.6\times10^7}\right)^2$
$= 10^8 \times \left(\frac{0.075}{4.524\times10^8}\right)^2 = 10^8 \times (1.658\times10^{-10})^2 = 10^8\times2.75\times10^{-20}$
$= 2.75\times10^{-12}$ W $= -85.6$ dBm
(c) $T_a = 0.95\times30 + 0.05\times290 = 28.5 + 14.5 = 43$ K
2.2.8 G/T Ratio
G/T ratio is a useful figure of merit for a receive antenna to evaluate the resulting SNR at the input to a receiver.
where $G_r$ is the gain of the antenna, and $T_a$ is the antenna noise temperature.
To see why $G_r/T_a$ governs the input SNR, combine the Friis transmission equation with the input noise:
For a given link (fixed EIRP, frequency, distance, bandwidth), every other factor is constant — only $G_r/T_a$ depends on the receive antenna. A higher G/T means better sensitivity. When the receiver noise is included:
What is dBi?
dBi = decibels relative to an isotropic radiator. An isotropic antenna is a hypothetical reference that radiates equally in all directions (linear gain $= 1$, i.e. $0$ dBi). Real antennas concentrate energy in some direction, so their gain expressed in dBi tells you how much stronger their peak radiation is compared to isotropic.
$$G(\text{dBi}) = 10\log_{10}\!\left(\frac{G_{\text{antenna}}}{G_{\text{isotropic}}}\right) = 10\log_{10}(G)$$Typical values:
- $0$ dBi — isotropic (reference, theoretical only)
- $2.15$ dBi — half-wave dipole
- $8$–$12$ dBi — Yagi or patch array
- $25$ dBi — small parabolic dish (e.g. point-to-point microwave link)
- $33$–$45$ dBi — DBS / VSAT terminal
- $60$+ dBi — deep-space ground stations
vs. dBd: Some antennas (e.g. broadcast) are quoted in dBd, referenced to a half-wave dipole. Conversion: $G(\text{dBi}) = G(\text{dBd}) + 2.15$.
Why dBi here? The Friis equation uses the absolute gain $G_r$ over an isotropic reference, so $G_r(\text{dBi})$ plugs directly into the link budget. In the G/T formula above, since $T$ is on a log scale ($10\log T$ in dBK), $G_r$ must also be on a log scale — and dBi is the natural choice.
Problem. A DBS system operates at 12.2–12.7 GHz (use $f = 12.45$ GHz, $\lambda = 3\times10^8/12.45\times10^9 = 0.0241$ m). It has:
- Transmit carrier power $P_t = 120$ W, transmit antenna gain $G_t = 34$ dB, IF bandwidth $B = 20$ MHz
- Worst-case slant range (30° elevation) from geosynchronous orbit to ground: $R = 39{,}000$ km
- 18-inch receiving dish: $G_r = 33.5$ dB, average background brightness $T_b = 50$ K
- LNB (low-noise block downconverter) noise figure $F_{LNB} = 1.1$ dB
Find: (a) EIRP of transmitter; (b) $G/T$ of the receive antenna w/o and w/ the LNB; (c) CNR at the input and output of the LNB. Then explain why $\text{CNR}_{in} - \text{CNR}_{out} \neq F$.
Click to reveal solution
Linear conversions: $34$ dB $= 2512$, $33.5$ dB $= 2239$, $1.1$ dB $= 1.29$.
(a) EIRP. By definition $\text{EIRP} = G_t P_t$:
$$\text{EIRP} = 2512 \times 120 = 3.01\times10^{5}\text{ W} = 54.8\text{ dBW} \;(=84.8\text{ dBm})$$(b) G/T of receive antenna. For a very-high-gain dish pointed at the cold sky, $T_a \approx T_b$ (sidelobe ground-pickup is negligible).
Without LNB (only antenna noise):
$$G/T \;=\; 10\log\frac{G_r}{T_a} \;=\; 10\log\frac{2239}{50} \;=\; 16.5\text{ dB/K}$$With LNB. Convert NF to noise temperature using Eq. (2): $T_e = (F-1)T_0$:
$$T_e = (1.29 - 1)\times 290 = 0.29\times 290 \approx 84\text{ K}$$The system noise temperature seen at the LNB input is $T_a + T_e = 50 + 84 = 134$ K, so:
$$G/T \;=\; 10\log\frac{G_r}{T_a + T_e} \;=\; 10\log\frac{2239}{134} \;=\; 12.2\text{ dB/K}$$Adding the LNB drops $G/T$ by $\Delta = 16.5 - 12.2 = 4.3$ dB — this is the receiver's contribution to the figure of merit.
(c) CNR at LNB input and output. Use the link equation (Friis transmission combined with $N = kT_{sys}B$):
$$\text{CNR} \;=\; \frac{S_i}{N_i} \;=\; \frac{G_r}{T_{sys}} \cdot \frac{\text{EIRP}\;\lambda^2}{kB\,(4\pi R)^2}$$Input of LNB ($T_{sys} = T_a$, no receiver noise yet):
$$\text{CNR}_{in} = \frac{2239}{50}\cdot\frac{3.01\times10^5\,(0.0241)^2}{1.38\times10^{-23}\times 20\times10^6\times(4\pi\times3.9\times10^7)^2} = 118.1 = 20.7\text{ dB}$$Output of LNB — gain cancels, but the noise floor is now $kB(T_a+T_e)$:
$$\text{CNR}_{out} = \frac{G_{LNB}\,S_i}{G_{LNB}\,kB(T_a+T_e)} = \frac{G_r}{T_a+T_e}\cdot\frac{\text{EIRP}\;\lambda^2}{kB(4\pi R)^2} = 44.1 = 16.4\text{ dB}$$Why is $\text{CNR}_{in} - \text{CNR}_{out} = 4.3$ dB, not $F = 1.1$ dB?
The textbook identity "SNR drop = noise figure" assumes $N_i = kT_0B$ (i.e. source at the standard 290 K) — that is the very definition of $F$. Here the source noise is the cold sky $T_a = 50$ K, not 290 K, so $F$ no longer equals the SNR-degradation:
$$\text{CNR drop} = 10\log\frac{T_a + T_e}{T_a} = 10\log\frac{134}{50} = 4.3\text{ dB}$$ $$\text{vs. } F\text{ (defined at } T_0 = 290\text{ K)} = 10\log\frac{T_0 + T_e}{T_0} = 10\log\frac{374}{290} = 1.1\text{ dB}$$When $T_a < T_0$, the receiver's $T_e$ looks proportionally larger compared to the cold input noise, so the SNR penalty is bigger than $F$ would suggest. This is exactly why satellite/radio-astronomy receivers use $G/T$ and noise temperature, not $F$, as the figure of merit.
For a worked example where $T_a \neq T_0$ is handled correctly through a full receiver cascade (LNA → BPF → Mixer), see Example — Analysis of Wireless Receiver. It uses $T_a = 150$ K (not 290 K) and computes the output noise via $N_o = G\,kB(T_a + T_e)$ instead of the (incorrect-here) $N_o = G\,kT_0BF$.
What is an LNB?
LNB = Low-Noise Block downconverter — the small device mounted at the focal point of a satellite dish (the part the coax cable plugs into). It bundles three RF stages in one weather-sealed package:
| Stage | Purpose |
|---|---|
| LNA (low-noise amp) | Boost the very weak microwave signal ($\sim -100$ dBm) before any cable loss adds noise — this is what sets the system noise figure (Friis: first stage dominates). |
| BPF (band-pass filter) | Reject out-of-band signals and image frequencies before mixing. |
| Mixer + LO | Down-convert the satellite RF (Ku-band 12.2–12.7 GHz) to a lower IF (typically 950–2150 MHz) so it can travel down cheap coax to the indoor receiver without huge loss. |
Why "Block"? It down-converts an entire block of frequencies (the whole 500 MHz Ku slice) at once, not channel-by-channel — channel selection happens later in the indoor set-top box.
Why mounted at the dish? Two reasons, both about noise:
- At 12 GHz, a 30-m coax run would have ~30 dB of loss. Putting the LNA right at the feed means the signal is amplified before the cable loss, so the cable's loss is divided by $G_{LNA}$ in the Friis cascade — its noise contribution becomes negligible.
- Down-converting to ~1 GHz IF lets cheap RG-6 coax carry it tens of meters with manageable loss.
In this example: the LNB has $F = 1.1$ dB (very low — typical for a modern Ku-band LNB, giving $T_e \approx 84$ K). Because it is the first amplifier and has high gain, by Friis it is the dominant noise contributor — everything after it is suppressed by $G_{LNB}$.
CNR is the ratio of the received carrier (signal) power to the noise power at the receiver input (or demodulator input). It is the link-budget version of "SNR" used at RF/IF — i.e. before demodulation, while the signal is still a modulated carrier.
$$\text{CNR} \;=\; \frac{C}{N} \;=\; \frac{S_i}{k\,T_{sys}\,B}$$In dB form, with all link components added:
$$\text{CNR (dB)} \;=\; \underbrace{\text{EIRP}}_{\text{transmitter}} \;+\; \underbrace{G/T}_{\text{receiver}} \;-\; \underbrace{\text{FSPL}}_{\text{free-space loss}} \;-\; 10\log_{10}(kB)$$More details — symbol meanings, CNR vs SNR vs $E_b/N_0$, why it matters
Symbol meanings:
- $C$ (or $S_i$) — received carrier power at the antenna terminals (W or dBW).
- $N = kT_{sys}B$ — thermal noise floor; $k = 1.38\times10^{-23}$ J/K, $T_{sys} = T_a + T_e$, $B$ = receiver noise bandwidth.
- $\text{FSPL} = 20\log_{10}(4\pi R/\lambda)$ — free-space path loss in dB.
- $G/T$ — receive-antenna figure of merit (dB/K), couples gain and noise into one number.
CNR vs. SNR vs. $E_b/N_0$:
- CNR — measured at RF/IF, before demodulation (modulated carrier in noise).
- SNR — usually post-demodulation, baseband signal-to-noise ratio.
- $E_b/N_0$ — energy-per-bit over noise spectral density; the modulation-independent figure used to predict BER. Conversion: $\text{CNR} = (E_b/N_0)\cdot(R_b/B)$.
Why we care: CNR sets the demodulator's BER. For a fixed link (EIRP, frequency, distance, bandwidth), CNR depends only on the receive antenna's $G/T$ — which is exactly why G/T is the antenna figure of merit. A 3 dB increase in G/T directly raises CNR by 3 dB, allowing higher-order modulation, lower BER, or smaller transmit power.
Problem: A ground station has $G_r = 45$ dBi, $T_a = 25$ K, and a receiver with $T_e = 50$ K. Find G/T. If EIRP $= 60$ dBW, $f = 12$ GHz, $R = 40{,}000$ km, $B = 36$ MHz, find the CNR.
Click to reveal solution
$G/T = 45 - 10\log(25+50) = 45 - 18.75 = 26.25$ dB/K
$\lambda = 0.025$ m. Free-space loss $= 20\log(4\pi R/\lambda) = 20\log(4\pi\times4\times10^7/0.025) = 20\log(2.01\times10^{10}) = 206.1$ dB
$\text{CNR} = \text{EIRP} + G/T - \text{FSPL} - 10\log(kB)$
$10\log(kB) = 10\log(1.38\times10^{-23}\times36\times10^6) = 10\log(4.97\times10^{-16}) = -153.0$ dBW
$\text{CNR} = 60 + 26.25 - 206.1 - (-153.0) = 60 + 26.25 - 206.1 + 153.0 = 33.15$ dB
2.2.9 Full Receiver Analysis ($T_a \neq T_0$)
Correct approach: $N_{added} = (F-1)kT_0B$, then $N_o = G[kT_aB + N_{added}] = GkB[T_a + T_e]$
Problem: In the previous example for an IS-54 cellular system using QPSK modulation, the receive antenna has a noise temperature of 900 K and the receiver has a noise figure of 8 dB. Find the minimum detectable signal level (sensitivity).
Given $(S_o/N_o)_{min} = 11.7$ dB $= 14.8$, $\;\;F = 8$ dB $= 6.3$
Click to reveal solution
Step 1 — Identify the key formula. The minimum detectable signal requires that the output SNR meets the modulation threshold (the bracket $T_a + (F-1)T_0 = T_a + T_e$ uses Eq. (2)):
$$S_{i,min} = kB\bigl[T_a + (F-1)T_0\bigr]\left(\frac{S_o}{N_o}\right)_{min}$$Step 2 — Convert noise figure to linear.
$F = 8$ dB $= 10^{0.8} = 6.31$, so $F - 1 = 5.31$
Step 3 — Compute the effective noise temperature seen at the input.
$T_a + (F-1)T_0 = 900 + 5.31\times 290 = 900 + 1540 = 2440$ K
Step 4 — Noise power over the bandwidth.
$kB \cdot 2440 = 1.38\times10^{-23}\times 30\times10^3\times 2440 = 1.010\times10^{-15}$ W
Step 5 — Required output SNR.
$(S_o/N_o)_{min} = 11.7$ dB $= 10^{1.17} = 14.8$
Step 6 — Minimum detectable signal.
$S_{i,min} = 1.010\times10^{-15}\times 14.8 = 1.49\times10^{-14}$ W $= \boxed{-108.3\text{ dBm}}$
Where do $900$, $5.31$, and $290$ come from?
The bracket is the system noise temperature at the receiver input, $T_{sys} = T_a + (F-1)T_0$. Each number maps to one physical piece:
- $900$ K $= T_a$, the antenna noise temperature — noise picked up by the antenna from sky / ground / surroundings.
- $5.31$ $= F - 1$. The given noise figure is $F = 8$ dB, so linear $F = 10^{0.8} = 6.31$, hence $F - 1 = 5.31$.
- $290$ K $= T_0$, the IEEE reference temperature at which $F$ is defined.
The product $(F-1)T_0 = 5.31 \times 290 \approx 1540$ K is the equivalent noise temperature of the receiver, $T_e$. This conversion exists because $F$ is defined at $T_0 = 290$ K, but to add receiver noise to antenna noise on equal footing you need it expressed as a temperature:
$$T_e = (F-1)T_0$$Both $T_a$ and $T_e$ appear at the same input node, so they simply add:
$$T_{sys} = T_a + T_e = 900 + 1540 = 2440\text{ K}$$That total is what goes into $N = kB\,T_{sys}$ for the input-referred noise power in Step 4.
Why is QPSK essentially two orthogonal BPSK channels?
Setup. A QPSK symbol carries 2 bits. Split them into an in-phase bit $b_I\in\{\pm 1\}$ and a quadrature bit $b_Q\in\{\pm 1\}$. The transmitted waveform is
$$s(t)=b_I\cos(\omega_c t)-b_Q\sin(\omega_c t)$$That single line is already the answer — it is a sum of two BPSK signals, one riding $\cos$, the other riding $\sin$.
Why the two streams don't interfere — orthogonality. Over one symbol period $T_s$,
$$\int_0^{T_s}\cos(\omega_c t)\sin(\omega_c t)\,dt = 0$$so $\cos$ and $\sin$ are orthogonal basis functions. At the receiver:
- multiply $s(t)$ by $\cos(\omega_c t)$ and integrate → only $b_I$ survives,
- multiply $s(t)$ by $-\sin(\omega_c t)$ and integrate → only $b_Q$ survives.
The two bits decouple cleanly; each branch is a textbook BPSK detector.
Geometric view (constellation). Treat $\cos$ and $\sin$ as orthogonal axes. The 4 QPSK symbols sit at $(\pm 1,\pm 1)$ — two BPSK constellations on perpendicular axes. Nearest-neighbor distance along each axis equals the BPSK distance, so the per-axis decision is identical to BPSK.
Consequences that follow for free:
QPSK doubles spectral efficiency for free — the I and Q channels share the same RF bandwidth with no crosstalk, so orthogonality buys you the second BPSK channel inside the same spectrum.
Caveat — coherent demodulation required. The receiver must lock to the carrier phase so its $\cos$ and $\sin$ match the transmitter's. A 90° phase error swaps $I$ and $Q$; a 45° error mixes them and destroys the orthogonality. That is why QPSK receivers need a Costas loop (or pilot / preamble) for carrier recovery.
Explanation of $\;S_{i,min} = kB\bigl[T_a + (F-1)T_0\bigr]\!\left(\tfrac{S_o}{N_o}\right)_{\!min}$
Smallest input signal $S_{i,min}$ that still meets the demodulator's threshold $(S_o/N_o)_{min}$.
Build-up from $\text{SNR}_{out} = G\,S_i / N_o$:
- System input noise at the antenna terminal: $$N_{sys} = kB\,T_{sys}, \qquad T_{sys} = T_a + T_e$$ $T_a$ — antenna noise temperature (sky / surroundings); $T_e$ — receiver equivalent noise temperature.
- Convert $T_e$ to noise figure via Eq. (2): $T_e = (F-1)T_0$, so $T_{sys} = T_a + (F-1)T_0$. The bracket splits the noise into antenna contribution $T_a$ and receiver contribution $(F-1)T_0$.
