3.1 Design for Specified Gain
In many cases it is preferable to design for less than the maximum available gain to improve noise figure, output power, and bandwidth. Impedance matches are chosen to optimize gain, noise figure, output power, or bandwidth.
Maximum gain requires conjugate matching at both ports ($\Gamma_s = S_{11}^*$, $\Gamma_l = S_{22}^*$). This leaves no room to optimize noise figure, bandwidth, or output power. By deliberately accepting less gain, you free up the matching network to pursue other objectives.
Common confusion: $Z_L = Z_S$ vs $Z_L = Z_S^*$
Max power transfer (and hence max amplifier gain at one port) requires the conjugate match, not equality:
- If $Z_S$ is purely real ($X_S = 0$), then $Z_S^* = Z_S$ and the two conditions coincide.
- If $Z_S$ is reactive ($X_S \neq 0$), the conjugate cancels the reactance: $Z_L = R_S - jX_S$. Setting $Z_L = Z_S$ would add reactance instead of canceling it, and you'd lose power.
For a two-port amplifier, this must hold at both ports simultaneously:
This simultaneous conjugate match exists only when the device is unconditionally stable ($K > 1$, $|\Delta| < 1$). Otherwise you trade gain for stability via constant-gain circles.
Why does the simultaneous match require unconditional stability?
First, the S-parameters of a two-port amplifier. Each $S_{ij}$ is a ratio of an outgoing wave at port $i$ to an incoming wave at port $j$, with all other ports terminated in $Z_0$ (no reflection back):
Physical meaning of $S_{11}$ (click to expand)
$S_{11}$ is the input reflection coefficient measured at port 1 with port 2 terminated in the reference impedance $Z_0$ (typically 50 Ω).
You send a travelling wave $a_1$ into port 1. $S_{11}$ is the ratio of the wave $b_1$ that bounces back out of port 1 to the wave you put in:
$$S_{11} = \frac{b_1}{a_1}\bigg|_{a_2=0}$$The condition $a_2 = 0$ means port 2 is matched — no wave arriving from the load side. This is the standard VNA measurement setup.
| Quantity | Meaning |
|---|---|
| $|S_{11}|$ | fraction of incident power reflected. 0 = perfect match, 1 = open/short |
| $\angle S_{11}$ | phase of reflection — tells whether mismatch is capacitive, inductive, or resistive (where on Smith chart the input impedance sits) |
| $-20\log|S_{11}|$ dB | return loss. A well-matched LNA input: ${\approx}{-15}$ dB ($|S_{11}|\approx 0.18$) |
Note: $S_{11}$ equals the input reflection coefficient $\Gamma_{in}$ only when port 2 is terminated in $Z_0$. When a real load $\Gamma_l \neq 0$ is connected and $S_{12}\neq 0$, the actual $\Gamma_{in}$ differs from $S_{11}$ — see eq. (3.1-1).
Here $a_i$ is the incident wave and $b_i$ the reflected/transmitted wave at port $i$, both normalized so $|a_i|^2$ and $|b_i|^2$ are powers. Two more quantities derived from them:
- $\Gamma_{in}$, $\Gamma_{out}$ — actual reflection coefficients looking into the amplifier's input/output when real (not $Z_0$) terminations $\Gamma_l$, $\Gamma_s$ are connected on the other side. These differ from $S_{11}$, $S_{22}$ whenever $S_{12} \ne 0$.
- $\Delta = S_{11}S_{22} - S_{12}S_{21}$ — the determinant of the S-matrix; appears in the stability factor $K$ and the simultaneous-match formulas.
A device is unilateral when $S_{12} \approx 0$ (no reverse leakage). Real transistors are never truly unilateral — that small $S_{12}$ is exactly what makes input and output coupled, and is the root cause of everything below.
The coupling problem. Through $S_{12}$, the load impedance feeds back into the input: changing $\Gamma_l$ changes $\Gamma_{in}$, and changing $\Gamma_s$ changes $\Gamma_{out}$.
$\Gamma_{out}$ vs $\Gamma_l$ — what is the difference? (click to expand)
They describe opposite directions at the same physical node (port 2):
| Direction | Who sets it | |
|---|---|---|
| $\Gamma_{out}$ | looking into port 2 of the transistor from the load side | fixed once you choose $\Gamma_s$ — device property |
| $\Gamma_l$ | looking into the load network from port 2 | your choice — set by the output matching network |
Analogy: $\Gamma_{out}$ is the Thévenin impedance of the transistor's output port. $\Gamma_l$ is the impedance of the load you connect to it. They face each other across the same reference plane.
When $\Gamma_l = \Gamma_{out}^*$: conjugate match — load = complex conjugate of the transistor's output impedance. Maximum power transfer, $G_T = G_A$.
When $\Gamma_l \neq \Gamma_{out}^*$: reflected power at port 2, $G_T < G_A$.
So $\Gamma_s = \Gamma_{in}^{\,*}$ and $\Gamma_l = \Gamma_{out}^{\,*}$ form a coupled system of two equations in two unknowns. Solving simultaneously gives the closed-form $\Gamma_{Ms}$, $\Gamma_{Ml}$:
with $B_1 = 1 + |S_{11}|^2 - |S_{22}|^2 - |\Delta|^2$, $\ C_1 = S_{11} - \Delta S_{22}^{\,*}$, and $\Delta = S_{11}S_{22} - S_{12}S_{21}$.
Where stability comes in. A physically realizable solution requires $|\Gamma_{Ms}| < 1$ and $|\Gamma_{Ml}| < 1$ — both reflection coefficients must live inside the Smith chart (passive terminations). The condition for this is the Rollett stability factor:
Equivalently $B_1^2 - 4|C_1|^2 > 0$, so the square-root term is real and the solution doesn't escape the unit disk.
What goes wrong when $K \le 1$ (potentially unstable).
- The discriminant $B_1^2 - 4|C_1|^2$ can be negative ⇒ $\Gamma_{Ms}$ becomes complex with $|\Gamma_{Ms}| = 1$ on a circle, or both roots fall outside the unit circle ⇒ no passive match exists.
- Even if a formal solution sits inside the unit circle, the conjugate-match terminations push $|\Gamma_{in}|$ or $|\Gamma_{out}|$ toward 1, where infinitesimal mismatches trigger oscillation. Trying to extract maximum gain from a potentially unstable device makes it oscillate.
