微波系統中的雜訊與失真Noise and Distortion in Microwave Systemsマイクロ波システムにおける雑音と歪み

2.1 線性接收機中的雜訊與 BER2.1 Noise and BER in Linear Receivers2.1 線形受信機における雑音と BER

2.1.1 雜訊源的分類2.1.1 Classification of Noise Sources2.1.1 雑音源の分類

2.1.2 熱雜訊電壓與功率2.1.2 Thermal Noise Voltage and Power2.1.2 熱雑音電圧と電力

普朗克表示式Planck Expressionプランクの式

其中各變數為：where each variable is:ここで各変数は次のとおり：

瑞立–金斯近似 ($hf \ll kT$)Rayleigh–Jeans Approximation ($hf \ll kT$)レイリー–ジーンズ近似 ($hf \ll kT$)

泰勒展開：$e^{hf/kT}-1 \approx hf/kT$，可得：Taylor expand: $e^{hf/kT}-1 \approx hf/kT$, yielding:テイラー展開：$e^{hf/kT}-1 \approx hf/kT$ より、次を得ます：

可用雜訊功率Available Noise Power利用可能雑音電力

匹配負載得到開路電壓的一半 → 傳送的功率為：Matched load receives half the open-circuit voltage → power delivered:整合負荷は開放電圧の半分を受け取ります → 供給される電力は：

戴維寧模型與最大功率傳輸Thévenin Model & Maximum Power Transferテブナン等価モデルと最大電力伝達

含雜訊的電阻 $R$ 可模型化為A noisy resistor $R$ is modeled as a雑音を持つ抵抗 $R$ は 戴維寧等效電路Thévenin equivalentテブナン等価回路：一個承載全部雜訊的理想電壓源 $v_n^s$，與無雜訊的內電阻 $R_s = R$ 串聯。這與任何其他戴維寧電源完全相同——內電阻就是往電源方向看進去的阻抗。: 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.としてモデル化されます。すなわち、すべての雑音を担う理想電圧源 $v_n^s$ と、雑音のない内部抵抗 $R_s = R$ の直列接続です。これは他のテブナン電源とまったく同じで、内部抵抗とは電源側を見込んだインピーダンスにほかなりません。

當接上負載 $R_L$ 時，會形成When the load $R_L$ is connected, a負荷 $R_L$ を接続すると、 分壓器voltage divider分圧回路 。負載上的電壓為：forms. The voltage across the load is:が形成されます。負荷にかかる電圧は次のとおりです：

一般情況下傳送到負載的功率為：The power delivered to the load in general:一般に負荷へ供給される電力は次のとおりです：

也可以用比值 $r = R_L/R_s$ 來表示：This can also be written in terms of the ratio $r = R_L/R_s$:比 $r = R_L/R_s$ を用いて次のように書くこともできます：

根據By theいわゆる 最大功率傳輸定理maximum power transfer theorem最大電力伝達の定理，$P_{load}$ 在 $R_L = R_s$ 時（即 $r = 1$）達到最大。取導數 $\frac{d}{dr}\!\left[\frac{r}{(1+r)^2}\right] = \frac{1-r}{(1+r)^3} = 0$ 可確認峰值位於 $r=1$。, $P_{load}$ 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 $r=1$.により、$P_{load}$ は $R_L = R_s$ のとき（すなわち $r = 1$）最大になります。微分 $\frac{d}{dr}\!\left[\frac{r}{(1+r)^2}\right] = \frac{1-r}{(1+r)^3} = 0$ を計算すれば、ピークが $r=1$ にあることが確認できます。

$1/4$ 是怎麼來的？Where Does the $1/4$ Come From?$1/4$ はどこから来るのか？

在一般功率表示式中令 $R_L = R_s = R$（匹配）：Set $R_L = R_s = R$ (matched) in the general power expression:一般的な電力の式で $R_L = R_s = R$（整合）とおきます：

現在代入 $v_n^s = \sqrt{4kTBR}$（來自瑞立–金斯的結果）：Now substitute $v_n^s = \sqrt{4kTBR}$ (from the Rayleigh–Jeans result):ここで $v_n^s = \sqrt{4kTBR}$（レイリー–ジーンズの結果）を代入します：

分子的因數 4（來自普朗克／瑞立–金斯）與分母的因數 4（來自匹配分壓）恰好相消——這就是為什麼最大可用雜訊功率 $kTB$ 與電阻 $R$ 無關。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 $R$.分子の係数 4（プランク／レイリー–ジーンズ由来）と分母の係数 4（整合した分圧回路由来）はちょうど打ち消し合います。これが、最大利用可能雑音電力 $kTB$ が抵抗 $R$ に依存しない理由です。

下圖顯示這個關係。將滑鼠移到曲線上可讀出精確數值。紅點標示匹配條件 $R_L = R_s$，此時傳送的功率最大（$= kTB$）：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:下のグラフはこの関係を示します。曲線上にカーソルを合わせると正確な値が読み取れます。赤い点は整合条件 $R_L = R_s$ を示し、このとき供給電力が最大（$= kTB$）になります：