- Output SNR through gain $G$: $$\frac{S_o}{N_o} = \frac{G\,S_i}{G\,kB[T_a+(F-1)T_0]} = \frac{S_i}{kB[T_a+(F-1)T_0]}$$ Gain cancels — sensitivity does not depend on $G$, only on the noise temperature looking into the receiver.
- Solve for the minimum $S_i$ that meets the threshold $(S_o/N_o)_{min}$: $$S_{i,min} = kB[T_a+(F-1)T_0]\,(S_o/N_o)_{min}$$
What each factor controls:
| Term | Meaning | How to reduce it |
|---|---|---|
| $kB$ | Thermal-noise floor over bandwidth $B$ | narrower $B$ (matched filter) |
| $T_a$ | Antenna / sky noise | point at cold sky, low side-lobes |
| $(F-1)T_0$ | Receiver-added noise | low-NF LNA first (Friis) |
| $(S_o/N_o)_{min}$ | Modulation requirement | better coding, lower-order modulation |
Intuition: sensitivity = (total input-referred noise power) × (required SNR margin). Halve the noise temperature → 3 dB better sensitivity. Halve the bandwidth → 3 dB better. Drop the required SNR by 3 dB (better FEC) → 3 dB better.
Where does $B = 30$ kHz come from?
The bandwidth $B = 30\times10^3$ Hz in Step 4 is not introduced in this example — it is inherited from the prior IS-54 QPSK lecture example, where $R_b = 48.6$ kbps and $B = 30$ kHz are stated up-front. The current problem opens with "In the previous example for an IS-54 cellular system using QPSK modulation...", so $B$ carries over.
Why specifically 30 kHz? IS-54 (a.k.a. D-AMPS / TDMA-54) is the early-1990s North-American digital cellular standard. It re-uses the analog AMPS 30 kHz channel grid mandated by the FCC — each RF carrier occupies a 30 kHz slot. Three users share that carrier via TDMA, but the per-carrier RF bandwidth stays fixed at 30 kHz.
Putting numbers into $kB\,T_{sys}$:
- $k = 1.38\times10^{-23}$ J/K (Boltzmann's constant)
- $B = 30\times10^3$ Hz (IS-54 channel bandwidth)
- $T_{sys} = T_a + (F-1)T_0 = 900 + 5.31\times290 = 2440$ K
For other standards just swap $B$: GSM uses 200 kHz, LTE uses 1.4–20 MHz, Wi-Fi 802.11 uses 20/40/80 MHz — and the noise floor $kBT_{sys}$ scales linearly with whichever channel bandwidth the radio actually filters.
Problem: In a simplified wireless receiver, the input noise power from a feeding antenna is $N_i = k\,T_a\,B$, where $T_a = 150$ K. Assume the system is at temperature $T_0 = 290$ K, with a characteristic impedance of 50 $\Omega$, and an IF bandwidth of $B = 10$ MHz.
$L$ is insertion loss (about transmission line)
→ For the full derivation of $F = L$ on a lossy line, see §2.2.4 Noise Figure of a Lossy Line.
$L$ = Insertion loss (in dB) — the power lost as the signal passes through a passive stage:
$$L_{\text{dB}} = 10\log_{10}\!\left(\frac{P_{\text{in}}}{P_{\text{out}}}\right) \;\;\Longleftrightarrow\;\; G = \frac{1}{L} = 10^{-L_{\text{dB}}/10}$$For passive components (filters, attenuators, cables, passive mixers), gain $G < 1$, so it's more natural to quote the loss $L > 1$ instead. The two are reciprocals.
In this diagram:
| Stage | Loss / Gain |
|---|---|
| BPF: $L = 1$ dB | $G_2 = 10^{-0.1} = 0.79$ — loses 21% of signal power |
| Mixer: $L = 3$ dB | $G_3 = 10^{-0.3} = 0.50$ — loses 50% (typical of a passive diode mixer) |
Why both $L$ and $F$ for the mixer? $L$ tells you how much signal is lost; $F$ tells you how much extra noise is added on top of the thermal floor. They are independent specs — a mixer can have low loss but high noise figure if it has internal noise sources (LO leakage, port mismatch, etc.).
Key relation for passive lossy stages at room temperature ($T = T_0$):
$$F = L \quad\text{(in dB: NF = insertion loss)}$$That's why for the BPF only $L$ is given — its noise figure is implicitly $F = L = 1$ dB. The mixer breaks this rule because it's an active/non-equilibrium component (LO drives it), so its $F = 4$ dB exceeds its $L = 3$ dB.
For the Friis cascade above, each $L$ is converted to gain: $G_2 = 1/L_2 = 0.79$, $G_3 = 1/L_3 = 0.50$ — that's where the linear values in part (a) come from.
(a) Find the cascaded parameters $F$, $T_e$, and $G_{total}$.
(b) Find the output noise power and two-sided power spectral density.
(c) Find the minimum detectable power and voltage for the input signal under a minimum detectable SNR of 20 dB at the output of the receiver.
Click to reveal solution
(a) Cascaded parameters $F$, $T_e$, $G_{total}$
Convert each stage to linear:
LNA: $F_1 = 2$ dB $= 1.58$, $G_1 = 10$ dB $= 10$
BPF: $F_2 = L = 1$ dB $= 1.26$, $G_2 = -1$ dB $= 0.79$
Mixer: $F_3 = 4$ dB $= 2.51$, $G_3 = -3$ dB $= 0.5$
Cascade noise figure (Friis formula):
$$F = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2} = 1.58 + \frac{0.26}{10} + \frac{1.51}{10\times 0.79} = 1.58 + 0.026 + 0.191 = \boxed{1.80\;(2.55\text{ dB})}$$From Eq. (1): $T_e = (F-1)\,T_0 = 0.80\times 290 = \boxed{232\text{ K}}$
$G_{total} = G_1 \cdot G_2 \cdot G_3 = 10\times 0.79\times 0.5 = \boxed{3.95\;(5.97\text{ dB})}$
(b) Output noise power and two-sided PSD
Input noise: $N_i = k\,T_a\,B = 1.38\times10^{-23}\times 150\times 10^7 = 2.07\times10^{-14}$ W (where $B = 10\text{ MHz} = 10\times 10^6 = 10^7$ Hz, from the problem statement)
Output noise includes amplified input noise plus receiver-added noise:
$$N_o = G_{total}\,k\,B\,(T_a + T_e) = 3.95\times 1.38\times10^{-23}\times 10^7\times(150+232)$$$= 3.95\times 1.38\times10^{-16}\times 382 = \boxed{2.08\times10^{-13}\text{ W}\;(-96.8\text{ dBm})}$
Two-sided PSD (noise is symmetric about $f = 0$, so divide by $2B$):
$$S_{no} = \frac{N_o}{2B} = \frac{2.08\times10^{-13}}{2\times10^7} = \boxed{1.04\times10^{-20}\text{ W/Hz}}$$(c) Minimum detectable signal for $\text{SNR}_{out} = 20$ dB
$\text{SNR}_{out,min} = 20$ dB $= 100$ (linear)
At the output: $\text{SNR}_{out} = G_{total}\,S_{i,min} / N_o$, so:
$$S_{i,min} = \frac{\text{SNR}_{out,min}\cdot N_o}{G_{total}} = \frac{100\times 2.08\times10^{-13}}{3.95} = \boxed{5.27\times10^{-12}\text{ W}\;(-82.8\text{ dBm})}$$Minimum detectable voltage (across $Z_0 = 50\;\Omega$):
$$V_{i,min} = \sqrt{S_{i,min}\cdot Z_0} = \sqrt{5.27\times10^{-12}\times 50} = \sqrt{2.63\times10^{-10}} = \boxed{16.2\;\mu\text{V}}$$Problem: For the following wireless receiver, if the received power at the antenna terminal is $S_i = -80$ dBm, calculate the input and output SNR. This is the worked example from the Chap 2 slides; the condensed slide derivation and the full step-by-step solution are merged below.
The receiver chain includes: Background noise ($T_b$) → Antenna ($G_A$, $\eta_{rad}$, $T_P$) → Transmission line ($L_T$, $T_P$) → RF Amp. ($G_{RF}$, $T_{RF}$) → Mixer with LO ($L_M$, $T_M$) → IF Amp. ($G_{IF}$, $T_{IF}$) → Output ($N_o$, $S_o$). The input signal and noise at the antenna terminal are $N_i$, $S_i$.
Given: $f = 4$ GHz, $B = 1$ MHz, $\eta_{rad}=0.9$, $T_p=300$ K, $T_b=200$ K, $L_T=1.5$ dB, $G_{RF}=20$ dB, $F_{RF}=3$ dB, $L_M=6$ dB, $F_M=7$ dB, $G_{IF}=30$ dB, $F_{IF}=1.1$ dB.
Click to reveal solution
Step 1 — Antenna noise temperature.
$T_A = \eta_{rad}\,T_b + (1-\eta_{rad})\,T_P = 0.9\times 200 + 0.1\times 300 = 180 + 30 = 210$ K
Step 2 — Transmission line noise contribution.
$L_T = 1.5$ dB $= 10^{0.15} = 1.413$
$T_{TL} = (L_T - 1)\,T_P = (1.413 - 1)\times 300 = 0.413\times 300 = 123$ K
Step 3 — Receiver cascade noise temperature.
Convert component values to linear:
$G_{RF} = 10^{2.0} = 100$, $\;F_{RF} = 10^{0.3} = 2.0$, $\;L_M = 10^{0.6} = 4$, $\;F_M = 10^{0.7} = 5.01$, $\;G_{IF} = 10^{3.0} = 1000$, $\;F_{IF} = 10^{0.11} = 1.29$
Individual noise temperatures:
$T_{RF} = (F_{RF}-1)\,T_0 = (2.0-1)\times 290 = 289$ K
$T_M = (F_M - 1)\,T_0 = (5.01-1)\times 290 = 1163$ K (where $F_M = 7$ dB $\Rightarrow 10^{7/10} = 10^{0.7} = 5.01$ linear, from the given values)
$T_{IF} = (F_{IF}-1)\,T_0 = (1.29-1)\times 290 = 84$ K
Friis cascade (RF amp → Mixer → IF amp):
$\color{#b44a2f}{T_{REC} = T_{RF} + \dfrac{T_M + T_{IF}\cdot L_M}{G_{RF}} = 289 + \dfrac{1163 + 84\times 4}{100} = 289 + \dfrac{1499}{100} = 289 + 15 = 304}$ K
How does the Friis cascade reduce to this form?
The general Friis cascade for noise temperature (see eq.) is:
$$T_e = T_{e1} + \dfrac{T_{e2}}{G_1} + \dfrac{T_{e3}}{G_1\,G_2} + \cdots$$Map the three stages of this receiver onto the formula:
| Stage | Role | Gain $G_i$ | Noise temp $T_{ei}$ |
|---|---|---|---|
| 1 | RF amp | $G_1 = G_{RF}$ | $T_{e1} = T_{RF}$ |
| 2 | Mixer (passive lossy) | $G_2 = 1/L_M$ | $T_{e2} = T_M$ |
| 3 | IF amp | $G_3 = G_{IF}$ | $T_{e3} = T_{IF}$ |
Substitute:
$$T_{REC} = T_{RF} + \dfrac{T_M}{G_{RF}} + \dfrac{T_{IF}}{G_{RF}\cdot(1/L_M)} = T_{RF} + \dfrac{T_M}{G_{RF}} + \dfrac{T_{IF}\cdot L_M}{G_{RF}}$$The mixer's loss $L_M$ pops up to the numerator because dividing by $G_2 = 1/L_M$ is the same as multiplying by $L_M$. Combine the last two terms over the common denominator $G_{RF}$:
$$T_{REC} = T_{RF} + \dfrac{T_M + T_{IF}\cdot L_M}{G_{RF}}$$Physical meaning: the IF amp's noise is referred back to the input through the RF amp's gain and through the mixer's loss — a passive lossy stage hurts everything downstream because it attenuates the signal but the noise still has to be referred back through it.
Step 4 — System noise temperature.
$T_{SYS} = T_A + T_{TL} + L_T\cdot T_{REC} = 210 + 123 + 1.413\times 304 = 210 + 123 + 429 = 762$ K
Why is $T_{REC}$ multiplied by $L_T$?
$T_{SYS}$ is being expressed referred to the antenna terminal — i.e., as if all noise sources were located at the antenna input. The transmission line sits between the antenna and the receiver, with gain $G_T = 1/L_T$ (loss $L_T \ge 1$).
Apply the Friis cascade for noise temperature with the line as stage 1 and the receiver as stage 2:
$$T_{e,\text{cascade}} = T_{TL} + \dfrac{T_{REC}}{G_T} = T_{TL} + \dfrac{T_{REC}}{1/L_T} = T_{TL} + L_T\cdot T_{REC}$$The antenna noise $T_A$ is already at the input, so it just adds:
$$\boxed{\,T_{SYS} = T_A + T_{TL} + L_T\cdot T_{REC}\,}$$Intuition: the line attenuates the signal by $1/L_T$ on its way to the receiver. So when we push the receiver's noise back through the line (toward the antenna), it has to be scaled up by $L_T$ to keep the same SNR — a small amount of noise generated past a 3 dB attenuator looks like twice that much noise at the attenuator's input.
Numerical check: $L_T = 1.5\text{ dB} = 1.413$, $T_{REC} = 304$ K $\Rightarrow L_T \cdot T_{REC} = 429$ K. The line itself contributes only 123 K, but it inflates the receiver's effective contribution from 304 K to 429 K — a 41% penalty just for sitting in front of the receiver.
Step 5 — Input SNR.
$N_i = k\,B\,T_A = 1.38\times10^{-23}\times 10^6\times 210 = 2.90\times10^{-15}$ W $= -115.4$ dBm
$\text{SNR}_{in} = S_i - N_i = -80 - (-115.4) \approx -80 - (-115) = 35$ dB
Step 6 — Output SNR.
$N_{SYS} = k\,B\,T_{SYS} = 1.38\times10^{-23}\times 10^6\times 762 = 1.05\times10^{-14}$ W $= -109.8$ dBm
$\text{SNR}_{out} = \dfrac{S_i}{k\,B\,T_{SYS}} = -80 - (-109.8) \approx -80 - (-110) = 30$ dB
Result: $\text{SNR}_{in} \approx \boxed{35\text{ dB}}$, $\;\text{SNR}_{out} \approx \boxed{30\text{ dB}}$
The SNR degrades by about 5 dB through the system. The dominant noise contributors are the receiver chain ($L_T \cdot T_{REC} = 429$ K, 56%) and the antenna ($T_A = 210$ K, 28%), with the transmission line adding $T_{TL} = 123$ K (16%).
Problem: Repeat the Minimum Detectable Signal Level example (IS-54 system, $T_a = 900$ K, $F = 8$ dB, $B = 30$ kHz) for (a) BPSK with $R_b = 48.6$ kbps, and (b) 64-QAM with $R_b = 48.6$ kbps. Find the minimum detectable input signal power for each. Assume $\color{#b44a2f}{\text{BER} = 10^{-5}}$ for both.
Pozar Table 14.1 reference values: For BER $= 10^{-5}$: BPSK requires $E_b/n_0 = 9.6$ dB; 64-QAM requires $E_b/n_0 \approx 22.5$ dB. BPSK spectral efficiency $R_b/B = 1$ bps/Hz; 64-QAM spectral efficiency $R_b/B = 6$ bps/Hz.
Click to reveal full solution
Common parameters: $F = 8$ dB $= 6.31$, $T_a = 900$ K, $B = 30$ kHz, $R_b = 48.6$ kbps.
$S_{i,min} = kB[T_a + (F-1)T_0] \cdot (S_o/N_o)_{min}$, where $(S_o/N_o)_{min} = (E_b/n_0)\cdot(R_b/B)$.
Where do $900$, $5.31$, and $290$ come from?
The bracket $T_a + (F-1)T_0$ is the system noise temperature at the receiver input, $T_{sys}$. Each number maps to one physical piece:
- $900$ K $= T_a$, antenna noise temperature — noise picked up from sky / ground / surroundings.
- $5.31$ $= F - 1$. Given $F = 8$ dB → linear $F = 10^{0.8} = 6.31$, hence $F - 1 = 5.31$.
- $290$ K $= T_0$, the IEEE reference temperature at which $F$ is defined.
The product $(F-1)T_0 = 5.31 \times 290 \approx 1540$ K is the equivalent noise temperature of the receiver, $T_e$. This conversion exists because $F$ is defined at $T_0 = 290$ K, but to add receiver noise to antenna noise on equal footing it must be expressed as a temperature:
$$T_e = (F-1)T_0$$Both $T_a$ and $T_e$ appear at the same input node, so they simply add:
$$T_{sys} = T_a + T_e = 900 + 1540 = 2440\text{ K}$$That total is what goes into $N = kB\,T_{sys}$ for the input-referred noise power — the bracket $[900 + 5.31\times 290]$ in the BPSK and 64-QAM lines below is exactly this $T_{sys}$.