- The "stability circles" on the Smith chart bound the region of $\Gamma_s$ (or $\Gamma_l$) that keeps $|\Gamma_{out}| < 1$ (or $|\Gamma_{in}| < 1$). When $K \le 1$, those circles cut into the unit disk and the simultaneous-match point lies outside the safe region.
Workaround when $K \le 1$. Don't aim for max gain. Pick $\Gamma_s$, $\Gamma_l$ on constant-gain circles well inside the stable region — accepting some gain reduction in exchange for $|\Gamma_{in}|, |\Gamma_{out}| < 1$ with margin. Or apply resistive loading / source degeneration to push $K$ above 1, then conjugate-match the now-unconditionally-stable device.
One-line summary: unconditional stability ensures the conjugate-match equations have a passive solution; without it, the same equations either have no valid solution or land on terminations that make the amplifier oscillate.
Proof: why $Z_L = Z_S^*$ maximizes power transfer
Source EMF $V_s$ with internal impedance $Z_S = R_S + jX_S$ drives a load $Z_L = R_L + jX_L$. The loop current is
$$I = \frac{V_s}{Z_S + Z_L}, \qquad |I|^2 = \frac{|V_s|^2}{(R_S+R_L)^2 + (X_S+X_L)^2}$$Only the resistive part of the load dissipates real power:
$$P_L = \tfrac{1}{2}|I|^2 R_L = \frac{|V_s|^2}{2}\cdot\frac{R_L}{(R_S+R_L)^2 + (X_S+X_L)^2}$$Step 1 — choose $X_L$. $X_L$ appears only in the denominator as $(X_S+X_L)^2 \ge 0$. To minimize the denominator (= maximize $P_L$):
Step 2 — choose $R_L$. With $X_L = -X_S$:
$$P_L = \frac{|V_s|^2}{2}\cdot\frac{R_L}{(R_S+R_L)^2}$$Differentiate w.r.t. $R_L$ and set to zero:
$$\frac{dP_L}{dR_L} = \frac{|V_s|^2}{2}\cdot\frac{(R_S+R_L)^2 - 2R_L(R_S+R_L)}{(R_S+R_L)^4} = \frac{|V_s|^2}{2}\cdot\frac{R_S - R_L}{(R_S+R_L)^3} = 0$$Combine both steps:
Maximum deliverable power (the available power of the source):
$$P_{L,\max} = \frac{|V_s|^2}{8\,R_S}$$Intuition. $X_L = -X_S$ tunes out reactive energy storage so the loop becomes purely resistive (resonance). $R_L = R_S$ then balances the resistive divider — half the source voltage drops across $R_L$ — delivering the most real power for a given $R_S$.
3.1.1 Interactive Tutorial: Reading & Using the Smith Chart
The Smith chart is a coordinate transform of the impedance plane onto the unit disk via $\Gamma = (Z-Z_0)/(Z+Z_0)$. Use the tabs below to learn one concept at a time. Click anywhere on the chart to drop / move a marker — the readout updates in real time.
What is VSWR?
VSWR — Voltage Standing Wave Ratio. When the load isn't matched, part of the incident wave reflects back. Forward + reflected waves interfere along the line, producing a standing-wave pattern with peaks $|V|_{\max}$ and nulls $|V|_{\min}$.
$$\text{VSWR} = \frac{|V|_{\max}}{|V|_{\min}} = \frac{1+|\Gamma|}{1-|\Gamma|}$$| $|\Gamma|$ | VSWR | Return loss | Meaning |
|---|---|---|---|
| 0 | 1.0 | $\infty$ dB | perfect match |
| 0.2 | 1.5 | 14 dB | very good |
| 0.33 | 2.0 | 9.5 dB | acceptable |
| 0.5 | 3.0 | 6 dB | poor |
| 1 | $\infty$ | 0 dB | total reflection (short / open) |
- Real number $\geq 1$ — depends only on $|\Gamma|$, not its phase.
- Constant along a lossless line (sliding the reference plane just rotates $\Gamma$ on its constant-$|\Gamma|$ circle).
- VSWR $=1$ $\Leftrightarrow$ matched $\Leftrightarrow$ no standing wave, just a traveling wave.
What is return loss?
Return loss (RL) — how much of the incident power is not reflected back, expressed in dB. A measure of match quality.
$$\text{RL} = -20\log_{10}|\Gamma|\;\;[\text{dB}] \;=\; -10\log_{10}\!\frac{P_{\text{ref}}}{P_{\text{inc}}}$$Defined as a positive number (the leading minus sign cancels the fact that $|\Gamma|\leq 1$ makes the log negative). Bigger RL = better match.
| $|\Gamma|$ | Reflected power | RL |
|---|---|---|
| 0 | 0% | $\infty$ dB (perfect match) |
| 0.1 | 1% | 20 dB |
| 0.2 | 4% | 14 dB |
| 0.316 | 10% | 10 dB |
| 0.5 | 25% | 6 dB |
| 1.0 | 100% | 0 dB (short/open) |
Physical meaning — what is the "loss"?
If you send 1 W of incident power into a port:
- RL = 10 dB → 0.1 W reflects back, 0.9 W actually enters the device
- RL = 20 dB → 0.01 W reflects, 0.99 W enters
- RL = 5 dB → 0.32 W reflects, only 0.68 W enters (≈ 32% wasted as reflection)
From the source's perspective, the reflected power never delivered useful work to the load — it is "lost" to the system, even though it isn't dissipated as heat. That's why it's called return loss: signal power wasted by bouncing back instead of going through.
This is exactly what is happening in the amplifier design below — by deliberately mismatching to back off from $G_{S,\max}/G_{L,\max}$ to hit a lower target gain, RL drops to ~5 dB, meaning about a third of the incident power is reflected at the port instead of being amplified.
- Same information as VSWR or $|\Gamma|$ — just a different unit. Engineers prefer dB because matching specs are usually written that way (e.g. "RL $>$ 15 dB across the band").
- "10 dB return loss" is a common minimum spec; 15–20 dB is good for most RF designs.
- Not the same as insertion loss (power lost going through a component, including reflection + dissipation).