2.1.3 白高斯雜訊與功率頻譜密度2.1.3 White Gaussian Noise and PSD2.1.3 白色ガウス雑音と PSD

其withであり、 均值為零zero mean平均は零 （鐘形曲線以 $v_n = 0$ 為中心）。(the bell curve is centered at $v_n = 0$).です（釣鐘型曲線は $v_n = 0$ を中心とします）。

Key Parameters of the Gaussian PDF

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:

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:

This is a Dirac delta function — infinite at $\tau = 0$ and zero everywhere else:

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:

AWGN — The Standard Channel Model

Combining "white" (flat PSD) + "Gaussian" (amplitude PDF) gives AWGN — Additive White Gaussian Noise:

2.1.3b Why Integrating PSD Gives Power — 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:

For an ergodic stationary random process — which thermal noise is — all three give the same numerical result.

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:

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.

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:

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

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 $1/2$:

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:

SSB vs DSB-SC: Signal Folding Comparison

The noise folds identically in both cases. The difference is what happens to the signal:

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:

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).

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:

Full Three-Way Comparison

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:

The two signals are simply negatives of each other — same amplitude, same frequency, opposite phase. Written compactly: $s(t) = \pm v_0\cos(\omega_c t)$.

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:

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.

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.

那對 $-v_0$（sent "0"）而言呢？And what about $-v_0$ (sent "0")?$-v_0$（"0" を送った場合）ではどうか？

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When a "0" is sent the signal sits at $-v_0$ and the whole picture is mirrored left-to-right. An error now occurs when the noise pushes the signal "0" を送るとき信号は $-v_0$ にあり、図全体が左右反転します。誤りが起こるのは、雑音が信号を往右推過 0to the right, past 0右へ押しやって 0 を越えさせた，誤判為 "1"：, so it is misread as a "1":とき、つまり "1" と誤判定されるときです：

為什麼不直接把門檻移開，讓錯誤不再發生？Why not just move the threshold so the errors stop happening?しきい値をずらして誤りが起きないようにすればよいのでは？

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：It sounds reasonable — move the threshold next to $+v_0$ and the "1" side will hardly ever be wrong. But there is a fatal cost:もっともらしく聞こえます — しきい値を $+v_0$ のそばへ動かせば "1" 側はほとんど誤らなくなります。しかしそこには致命的な代償があります：

真正的解法：拉開兩個訊號之間的距離The real fix: pull the two signals further apart本当の解決策：2 つの信号の間隔を広げる

縮小尾巴的正確方法不是移門檻，而是The right way to shrink the tails is not to move the threshold but to 裾を小さくする正しい方法はしきい値を動かすことではなく、增加 $+v_0$ 和 $-v_0$ 之間的間距increase the separation between $+v_0$ and $-v_0$$+v_0$ と $-v_0$ の間隔を広げる，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：, that is, to raise the signal power. Both distributions then move away from the threshold at once and both tails shrink:こと、つまり信号電力を上げることです。そうすれば両方の分布が同時にしきい値から遠ざかり、両側の裾がともに小さくなります：

Key erfc Values

The Two Forms — Why They Are Equal

The two expressions in the equation above are related by:

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.

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:

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:

2.1.6 BER Improvement by Integrate-and-Dump

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:

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:

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:

Plugging in:

2.1.7 Comparison of Modulation Methods

BPSK vs QPSK

Comparison of bit error rates for ASK, FSK, and PSK detection. PSK achieves the lowest $E_b/n_0$ for a given BER. (Pozar Fig. 14.4)

BER vs $E_b/n_0$ for QPSK. At $P_e = 10^{-5}$, the required $E_b/n_0 = 9.6$ dB — same as BPSK.

2.1 Noise and BER in Linear Receivers (continued)

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.

  Q
 +3  •  •  •  •
 +1  •  •  •  •
 -1  •  •  •  •
 -3  •  •  •  •
     -3 -1 +1 +3  → I

 Each • = one unique 4-bit symbol

Transmitter

1. Split the bitstream into groups of 4 bits
2. Map 2 bits → I amplitude level (one of $\pm1, \pm3$)
3. Map 2 bits → Q amplitude level (one of $\pm1, \pm3$)
4. Modulate: $s(t) = I\cos(\omega_c t) - Q\sin(\omega_c t)$

Receiver

1. I/Q demodulation — multiply by $\cos(\omega_c t)$ and $\sin(\omega_c t)$, integrate over $T_s$
2. Get two numbers: one I value and one Q value
3. Decision — compare each against 3 thresholds (not 1 like BPSK/QPSK): $-2, 0, +2$
4. Map each detected level back to 2 bits → combine for 4 bits total

2.2 Noise Temperature and Noise Figure

2.2.1 Equivalent Noise Temperature

One-port: A white noise source delivers power $N_o$ into a matched load $R$. The equivalent noise temperature is $T_e = N_o/(kB)$.

Two-port: A noisy amplifier (left) is modeled as a noiseless amplifier with a noise source $T_e$ at its input (right). With $T_s = 0$ K input, $T_e = N_o/(GkB)$.