(a) BPSK:
$E_b/n_0 = 9.6$ dB $= 9.12$ (linear). Spectral efficiency $R_b/B = 48.6/30 = 1.62$.
$(S_o/N_o)_{min} = 9.12 \times 1.62 = 14.8$ → $= 11.7$ dB (same as QPSK!)
$S_{i,min} = 1.38\times10^{-23}\times30\times10^3\times[900 + 5.31\times290]\times14.8$
$= 4.14\times10^{-19}\times2440\times14.8 = 1.49\times10^{-14}$ W $= \boxed{-108.3\text{ dBm}}$
Note: BPSK and QPSK have the same $E_b/n_0$ requirement and (for the same $R_b/B$ ratio) the same sensitivity. This makes sense because QPSK is essentially two orthogonal BPSK channels.
(b) 64-QAM:
$E_b/n_0 = 22.5$ dB $= 177.8$ (linear). Spectral efficiency $R_b/B = 48.6/30 = 1.62$.
$(S_o/N_o)_{min} = 177.8 \times 1.62 = 288.0$ → $= 24.6$ dB
$S_{i,min} = 1.38\times10^{-23}\times30\times10^3\times[900 + 5.31\times290]\times288.0$
$= 4.14\times10^{-19}\times2440\times288.0 = 2.91\times10^{-13}$ W $= \boxed{-95.4\text{ dBm}}$
64-QAM requires about 12.9 dB more signal power than BPSK/QPSK to achieve the same BER. This is the cost of higher spectral efficiency — 64-QAM packs 6 bits per symbol, so constellation points are much closer together and more vulnerable to noise.
Summary comparison:
| Modulation | $E_b/n_0$ (dB) | $\eta = R_b/B$ | $S_o/N_o$ (dB) | $S_{i,min}$ (dBm) |
|---|---|---|---|---|
| BPSK | 9.6 | 1.62 | 11.7 | -108.3 |
| QPSK | 9.6 | 1.62 | 11.7 | -108.3 |
| 64-QAM | 22.5 | 1.62 | 24.6 | -95.4 |
Why is QPSK essentially two orthogonal BPSK channels?
Setup. A QPSK symbol carries 2 bits. Split them into an in-phase bit $b_I\in\{\pm 1\}$ and a quadrature bit $b_Q\in\{\pm 1\}$. The transmitted waveform is
$$s(t)=b_I\cos(\omega_c t)-b_Q\sin(\omega_c t)$$That single line is already the answer — it is a sum of two BPSK signals, one riding $\cos$, the other riding $\sin$.
Why the two streams don't interfere — orthogonality. Over one symbol period $T_s$,
$$\int_0^{T_s}\cos(\omega_c t)\sin(\omega_c t)\,dt = 0$$so $\cos$ and $\sin$ are orthogonal basis functions. At the receiver:
- multiply $s(t)$ by $\cos(\omega_c t)$ and integrate → only $b_I$ survives,
- multiply $s(t)$ by $-\sin(\omega_c t)$ and integrate → only $b_Q$ survives.
The two bits decouple cleanly; each branch is a textbook BPSK detector.
Geometric view (constellation). Treat $\cos$ and $\sin$ as orthogonal axes. The 4 QPSK symbols sit at $(\pm 1,\pm 1)$ — two BPSK constellations on perpendicular axes. Nearest-neighbor distance along each axis equals the BPSK distance, so the per-axis decision is identical to BPSK.
Consequences that follow for free:
QPSK doubles spectral efficiency for free — the I and Q channels share the same RF bandwidth with no crosstalk, so orthogonality buys you the second BPSK channel inside the same spectrum.
Caveat — coherent demodulation required. The receiver must lock to the carrier phase so its $\cos$ and $\sin$ match the transmitter's. A 90° phase error swaps $I$ and $Q$; a 45° error mixes them and destroys the orthogonality. That is why QPSK receivers need a Costas loop (or pilot / preamble) for carrier recovery.
Problem: Use the wireless receiver from the full receiver analysis example (Section 2.9: $f=4$ GHz, $B=1$ MHz, $\eta_{rad}=0.9$, $T_p=300$ K, $T_b=200$ K, $L_T=1.5$ dB, $G_{RF}=20$ dB, $F_{RF}=3$ dB, $L_M=6$ dB, $F_M=7$ dB, $G_{IF}=30$ dB, $F_{IF}=1.1$ dB) in a point-to-point microwave link. The transmitter has EIRP $= 30$ dBW, $R = 30$ km, and the receive antenna gain is $G_r = 25$ dBi. Find (a) the received signal power $S_i$, (b) the system noise temperature $T_{SYS}$, (c) the input and output SNR.
Click to reveal full solution
(a) Received signal power via Friis equation:
$\lambda = c/f = 3\times10^8/(4\times10^9) = 0.075$ m
Free-space path loss: $\text{FSPL} = 20\log(4\pi R/\lambda) = 20\log(4\pi\times3\times10^4/0.075) = 20\log(5.03\times10^6) = 134.0$ dB
$S_i = \text{EIRP} + G_r - \text{FSPL} = 30 + 25 - 134.0 = -79.0$ dBW $= -49.0$ dBm
(b) System noise temperature (from the Section 2.9 example, already computed):
$T_A = 0.9\times200 + 0.1\times300 = 210$ K
$T_{TL} = (L_T - 1)T_p = (1.413-1)\times300 = 123$ K
$T_{REC} = T_{RF} + \frac{T_M + T_{IF}\cdot L_M}{G_{RF}} = 289 + \frac{1163 + 84\times4}{100} = 289 + 15 = 304$ K (Friis cascade)
$T_{SYS} = T_A + T_{TL} + L_T\cdot T_{REC} = 210 + 123 + 1.413\times304 = 762$ K
(c) Input SNR:
$N_i = kBT_A = 1.38\times10^{-23}\times10^6\times210 = 2.90\times10^{-15}$ W $= -115.4$ dBm
$\text{SNR}_{in} = S_i/N_i = -49.0 - (-115.4) = 66.4$ dB
Output SNR:
$N_{SYS} = kBT_{SYS} = 1.38\times10^{-23}\times10^6\times762 = 1.05\times10^{-14}$ W $= -109.8$ dBm
$\text{SNR}_{out} = S_i/kBT_{SYS} = -49.0 - (-109.8) = \boxed{60.8\text{ dB}}$
The SNR degradation from input to output is $66.4 - 60.8 = 5.6$ dB, which reflects the total system noise contributions from the transmission line and receiver chain. This is a very healthy link margin — even 64-QAM (requiring ~24.6 dB SNR) would work comfortably.
Why not use the $G/T$ figure of merit here? Would it be more correct?
Short answer: using $G/T$ wouldn't be more correct, only a different bookkeeping for the same equation. In this problem $S_i$ is already given at the receiver input port, so the antenna gain has already been "spent" — folding it back in via $G/T$ would double-count it.
Where $G/T$ comes from. The full link-budget form for the carrier-to-noise ratio is
$$\frac{C}{N_0}\;=\;\frac{\mathrm{EIRP}}{L_{path}}\cdot\frac{G_r}{T_{sys}}\cdot\frac{1}{k}$$The receiver-side ratio $G_r/T_{sys}$ — the figure of merit — is grouped together because both terms belong to the receiving station:
- $G_r$ — receive-antenna gain (boosts wanted signal),
- $T_{sys}$ — total system noise temperature (sets noise floor).
A station with higher $G/T$ collects more signal and adds less noise, captured in one number.
Why this problem doesn't need it. Here $S_i = -49$ dBm is the signal power at the receiver input port, after the antenna. Tracing the chain:
$$S_i=\frac{\mathrm{EIRP}\cdot G_r}{L_{path}}\quad\Longrightarrow\quad \frac{S_i}{N_0}=S_i\cdot\frac{1}{kT_{sys}}$$The $G_r$ has already been absorbed into $S_i$ (see Step (a), where $G_r = 25$ dBi was added to EIRP and FSPL was subtracted to get $S_i$). Multiplying by $G/T$ now would apply the antenna gain twice. So the correct formula given a known $S_i$ is exactly what Step (c) uses:
$$\mathrm{SNR}_{out}=\frac{S_i}{kBT_{sys}}$$When $G/T$ is the right tool:
Sanity check. If you wanted to back out an effective $G/T$ from this problem:
$$G/T\;[\text{dB/K}]=G_r\,[\text{dBi}]-10\log_{10}T_{sys}=25-10\log_{10}(762)=25-28.8=-3.8\ \text{dB/K}$$The 60.8 dB output SNR is unchanged — same physics, two different ways of grouping the same factors. So Step (c) is correct as written; $G/T$ would just be a re-derivation route, not a more accurate one.
2.3 Gain Compression and Intermodulation Distortion
2.3.1 Nonlinear Model (Taylor Series)
$a_1$ = small-signal gain, $a_2$ → second harmonics/DC offset, $a_3$ → gain compression and IM3.
➤For an arbitrary input signal of certain bandwidth, the resulting in-band intermodulation products from a similar mechanism will cause distortion of the output signal. This effect is called third-order intermodulation distortion.
2.3.2 Gain Compression & 1-dB Compression Point
Assume that the input signal is a single frequency sinusoid (one tone):
Effective voltage gain at $\omega_0$:
Derivation of $G_v = a_1 + \tfrac{3}{4}a_3 V_0^2$ from the single-tone Taylor series
The two identities used throughout this derivation:
$$\cos^2 x = \tfrac{1}{2}\bigl(1 + \cos 2x\bigr) \qquad\text{and}\qquad \cos^3 x = \tfrac{3}{4}\cos x + \tfrac{1}{4}\cos 3x$$These convert powers of a cosine into a sum of cosines at integer multiples of the frequency — the mechanism by which an $n$th-order nonlinearity produces the $n$th harmonic plus a back-contribution to the fundamental (gain compression).
Substitute $v_i = V_0\cos\omega_0 t$ into the Taylor series and apply the identities above:
$$\begin{aligned} v_o(t) &= a_0 + a_1 V_0\cos\omega_0 t + a_2 V_0^2\cos^2\omega_0 t + a_3 V_0^3\cos^3\omega_0 t + \cdots \\[4pt] &= \underbrace{\left(a_0+\tfrac{1}{2}a_2 V_0^2\right)}_{\text{DC}} + \underbrace{\left(a_1 V_0+\tfrac{3}{4}a_3 V_0^3\right)\cos\omega_0 t}_{\text{at }\omega_0} \\ &\quad + \underbrace{\tfrac{1}{2}a_2 V_0^2\cos 2\omega_0 t}_{\text{2nd harmonic}} + \underbrace{\tfrac{1}{4}a_3 V_0^3\cos 3\omega_0 t}_{\text{3rd harmonic}} + \cdots \end{aligned}$$Keeping only Taylor terms up to third order, the component at the input frequency $\omega_0$ has amplitude $a_1 V_0 + \tfrac{3}{4}a_3 V_0^3$. The voltage gain at $\omega_0$ is the ratio of that output amplitude to $V_0$:
$$G_v \;=\; \frac{v_o^{\omega_0}}{v_i^{\omega_0}} \;=\; \frac{a_1 V_0 + \tfrac{3}{4}a_3 V_0^3}{V_0} \;=\; a_1 + \tfrac{3}{4}a_3 V_0^2$$Two observations follow directly:
- The small-signal gain $a_1$ is recovered as $V_0\to 0$ (the cubic correction vanishes).
小訊號增益 $a_1$ 在 $V_0\to 0$ 時還原（三次項修正消失）。 - For real amplifiers $a_3 < 0$, so the correction $\tfrac{3}{4}a_3 V_0^2$ is negative — the gain drops as drive level rises. That drop is gain compression, and the level at which it reaches 1 dB defines $P_{1\text{dB}}$.
實際放大器中 $a_3 < 0$，因此修正項 $\tfrac{3}{4}a_3 V_0^2$ 為負——驅動電平升高時增益下降。此下降即為增益壓縮，當下降量達到 1 dB 時所對應的電平即為 $P_{1\text{dB}}$。
$P_{1\text{dB}}$ 壓縮特性：實線為實際輸出，虛線為理想線性外推（斜率 = 1）；兩線在 $IP_{1\text{dB}}$ 處相差 1 dB，對應之輸出電平即為 $OP_{1\text{dB}}$。
Since $a_3 < 0$ (typically), gain decreases with increasing $V_0$ → gain compression.
The 1 dB compression point $P_{1\text{dB}}$ is where actual output is 1 dB below the ideal linear extension.
Problem: An amplifier has a small-signal gain of 15 dB. The datasheet specifies $OP_{1\text{dB}} = 22$ dBm. Find (a) $IP_{1\text{dB}}$, and (b) the actual output power at the 1 dB compression point compared to an ideal amplifier.
Click to reveal solution
(a) $IP_{1\text{dB}} = OP_{1\text{dB}} - G = 22 - 15 = 7$ dBm
Why $IP_{1\text{dB}} = OP_{1\text{dB}} - G$?
In the linear region of the amplifier, gain in dB simply adds to input power (in dBm):
which is just the dB-form of the linear power law $P_{\text{out}} = G\cdot P_{\text{in}}$. Solving for the input:
$$IP_{1\text{dB}} = OP_{1\text{dB}} - G = 22 - 15 = 7\ \text{dBm}$$Why no $+1$ dB correction here? Strictly, at the compression point the actual gain is $G - 1$ dB (that's the definition of 1 dB compression), so the exact relation is
$$OP_{1\text{dB}} = IP_{1\text{dB}} + (G - 1) \;\Longrightarrow\; IP_{1\text{dB}} = OP_{1\text{dB}} - G + 1 = 8\ \text{dBm}.$$This problem uses the common simplified form $IP_{1\text{dB}} = OP_{1\text{dB}} - G$ (ignoring the 1 dB compression of the gain itself). Both conventions appear in textbooks — the 1 dB discrepancy is usually small compared to other approximations. See part (b) for the convention-aware version.
中文：在線性區，輸出功率 (dBm) = 輸入功率 (dBm) + 增益 (dB)。在 1 dB 壓縮點將同一線性關係延用，便得 $IP_{1\text{dB}} = OP_{1\text{dB}} - G$。若嚴格考慮壓縮處實際增益為 $G - 1$ dB，正確結果應為 $OP_{1\text{dB}} - G + 1 = 8$ dBm；此處採用簡化慣例不加 1。
(b) An ideal linear amplifier would produce $7 + 15 = 22$ dBm at that input. The actual output is 1 dB below: $22 - 1 = 21$ dBm. Wait — the datasheet says $OP_{1\text{dB}} = 22$ dBm. This means the ideal output at the compression input would be $22 + 1 = 23$ dBm, so $IP_{1\text{dB}} = 23 - 15 = 8$ dBm.
Convention note: $OP_{1\text{dB}}$ is the actual compressed output, which is 1 dB below the ideal. So ideal output = 23 dBm; actual = 22 dBm at $P_{in} = 8$ dBm.
2.3.3 Intermodulation Products
Intermodulation distortion — full two-tone Taylor expansion
Starting from the Taylor series with a two-tone input $v_i = V_0(\cos\omega_1 t + \cos\omega_2 t)$, expand each power and apply the product-to-sum identities (with $\cos^2 x = \tfrac{1}{2}(1+\cos 2x)$, $\cos x\cos y = \tfrac{1}{2}[\cos(x-y)+\cos(x+y)]$, $\cos^3 x = \tfrac{3}{4}\cos x + \tfrac{1}{4}\cos 3x$):
$$\begin{aligned} v_o &= a_0 + a_1 V_0(\cos\omega_1 t + \cos\omega_2 t) + a_2 V_0^2(\cos\omega_1 t + \cos\omega_2 t)^2 \\ &\quad + a_3 V_0^3(\cos\omega_1 t + \cos\omega_2 t)^3 + \cdots \\[6pt] &= a_0 + a_1 V_0\cos\omega_1 t + a_1 V_0\cos\omega_2 t + \tfrac{1}{2}a_2 V_0^2(1+\cos 2\omega_1 t) \\ &\quad + \tfrac{1}{2}a_2 V_0^2(1+\cos 2\omega_2 t) + a_2 V_0^2\cos(\omega_1-\omega_2)t + a_2 V_0^2\cos(\omega_1+\omega_2)t \\ &\quad + a_3 V_0^3\!\left(\tfrac{3}{4}\cos\omega_1 t + \tfrac{1}{4}\cos 3\omega_1 t\right) + a_3 V_0^3\!\left(\tfrac{3}{4}\cos\omega_2 t + \tfrac{1}{4}\cos 3\omega_2 t\right) \\ &\quad + a_3 V_0^3\!\left[\tfrac{3}{2}\cos\omega_2 t + \tfrac{3}{4}\cos(2\omega_1-\omega_2)t + \tfrac{3}{4}\cos(2\omega_1+\omega_2)t\right] \\ &\quad + a_3 V_0^3\!\left[\tfrac{3}{2}\cos\omega_1 t + \tfrac{3}{4}\cos(2\omega_2-\omega_1)t + \tfrac{3}{4}\cos(2\omega_2+\omega_1)t\right] + \cdots \end{aligned}$$Group the output products by frequency:
| Frequency | Coefficient | Origin / comment |
|---|---|---|
| DC | $a_0 + a_2 V_0^2$ | $a_2$ rectifies → DC offset |
| $\omega_1,\ \omega_2$ (fundamentals) | $\bigl(a_1 V_0 + \tfrac{9}{4}a_3 V_0^3\bigr)$ | $\tfrac{3}{4}+\tfrac{3}{2}=\tfrac{9}{4}$ — the $\tfrac{3}{4}a_3 V_0^3$ term is what causes gain compression |
| $2\omega_1,\ 2\omega_2$ (2nd harmonics) | $\tfrac{1}{2}a_2 V_0^2$ | far out of band — easy to filter |
| $\omega_1\pm\omega_2$ (2nd-order intermod) | $a_2 V_0^2$ | out of band for narrow-band RF — easy to filter |
| $3\omega_1,\ 3\omega_2$ (3rd harmonics) | $\tfrac{1}{4}a_3 V_0^3$ | far out of band — easy to filter |
| $2\omega_1-\omega_2,\ 2\omega_2-\omega_1$ (IM3) | $\tfrac{3}{4}a_3 V_0^3$ | in-band — cannot be filtered. This is the IM3 product. |
| $2\omega_1+\omega_2,\ 2\omega_2+\omega_1$ | $\tfrac{3}{4}a_3 V_0^3$ | far out of band — easy to filter |
The IM3 amplitude:
$$\boxed{\,|v_{\text{IM3}}|=\tfrac{3}{4}\,|a_3|\,V_0^3\,}$$Two facts make IM3 the dominant linearity worry:
- It scales as $V_0^3$. Doubling the drive amplitude multiplies IM3 by 8 — a 6 dB rise in input gives 18 dB more IM3, so the spurious-free dynamic range collapses fast.