What are conductance & susceptance? (admittance Smith chart)
Admittance is the reciprocal of impedance: $Y = 1/Z$. Splitting into real and imaginary parts:
$$Y = G + jB \;\;[\text{S, siemens}] \qquad y = Y\cdot Z_0 = g + jb \;\;\text{(normalized)}$$- $G$ — conductance: real part of $Y$. How much current flows in phase with the voltage (resistive part of the load). $G = 1/R$ only when $X=0$; in general $G = R/(R^2+X^2)$.
- $B$ — susceptance: imaginary part of $Y$. How much current flows $90°$ out of phase with the voltage (reactive part). $B = -X/(R^2+X^2)$.
Sign convention is opposite to reactance (because of the $1/Z$ flip):
| Element | Reactance $X$ | Susceptance $B$ |
|---|---|---|
| Inductor $L$ | $+\omega L$ (positive) | $-1/(\omega L)$ (negative) |
| Capacitor $C$ | $-1/(\omega C)$ (negative) | $+\omega C$ (positive) |
Why use admittance? For elements in series, impedances add: $Z_{tot} = Z_1 + Z_2$. For elements in parallel (shunt), admittances add: $Y_{tot} = Y_1 + Y_2$. So when you tack on a shunt L or C in a matching network, the math is clean in admittance — and on the Smith chart the move is along a constant-$g$ circle (resistance‑equivalent: constant-$r$ stays put under series additions; constant-$g$ stays put under shunt additions).
The admittance Smith chart is the impedance chart rotated $180°$ (equivalently, $\Gamma \to -\Gamma$). Constant-$g$ circles now pass through the left edge ($\Gamma=-1$, $y=\infty$ = short), and the upper/lower halves swap reactive sense:
● blue circles = constant $g$ (conductance) · ● orange arcs = constant $b$ (susceptance) · centre = $y=1$ (perfect match).
Compared with the impedance chart:
| Z-Smith chart | Y-Smith chart | |
|---|---|---|
| Right edge ($\Gamma=+1$) | $z=\infty$ (open) | $y=0$ (open) |
| Left edge ($\Gamma=-1$) | $z=0$ (short) | $y=\infty$ (short) |
| Upper half | $x>0$ inductive | $b<0$ inductive |
| Lower half | $x<0$ capacitive | $b>0$ capacitive |
| Series element moves along | constant-$r$ circle | — (use Z view) |
| Shunt element moves along | — (use Y view) | constant-$g$ circle |
3.1.2 Constant Gain Circles — Unilateral Case
When $S_{12} \approx 0$ (unilateral assumption), the transducer gain factors into three independent pieces:
where
$$G_S(\Gamma_s) = \frac{1-|\Gamma_s|^2}{|1-S_{11}\Gamma_s|^2}, \qquad G_L(\Gamma_l) = \frac{1-|\Gamma_l|^2}{|1-S_{22}\Gamma_l|^2}$$$G_S$ depends only on $\Gamma_s$ and $G_L$ depends only on $\Gamma_l$. This means input and output matching networks can be designed independently.
Maximum values
Normalized gain factors
Define $g_S = G_S / G_{S,\max}$ and $g_L = G_L / G_{L,\max}$, so $0 \le g_S \le 1$ and $0 \le g_L \le 1$.
$$g_S = \frac{(1-|\Gamma_s|^2)(1-|S_{11}|^2)}{|1-S_{11}\Gamma_s|^2}, \qquad g_L = \frac{(1-|\Gamma_l|^2)(1-|S_{22}|^2)}{|1-S_{22}\Gamma_l|^2}$$Constant gain circles on the Smith chart
Setting $g_S = \text{const}$ traces a circle on the $\Gamma_s$ plane:
Similarly for the output:
- At maximum gain ($g_S=1$): $C_S = S_{11}^*$, $R_S = 0$ — the circle collapses to a point at the conjugate match.
- At 0 dB gain ($G_S=1$, i.e. $\Gamma_s=0$): the circle always passes through the center of the Smith chart.
- Centers lie along a line from the origin toward $S_{11}^*$ (or $S_{22}^*$) on the Smith chart.
Drag $S_{11}$ on the Smith chart (or use the sliders) and sweep $g_S$. The dashed orange ray from origin through $S_{11}^*$ is the locus of every $C_S$. As $g_S$ runs from $0$ to $1$, the highlighted circle slides along that line from the unit-circle limit down to the single point $S_{11}^*$.
Why? $C_S = \dfrac{g_S\,S_{11}^{\,*}}{1-(1-g_S)|S_{11}|^2}$. The denominator is a positive real number for $0 \le g_S \le 1$, and $g_S$ is real, so $C_S$ has the same phase as $S_{11}^{\,*}$ — only its magnitude changes with $g_S$.
S11* =
GS,max= dB
GS = dB
CS =
RS =
Try the limits:
• $g_S \to 1$ → red circle shrinks to a point at $S_{11}^*$ (max gain).
• $g_S$ small → red circle bulges out and passes through the chart origin (0 dB gain).
• Drag $S_{11}$ around → the dashed ray pivots; centers always live on it.
Show derivation of the constant gain circle formula
Start from $g_S|1-S_{11}\Gamma_s|^2 = (1-|\Gamma_s|^2)(1-|S_{11}|^2)$.
Expand both sides:
$$g_S(1 - S_{11}\Gamma_s - S_{11}^*\Gamma_s^* + |S_{11}|^2|\Gamma_s|^2) = (1-|S_{11}|^2) - (1-|S_{11}|^2)|\Gamma_s|^2$$Collect $|\Gamma_s|^2$ terms and $\Gamma_s$ terms, then complete the square:
$$\left|\Gamma_s - \frac{g_S S_{11}^*}{1-(1-g_S)|S_{11}|^2}\right|^2 = \left[\frac{\sqrt{1-g_S}(1-|S_{11}|^2)}{1-(1-g_S)|S_{11}|^2}\right]^2$$This is a circle with center $C_S$ and radius $R_S$.
GaAs MESFET: $S_{11}=0.75\angle{-120°}$, $S_{21}=2.5\angle{80°}$, $S_{12}\approx 0$, $S_{22}=0.6\angle{-70°}$.