2.2.2 Y-Factor Measurement

Y-factor method: switch between a hot source ($T_1$) and cold source ($T_2$), measure output powers $N_1$ and $N_2$, then compute $T_e = (T_1 - YT_2)/(Y-1)$ where $Y = N_1/N_2$.

2.2.3 Noise Figure

Noise figure setup: source at $T_0 = 290$ K with signal $S_i$ drives a noisy network ($G, B, T_e$). Input power $P_i = S_i + N_i$ where $N_i = kT_0B$. Output power $P_o = S_o + N_o$.

2.2.4 Noise Figure of a Lossy Line

A lossy line with $G = 1/L$, physical temperature $T$, and characteristic impedance $Z_0 = R$. Due to thermodynamic equilibrium, $N_o = N_i = kTB$ when the source is at the same temperature $T$.

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.

2.2.5 Cascaded Noise Figure (Friis Formula)

Two cascaded stages ($G_1, F_1, T_{e1}$) and ($G_2, F_2, T_{e2}$) are equivalent to a single stage with combined gain $G_1G_2$, overall noise figure $F$, and equivalent noise temperature $T_e$.

A simplified wireless receiver: LNA ($G=10$ dB, $F=2$ dB) → Bandpass filter ($L=1$ dB) → Mixer ($L=3$ dB, $F=4$ dB). Input $S_i, N_i$ at antenna, output $S_o, N_o$ after mixer.

2.2.6 Noise Figure of Mismatched Networks & Wilkinson Divider

Mismatched Passive Two-Port Network

A passive two-port network with S-parameters $[S]$, source impedance $Z_g$ ($\Gamma_s, \Gamma_{in}$), and load $Z_L$ ($\Gamma_{out}, \Gamma_L$). Noise $N_1 = kTB$ enters port 1, noise $N_2$ exits port 2.

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:

A voltage wave of initial amplitude $V^+$, after traveling distance $l$:

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:

For a two-port network:

Key properties

S-matrices of common devices

Application to a Mismatched Lossy Line

Lossy line with parameters $L, T, Z_0, \beta$ and length $l$. Source has reflection coefficient $\Gamma_s$, load is matched ($\Gamma_L = 0$). Mismatch at port 1 increases the noise figure because reflected noise re-enters the line.

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.

Wilkinson Power Divider (lossless, port 3 terminated)

2.2.7 Antenna Noise Temperature and Friis Transmission

Friis transmission: transmit antenna ($P_t$, gain $G_t$) and receive antenna (gain $G_r$, received power $P_r$) separated by distance $R$. EIRP $= G_t P_t$.

$P_t$: 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:

The power density radiated by an arbitrary transmit antenna with an antenna gain of $G_t$ at a distance $R$ is:

where $EIRP = G_t P_t$, representing "effective isotropic radiated power".

The power received by a receive antenna with an effective aperture area is:

The Friis equation can be derived from the above two expressions as:

Left: Matched source resistance at $T_p$ driving a noisy receiver ($N_i = kT_pB$). Right: Receive antenna with Friis-received signal $S_i$ and antenna noise $N_i = kT_aB$. In both cases $N_o = GkB(T_{source} + T_e)$.

Background noise temperature $T_B$ vs frequency for various elevation angles $\theta$. Below 10 GHz: zenith ($\theta=90°$) ≈ 3–5 K, horizon ($\theta=0°$) ≈ 290 K. Absorption peaks at H₂O (~22 GHz) and O₂ (~60 GHz).

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.

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:

2.2.9 Full Receiver Analysis ($T_a \neq T_0$)

Lecture Example — Full Receiver Analysis: Antenna + Transmission Line + Cascade (Chap 2 slides 40–42)

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.

Complete receiver chain: background $T_b$ → antenna ($G_A, \eta_{rad}, T_P$) → transmission line ($L_T, T_P$) → RF amplifier ($G_{RF}, T_{RF}$) → mixer with LO ($L_M, T_M$) → IF amplifier ($G_{IF}, T_{IF}$). Signal and noise reference planes: $N_i, S_i$ at the antenna-line junction, $N_o, S_o$ at the IF output.

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%).

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 $\omega_0$:

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.

2.3.3 Intermodulation Products

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:

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 Extrapolate both lines until they cross. The crossing input level is 2 本の直線を交わるまで外挿します。交点の入力レベルが 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).

On a dBm-vs-dBm plot: fundamental has slope 1, IM3 has slope 3. Their idealized extensions intersect at IP3.

2.3.5 Relation Between $P_{1\text{dB}}$ and IP3

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:

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.

2.3.7 Intercept Point of Cascaded Components

Two-stage cascade and its equivalent single-block representation:

Lecture Example — Cascaded IP3 (Amplifier + Mixer)

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!)

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.

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

Part 2 — Noise Figure, Temperature, Antenna & Link Budget

Part 3 — Distortion (IP3, P1dB, DR)

Quick Reference Summary

Important Symbols

Part 1 — Random Processes, PSD & Modulation

Part 2 — Noise Figure, Antenna & Distortion