- It lands in-band. $2\omega_1-\omega_2$ sits next to $\omega_1$, and $2\omega_2-\omega_1$ sits next to $\omega_2$ — a band-pass filter cannot remove them without also removing the wanted signals.
Every other 3rd-order product either falls out-of-band (filterable) or is mathematically identical in amplitude but spectrally harmless. That is why IM3 — and its figure of merit IIP3 — gets the spotlight in receiver linearity specs.
Two-tone input: $v_i = V_0(\cos\omega_1 t + \cos\omega_2 t)$. The Taylor expansion generates products at $m\omega_1 + n\omega_2$; the order is $|m|+|n|$.
Output spectrum of two-tone intermodulation products:
Problem: Two signals at $f_1 = 1000$ MHz and $f_2 = 1001$ MHz are applied to a nonlinear amplifier. List the frequencies of all IM3 products and identify which ones fall in-band.
Click to reveal solution
IM3 products: $3f_1 = 3000$, $3f_2 = 3003$, $2f_1+f_2 = 3001$, $f_1+2f_2 = 3002$ — all far from passband.
In-band IM3: $2f_1-f_2 = 2000-1001 = 999$ MHz, $2f_2-f_1 = 2002-1000 = 1002$ MHz.
These fall only 1 MHz away from the fundamentals — directly in the passband and impossible to filter!
2.3.4 Third-Order Intercept Point (IP3)
What is IM3 (third-order intermodulation), and why does it matter?
Setup — two-tone test. Drive the device with two equal-amplitude carriers close in frequency:
$$v_i = V_0\bigl[\cos\omega_1 t + \cos\omega_2 t\bigr]$$The cubic Taylor term $a_3 v_i^3$ produces every combination $\pm m\omega_1 \pm n\omega_2$ with $m+n\le 3$. Most fall far outside the RF band and are removed by a filter — the dangerous ones are $2\omega_1-\omega_2$ and $2\omega_2-\omega_1$, sitting symmetrically just outside $\omega_1$ and $\omega_2$. They land inside the RF band, where no filter can help.
Amplitude — the 3:1 slope. The IM3 component at $2\omega_1-\omega_2$ has amplitude
$$|v_{\text{IM3}}|=\tfrac{3}{4}|a_3|\,V_0^3$$The wanted output grows as $V_0$, but IM3 grows as $V_0^3$. On a log-log scale that's a 1:1 line for the fundamental and a 3:1 line for IM3 — every 1 dB more drive gives 3 dB more IM3, so spurious-free dynamic range gets worse by 2 dB per dB of drive.
Intercept point. Extrapolate（外推）both lines until they cross. The crossing input level is IIP3; the crossing output is OIP3 = IIP3 + Gain. The device never actually reaches IP3 (it compresses long before), but the imaginary intercept is a clean linearity figure of merit. Rule of thumb:
$$\mathrm{OIP3}\;\approx\;P_{1\mathrm{dB}}+10\ \text{dB}$$Why IM3 is the key linearity spec — not IM2 or IM5:
- Falls in-band. $2f_1-f_2$ sits next to $f_1$, can't be filtered. IM2 products ($f_1\pm f_2$, $2f_1$, $2f_2$) are far away and a band filter removes them.
- Strongest in-band term. IM5 also lands in-band, but $V_0^5$ scaling makes it much weaker than IM3 unless you over-drive hard.
- Sets receiver dynamic range. IIP3 directly determines how strong an interferer the receiver can tolerate before its own intermodulation masks the wanted signal.
Cascade rule (input-referred). For a chain of stages with individual IIP3:
$$\frac{1}{\mathrm{IIP3}_{\text{tot}}}=\frac{1}{\mathrm{IIP3}_1}+\frac{G_1}{\mathrm{IIP3}_2}+\frac{G_1 G_2}{\mathrm{IIP3}_3}+\cdots$$Opposite of Friis noise figure: in noise the first stage dominates; in IIP3 the last stage dominates because the signal is largest there. So the LNA sets the noise floor; the mixer or PA sets the linearity ceiling.
➤As the input voltage $V_0$ increases, the voltage associated with the third-order products increases as $V_0^{\,3}$. Since power is proportional to the square of the voltage, the output power of the third-order products must increase as the cube of the input power. So for small input powers the third-order products must be very small, but will increase quickly as input power increases.
➤On a log-log ($P_{ ext{out}}$ vs $P_{ ext{in}}$ in dBm) scale the linear (1st-order) curve has slope 1 and the IM3 curve has slope 3; both compress at high drive. Extending the two idealized lines, they intersect at the hypothetical point where the 1st- and 3rd-order powers would be equal — that point is the third-order intercept point (IP3).
Problem: Research and compare Adjacent Channel Power Ratio (ACPR) and Intermodulation Ratio (IMR) as measures of amplifier/system linearity. Explain: (a) the definition of each, (b) how they differ in test signal and measurement method, (c) when each is preferred, and (d) the approximate relationship between them.
Click to reveal full solution
(a) Definitions:
IMR (Intermodulation Ratio) — also called IMD or IM3 — is measured using a two-tone CW test. It is defined as the ratio of the output power of a fundamental tone to the power of the third-order intermodulation product (e.g., $2\omega_1-\omega_2$):
$$\text{IMR} = \frac{P_{\omega_1}}{P_{2\omega_1-\omega_2}} \quad\text{(dBc, always positive)}$$
From the IP3 analysis: $\text{IMR (dB)} = 2[\text{OIP}_3\text{(dBm)} - P_{out}\text{(dBm)}]$
ACPR (Adjacent Channel Power Ratio) is measured using a modulated wideband signal (e.g., W-CDMA, LTE, 5G NR). It is defined as the ratio of power leaked into an adjacent channel to the total power in the main channel:
$$\text{ACPR} = \frac{P_{\text{adjacent channel}}}{P_{\text{main channel}}} \quad\text{(dBc, negative value = better)}$$
(b) Key differences:
| Feature | IMR (Two-Tone) | ACPR (Modulated Signal) |
|---|---|---|
| Test signal | Two CW tones (simple) | Digitally modulated signal (realistic) |
| What it captures | Only 3rd-order products at specific frequencies | All spectral regrowth including 3rd, 5th, 7th order and memory effects |
| Ease of measurement | Simple spectrum analyzer | Requires signal generator with IQ modulation + spectrum analyzer |
| Accuracy for real signals | Approximation only | Directly represents real-world interference |
| Relates to | IP3 (directly calculable) | IP3 (approximately, but also depends on signal statistics) |
(c) When each is preferred:
IMR/IP3 is preferred for: component-level characterization (LNAs, mixers), quick bench testing, comparing datasheets, and cascaded system analysis (the Friis-like IP3 cascade formula uses two-tone IP3).
ACPR is preferred for: system-level compliance testing (3GPP, IEEE standards mandate ACPR specs), PA characterization under realistic signal conditions, and evaluating digital predistortion (DPD) effectiveness. Modern standards (LTE, 5G NR, W-CDMA) all specify ACPR limits.
(d) Approximate relationship:
For a memoryless weakly nonlinear system with $N$ equally-spaced tones of equal power (approximating a modulated signal), analysis shows:
$$\text{ACPR (dBc)} \approx \text{IMR}_2\text{(dBc)} - C(N) \text{ dB}$$
where $\text{IMR}_2$ is the two-tone IMR, and $C(N)$ is a correction factor that depends on the number of tones. For large $N$ (good approximation of a continuous spectrum), the ACPR is approximately 6 dB worse (higher leakage) than the two-tone IMR would predict, because many intermodulation products from different tone pairs accumulate in the adjacent channel.
Rule of thumb: $\text{ACPR} \approx \text{IMR}_2 + 6$ dB (i.e., ACPR is about 6 dB worse than two-tone IMR for the same output power backoff from IP3).
Bottom line: IP3/IMR is a convenient single-number metric, but ACPR gives a more realistic picture of linearity under real modulated signals. Always measure ACPR for final system compliance.
On a dBm-vs-dBm plot: fundamental has slope 1, IM3 has slope 3. Their idealized extensions intersect at IP3.
P_IM3 與 OIP3 的差別？（點擊展開）
P_IM3（dBm） — 在某個操作點下，三階互調失真產物的實際輸出功率。隨驅動電平而改變：P_in 上升 1 dB，P_IM3 上升 3 dB（斜率 = 3）。可在頻譜分析儀上量測到的真實功率。
OIP3 — 固定的元件線性度指標。是外推的基頻直線（斜率 1）與 IM3 直線（斜率 3）的假想交點。元件在到達 OIP3 之前就已壓縮，無法真正達到此功率。不隨驅動電平改變，是元件特性值。
| P_IM3 | OIP3 | |
|---|---|---|
| 本質 | 實際失真功率 | 外推的假想交點 |
| 隨 P_in 改變？ | 是（斜率 3） | 否（固定值） |
| 在圖上 | 斜率 3 曲線上的一點 | 固定的 y 座標 |
| 用途 | 當前失真量 | 線性度規格 / 比較基準 |
一句話：OIP3 是 P_IM3 在完全線性假設下「會到達」的點——是標準值；P_IM3 才是實際輸出的失真量。
這兩條公式怎麼來的？（點擊展開）
縮寫說明： IIP = Input Intercept Point（輸入截距點）；OIP = Output Intercept Point（輸出截距點）。下標 3 代表三階。
從 dBm 圖上的兩條直線出發。
- 基頻輸出：斜率 1 → $P_{out} = P_{in} + G$
- IM3 輸出：斜率 3（振幅正比於 $V_0^3$） → $P_{IM3,out} = 3\,P_{in} + c$（$c$ 為常數）
利用截距定義求 $c$。 IIP3 定義為兩條外推線的交點輸入功率。在 $P_{in} = \text{IIP}_3$ 時，兩線都等於 OIP3：
$$\text{OIP}_3 = 3\,\text{IIP}_3 + c \;\Rightarrow\; c = \text{OIP}_3 - 3\,\text{IIP}_3$$代回：
$$P_{IM3,out} = 3\,P_{in} + \text{OIP}_3 - 3\,\text{IIP}_3 \;\stackrel{\text{OIP}_3=\text{IIP}_3+G}{=}\; 3\,P_{in} + \text{IIP}_3 + G - 3\,\text{IIP}_3 = 3\,P_{in} - 2\,\text{IIP}_3 + G$$輸入端等效 IM3（減去增益 $G$，折算回輸入）：
$$P_{IM3}\text{(dBm)} = P_{IM3,out} - G = 3\,P_{in} - 2\,\text{IIP}_3 \;\checkmark$$OIP3 = IIP3 + G。 IIP3 與 OIP3 是同一個截距點，分別從輸入軸和輸出軸讀出。把 $P_{in} = \text{IIP}_3$ 代入基頻輸出線 $P_{out} = P_{in} + G$，直接得到輸出截距：
$$\text{OIP}_3 = \underbrace{\text{IIP}_3}_{P_{in}\text{ at intercept}} + G \;\checkmark$$圖中的 G = 小訊號功率增益（small-signal power gain）。在 Po vs Pi 圖的線性區，基頻輸出線（slope = 1）與 45° 線之間有 G dB 的垂直距離——即 Pin = 0 時 Pout = G 的截距。它同時也等於 OIP3 與 IIP3 之間的縱軸差：$G = \text{OIP3} - \text{IIP3}$。圖中雖然不一定標示箭頭，但壓縮曲線與直線外推的分歧，以及 OIP3/IIP3 的相對位置，都隱含了 G 的大小。
注意：元件在實際上遠未達到 IP3 就已經壓縮了（通常 OIP3 ≈ $P_{1\text{dB}}$ + 10 dB）。IP3 是純粹的外推交點，作為線性度指標使用，並非元件真實工作點。
Problem: An amplifier has $G = 12$ dB and $\text{OIP}_3 = 30$ dBm. Two equal-power tones at $P_{in} = -20$ dBm each are applied. Find (a) IIP3, (b) the output power of each fundamental, and (c) the output IM3 power.
Click to reveal solution
(a) $\text{IIP}_3 = \text{OIP}_3 - G = 30 - 12 = 18$ dBm
(b) $P_{out,fund} = P_{in} + G = -20 + 12 = -8$ dBm
(c) $P_{IM3,out} = 3\,P_{out} - 2\,\text{OIP}_3 = 3(-8) - 2(30) = -24 - 60 = -84$ dBm
Alternatively: $P_{IM3,in} = 3\,P_{in} - 2\,\text{IIP}_3 = 3(-20) - 2(18) = -60 - 36 = -96$ dBm, then $P_{IM3,out} = -96 + 12 = -84$ dBm ✓
The IM3 products are 76 dB below the fundamentals.
2.3.5 Relation Between $P_{1\text{dB}}$ and IP3
Derivation of $\text{OIP}_3 \approx OP_{1\text{dB}} + 10.64$ dB (from Taylor coefficients)
Setup. From the single-tone Taylor expansion (keeping terms up to third order), the small-signal voltage gain at $\omega_0$ is
$$G_v(V_0) = a_1 + \tfrac{3}{4}\,a_3\,V_0^{\,2}$$Step 1 — Solve for $V_{1\text{dB}}$ at the 1 dB compression point. By definition, the gain at $V_{1\text{dB}}$ has dropped by 1 dB:
$$20\log G_{1\text{dB}} \;=\; 20\log\!\left(a_1 + \tfrac{3}{4}a_3 V_{1\text{dB}}^{\,2}\right) \;=\; 20\log a_1 - 1$$so $G_{1\text{dB}} = a_1\cdot 10^{-0.05}$. Solving the equality for $V_{1\text{dB}}$:
$$V_{1\text{dB}} \;=\; \sqrt{\frac{4\,(10^{-0.05}-1)\,a_1}{3\,a_3}}$$Step 2 — Compute $OP_{1\text{dB}}$. The output power at compression is $(G_{1\text{dB}}V_{1\text{dB}})^2 / 2Z_0$:
$$\begin{aligned} OP_{1\text{dB}} &= \frac{(G_{1\text{dB}}\,V_{1\text{dB}})^2}{2Z_0} = \frac{1}{2Z_0}\,(a_1\cdot 10^{-0.05})^2 \cdot \frac{4(1-10^{-0.05})\,|a_1|}{3\,|a_3|} \\[3pt] &= \frac{2\times 10^{-0.1}\,(1-10^{-0.05})}{3\,Z_0}\,\left|\frac{a_1^{\,3}}{a_3}\right| \quad\text{(W)} \end{aligned}$$For $Z_0 = 50\ \Omega$ and converting to dBm:
$$\boxed{\,OP_{1\text{dB}}\,(\text{dBm}) \;=\; 10\log\!\left|\frac{a_1^{\,3}}{a_3}\right| + 0.61\,}$$Step 3 — In the linear region, find $\text{OIP}_3$. The fundamental and IM3 output powers are
$$P_{\omega_1} = P_{\omega_2} = \frac{a_1^{\,2}\,V_0^{\,2}}{2Z_0},\qquad P_{2\omega_1-\omega_2} = P_{2\omega_2-\omega_1} = \frac{1}{2Z_0}\left(\tfrac{3}{4}a_3\,V_0^{\,3}\right)^2 = \frac{9\,a_3^{\,2}\,V_0^{\,6}}{32\,Z_0}$$At the intercept $V_{\text{IP}_3}$, the two are equal: $P_{\omega_1}(V_{\text{IP}_3}) = P_{2\omega_1-\omega_2}(V_{\text{IP}_3})$. Solving:
$$V_{\text{IP}_3} \;=\; \sqrt{\frac{4\,a_1}{3\,|a_3|}}$$Substituting back gives the output intercept power:
$$\text{OIP}_3 \;=\; \frac{a_1^{\,2}\,V_{\text{IP}_3}^{\,2}}{2Z_0} \;=\; \frac{2\,a_1^{\,3}}{3\,Z_0\,|a_3|}\quad\text{(W)}$$For $Z_0 = 50\ \Omega$ in dBm:
$$\boxed{\,\text{OIP}_3\,(\text{dBm}) \;=\; 10\log\!\left|\frac{a_1^{\,3}}{a_3}\right| + 11.25\,}$$Step 4 — Subtract. Both expressions share the same $10\log|a_1^{\,3}/a_3|$ term, so
$$\boxed{\,\text{OIP}_3\,(\text{dBm}) - OP_{1\text{dB}}\,(\text{dBm}) \;=\; 11.25 - 0.61 \;=\; 10.64\ \text{dB}\,}$$Bonus — the 3:1 slope rule for IM3. Take the ratio
$$\frac{P_{2\omega_1-\omega_2}}{(P_{\omega_1})^3} \;=\; \frac{9\,Z_0^{\,2}\,a_3^{\,2}}{4\,a_1^{\,6}} \;=\; \left(\frac{1}{\text{OIP}_3}\right)^{\!2}$$which gives the textbook IM3-from-fundamental formula in dB:
$$\boxed{\,P_{2\omega_1-\omega_2}\,(\text{dBm}) \;=\; 3\,P_{\omega_1}\,(\text{dBm}) - 2\,\text{OIP}_3\,(\text{dBm})\,}$$Every 1 dB rise in $P_{\omega_1}$ adds 3 dB to the IM3 product, so the gap shrinks by 2 dB per dB of drive — that is the dynamic-range squeeze you fight with cascaded IIP3 budgeting.