▶ Click to show full solution
Step 1: Compute maximum gains
$$G_0 = |S_{21}|^2 = 6.25 = 8.0\text{ dB}$$ $$G_{S,\max} = \frac{1}{1-0.75^2} = 2.286 = 3.6\text{ dB}$$ $$G_{L,\max} = \frac{1}{1-0.6^2} = 1.563 = 1.9\text{ dB}$$ $$G_{TU,\max} = 3.6+8.0+1.9 = 13.5\text{ dB}$$Step 2: Choose gain allocation for 11 dB total
$G_S = 2$ dB, $G_L = 1$ dB, $G_0 = 8$ dB → total $= 11$ dB.
Step 3: Compute circle parameters
$G_S = 2$ dB: $g_S = 2/2.286 = 0.691$ → $C_S = 0.627\angle{120°}$, $R_S = 0.294$
How is $C_S = 0.627\angle 120°$, $R_S = 0.294$ obtained from $g_S = 0.691$?
Use the unilateral constant-gain circle formulas:
$$C_S = \frac{g_S\, S_{11}^{\,*}}{1 - (1-g_S)|S_{11}|^2}, \qquad R_S = \frac{\sqrt{1-g_S}\,(1-|S_{11}|^2)}{1 - (1-g_S)|S_{11}|^2}.$$With $S_{11} = 0.75\angle{-120°}$ ⇒ $|S_{11}| = 0.75$, $|S_{11}|^2 = 0.5625$, $S_{11}^{\,*} = 0.75\angle{+120°}$, and $g_S = 0.691$, $1-g_S = 0.309$:
Common denominator $D = 1 - (1-g_S)|S_{11}|^2$:
$$D = 1 - 0.309 \times 0.5625 = 1 - 0.1738 = 0.8262.$$Center (phase inherited from $S_{11}^{\,*}$ since denominator is real positive):
$$C_S = \frac{0.691 \times 0.75\angle{120°}}{0.8262} = \frac{0.518\angle{120°}}{0.8262} = 0.627\angle{120°}.$$Radius:
$$\sqrt{1-g_S} = \sqrt{0.309} = 0.556, \quad 1-|S_{11}|^2 = 0.4375$$ $$R_S = \frac{0.556 \times 0.4375}{0.8262} = \frac{0.2432}{0.8262} = 0.294.$$The center sits on the radial line from origin through $S_{11}^{\,*}$; the far edge of the circle (distance $C_S + R_S = 0.921 \approx |S_{11}|$) approaches the conjugate-match point as $g_S \to 1$.
$G_L = 1$ dB: $g_L = 1.259/1.563 = 0.806$ → $C_L = 0.520\angle{70°}$, $R_L = 0.303$
How is $C_L = 0.520\angle 70°$, $R_L = 0.303$ obtained from $g_L = 0.806$?
Same formulas with $S_{22}$ instead of $S_{11}$:
$$C_L = \frac{g_L\, S_{22}^{\,*}}{1 - (1-g_L)|S_{22}|^2}, \qquad R_L = \frac{\sqrt{1-g_L}\,(1-|S_{22}|^2)}{1 - (1-g_L)|S_{22}|^2}.$$With $S_{22} = 0.6\angle{-70°}$ ⇒ $|S_{22}| = 0.6$, $|S_{22}|^2 = 0.36$, $S_{22}^{\,*} = 0.6\angle{+70°}$, and $g_L = 0.806$, $1-g_L = 0.194$:
Common denominator $D = 1 - (1-g_L)|S_{22}|^2$:
$$D = 1 - 0.194 \times 0.36 = 1 - 0.0698 = 0.9302.$$Center:
$$C_L = \frac{0.806 \times 0.6\angle{70°}}{0.9302} = \frac{0.4836\angle{70°}}{0.9302} = 0.520\angle{70°}.$$Radius:
$$\sqrt{1-g_L} = \sqrt{0.194} = 0.440, \quad 1-|S_{22}|^2 = 0.64$$ $$R_L = \frac{0.440 \times 0.64}{0.9302} = \frac{0.2816}{0.9302} = 0.303.$$Step 4: Choose $\Gamma_s$, $\Gamma_l$
Pick points on the gain circles closest to the Smith chart center (smaller matching elements, wider bandwidth):
Why "smaller matching elements, wider bandwidth"?
Both follow from being closer to the Smith chart center ($\Gamma = 0$, i.e. $Z = Z_0$):
1. Smaller matching elements. The matching network has to transform $Z_0$ into the impedance corresponding to your chosen $\Gamma$. The further $|\Gamma|$ is from the center, the more reactance the network must add:
| $|\Gamma|$ | Normalized $z$ | Network burden (stub = 短截線) |
|---|---|---|
| 0 | 1 (matched) | no element needed |
| 0.33 | $\sim 2$ or $\sim 0.5$ | modest L/C, short stub |
| 0.7 | $\sim 5$ or $\sim 0.2$ | large reactance, long stub |
| $\to 1$ | highly reactive | stub length $\to \lambda/4$, very high L/C |
2. Wider bandwidth. The loaded $Q$ of an L-section (or any reactive matching network) is set by the impedance transformation ratio:
How to get $R_{\text{high}}$ and $R_{\text{low}}$ from a chosen $\Gamma$
For an L-section matching $Z_0$ (the 50 Ω reference) to the impedance at your chosen $\Gamma$:
$$R_{\text{high}} = \max(Z_0,\; R_\Gamma), \qquad R_{\text{low}} = \min(Z_0,\; R_\Gamma)$$where $R_\Gamma = \mathrm{Re}(Z_\Gamma)$ is the real part of the impedance at point $\Gamma$:
$$Z_\Gamma = Z_0\,\frac{1+\Gamma}{1-\Gamma}, \qquad R_\Gamma = Z_0\,\frac{1-|\Gamma|^2}{|1-\Gamma|^2}.$$The reactive part $X_\Gamma$ is absorbed by the L-section and does not enter the resistance ratio.
Example with this homework. With $\Gamma_s = 0.333\angle 120°$ ($Z_0 = 50\,\Omega$):
- $z_s = (1+\Gamma_s)/(1-\Gamma_s) \approx 0.616 + j\,0.399$
- $R_\Gamma = 0.616 \times 50 = 30.8\,\Omega$
- $R_{\text{high}}/R_{\text{low}} = 50/30.8 = 1.62$
- $Q = \sqrt{1.62 - 1} \approx 0.79$ → wide bandwidth
Compare with the conjugate-match point $\Gamma_s = S_{11}^* = 0.75\angle 120°$ on the outer edge of the same gain circle:
- $z_s \approx 0.189 + j\,0.562$, $R_\Gamma = 9.46\,\Omega$
- $Q = \sqrt{50/9.46 - 1} \approx 2.07$ → ~2.6× narrower bandwidth
Same gain (both points lie on the $G_S = 2$ dB circle), but the center-side point has a much lower $Q$. That is the entire reason you pick the closest-to-center option.
- $|\Gamma|$ close to $0$ → $R_{\text{high}}/R_{\text{low}} \approx 1$ → $Q \approx 0$ → wide bandwidth.
- $|\Gamma|$ close to $1$ → huge resistance ratio → high $Q$ → narrow bandwidth.
Since bandwidth $\sim f_0/Q$, low-$Q$ networks pass a wider frequency range without significant gain droop or return-loss degradation.
The trade-off: the only hard constraint is "stay on the gain circle." Anywhere on it gives the target gain, so within that freedom you pick the point closest to the center for the smallest network and best bandwidth. Other reasons to leave the center (e.g. hitting a noise-figure target) only matter when they conflict with the gain constraint.
中文摘要：越靠近 Smith 圖中心 ($\Gamma=0$ 即 $Z=Z_0$)，匹配網路需要做的阻抗轉換越小：所需電感、電容或短截線長度都較小，且網路 $Q$ 值低 ($Q=\sqrt{R_{\text{high}}/R_{\text{low}}-1}$)，因此頻寬較寬。增益圓上任何點都能達到目標增益，因此在圓上挑最靠近中心的點，是免費換來最小元件與最寬頻寬的選擇。
術語：「stub」中文為短截線（亦稱支線、短路線）。常見組合：open stub＝開路短截線，shorted stub＝短路短截線，quarter-wave stub＝四分之一波長短截線（$\lambda/4$ 短截線）；single / double-stub matching 即單／雙短截線匹配。
Step 5 — Realize the matching networks. With $\Gamma_s = 0.333\angle{120°}$ and $\Gamma_l = 0.217\angle{70°}$, design open-circuited single-stub matching on each side. The completed schematic (line lengths in $\lambda$, all $Z_0 = 50\,\Omega$):
Input network: 0.179 λ — series LINE length (ℓ₁) + 0.100 λ — shunt open STUB length (ℓₛ). Output network: 0.045 λ — series LINE (ℓ₁) + 0.432 λ — shunt open STUB (ℓₛ).
- Horizontal segments between blocks (along the signal path) = series LINE of length ℓ₁ — provides the rotation along the constant-|Γ| circle on the Smith chart.
- Vertical pieces branching to ground = shunt open STUB of length ℓₛ — provides the pure susceptance that cancels the residual b, finishing the match.
Reading the schematic: how each element realizes the chosen $\Gamma_s$ / $\Gamma_l$
Topology (left → right).
| Block | Components | Purpose |
|---|---|---|
| Source | 50 Ω generator + ground | provides the incident wave |
| Input match | 0.179 λ series line (50 Ω) + 0.100 λ shunt open stub (50 Ω) | transforms $Z_0$ into $\Gamma_s = 0.333\angle 120°$ seen at the device input |
| Active device | FET (uses the given S-parameters) | provides the gain $G_T = 11$ dB |
| Output match | 0.045 λ series line (50 Ω) + 0.432 λ shunt open stub (50 Ω) | presents $\Gamma_l = 0.217\angle 70°$ to the device output, then matches to 50 Ω load |
| Load | 50 Ω | sinks the amplified signal |
Why each piece works.
- All lines are 50 Ω. The characteristic impedance is fixed at $Z_0$. Impedance transformation comes from the line's electrical length $\beta\ell$, not from changing $Z_0$.
- Series line (e.g. 0.179 λ): rotates the impedance point clockwise on the Smith chart along a constant-$|\Gamma|$ circle by $2\beta\ell$ (0.179 λ ≈ 129° round-trip phase).
- Shunt open stub (e.g. 0.100 λ): looks like a pure susceptance $jB = jY_0\tan(\beta\ell)$ from the main line — open at the far end, so any $B$ is reachable by choosing $\ell \in (0, \lambda/4)$. It slides the operating point along a constant-conductance ($g$) circle on the Y-Smith chart.
- "Open" rather than shorted: at microwave frequencies an open termination is easier on a PCB than a shorted one — a shorted stub would need a plated-through via to ground.
Recipe (input side) to realize $\Gamma_s = 0.333\angle 120°$:
- Start at the chart center ($Z_0 = 50\,\Omega$, $\Gamma = 0$, coming from the source).
- Add the 0.100 λ shunt open stub at the source-side junction — jumps from the center onto the $g=1$ circle ($g$ = normalized conductance, $g = G/Y_0 = G\cdot Z_0$). The stub's susceptance $b=\tan(\beta\ell_s)\approx 0.727$ sends the operating point to $\Gamma \approx 0.333\angle{-110°}$.
- Walk through the 0.179 λ series line toward the device — rotates clockwise (toward generator, in the look-back convention) along the constant-$|\Gamma|=0.333$ circle by $2\beta\ell_1 \approx 129°$, landing on $\Gamma_s = 0.333\angle 120°$ at the device input.
Same two-element procedure on the output side: 0.432 λ stub + 0.045 λ series → $\Gamma_l = 0.217\angle 70°$.
▸ Why two elements (series + shunt)? (click to expand)
$\Gamma_s$ is a complex number — magnitude and phase, i.e. two real parameters — so the matching network needs two independently adjustable knobs to hit an arbitrary target. The series line and the shunt stub provide exactly those two degrees of freedom:
| Element | Effect on the Smith chart | Adjusts |
|---|---|---|
| Shunt open stub ($\ell_s$) | Adds $jb = j\tan(\beta\ell_s)$ in parallel — slides along a constant-$g$ circle | Lifts the point off the center onto the $g=1$ circle (sets susceptance) |
| Series 50 Ω line ($\ell_1$) | Rotates CW along constant-$|\Gamma|$ circle by $2\beta\ell_1$ | Sets the angle of $\Gamma_s$ on the chart |
Two unknowns $(\ell_s,\ell_1)$ ↔ two real constraints $(|\Gamma_s|,\;\angle\Gamma_s)$ ⇒ unique solution (modulo the two-branch / periodicity choice). Other topologies that also provide ≥ 2 adjustable parameters reach the same $\Gamma_s$ with different lengths:
- Shorted shunt stub + series line — same idea, but the stub needs a via to ground (harder on PCB at microwave).