$G=10$ dB, $F=5$ dB, $\text{OIP}_3 = 20$ dBm, $T_0=290$ K, $B=1$ MHz.
$OP_{1\text{dB}} = 20 - 10.64 = 9.36$ dBm
$N_o = -99$ dBm. $OP_{sf} = \frac{2}{3}(9.36) + 7.09 + \frac{1}{3}(-99) = -19.67$ dBm
Problem: A mixer has $\text{IIP}_3 = +5$ dBm and conversion loss $L = 7$ dB. Estimate $IP_{1\text{dB}}$ and $OP_{1\text{dB}}$.
Click to reveal solution
$\text{OIP}_3 = \text{IIP}_3 + G = 5 + (-7) = -2$ dBm
$OP_{1\text{dB}} = \text{OIP}_3 - 10.64 = -2 - 10.64 = -12.64$ dBm
$IP_{1\text{dB}} = OP_{1\text{dB}} - G = -12.64 - (-7) = -5.64$ dBm
(Rule of thumb check: $\text{IIP}_3 - IP_{1\text{dB}} \approx 5 - (-5.6) = 10.6$ dB ✓)
2.3.6 Dynamic Range
Linear dynamic range ($DR_l$)
➤A power range limited at the low end by noise and at the high end by the compression point. This is usually considered the usable power range for a linear power amplifier.
Spurious-free dynamic range ($DR_f$)
➤A power range limited at the low end by the third-order intermodulation product equal to the noise level and at the high end by the corresponding output signal power. This is usually considered the usable power range for low-noise amplifiers or mixers in a receiver.
$P_{\text{out}}$ vs $P_{\text{in}}$ — the two dynamic ranges $DR_l$ (linear) and $DR_f$ (spurious-free) sit between the noise floor $N_o$ and the compression / IM3 corners:
Derivation — $DR_f$, $DR_l$, and $OP_{sf}$ in dB form
Setup. From the 3:1 IM3 slope rule (§3.5), the fundamental output power can be written in terms of IM3 and OIP3:
$$P_{\omega_1} \;=\; \bigl(P_{2\omega_1-\omega_2}\bigr)^{\!1/3}\,(\text{OIP}_3)^{2/3}$$Step 1 — $DR_f$ in linear form. Substitute into the $DR_f$ definition with the constraint $P_{2\omega_1-\omega_2}=N_o$:
$$DR_f \;=\; \left.\frac{P_{\omega_1}}{P_{2\omega_1-\omega_2}}\right|_{P_{2\omega_1-\omega_2}=N_o} \;=\; \left.\frac{(P_{2\omega_1-\omega_2})^{1/3}(\text{OIP}_3)^{2/3}}{P_{2\omega_1-\omega_2}}\right|_{P_{2\omega_1-\omega_2}=N_o} \;=\; \left(\frac{\text{OIP}_3}{N_o}\right)^{\!2/3} \;=\; \left(\frac{\text{OIP}_3}{kT_0\,B\cdot G_A\cdot F}\right)^{\!2/3}$$In dB:
$$\boxed{\,DR_f\,(\text{dB}) \;=\; \tfrac{2}{3}\,\text{OIP}_3\,(\text{dBm}) - \tfrac{2}{3}\,N_o\,(\text{dBm})\,}$$Step 2 — $DR_l$ in dB form. Directly from the definition $DR_l = OP_{1\text{dB}}/N_o$:
$$\boxed{\,DR_l\,(\text{dB}) \;=\; OP_{1\text{dB}}\,(\text{dBm}) - N_o\,(\text{dBm})\,}$$Step 3 — relate $DR_l$ and $DR_f$. Solving Step 1 for $\text{OIP}_3$ and substituting into Step 2:
$$DR_l\,(\text{dB}) \;=\; \tfrac{3}{2}\,DR_f\,(\text{dB}) + OP_{1\text{dB}}\,(\text{dBm}) - \text{OIP}_3\,(\text{dBm})$$Using §3.5's empirical rule $\text{OIP}_3 - OP_{1\text{dB}} = 10.64$ dB:
$$\boxed{\,DR_l\,(\text{dB}) \;=\; \tfrac{3}{2}\,DR_f\,(\text{dB}) - 10.64\ \text{dB}\,}$$Comparing the two ranges:
- $DR_f > 21.28$ dB ⇒ $DR_l > DR_f$
- $DR_f = 21.28$ dB ⇒ $DR_l = DR_f$ (crossover)
- $DR_f < 21.28$ dB ⇒ $DR_l < DR_f$
Step 4 — output spurious-free power $OP_{sf}$. The fundamental output level at which the IM3 product equals the noise floor:
$$\begin{aligned} OP_{sf}\,(\text{dBm}) &\equiv \left.P_{\omega_1}\right|_{P_{2\omega_1-\omega_2}=N_o} \;=\; DR_f\,(\text{dB}) + N_o\,(\text{dBm}) \\[3pt] &=\; \tfrac{2}{3}\,\text{OIP}_3\,(\text{dBm}) + \tfrac{1}{3}\,N_o\,(\text{dBm}) \\[3pt] &=\; \tfrac{2}{3}\,OP_{1\text{dB}}\,(\text{dBm}) + 7.09\,(\text{dBm}) + \tfrac{1}{3}\,N_o\,(\text{dBm}) \end{aligned}$$The constant $7.09 = \tfrac{2}{3}\times 10.64$ dB carries the OIP3-vs-P1dB offset into the spur-free output expression.
With minimum detectable SNR specified: $DR_f = \tfrac{2}{3}[\text{OIP}_3 - N_o] - \text{SNR}_{min}$.
Always: $DR_l - DR_f \approx 21.3$ dB.
$F=7$ dB, $OP_{1\text{dB}}=25$ dBm, $G=40$ dB, $\text{OIP}_3=35$ dBm, $T_a=150$ K, SNR$_{min}$=10 dB, $B=100$ MHz.
$N_o = GkB[T_a+(F-1)T_0] = 10^4\times1.38\times10^{-23}\times10^8\times[150+4.01\times290] = -47.4$ dBm
$DR_l = 25 - (-47.4) = 72.4$ dB
$DR_f = \frac{2}{3}[35-(-47.4)] - 10 = \frac{2}{3}(82.4) - 10 = 54.9 - 10 = 44.9$ dB
Problem: A receiver has $\text{OIP}_3 = 25$ dBm, $G = 30$ dB, $F = 4$ dB, $B = 10$ MHz, $T_0 = 290$ K. Find $DR_l$ and $DR_f$ (no minimum SNR constraint).
Click to reveal solution
$N_o = kT_0B\cdot G\cdot F$ (since source is at $T_0$)
$= 1.38\times10^{-23}\times290\times10^7\times10^3\times10^{0.4}$
$= 4\times10^{-14}\times1000\times2.512 = 1.005\times10^{-10}$ W $= -70.0$ dBm
$OP_{1\text{dB}} = \text{OIP}_3 - 10.64 = 25 - 10.64 = 14.36$ dBm
$DR_l = 14.36 - (-70.0) = 84.4$ dB
$DR_f = \frac{2}{3}[25 - (-70.0)] = \frac{2}{3}\times95 = 63.3$ dB
Check: $DR_l - DR_f = 84.4 - 63.3 = 21.1$ dB $\approx 21.3$ dB ✓
2.3.7 Intercept Point of Cascaded Components
Two-stage cascade and its equivalent single-block representation:
At the output of the first stage:
$$P'_{2\omega_1-\omega_2} \;=\; \frac{(P'_{\omega_1})^3}{(\text{OIP}_3')^2}\;\Longrightarrow\;V'_{2\omega_1-\omega_2} \;=\; \sqrt{2 Z_0\cdot P'_{2\omega_1-\omega_2}} \;=\; \frac{\sqrt{2 Z_0\,(P'_{\omega_1})^3}}{\text{OIP}_3'}$$At the output of the second stage:
$$V''_{2\omega_1-\omega_2} \;=\; \sqrt{G_2}\cdot V'_{2\omega_1-\omega_2} \;+\; \frac{\sqrt{2 Z_0\,(P''_{\omega_1})^3}}{\text{OIP}_3''} \;=\; \frac{\sqrt{2 G_2 Z_0\,(P'_{\omega_1})^3}}{\text{OIP}_3'} \;+\; \frac{\sqrt{2 Z_0\,(P''_{\omega_1})^3}}{\text{OIP}_3''}$$Since $P''_{\omega_1} = G_2\,P'_{\omega_1}$:
$$V''_{2\omega_1-\omega_2} \;=\; \frac{\sqrt{2 G_2 Z_0\,(P''_{\omega_1}/G_2)^3}}{\text{OIP}_3'} \;+\; \frac{\sqrt{2 Z_0\,(P''_{\omega_1})^3}}{\text{OIP}_3''} \;=\; \left(\frac{1}{G_2\cdot\text{OIP}_3'} + \frac{1}{\text{OIP}_3''}\right)\sqrt{2 Z_0\,(P''_{\omega_1})^3}$$ $$P''_{2\omega_1-\omega_2} \;=\; \frac{(V''_{2\omega_1-\omega_2})^2}{2Z_0} \;=\; \left(\frac{1}{G_2\cdot\text{OIP}_3'} + \frac{1}{\text{OIP}_3''}\right)^{\!2}\!(P''_{\omega_1})^3 \;\equiv\; \frac{(P''_{\omega_1})^3}{(\text{OIP}_3)^2}$$where $\text{OIP}_3$ is the equivalent output third-order intercept point of the cascaded network.
$$\Longrightarrow\quad \frac{1}{\text{OIP}_3} \;=\; \frac{1}{G_2\cdot\text{OIP}_3'} \;+\; \frac{1}{\text{OIP}_3''}$$In general, for $N$ networks that are cascaded:
$$\frac{1}{\text{OIP}_3} \;=\; \frac{1}{G_2 G_3\cdots G_N\cdot\text{OIP}_3^{(1)}} \;+\; \frac{1}{G_3\cdots G_N\cdot\text{OIP}_3^{(2)}} \;+\; \cdots \;+\; \frac{1}{G_N\cdot\text{OIP}_3^{(N-1)}} \;+\; \frac{1}{\text{OIP}_3^{(N)}}$$With overall gain $G = G_1 G_2 \cdots G_{N-1}\,G_N$:
$$\frac{1}{\text{OIP}_3} \;=\; \frac{1}{G}\!\left(\frac{1}{\text{OIP}_3^{(1)}/G_1} + \frac{1}{\text{OIP}_3^{(2)}/(G_1 G_2)} + \cdots + \frac{1}{\text{OIP}_3^{(N-1)}/(G_1 G_2\cdots G_{N-1})} + \frac{1}{\text{OIP}_3^{(N)}/(G_1 G_2\cdots G_{N-1}\,G_N)}\right)$$The above expression shows that the third-order intercept point of a cascaded system is dominated by the last few stages, since the later stages divided by a larger product of the gains reduce $\text{IP}_3$ more significantly than the earlier stages.
Noise figure: first stage dominates (later stages divided by preceding gain).
IP3: last stage dominates (later stages multiplied by preceding gain).
This creates a fundamental design tension in receiver chains.
Amplifier: $G=20$ dB, $\text{OIP}'_3=22$ dBm $=158$ mW. Mixer: $L=6$ dB ($G_2=0.25$), $\text{IIP}''_3=13$ dBm → $\text{OIP}''_3 = 7$ dBm $=5$ mW.
$\frac{1}{\text{OIP}_3} = \frac{1}{0.25\times158} + \frac{1}{5} = \frac{1}{39.5} + \frac{1}{5} = 0.0253 + 0.200 = 0.2253$
$\text{OIP}_3 = 4.4$ mW $= 6.4$ dBm. (Dominated by the mixer's OIP3!)
Problem: Three cascaded stages: LNA ($G_1=15$ dB, $\text{OIP}_3^{(1)}=25$ dBm), Filter ($G_2=-2$ dB, $\text{OIP}_3^{(2)}=40$ dBm), Mixer ($G_3=-8$ dB, $\text{OIP}_3^{(3)}=10$ dBm). Find the overall OIP3.
Click to reveal solution
Convert: $G_1=31.62$, $G_2=0.631$, $G_3=0.158$. $G_{total} = 31.62\times0.631\times0.158 = 3.15 = 4.98$ dB.
$\text{OIP}_3^{(1)}=316.2$ mW, $\text{OIP}_3^{(2)}=10{,}000$ mW, $\text{OIP}_3^{(3)}=10$ mW.
Using the formula (all powers in mW, referred to overall output):
$\frac{1}{\text{OIP}_3} = \frac{1}{\text{OIP}_3^{(1)}\cdot G_2 G_3} + \frac{1}{\text{OIP}_3^{(2)}\cdot G_3} + \frac{1}{\text{OIP}_3^{(3)}}$
$= \frac{1}{316.2\times0.631\times0.158} + \frac{1}{10000\times0.158} + \frac{1}{10}$
$= \frac{1}{31.52} + \frac{1}{1580} + \frac{1}{10} = 0.0317 + 0.000633 + 0.100 = 0.1324$
$\text{OIP}_3 = 7.55$ mW $= 8.78$ dBm
(Dominated by the mixer — the last-stage term $1/10 = 0.100$ is the largest contributor.)
2.3.8 Passive Intermodulation (PIM)
➤Intermodulation products can also be generated by passive nonlinear effects — anywhere there is a metal-to-metal contact (connectors, cables, antennas, etc.). The result is called passive intermodulation (PIM).
Common causes of PIM
- Poor mechanical contact at RF junctions
- Oxidation between ferrous-based metals
- Contamination of conducting surfaces at RF junctions
- Use of nonlinear materials such as carbon-fiber composites or ferromagnetic metals
- Thermal effects when high powers heat junctions (adds to overall nonlinearity)
When PIM matters
- Third-order coefficients of passive components are normally very small — PIM is usually only significant when input powers are large.
- Cellular base-station transmitters are the canonical case: per-channel powers of 30–40 dBm with many closely spaced RF channels.
- Typical PIM target: keep PIM below −125 dBm with two simultaneous 40-dBm transmit signals — a dynamic range of $\sim$165 dB.
- Receivers usually do not have a PIM problem because power levels are far lower.
Design and maintenance implications
- Predicting PIM from first principles is very difficult — measurement is the standard approach.
- Carefully select cables, connectors, and antenna components used in the high-power portions of the transmitter.
- Outdoor exposure (oxidation, vibration, sunlight) degrades passive components over time, so a regular maintenance program is essential.
Problems & Exercises
A problem bank covering every topic in this chapter — noise sources & thermal noise, PSD & autocorrelation, modulation/BER, noise figure & temperature, Friis cascade, antenna noise, link budgets, and nonlinear distortion (P1dB, IP3, dynamic range). Click ? to flag a problem for review or ✓ to mark it done — state persists in your browser.
(a) clean: antenna → LNA → mixer → IF;
(b) with cable: antenna → 1.5 dB cable → LNA → mixer → IF.
By how many dB does the cable degrade the system NF? State the design rule it illustrates.
$\text{CNR} = \text{EIRP} + G/T - \text{FSPL} - 10\log(kB)$.