- Series stub + shunt line — uncommon on PCB but possible in coax / waveguide.
- Lumped L-network (L + C) — low frequency, narrowband, no transmission line needed.
- Two-stub / three-stub tuner — extra DOF for broader bandwidth or fixed line lengths.
All of them hit the same $\Gamma_s$ because the constraint is on the final complex value, not on the topology.
| after stub (Γ) | 0.000 ∠ 0.0° |
| y at junction | 1 + j0.000 |
| after line (Γ) | 0.000 ∠ 0.0° |
| distance to Γ_s | 0.333 |
▸ How to design $\ell_s$ and $\ell_1$ from a target $\Gamma_s$ (click to expand)
Step A — Convert the target $\Gamma_s$ into a target impedance, then normalize to $Z_0$. The Smith chart is always normalized — you plot $z = Z/Z_0$, not raw ohms:
$$Z_s \;=\; Z_0 \cdot \frac{1+\Gamma_s}{1-\Gamma_s} \;=\; 50 \cdot \frac{1 + 0.333\angle 120°}{1 - 0.333\angle 120°} \;\approx\; 30.8 \;+\; j\,20.0\;\;\Omega$$ $$\Longrightarrow\;\; z_s \;=\; \frac{Z_s}{Z_0} \;=\; 0.616 \;+\; j\,0.400 \quad\text{(this is what you mark on the chart).}$$$z_s$, $\Gamma_s$, and $y_s$ are three equivalent labels for the same point:
| Form | Value | Read off chart by |
|---|---|---|
| normalized impedance $z_s$ | 0.616 + j0.400 | intersection of $r=0.616$ and $x=+0.400$ circles |
| reflection coefficient $\Gamma_s$ | 0.333 ∠ 120° | polar coords from center |
| normalized admittance $y_s$ | 1.143 − j0.742 | rotate the $z_s$ point 180° |
You only de-normalize at the very end: $Z_s = z_s \cdot Z_0 = 50 \cdot (0.616 + j0.400) = 30.8 + j20.0\,\Omega$.
Step B — Adjust $\ell_s$ and $\ell_1$ until the network's normalized input impedance equals $z_s$. With the 50 Ω source attached at the far end, the impedance looking from the FET back into the network is a function of both lengths:
$$z_{\text{in,\,FET}}(\ell_s,\,\ell_1) \;\stackrel{!}{=}\; z_s \;=\; 0.616 + j\,0.400.$$One complex equation = two real equations (real and imaginary parts), matched by the two unknowns $(\ell_s,\,\ell_1)$. Solving gives:
- $\ell_s = 0.100\,\lambda\;\Rightarrow\; b = \tan(\beta\ell_s) = \tan 36° \approx 0.727$ — lifts the operating point from the chart center onto the $g=1$ circle.
- $\ell_1 = 0.179\,\lambda\;\Rightarrow\; 2\beta\ell_1 \approx 129°$ — rotates that point CW along the constant-$|\Gamma|$ circle until $z_{\text{in,\,FET}} \approx 0.616 + j0.400$. ✓
What the interactive does. The Smith chart procedure is a graphical solver for the above equation in normalized units. Each slider position corresponds to one trial $(\ell_s,\ell_1)$; the orange marker shows the resulting $\Gamma_{\text{in,\,FET}}$ (equivalent to $z_{\text{in,\,FET}}$ via the chart's $r$/$x$ grid), and the design is finished when the orange marker lands on the red $\Gamma_s$ / $z_s$ target.
Equivalent statements: "present $\Gamma_s$ at the FET input" ⇔ "make $z_{\text{in,\,FET}} = 0.616 + j0.400$" ⇔ "make $Z_{\text{in,\,FET}} = 30.8 + j20.0\,\Omega$" ⇔ "the orange marker lands on the red target on the Smith chart". All four describe the same condition.
Why these specific lengths? They are read off the Smith chart — choose the intersection of the constant-$|\Gamma|$ circle with the appropriate constant-$g$ circle, then convert angular positions to electrical line length using $\lambda$. Alternative realizations (shorted stub, series stub, two-stub network, etc.) yield different lengths but the same final $\Gamma_s, \Gamma_l$.
中文摘要：本電路用「串聯傳輸線 + 並聯開路短截線」實現所選的 $\Gamma_s, \Gamma_l$。所有線都是 50 Ω 特性阻抗，只靠長度（單位 $\lambda$）做阻抗轉換：串聯線把運作點沿等 $|\Gamma|$ 圓旋轉，開路短截線提供純電納（susceptance $B$），沿等 $g$ 圓移動。「開路」短截線在 PCB 上比短路（要打 via 接地）容易實作，因此微波電路常用。輸入端 0.179 λ + 0.100 λ、輸出端 0.045 λ + 0.432 λ 是從 Smith 圖讀出 $|\Gamma|$ 圓與 $g$ 圓的交點所得到的長度；換成短路短截線或其他拓樸雖長度不同，但最終 $\Gamma$ 值仍然一致。
Step 6 — Simulation results. Sweep transducer gain $G_T$ and return loss $R_L$ across 3 – 5 GHz:
It can be seen that the desired gain of 11 dB is achieved at 4 GHz. The return loss, however, is not very good — only about 5 dB at the design frequency. This is due to the deliberate mismatch introduced into the matching sections to back off from $G_{S,\max}$ and $G_{L,\max}$ in order to achieve the specified (lower) gain.