Recall $10\log(kB) = -228.6 + 10\log B$. Is the CNR sufficient for QPSK at BER $= 10^{-6}$?
$\dfrac{1}{\text{OIP3}} = \dfrac{1}{G_2 \cdot \text{OIP3}_{,1}} + \dfrac{1}{\text{OIP3}_{,2}}$.
Which stage dominates? Contrast with noise figure where the first stage dominates.
Worked Solutions
Detailed worked solutions to all 42 problems, with formulas, substitutions, unit checks, and design interpretation. Double-click any problem above to jump to its solution; double-click any solution to jump back to the question.
Part 1 — Noise & BER
Thermal — thermal vibration of bound charges in any resistive element (resistors, lossy lines, atmosphere); white spectrum.
Shot — discrete charge carriers crossing a junction; $\propto I_{DC}$; dominant in diodes/transistors.
Flicker (1/f) — surface defects in semiconductors; sets oscillator phase-noise floor.
Plasma — random motion in ionized gas (atmosphere, sparking contacts).
Quantum — quantized photons; negligible at microwave because $hf \ll kT$.
(a) $v_n^s = \sqrt{4kTBR} = \sqrt{4(1.38\!\times\!10^{-23})(290)(10^7)(50)} = \sqrt{8.0\!\times\!10^{-12}} \approx 2.83\;\mu\text{V rms}$.
(b) $P_n = kTB = (1.38\!\times\!10^{-23})(290)(10^7) = 4.0\!\times\!10^{-14}\;\text{W}$.
In dBm: $10\log(4.0\!\times\!10^{-11}\;\text{mW}) = -104\;\text{dBm}$. Equivalently $-174 + 10\log(10^7) = -174+70 = -104$ dBm.
$kT_0 = (1.38\!\times\!10^{-23})(290) = 4.00\!\times\!10^{-21}$ W/Hz $= 4.00\!\times\!10^{-18}$ mW/Hz.
$10\log(4.00\!\times\!10^{-18}) = -174$ dBm/Hz. ✓
Floor $= -174 + 10\log B$:
- $B=1$ kHz → $-144$ dBm
- $B=1$ MHz → $-114$ dBm
- $B=1$ GHz → $-84$ dBm
Two-sided $S(f) = n_0/2$ for all $f$. An ideal BPF passes $[-f_c-B/2, -f_c+B/2]$ and $[f_c-B/2, f_c+B/2]$.
$P_n = \displaystyle\int_{-\infty}^{\infty} S(f)\,|H(f)|^2 df = \tfrac{n_0}{2}(B) + \tfrac{n_0}{2}(B) = n_0 B = kTB$.
The two halves account for negative and positive frequencies — both are physical (real-signal symmetry).
Method
Use the Wiener-Khinchin theorem: autocorrelation and PSD are Fourier transform pairs. For ideal white noise the PSD is constant, so its inverse Fourier transform is an impulse.
$R(\tau) = \mathcal{F}^{-1}\{S(f)\} = \mathcal{F}^{-1}\{n_0/2\} = (n_0/2)\,\delta(\tau)$.
For $\tau \neq 0$: $R(\tau)=E[n(t)n(t+\tau)] = 0$ — distinct samples are uncorrelated.
At $\tau = 0$: $R(0) = \int S(f)\,df = \infty$ — infinite total power. White noise is a mathematical idealization; real noise is band-limited.
Method
Start from the finite-energy Parseval relation, then use the power-signal version for stationary random noise.
For finite-energy signals: $\int|x(t)|^2 dt = \int|X(f)|^2 df$.
For random/power signals, $S(f)$ is the time-averaged $|X_T(f)|^2/T$ as $T\to\infty$.
Setting $\tau=0$: $R(0) = E[|n(t)|^2]$ — the mean-square value, i.e. the time-average power.
Wiener–Khinchin: $R(0) = \int S(f)\,df$. Hence $\int S(f)\,df$ is the total power.
For a receiver filter, include the filter response: $P_n=\int S(f)|H(f)|^2df$. With an ideal bandpass of physical bandwidth $B$ and two-sided $S(f)=n_0/2$, the positive and negative passbands contribute $n_0B/2+n_0B/2=n_0B$.
Coherent demod: multiply by $2\cos\omega_c t$, lowpass.
SSB: Signal occupies one sideband only. Multiplication recovers baseband; noise in that single sideband is shifted to baseband. Both signal power and noise power transfer 1:1 → SNR$_{out}$ = SNR$_{in}$.
DSB-SC: Signal in both sidebands adds coherently after demod (signal voltage doubles → power $\times 4$). Noise in upper and lower sidebands is uncorrelated (white) so noise powers add (noise $\times 2$). Net SNR gain $= 4/2 = 2 = +3$ dB.
Solve $P_e = \tfrac12\mathrm{erfc}(\sqrt{E_b/n_0})$:
- $P_e=10^{-3}$ → $\sqrt{E_b/n_0}\!\approx\!2.18$ → $E_b/n_0 \approx 4.75 = 6.8$ dB
- $P_e=10^{-5}$ → $\sqrt{E_b/n_0}\!\approx\!3.0$ → $E_b/n_0 = 9.0 \approx 9.6$ dB
- $P_e=10^{-9}$ → $\sqrt{E_b/n_0}\!\approx\!4.27$ → $E_b/n_0 \approx 18.2 = 12.6$ dB
Input: $\pm V\cos\omega_c t + n(t)$ over $[0,T]$. Multiply by $\cos\omega_c t$, integrate.
(a) Signal: $\int_0^T (\pm V)\cos^2\omega_c t\,dt = \pm V T/2$ (peak $\omega_c T \gg 1$ assumption).
(b) Noise variance: $\sigma^2 = E\!\left[\left(\!\int_0^T n(t)\cos\omega_c t\,dt\right)^{\!2}\right] = \int_0^T\!\!\int_0^T R_n(t-s)\cos\omega_c t\cos\omega_c s\,ds\,dt$.
With $R_n(\tau)=(n_0/2)\delta(\tau)$: $\sigma^2 = (n_0/2)\int_0^T\cos^2\omega_c t\,dt = n_0 T/4$.
Some books use the convention $\sigma^2_{out} = n_0/T$ after normalizing by $T$.
(c) SNR: Signal power $(VT/2)^2 = V^2T^2/4$; noise power $n_0 T/4$. SNR $= V^2 T / n_0 = 2E_b/n_0$ where $E_b = V^2T/2$.
$\eta = m^2/(2+m^2)$:
- $m=0.3$: $0.09/2.09 = 4.3\%$
- $m=0.5$: $0.25/2.25 = 11.1\%$
- $m=0.7$: $0.49/2.49 = 19.7\%$
- $m=1.0$: $1/3 = 33.3\%$
$E_b/n_0$ (dB) at BER $= 10^{-5}$:
- Coh BPSK: 9.6 ← best
- Coh FSK: 12.6
- Noncoh FSK: 13.4
- Coh ASK: 15.6
- Noncoh ASK: 16.4 ← worst
$\eta_s = \log_2 M$ (bps/Hz, ideal Nyquist):
- BPSK: 1, QPSK: 2, 8-PSK: 3, 16-QAM: 4, 64-QAM: 6
Need $\eta_s \ge 24/6 = 4$ bps/Hz.
$S/N|_{dB} = (E_b/n_0)|_{dB} + 10\log(R_b/B)$.
$10\log(48.6/30) = 10\log(1.62) = 2.10$ dB.
(a) For 16-QAM at $R_b/B = 12/4 = 3$:
$S/N = 14.4 + 10\log 3 = 14.4 + 4.77 = 19.2$ dB.
(b) QPSK at the same $R_b=12$ Mbps needs $B = 6$ MHz (since $\eta_s = 2$):
$S/N = 9.6 + 10\log(12/6) = 9.6 + 3.01 = 12.6$ dB.
(a) GaAs HEMT LNA at 4 GHz: thermal (Johnson-Nyquist) noise from the channel resistance dominates; shot noise of the gate is negligible because gate current is essentially zero in a HEMT.
(b) Si BJT mixer at 100 MHz, low $I_C$: shot noise of the base/emitter junctions dominates because thermal noise of the small-signal $r_e$ is comparable but the $2qI$ shot term wins at low currents and low frequency.
(c) Varactor multiplier at 8 GHz: thermal noise of the series resistance and the lossy substrate; the device is essentially passive (no DC bias current → no shot noise).
(a) $v_n^{rms} = \sqrt{4kTRB} = \sqrt{4(1.38\!\times\!10^{-23})(300)(2200)(5\!\times\!10^6)} \approx 6.0\ \mu$V.
(b) Available noise power $P_n = kTB = (1.38\!\times\!10^{-23})(300)(5\!\times\!10^6) = 2.07\!\times\!10^{-14}$ W $= -106.8$ dBm.
(c) At 77 K, $v_n^{rms}$ scales as $\sqrt{T}$: factor $\sqrt{77/300} = 0.507$, so new value $\approx 3.0\ \mu$V (and noise power drops by $10\log(77/300) = -5.9$ dB).
For an ideal first-order RC LPF, the noise bandwidth is $B_n = (\pi/2) f_c = 1.571\times f_c = 157.1$ MHz.
Total output noise voltage variance: $\sigma_v^2 = 2 S_n B_n$ (factor 2 because two-sided integrates positive + negative frequencies):
$\sigma_v^2 = 2 \times 2\!\times\!10^{-18} \times 1.571\!\times\!10^8 = 6.28\!\times\!10^{-10}$ V$^2$.
$v_{rms} = \sqrt{\sigma_v^2} = 25.1\ \mu$V across 50 $\Omega$.
Noise power into 50 $\Omega$: $P_n = v_{rms}^2 / 50 = 1.26\!\times\!10^{-11}$ W $= -79.0$ dBm.
Signal power: $S_i = (1\!\times\!10^{-3})^2 / 50 = 2\!\times\!10^{-8}$ W $= -47.0$ dBm.
Noise power across 30 kHz: $N_i = -174 + 10\log(30\!\times\!10^3) = -174 + 44.8 = -129.2$ dBm.
(a) SNR$_i = -47 - (-129.2) = 82.2$ dB.
(b) For a coherent SSB demodulator the output SNR equals the input SNR within a constant; SNR$_o = 82.2$ dB.
(c) Improvement = 0 dB. SSB is a linear demodulator — no SNR gain, but also no penalty (unlike envelope-detected AM, which has a threshold).
BPSK at $E_b/n_0 = 7$ dB: linear $E_b/n_0 = 5.01$. $P_b = Q(\sqrt{2\times 5.01}) = Q(3.17) \approx 7.6\!\times\!10^{-4}$.
64-QAM at the same $E_b/n_0$: required $E_b/n_0$ for $P_b = 10^{-6}$ is $\approx 18.5$ dB; at 7 dB the BER is essentially $> 10^{-1}$ (deep below threshold).
Quantitatively, for square M-QAM with Gray coding: $P_b \approx \frac{4}{\log_2 M}\Big(1-\frac{1}{\sqrt{M}}\Big) Q\!\left(\sqrt{\frac{3\log_2 M}{M-1}\,\frac{E_b}{n_0}}\right)$. Plug $M=64$, $E_b/n_0 = 5.01$ → argument $\sqrt{(3\cdot 6/63)\cdot 5.01} = \sqrt{1.43} = 1.20$, so $Q(1.2) \approx 0.115$ and $P_b \approx 0.115$.
(a) Matched filter (integrate-and-dump): noise bandwidth equals signal bandwidth; $P_b = Q(\sqrt{2\, E_b/n_0}) = Q(\sqrt{2\times 7.94}) = Q(3.99) \approx 3.4\!\times\!10^{-5}$.
(b) Brick-wall LPF at $B = 1.5$ MHz (excess bandwidth $50\%$): noise power is $1.5\times$ that of the matched filter, so effective $E_b/n_0$ drops by $10\log(1.5) = 1.76$ dB. New $E_b/n_0 = 9 - 1.76 = 7.24$ dB → $P_b = Q(\sqrt{2\times 5.30}) = Q(3.26) \approx 5.6\!\times\!10^{-4}$.
| Modulation | $E_b/n_0$ for $10^{-6}$ | $\eta_s$ (max bps/Hz) | Penalty vs BPSK | Used in |
|---|---|---|---|---|
| BPSK | 10.5 dB | 1 | 0 dB | satellite TT&C |
| QPSK | 10.5 dB | 2 | 0 dB | DVB-S, GPS |
| 8-PSK | 14.0 dB | 3 | +3.5 dB | DVB-S2 (some MODCODs) |
| 16-QAM | 14.4 dB | 4 | +3.9 dB | Wi-Fi, LTE, 5G NR |
| 64-QAM | 18.6 dB | 6 | +8.1 dB | Wi-Fi 5/6, LTE, 5G NR |
| 256-QAM | 23.5 dB | 8 | +13.0 dB | Wi-Fi 6/7, 5G mmWave |
(a) Symbol rate $R_s = R_b / \log_2 M = 80 / 4 = 20$ Msym/s. Spectral efficiency $\eta_s = R_b / B = 80 / 25 = 3.2$ bps/Hz.
(b) From Pozar Table 14.1 / standard curves: 16-QAM BER $= 10^{-5}$ → $E_b/n_0 \approx 13.4$ dB.
(c) $S/N = E_b/n_0 + 10\log(R_b/B) = 13.4 + 10\log 3.2 = 13.4 + 5.05 = 18.5$ dB.
(d) Noise floor $N = kT_{sys}B = (1.38\!\times\!10^{-23})(400)(25\!\times\!10^6) = 1.38\!\times\!10^{-13}$ W $= -98.6$ dBm. Required $P_r = N + (S/N) = -98.6 + 18.5 = -80.1$ dBm.
Part 2 — Noise Figure, Temperature, Antenna & Link Budget
$T_e = (F-1)T_0$, $T_0 = 290$ K:
- $F = 1$ dB ($1.259$) → $T_e = 75.1$ K
- $F = 3$ dB ($1.995$) → $T_e \approx 289$ K
- $F = 6$ dB ($3.98$) → $T_e \approx 864$ K
- $F = 10$ dB ($10$) → $T_e = 2610$ K
$F = 1 + T_e/T_0$:
- $T_e=50$ → $F=1.172 = 0.69$ dB
- $T_e=170$ → $F=1.586 = 2.00$ dB
- $T_e=580$ → $F=3.0 = 4.77$ dB
- $T_e=2610$ → $F=10 = 10$ dB
Output noise added (source at 0 K) $= G\,k T_e B$. Solve:
$T_e = N_{add}/(GkB) = (4\!\times\!10^{-13})/(100 \cdot 1.38\!\times\!10^{-23} \cdot 10^8) = 4\!\times\!10^{-13}/1.38\!\times\!10^{-13} \approx 2.9$ K.
$F = 1 + T_e/T_0 = 1 + 2.9/290 \approx 1.010 \;\Rightarrow\; 0.04$ dB.
$T_e = (T_h - YT_c)/(Y-1) = (873 - 2.5\!\cdot\!77)/(2.5-1) = (873 - 192.5)/1.5 = 680.5/1.5$.
$T_e = 454$ K. $F = 1 + 454/290 = 2.56$.
$L = 10^{0.25} = 1.778$. At $T_p = T_0$: $F = L = 2.5$ dB exactly.
$T_e = (L-1)T_p = 0.778 \cdot 290 = 226$ K.
Why $F = L$: at $T_p = T_0$, both source noise and added noise are $kT_0B$. Power gets attenuated by $L$ but added noise refills to keep output noise at $kT_0B$ → SNR drops by exactly $L$.
Equivalently, use the temperature form: $F=1+T_e/T_0=1+(L-1)T_0/T_0=L$. This is why a passive room-temperature loss has noise figure equal to insertion loss.
$L = 10^{0.3} = 2.0$. At $T_p = 200$ K: $T_e = (L-1)T_p = 1\cdot 200 = 200$ K.
If naively $T_p = 290$: $T_e = 1\cdot 290 = 290$ K.
Linear: $F_1=1.585$, $G_1=31.62$; $F_2=5.01$, $G_2=0.251$; $F_3=2.512$.
$F = F_1 + (F_2-1)/G_1 + (F_3-1)/(G_1 G_2)$
$= 1.585 + 4.01/31.62 + 1.512/(31.62\cdot 0.251)$
$= 1.585 + 0.127 + 1.512/7.93 = 1.585 + 0.127 + 0.190 = 1.902$.
(a) Clean: $F_1=1.585\;(2\,\text{dB})$, $G_1=100$; $F_2=6.31\;(8\,\text{dB})$, $G_2=0.251$; $F_3=2.512\;(4\,\text{dB})$.
$F_a = 1.585 + 5.31/100 + 1.512/(100\cdot 0.251) = 1.585 + 0.053 + 0.060 = 1.698 \Rightarrow 2.30$ dB.
(b) With cable: add 1.5 dB cable ($F_0 = L = 1.413$, $G_0 = 0.708$) at front:
$F_b = 1.413 + (1.585-1)/0.708 + 5.31/(0.708\cdot 100) + 1.512/(0.708\cdot 100\cdot 0.251)$
$= 1.413 + 0.826 + 0.075 + 0.085 = 2.40 \Rightarrow 3.80$ dB.
A matched, lossless Wilkinson splits input power equally to two output ports — each port sees $P_{in}/2$, i.e. 3 dB lower.
From one output port's perspective, the divider looks like a 3 dB lossy two-port (input → that output), so by the lossy-line rule, $F = L = 3$ dB at $T_p = T_0$.
No contradiction: the "missing" half-power didn't dissipate — it went out the other output port. The receiver at port 2 still sees its input SNR halved.
Detailed SNR view: for one branch, the desired signal available at the receiver is halved. The source noise delivered into that branch is also halved, but the branch still sees a matched $T_0$ environment. Referred to a single output path, the available SNR is degraded by the same factor as a 3 dB attenuator.
$T_a = \eta_{rad}T_b + (1-\eta_{rad})T_p = 0.9\cdot 50 + 0.1\cdot 290 = 45 + 29 = 74$ K.
Pointed at the ground ($T_b = 290$): $T_a = 0.9\cdot 290 + 0.1\cdot 290 = 290$ K.
(a) EIRP $= P_t + G_t = 10\log(100) + 35 = 20 + 35 = 55$ dBW $= 85$ dBm.
(b) $\lambda = 3\!\times\!10^8/12\!\times\!10^9 = 0.025$ m.
FSPL $= 32.4 + 20\log(12000) + 20\log(36000) = 32.4 + 81.6 + 91.1 \approx 205$ dB.
(c) $P_r = $ EIRP $- $ FSPL $+ G_r = 55 - 205 + 30 = -120$ dBW $= -90$ dBm.
$G/T = G_r - 10\log(T_a + T_e) = 30 - 10\log(80+120) = 30 - 10\log 200 = 30 - 23.0 = 7.0$ dB/K.
With $T_e \to 60$ K: $G/T = 30 - 10\log(140) = 30 - 21.46 = 8.54$ dB/K.
$10\log(kB) = -228.6 + 10\log(2.7\!\times\!10^7) = -228.6 + 74.3 = -154.3$ dBW/Hz·Hz.
CNR $= $ EIRP $+ G/T - $ FSPL $- 10\log(kB)$
$= 55 + 12 - 205 - (-154.3) = 55 + 12 - 205 + 154.3 = 16.3$ dB.
QPSK at BER $=10^{-6}$ needs $E_b/n_0 \approx 10.5$ dB; with $R_b/B=2$ → $S/N_{req} \approx 13.5$ dB.
$F = 2$ ($3$ dB), so $(F-1)T_0 = 290$ K. Effective noise temp at antenna port $= T_a + (F-1)T_0 = 100 + 290 = 390$ K.
$S_{i,min} = kB\cdot 390 \cdot 10 = (1.38\!\times\!10^{-23})(10^6)(390)(10) = 5.4\!\times\!10^{-14}$ W $= -132.7$ dBm.
Why not $kT_0BF$? $kT_0BF \cdot (S_o/N_o) = (1.38e^{-23})(290)(10^6)(2)(10) = 8.0e^{-14}$ W $= -130.9$ dBm — about 1.8 dB too high. Because the antenna is cold ($T_a = 100$ K $\ne T_0$), the standard formula overstates the noise.
$L_T = 10^{0.15} = 1.413$.
(a) $T_{TL} = (L_T-1)T_p = 0.413\cdot 290 = 119.7$ K.
$T_{REC} = (F-1)T_0 = (1.585-1)(290) = 169.7$ K.
(b) $T_{SYS} = T_a + T_{TL} + L_T\cdot T_{REC} = 60 + 119.7 + 1.413\cdot 169.7 = 60 + 119.7 + 239.8 = 419.5$ K.
(c) Equivalent system NF: $F_{sys} = 1 + T_{SYS}/T_0 = 1 + 419.5/290 = 2.45 \Rightarrow 3.89$ dB.
$L_T = 10^{0.2} = 1.585$, $F = 10^{0.4} = 2.512$.
$T_{TL} = 0.585\cdot 270 = 158.0$ K; $T_{REC} = 1.512\cdot 290 = 438.5$ K.
$T_{SYS} = 200 + 158.0 + 1.585\cdot 438.5 = 200 + 158 + 695 = 1053$ K.
$F_{sys} = 1 + 1053/290 = 4.63 \Rightarrow 6.65$ dB.
Sanity check at $T_a = T_0 = 290$ K:
$T_{SYS} = 290 + 158 + 695 = 1143$ K, $F_{sys} = 1 + 1143/290 = 4.94 \Rightarrow 6.94$ dB.
Friis (line + receiver): $F_{line+rec} = L_T \cdot F = 1.585\cdot 2.512 = 3.98 \Rightarrow 6.0$ dB. With $T_p = 270 \neq T_0$ the equality is approximate; if we set $T_p = T_0$ also, $T_{SYS}$ would be exactly $T_0\cdot F_{tot}$.
Open-ended; one workable budget for 64-QAM, $R_b = 100$ Mbps, BER $=10^{-6}$ ($E_b/n_0 = 22.5$ dB):
- $\eta_s \approx 5$ bps/Hz (with rolloff) → $B \approx 20$ MHz
- Required $S/N = 22.5 + 10\log(100/20) = 22.5 + 7 = 29.5$ dB; aim for $\ge 32.5$ dB with 3 dB margin
- Try Ku-band: $f=20$ GHz, $R=36{,}000$ km → FSPL $\approx 209.6$ dB
- $P_t = 50$ W (47 dBm), $G_t = 40$ dBi → EIRP $= 87$ dBm $= 57$ dBW
- $G_r = 50$ dBi (dish), $T_a = 30$ K, $T_e = 70$ K (cooled LNA) → $G/T = 50 - 10\log(100) = 30$ dB/K
- $10\log(kB) = -228.6 + 73 = -155.6$ dBW/Hz
- CNR $= 57 + 30 - 209.6 + 155.6 = 33.0$ dB → margin $= 33.0 - 29.5 = 3.5$ dB ✓
(a) $F = 2.5$ dB → linear $F = 10^{0.25} = 1.778$. $T_e = (F-1)T_0 = 0.778 \times 290 = 225.6$ K.
(b) New source $T_s = 100$ K → $T_{sys} = 100 + 225.6 = 325.6$ K. Effective $F$ at this source temperature: $F_{eff} = 1 + T_e/T_s = 1 + 225.6/100 = 3.256 = 5.13$ dB.
$Y = (T_h + T_e) / (T_c + T_e) = 1.45$
$\Rightarrow T_e = (T_h - Y T_c) / (Y - 1) = (470 - 1.45 \times 80) / 0.45 = 354/0.45 = 786.7$ K.
$F = 1 + T_e/T_0 = 1 + 786.7/290 = 3.71 = 5.69$ dB.
(a) Input-referred noise: $kT_0BF = -174 + 10\log(10^7) + 4 = -174 + 70 + 4 = -100$ dBm.
(b) Output noise: $G + (-100) = 30 + (-100) = -70$ dBm.
(c) SNR$_i = -80 - (-100) = 20$ dB. Output: $S_o = -80 + 30 = -50$ dBm; SNR$_o = -50 - (-70) = 20$ dB.
(a) Passive line at $T_p = T_0$: $F_{line} = L = 2$ dB → linear $1.585$. $T_{line} = (L-1)T_p = 0.585 \times 290 = 169.7$ K.
(b) LNA: $F_{LNA} = 1.5$ dB = 1.413, $T_{LNA} = (1.413-1)\times 290 = 119.8$ K. Cascade with line first ($G_1 = 1/L = 0.631$):
$T_{cas} = T_{line} + T_{LNA}/G_1 = 169.7 + 119.8/0.631 = 169.7 + 189.8 = 359.5$ K.
$F_{cas} = 1 + 359.5/290 = 2.24 = 3.50$ dB.
(c) Sensitivity penalty: $F_{cas} - F_{LNA} = 3.50 - 1.5 = 2.0$ dB — exactly the line loss (this is the universal "every dB of pre-LNA loss costs 1 dB of NF" rule).
Linear gains/figures: $G_1 = 31.6$, $F_1 = 1.585$ (T₁ = 169.7 K); $L_2 = 3.98$, $G_2 = 1/3.98 = 0.251$, $F_2 = 6.31$ (T₂ = 1540 K); $G_3 = 1000$, $F_3 = 1.585$ (T₃ = 169.7 K).
$F_{cas} = F_1 + (F_2-1)/G_1 + (F_3-1)/(G_1 G_2) = 1.585 + 5.31/31.6 + 0.585/(31.6\times 0.251)$
$= 1.585 + 0.168 + 0.0738 = 1.827 = 2.62$ dB.
An ideal Wilkinson is lossless (3 dB split is just power division, not insertion loss). The 0.3 dB excess loss per arm is the actual insertion loss seen by signal and noise alike.
From one output port: looking back, the device behaves as a 3 dB attenuator (split) + 0.3 dB extra loss = 3.3 dB total loss at $T_p = 290$ K.
$T_e = (L - 1) T_p = (10^{0.33} - 1) \times 290 = (2.14 - 1) \times 290 = 330.5$ K.
$F = L = 3.3$ dB per arm.
(a) $T_a = \eta_{rad} T_{sky} + (1-\eta_{rad}) T_p = 0.85 \times 20 + 0.15 \times 290 = 17 + 43.5 = 60.5$ K.
(b) $T_{sys} = T_a + T_e = 60.5 + 60 = 120.5$ K.
(c) $N = kT_{sys}B = 1.38\!\times\!10^{-23} \times 120.5 \times 36\!\times\!10^6 = 5.99\!\times\!10^{-14}$ W $= -132.2$ dBm.
(a) $G/T = G_r - 10\log T_{sys} = 45 - 10\log 120 = 45 - 20.79 = 24.2$ dB/K.
(b) Link budget in dB:
$C/N_0 = $ EIRP $-$ FSPL $+ G/T - 10\log k$
$= 50 - 205 + 24.2 - (-228.6) = 97.8$ dB-Hz.
(c) For e.g. a 36 MHz transponder: $C/N = C/N_0 - 10\log B = 97.8 - 75.6 = 22.2$ dB — comfortably enough for 16-APSK / DVB-S2.
$T_{TL} = (L_T - 1)T_p$ with $L_T = 10^{0.05} = 1.122$ → $T_{TL} = 0.122 \times 290 = 35.4$ K.
$T_{SYS} = T_a + T_{TL} + L_T \cdot T_{REC} = 200 + 35.4 + 1.122 \times 180 = 200 + 35.4 + 202 = 437.4$ K.
$N_{SYS} = kT_{SYS}B = 1.38\!\times\!10^{-23}\times 437.4 \times 2\!\times\!10^7 = 1.21\!\times\!10^{-13}$ W $= -99.2$ dBm.
$\text{SNR}_{out} = S_i - N_{SYS} = -80 - (-99.2) = 19.2$ dB.
Part 3 — Distortion (IP3, P1dB, DR)
Substitute $v_i = A\cos\omega t$:
- $a_1 v_i = a_1 A\cos\omega t$ → linear gain
- $a_2 v_i^2 = a_2 A^2(\tfrac12 + \tfrac12\cos 2\omega t)$ → DC offset + 2nd harmonic
- $a_3 v_i^3 = a_3 A^3(\tfrac34\cos\omega t + \tfrac14\cos 3\omega t)$ → fundamental gets a contribution $\tfrac{3a_3 A^3}{4}\cos\omega t$ (compression / expansion) and $\tfrac{a_3 A^3}{4}\cos 3\omega t$ (3rd harmonic)
Method
First compute the output predicted by the small-signal straight-line gain, then subtract the 1 dB compression error.
Small-signal extrapolated output at $P_{in}=-10$ dBm is $P_{out,ideal}=P_{in}+G=-10+20=10$ dBm.
At the 1 dB compression point, the actual output is 1 dB lower than this extrapolated value: $\text{OP1dB}=10-1=9$ dBm.
Set $\Delta = f_2 - f_1 = 1$ MHz.
2nd-order: $2f_1 \approx 2$ GHz, $2f_2 \approx 2$ GHz, $f_1 + f_2 \approx 2$ GHz, $f_2 - f_1 = 1$ MHz — all out-of-band.
3rd-order: $3f_1, 3f_2 \approx 3$ GHz (out); $2f_1 \pm f_2$ → $0.999, 3.001$ GHz; $2f_2 \pm f_1$ → $1.002, 3.002$ GHz.
$\Delta P = P_{fund} - P_{IM3} = -10 - (-70) = 60$ dB.
OIP3 $= P_{fund} + \Delta P/2 = -10 + 30 = 20$ dBm.
Slope rule: on a $\log P_{out}$ vs $\log P_{in}$ plot, fund grows 1 dB / dB; IM3 grows 3 dB / dB. They intersect at IP3. At $\Delta$ dB backoff from IP3, fund is $\Delta$ dB below IP3 and IM3 is $3\Delta$ dB below — gap $= 2\Delta$.
Here the measured gap is 60 dB, so the fundamental is $60/2=30$ dB below the intercept. Since the measured fundamental is $-10$ dBm, the intercept is $-10+30=20$ dBm.
(a) $P_{out,fund} = P_{in} + G = -30 + 20 = -10$ dBm/tone.
(b) $P_{IM3} = 3 P_{out} - 2\,\text{OIP3} = 3(-10) - 2(30) = -30 - 60 = -90$ dBm.
Backoff check: the fundamental is $30-(-10)=40$ dB below OIP3. IM3 is twice that gap below the fundamental, so IM3 should be $2\cdot40=80$ dBc. $-10-80=-90$ dBm, matching the formula.
(a) OP1dB $= 22$ → OIP3 $\approx 22 + 10.6 = 32.6$ dBm.
(b) OIP3 $= 35$ → OP1dB $\approx 35 - 10.6 = 24.4$ dBm.
(c) Breakdown cases:
- Saturating PAs (Class C/E/F): output clips abruptly, OP1dB much closer to OIP3.
- Strong even-order distortion: 10.6 dB rule assumes pure 3rd-order — fails when $a_2$-driven products dominate.
- Heavy back-off / pre-distortion linearizers can push the gap to 12–15 dB.
(a) $DR_l = $ OP1dB $- N_o = 20 - (-90) = 110$ dB.
(b) $DR_f = (2/3)($ OIP3 $- N_o) = (2/3)(120) = 80$ dB.
(c) $DR_l - DR_f = 30$ dB. The "$\approx 21.3$ dB" rule from §3.6 holds when OIP3 $-$ OP1dB $= 10.6$ dB and $N_o$ tracks the canonical $kT_0 B F$ floor; here OIP3 $-$ OP1dB $= 10$ dB and $N_o$ was specified independently, so the gap differs slightly from the rule.
Convert: OIP3$_{,1} = 20$ dBm $= 100$ mW, $G_1 = 10$ dB $=10$, $G_2 = 20$ dB $=100$, OIP3$_{,2} = 25$ dBm $= 316$ mW.
$\dfrac{1}{\text{OIP3}} = \dfrac{1}{G_2 \cdot \text{OIP3}_{,1}} + \dfrac{1}{\text{OIP3}_{,2}} = \dfrac{1}{100\cdot 100} + \dfrac{1}{316} = 10^{-4} + 3.16\!\times\!10^{-3} = 3.26\!\times\!10^{-3}$.
OIP3 $= 307$ mW $= 24.9$ dBm. Stage 2 dominates ($1/316$ term is 31× larger).
Linear: $G_1=31.62$, $G_2=0.200$, $G_3=316.2$. OIP3$_{,1}=31.6$ mW, OIP3$_{,2}=63.1$ mW, OIP3$_{,3}=631$ mW.
$\dfrac{1}{\text{OIP3}} = \dfrac{1}{\text{OIP3}_{,1} G_2 G_3} + \dfrac{1}{\text{OIP3}_{,2} G_3} + \dfrac{1}{\text{OIP3}_{,3}}$
$= \dfrac{1}{31.6 \cdot 0.2 \cdot 316.2} + \dfrac{1}{63.1 \cdot 316.2} + \dfrac{1}{631}$
$= \dfrac{1}{1998} + \dfrac{1}{19952} + \dfrac{1}{631} = 5.0\!\times\!10^{-4} + 5.0\!\times\!10^{-5} + 1.585\!\times\!10^{-3} = 2.14\!\times\!10^{-3}$.
OIP3 $= 467$ mW $= 26.7$ dBm.
IIP3 $= $ OIP3 $- G = 24 - 12 = 12$ dBm.
Input-referred: $P_{IM3,in} = 3 P_{in} - 2\,\text{IIP3} = 3(-20) - 2(12) = -60 - 24 = -84$ dBm.
(Sanity-check at output: $P_{out,fund} = -8$ dBm, $P_{out,IM3} = 3(-8) - 2(24) = -72$ dBm = $-84 + G$ ✓.)
ACPR = $10\log\dfrac{P_{adj\,channel}}{P_{main\,channel}}$ for a digitally modulated test signal — power that "leaks" into a defined adjacent channel BW.