可以看到設計頻率 4 GHz 處達到了目標增益 11 dB。但回波損失並不理想 — 在設計頻率僅約 5 dB。這是因為匹配網路刻意引入失配（mismatch），從 $G_{S,\max}$ 與 $G_{L,\max}$ 後退（back off），以換取指定的（較低的）增益。
3.1.3 Unilateral Figure of Merit （單向化品質因數 $U$）
用來判斷將雙向 (bilateral) 放大器近似為單向 (unilateral, $S_{12}pprox 0$) 時所引入的誤差大小。當 $U$ 很小時，可以放心使用單向公式設計；若 $U$ 顯著，必須改用雙向 (bilateral) 方法。
The unilateral assumption ($S_{12}=0$) introduces error. The actual bilateral gain $G_T$ is bounded by:
Derivation of the bound (click to expand)
Step 1 — Write out both gains. The exact (bilateral) transducer gain is
Setting $S_{12} = 0$ gives the unilateral gain
$$G_{TU} = \frac{(1-|\Gamma_s|^2)\,|S_{21}|^2\,(1-|\Gamma_l|^2)}{|1-S_{11}\Gamma_s|^2\,|1-S_{22}\Gamma_l|^2}.$$Step 2 — Take the ratio. Numerators are identical, so they cancel:
$$\frac{G_T}{G_{TU}} \;=\; \frac{|1-S_{11}\Gamma_s|^2\,|1-S_{22}\Gamma_l|^2}{\bigl|(1-S_{11}\Gamma_s)(1-S_{22}\Gamma_l) - S_{12}S_{21}\Gamma_s\Gamma_l\bigr|^2}.$$Step 3 — Factor out the unilateral product. Let
$$A \;=\; (1-S_{11}\Gamma_s)(1-S_{22}\Gamma_l), \qquad B \;=\; S_{12}S_{21}\Gamma_s\Gamma_l.$$Then $G_T/G_{TU} = |A|^2/|A-B|^2 = 1/|1 - B/A|^2$. Define the $X$ parameter:
Step 4 — Apply the triangle inequality. For any complex $X$:
$$\bigl|\,1 - |X|\,\bigr| \;\le\; |1 - X| \;\le\; 1 + |X|.$$Squaring (all quantities non-negative when $|X|<1$) and inverting (flips the inequalities):
$$\frac{1}{(1+|X|)^2} \;\le\; \frac{1}{|1-X|^2} \;\le\; \frac{1}{(1-|X|)^2}.$$Step 5 — Substitute back. Since $G_T/G_{TU} = 1/|1-X|^2$, this is exactly the claimed bound:
Step 6 — Worst-case $|X|$. At the unilateral conjugate match $\Gamma_s = S_{11}^*$, $\Gamma_l = S_{22}^*$:
- $|\Gamma_s| = |S_{11}|$ and $S_{11}\Gamma_s = |S_{11}|^2$ (real positive), so $|1-S_{11}\Gamma_s| = 1 - |S_{11}|^2$.
- Similarly $|1-S_{22}\Gamma_l| = 1 - |S_{22}|^2$.
- $|S_{12}S_{21}\Gamma_s\Gamma_l| = |S_{12}||S_{21}||S_{11}||S_{22}|$.
Putting it together gives the unilateral figure of merit:
$$U \;=\; |X|_{\max} \;=\; \frac{|S_{11}||S_{12}||S_{21}||S_{22}|}{(1-|S_{11}|^2)(1-|S_{22}|^2)}.$$Replacing $|X|$ by $U$ gives the conservative form used in practice:
$$\frac{1}{(1+U)^2} \;\le\; \frac{G_T}{G_{TU}} \;\le\; \frac{1}{(1-U)^2}.$$中文摘要：誤差比 $G_T/G_{TU} = 1/|1-X|^2$，其中 $X = S_{12}S_{21}\Gamma_s\Gamma_l / [(1-S_{11}\Gamma_s)(1-S_{22}\Gamma_l)]$。對 $|1-X|$ 套用三角不等式 $1-|X| \le |1-X| \le 1+|X|$，再倒數平方即得上下界。當 $\Gamma_s, \Gamma_l$ 取共軛匹配時，$|X|$ 達到最大值 $U$（單向化品質因數）。
where at conjugate match ($\Gamma_s = S_{11}^*$, $\Gamma_l = S_{22}^*$):
$$U = |X|_{\max} = \frac{|S_{12}||S_{21}||S_{11}||S_{22}|}{(1-|S_{11}|^2)(1-|S_{22}|^2)}$$$U$ is the unilateral figure of merit. When $U$ is small (say $U < 0.1$), the error bound is tight and the unilateral approximation is accurate. The error in dB is approximately $\pm 20\log_{10}\frac{1}{1 \mp U}$.
3.1.4 Constant Gain Circles — Bilateral Case
When $S_{12} \neq 0$, gain circles become more complex. Define:
$$\Delta = S_{11}S_{22} - S_{12}S_{21}$$ $$C_2 = S_{22} - \Delta S_{11}^*, \qquad C_1 = S_{11} - \Delta S_{22}^*$$ $$k = \text{stability factor}$$Constant operating power gain circles ($G_P$ = const, on $\Gamma_l$ plane)
Normalize: $g_P = G_P / |S_{21}|^2$
Derivation of the bilateral output gain circle (click to expand)
Step 1 — Start from the operating power gain. With $\Gamma_{in} = S_{11} + \dfrac{S_{12}S_{21}\Gamma_l}{1-S_{22}\Gamma_l}$, the operating power gain is
$$G_P \;=\; \frac{|S_{21}|^2\,(1-|\Gamma_l|^2)}{(1-|\Gamma_{in}|^2)\,|1-S_{22}\Gamma_l|^2}.$$A bit of algebra (using $\Delta = S_{11}S_{22} - S_{12}S_{21}$) simplifies $1 - |\Gamma_{in}|^2$ to
$$1 - |\Gamma_{in}|^2 \;=\; \frac{|1 - S_{22}\Gamma_l|^2 - |S_{11} - \Delta\Gamma_l|^2}{|1 - S_{22}\Gamma_l|^2}.$$Substituting back:
$$G_P \;=\; \frac{|S_{21}|^2\,(1-|\Gamma_l|^2)}{|1 - S_{22}\Gamma_l|^2 - |S_{11} - \Delta\Gamma_l|^2}.$$Step 2 — Normalize. Define $g_P = G_P / |S_{21}|^2$. The equation becomes
$$g_P \;=\; \frac{1-|\Gamma_l|^2}{|1 - S_{22}\Gamma_l|^2 - |S_{11} - \Delta\Gamma_l|^2}.$$Step 3 — Expand the denominator. Use $|a|^2 = a a^{\,*}$ and collect $\Gamma_l$ terms. After expansion:
$$|1-S_{22}\Gamma_l|^2 - |S_{11}-\Delta\Gamma_l|^2 \;=\; 1 - |S_{11}|^2 - (|S_{22}|^2 - |\Delta|^2)|\Gamma_l|^2 - 2\,\mathrm{Re}\{C_2\,\Gamma_l\},$$where
$$\boxed{\;C_2 \;\equiv\; S_{22} - \Delta\,S_{11}^{\,*}\;}$$is the bilateral-circle coefficient. Also use the identity $1 - |S_{11}|^2 - |S_{22}|^2 + |\Delta|^2 = 2k\,|S_{12}S_{21}|$ where $k$ is the stability factor.