For weakly-nonlinear amps the ACPR floor is set by 3rd-order products generated by all the in-band tones convolving with each other. Same physics as 2-tone IMR but with a continuum of tones.
Empirically, the ACPR test "sees" the integrated 3rd-order power vs 2-tone's peak product, so ACPR specs typically demand $\sim 3$–$4$ dB more OIP3 than the equivalent 2-tone IMR target.
(a) Input noise floor: $N_i = kT_0 B F = -174 + 60 + 4 = -110$ dBm. Output: $N_o = N_i + G = -110 + 60 = -50$ dBm.
(b) IIP3 $= $ OIP3 $- G = 30 - 60 = -30$ dBm.
(c) Input-referred SFDR $= (2/3)($ IIP3 $- N_i) = (2/3)(-30 - (-110)) = (2/3)(80) = 53.3$ dB.
Interferer scenario: $-30$ dBm interferer is right at IIP3 — IM3 generated will be $3(-30) - 2(-30) = -30$ dBm at input… that's the IP3 itself, completely overwhelming a $-90$ dBm desired signal. The receiver fails: gap $= -30 - (-90) = 60$ dB > SFDR (53.3 dB).
Use $\cos^2 = \tfrac{1}{2}(1+\cos 2\omega t)$ and $\cos^3 = \tfrac{3}{4}\cos\omega t + \tfrac{1}{4}\cos 3\omega t$. With $V_0 = 0.4$ V:
(a) DC: $\tfrac{1}{2}a_2 V_0^2 = 0.5 \times 0.3 \times 0.16 = 0.024$ V.
(b) Fundamental amplitude: $a_1 V_0 + \tfrac{3}{4}a_3 V_0^3 = 5 \times 0.4 + 0.75 \times (-0.05) \times 0.064 = 2.0 - 0.0024 = 1.998$ V.
(c) 2nd harmonic: $\tfrac{1}{2}a_2 V_0^2 = 0.024$ V at $2\omega_0$.
(d) 3rd harmonic: $\tfrac{1}{4}a_3 V_0^3 = 0.25 \times (-0.05) \times 0.064 = -8\!\times\!10^{-4}$ V (i.e. $0.8$ mV at $3\omega_0$).
(e) Effective gain at $\omega_0$: $G_v = a_1 + \tfrac{3}{4}a_3 V_0^2 = 5 - 0.006 = 4.994$ ≈ 14 dB. Compression $\approx 0.01$ dB.
(a) Linear region: $P_{out} = P_{in} + G = P_{in} + 18$. As $P_{in}$ rises, gain compresses; at $P_{1dB,out} = 10$ dBm, gain has dropped to 17 dB.
(b) $P_{1dB,in} = P_{1dB,out} - (G - 1) = 10 - 17 = -7$ dBm. At $P_{in} = -8$ dBm, the device is just below 1 dB compression — actual $G \approx 17.6$ dB → $P_{out} \approx -8 + 17.6 = 9.6$ dBm.
(c) $P_{1dB,in} = -7$ dBm.
$f_1 = 2400$, $f_2 = 2410$ MHz. Passband $2355\text{–}2455$ MHz.
IM2: $f_2 - f_1 = 10$ MHz (well below passband); $f_1 + f_2 = 4810$ MHz (well above); $2 f_1 = 4800$, $2 f_2 = 4820$ MHz (above). None fall in band.
IM3: $2f_1 - f_2 = 2390$ MHz, $2f_2 - f_1 = 2420$ MHz — both in-band (next to fundamentals). $3f_1 = 7200$ MHz, $3f_2 = 7230$ MHz, $2f_1 + f_2 = 7210$ MHz, $f_1 + 2f_2 = 7220$ MHz — far above passband.
The two in-band products $2f_1 - f_2$ and $2f_2 - f_1$ both come from the cubic $a_3$ term.
Slope-1 fundamental, slope-3 IM3 lines on a log-log $P_{out}$ vs $P_{in}$ chart.
Gap at $P_{in} = -10$ dBm: $\Delta = P_{out,fund} - P_{IM3} = 0 - (-50) = 50$ dB.
$\text{IIP}_3 = P_{in} + \Delta/2 = -10 + 25 = 15$ dBm.
$\text{OIP}_3 = \text{IIP}_3 + G = 15 + 10 = 25$ dBm.
IM3 ratio = 50 dBc — adequate for moderate-linearity systems but marginal for cellular base stations (which require > 70 dBc typically).
(a) $OP_{1dB} \approx \text{OIP}_3 - 10.6 = 28 - 10.6 = 17.4$ dBm.
(b) $IP_{1dB,in} = OP_{1dB} - G_{1dB} \approx OP_{1dB} - (G - 1) = 17.4 - 24 = -6.6$ dBm.
(c) $\text{IIP}_3 = \text{OIP}_3 - G = 28 - 25 = 3$ dBm.
Limitation: the 10.6 dB constant comes from the strict cubic Taylor model. Real devices with significant 5th-order content, AM-PM distortion, or memory effects can deviate by $\pm 2$ dB or more — always verify with two-tone measurement when accuracy matters.
(a) $N_o = -174 + 10\log B + F + G = -174 + 60 + 4 + 30 = -80$ dBm.
(b) $DR_l = OP_{1dB} - N_o = 10 - (-80) = 90$ dB.
(c) $DR_f = (2/3)(\text{OIP}_3 - N_o) = (2/3)(20 - (-80)) = (2/3)(100) = 66.7$ dB.
(d) $DR_l - DR_f = 90 - 66.7 = 23.3$ dB ≈ 21 dB ✓ (the $\sim 21.3$ dB rule).
Convert: $G_1 = 100$, $L = 4$ (so $G_2 = 0.25$), OIP3$_1 = 31.6$ mW, OIP3$_2 = 100$ mW.
$1/\text{OIP}_3 = 1/(G_2 \cdot \text{OIP}_3^{(1)}) + 1/\text{OIP}_3^{(2)} = 1/(0.25\times 31.6) + 1/100 = 1/7.9 + 1/100 = 0.1266 + 0.01 = 0.1366$.
OIP3$_{cas} = 7.32$ mW $= 8.65$ dBm.
Mixer dominates linearity ($1/(G_2\cdot \text{OIP}_3^{(1)}) = 1/7.9$ is 12.7× larger than $1/\text{OIP}_3^{(2)}$).
(a) PIM3 / single carrier: $-125 - 43 = -168$ dBc — a typical good base-station spec ($< -150$ dBc).
(b) PIM3 vs noise floor: $-125 - (-110) = -15$ dB → PIM3 is $15$ dB below the receiver noise floor. Just below the noise — would corrupt sensitivity if it rose any further.
(c) PIM is generated by passive nonlinearities (bad connectors, oxidation, ferromagnetic parts). It only matters when powers are large — i.e. on the transmit side. The receiver's small signal levels can't generate appreciable PIM, so receivers don't worry about it. But the same antenna often sees both Tx and Rx signals (FDD), so PIM generated by the Tx can fall into the Rx band.
Quick Reference Summary
Important Symbols
Part 1 — Random Processes, PSD & Modulation
| Symbol | Name | Units | Notes |
|---|---|---|---|
| $S(f)$, $S(\omega)$ | Power spectral density (PSD) | W/Hz | Two-sided. Total power $P = \int S(f)\,df$ |
| $R(\tau)$ | Autocorrelation function | W (or V²) | $R(\tau) = \mathcal{F}^{-1}\{S(f)\}$. $R(0)$ = total power |
| $n_0$ | Noise spectral density (one-sided) | W/Hz | $n_0 = kT$. Two-sided is $n_0/2$ |
| $\delta(\tau)$ | Dirac delta function | 1/s | White noise: $R(\tau) = (n_0/2)\delta(\tau)$ (uncorrelated) |
| $\sigma$, $\sigma^2$ | Noise standard deviation / variance | V / V² | Gaussian PDF: $p(v) = \tfrac{1}{\sqrt{2\pi}\sigma}e^{-v^2/2\sigma^2}$ |
| $V$ or $v_0$ | Peak signal voltage | V | PSK signal $s(t)=\pm V\cos\omega_c t$ |
| $T$ | Bit/symbol period | s | $T = 1/R_b$ for binary signaling |
| $E_b$ | Energy per bit | J | $E_b = V^2 T/2$ for sinusoidal PSK |
| $P_e$ | Bit error probability (BER) | unitless | Typical target $10^{-5}$ to $10^{-9}$ |
| $\mathrm{erfc}(x)$ | Complementary error function | unitless | $\mathrm{erfc}(x) = 1 - \mathrm{erf}(x)$, Gaussian-tail probability |
| $\omega_c$, $f_c$ | Carrier (angular) frequency | rad/s, Hz | $\omega_c = 2\pi f_c$ |
| $M$ | Modulation order | integer | M-QAM, M-PSK; bits/symbol $= \log_2 M$ |
Part 2 — Noise Figure, Antenna & Distortion
| Symbol | Name | Units | Notes |
|---|---|---|---|
| $k$ | Boltzmann constant | $1.38\times10^{-23}$ J/K | $10\log(k) = -228.6$ dBW/Hz/K |
| $T_0$ | Reference temperature | $290$ K | Used in noise-figure definition; $kT_0 = -174$ dBm/Hz |
| $B$ | Noise bandwidth | Hz | IF/channel bandwidth that defines noise power |
| $T_e$ | Equivalent noise temperature | K | Of receiver/component; $T_e = (F-1)T_0$ |
| $F$ | Noise figure (factor) | linear / dB | $F = \text{SNR}_{in}/\text{SNR}_{out}$ at $T_0=290$ K |
| $G$ | Available power gain | linear / dB | Antenna gain in linear or dBi |
| $L$ | Loss factor | $\ge 1$ / dB | Passive lossy line at $T_p$: $T_e=(L-1)T_p$, $F=L$ when $T_p=T_0$ |
| $T_b$ | Brightness (background) temperature | K | Beam-weighted: cold sky 3–50 K, ground ~290 K |
| $T_p$ | Physical temperature | K | Of antenna structure / lossy line |
| $T_a$ / $T_A$ | Antenna noise temperature | K | $T_a = \eta_{rad}T_b + (1-\eta_{rad})T_p$ |
| $\eta_{rad}$ | Radiation efficiency | $0$–$1$ | Fraction radiated; $1-\eta_{rad}$ dissipated |
| $T_{SYS}$ | System noise temperature | K | $T_{SYS} = T_a + T_e$ (referred to antenna) |
| EIRP | Equivalent Isotropic Radiated Power | W / dBW | $\text{EIRP} = G_t P_t$ |
| FSPL | Free-space path loss | dB | $20\log_{10}(4\pi R/\lambda)$ |
| $G/T$ | Antenna figure of merit | dB/K | $G_r(\text{dBi}) - 10\log_{10}(T_a + T_e)$ |
| $E_b/n_0$ | Energy-per-bit / noise PSD | dB | Sets BER independent of modulation |
| $R_b$ | Bit rate | bps | $\text{CNR} = (E_b/n_0)(R_b/B)$ |
| P1dB | 1 dB compression point | dBm | Input or output (IP1dB / OP1dB) |
| IP3 / OIP3 | 3rd-order intercept point | dBm | Input/output; OIP3 ≈ OP1dB + 10.6 dB |
| $DR_l$, $DR_f$ | Linear DR / Spurious-free DR | dB | $DR_l - DR_f \approx 21.3$ dB |
Important Formulas
Part 1 — Random Processes, PSD & Modulation
Power ↔ PSD
$P = \displaystyle\int_{-\infty}^{\infty} S(f)\,df$
Time avg = area under PSD
Wiener–Khinchin
$S(f) \;\longleftrightarrow\; R(\tau)$ (Fourier pair)
$R(0)$ = total power
White Noise PSD
$S(f) = n_0/2$ (two-sided)
$R(\tau) = (n_0/2)\,\delta(\tau)$
$n_0 = kT$
Band-Limited Noise
$P_n = n_0 B = kTB$
Integrated over physical $B$
Gaussian PDF
$p(v) = \dfrac{1}{\sqrt{2\pi}\,\sigma}\,e^{-v^2/2\sigma^2}$
Tail: $P(v>V)=\tfrac{1}{2}\mathrm{erfc}(V/\sqrt{2}\sigma)$
Energy per Bit
$E_b = \dfrac{V^2 T}{2}$ (PSK, sinusoidal)
$E_b = P_s \cdot T_b$
Integrate-and-Dump
Output signal: $VT$
$\sigma_{out}^2 = n_0/T$
$\Rightarrow E_b/n_0$ ratio
Universal BER (PSK)
$P_e = \tfrac{1}{2}\mathrm{erfc}\!\left(\sqrt{E_b/n_0}\right)$
$10^{-5}$ → $E_b/n_0 = 9.6$ dB
Spectral Efficiency
$\eta_s = R_b/B$ (bps/Hz)
BPSK: 1, QPSK: 2, 16-QAM: 4, 64-QAM: 6
SNR ↔ $E_b/n_0$ (Part 1)
$\dfrac{S}{N} = \dfrac{E_b}{n_0}\cdot\dfrac{R_b}{B}$
Bridges symbol-rate and bandwidth views
Modulation Thresholds (BER $10^{-5}$)
Coherent PSK: 9.6 dB
Coherent FSK: 12.6 dB; Noncoh: 13.4 dB
Coherent ASK: 15.6 dB; Noncoh: 16.4 dB
QAM thresholds (BER $10^{-5}$)
16-QAM: $E_b/n_0 \approx 14.4$ dB
64-QAM: $E_b/n_0 \approx 22.5$ dB
SSB vs DSB-SC
SSB: SNR unchanged after demod
DSB-SC (coherent): +3 dB SNR gain
AM (DSB-LC) Efficiency
$\eta = \dfrac{m^2}{2+m^2}$, $m \le 1$
Max 33% at $m=1$ (carrier wastes power)
Part 2 — Noise Figure, Antenna & Distortion
Thermal Noise
$P_n = kTB$
$v_n = \sqrt{4kTBR}$
$kT_0 = -174$ dBm/Hz
Noise Figure ↔ $T_e$
$F = 1 + T_e/T_0$
$T_e = (F-1)T_0$
$T_0 = 290$ K
Friis Cascade ($F$)
$F = F_1 + \dfrac{F_2-1}{G_1} + \dfrac{F_3-1}{G_1G_2} + \cdots$
1st stage dominates → low-NF, high-gain LNA first
Cascade ($T_e$)
$T_e = T_1 + \dfrac{T_2}{G_1} + \dfrac{T_3}{G_1G_2} + \cdots$
Lossy Line
$T_e = (L-1)T_p$
$F = L$ when $T_p = T_0$
$F$(dB) $=$ IL(dB)
Antenna Noise Temp.
$T_a = \eta_{rad}T_b + (1-\eta_{rad})T_p$
Lossless: $T_a = T_b$
System $T_{SYS}$ (with line)
$T_{SYS} = T_a + T_{TL} + L_T\,T_{REC}$
$T_{TL} = (L_T-1)T_p$
Friis Transmission
$P_r = P_t G_t G_r\!\left(\dfrac{\lambda}{4\pi R}\right)^{\!2}$
EIRP $= G_t P_t$
Free-Space Path Loss
FSPL $= 20\log_{10}\!\left(\dfrac{4\pi R}{\lambda}\right)$
$= 32.4 + 20\log f_{\text{MHz}} + 20\log R_{\text{km}}$
G/T Figure of Merit
$G/T$ (dB/K) $= G_r$(dBi) $- 10\log_{10}(T_a + T_e)$
CNR (Link Budget)
CNR $= \text{EIRP} + G/T - \text{FSPL} - 10\log(kB)$
$10\log(kB) = -228.6 + 10\log B$
Min. Detectable Signal
$S_{i,min} = kB[T_a + (F-1)T_0]\,(S_o/N_o)_{min}$
SNR ↔ $E_b/n_0$
$S/N = (E_b/n_0)(R_b/B)$
BPSK/QPSK: $E_b/n_0 = 9.6$ dB → BER $10^{-5}$
BER (BPSK/QPSK)
$P_e = \tfrac{1}{2}\mathrm{erfc}\!\left(\sqrt{E_b/n_0}\right)$
IP3 vs P1dB
OIP3 $\approx$ OP1dB $+ 10.6$ dB
IM3 slope = 3:1, fundamental = 1:1
Cascaded IP3 (output)
$\dfrac{1}{\text{OIP}_3} = \dfrac{1}{G_2\,\text{OIP}_{3,1}} + \dfrac{1}{\text{OIP}_{3,2}} + \cdots$
Last stage dominates (opposite of noise)
Dynamic Range
$DR_l = $ P1dB $- N_o$
$DR_f = \tfrac{2}{3}(\text{OIP}_3 - N_o)$
$DR_l - DR_f \approx 21.3$ dB
Y-factor (Noise Measurement)
$T_e = \dfrac{T_h - YT_c}{Y - 1}$
$Y = N_h/N_c$
Part 1 — Random Processes, PSD & Modulation
Part 2 — Noise Figure, Antenna & Distortion