Step 4 — Rearrange to a circle equation. Multiplying out and grouping terms in $|\Gamma_l|^2$ and $\Gamma_l$:
$$\bigl[1 + g_P(|S_{22}|^2 - |\Delta|^2)\bigr]\,|\Gamma_l|^2 \;-\; 2\,g_P\,\mathrm{Re}\{C_2\,\Gamma_l\} \;+\; \bigl[g_P(1-|S_{11}|^2) - 1\bigr]\cdot(-1) \;=\; 0.$$This is the equation of a circle in the $\Gamma_l$ plane. Completing the square produces
$$\Bigl|\,\Gamma_l - C_L\,\Bigr|^2 \;=\; R_L^{\,2}.$$Step 5 — Read off the center. The linear term gives the center directly:
The conjugate appears because the cross term $\mathrm{Re}\{C_2\Gamma_l\} = \tfrac{1}{2}(C_2\Gamma_l + C_2^*\Gamma_l^*)$ pairs $\Gamma_l$ with $C_2^*$.
Step 6 — Compute the radius. Collecting the constant terms after completing the square and simplifying with $1 - |S_{11}|^2 - |S_{22}|^2 + |\Delta|^2 = 2k|S_{12}S_{21}|$ and $|C_2|^2 = (|S_{22}|^2 - |\Delta|^2)(1-|S_{11}|^2) + |S_{12}S_{21}|^2$, the discriminant collapses to
$$R_L^{\,2}\!\bigl[1 + g_P(|S_{22}|^2 - |\Delta|^2)\bigr]^2 \;=\; 1 - 2k\,|S_{12}S_{21}|\,g_P + |S_{12}S_{21}|^2\,g_P^{\,2}.$$Taking the positive root:
Sanity checks.
- Unilateral limit $S_{12} \to 0$: $|S_{12}S_{21}| \to 0$, $\Delta \to S_{11}S_{22}$. Numerator under the root $\to 1$, $C_2 \to S_{22}(1-|S_{11}|^2)$. The circle reduces to the unilateral $G_L$ circle around $S_{22}^*$.
- Maximum gain $G_P = G_{T,\max}$: the radicand $1 - 2k|S_{12}S_{21}|g_P + |S_{12}S_{21}|^2 g_P^2$ touches zero, $R_L \to 0$ — the circle collapses to a point at $\Gamma_{Ml}$ (simultaneous-conjugate-match load).
- $G_P = 0$: $g_P = 0$, $R_L = 1$, $C_L = 0$ — the circle becomes the entire unit disk boundary (every $\Gamma_l$ gives zero gain, trivially).
中文摘要：從操作功率增益 $G_P$ 出發，將 $\Gamma_{in}$ 對 $\Gamma_l$ 的依賴帶入，化簡分母為關於 $|\Gamma_l|^2$ 與 $\mathrm{Re}\{C_2\Gamma_l\}$ 的線性式（其中 $C_2 = S_{22} - \Delta S_{11}^*$），即為圓的標準式。配方後可讀出圓心 $C_L = g_P C_2^*/[1+g_P(|S_{22}|^2-|\Delta|^2)]$ 與半徑 $R_L$；判別式中出現穩定因子 $k$ 與 $|S_{12}S_{21}|$，因此圓的大小與穩定性息息相關。輸入端的 $G_A$ 圓推導完全對稱（$1\leftrightarrow 2$）。
Constant available power gain circles ($G_A$ = const, on $\Gamma_s$ plane)
Normalize: $g_A = G_A / |S_{21}|^2$
Derivation of the bilateral input gain circle (click to expand)
The input-side derivation is the mirror image of the output side — swap subscripts $1\leftrightarrow 2$ and use the available gain $G_A$ in place of the operating gain $G_P$. The whole argument runs in the $\Gamma_s$ plane.
Step 1 — Start from the available power gain. With $\Gamma_{out} = S_{22} + \dfrac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s}$, the available gain is
Why $G_T \xrightarrow{\,\Gamma_l=\Gamma_{out}^*\,} G_A$ — denominator simplification (click to expand)
Start. The exact bilateral transducer gain (eq. (3.1.3-1)) is
$$G_T \;=\; \frac{(1-|\Gamma_s|^2)\,|S_{21}|^2\,(1-|\Gamma_l|^2)}{\bigl|(1-S_{11}\Gamma_s)(1-S_{22}\Gamma_l) - S_{12}S_{21}\Gamma_s\Gamma_l\bigr|^2}.$$Substitute $\Gamma_l = \Gamma_{out}^*$. Let $D \equiv 1 - S_{11}\Gamma_s$. By definition
$$\Gamma_{out} \;=\; S_{22} + \frac{S_{12}S_{21}\Gamma_s}{D} \;=\; \frac{D\,S_{22} + S_{12}S_{21}\Gamma_s}{D},$$so
$$D\,\Gamma_{out} \;=\; D\,S_{22} + S_{12}S_{21}\Gamma_s. \tag{$*$}$$Simplify the denominator factor.
$$\underbrace{(1-S_{11}\Gamma_s)}_{D}(1-S_{22}\Gamma_{out}^*) - S_{12}S_{21}\Gamma_s\cdot\Gamma_{out}^*$$ $$= D - \Gamma_{out}^*\!\underbrace{\bigl(D\,S_{22} + S_{12}S_{21}\Gamma_s\bigr)}_{\displaystyle= D\,\Gamma_{out}\;\text{by }(*)}$$ $$= D - \Gamma_{out}^*\cdot D\,\Gamma_{out} \;=\; D\bigl(1 - |\Gamma_{out}|^2\bigr).$$Therefore
$$\bigl|\cdots\bigr|^2 \;=\; |D|^2\,(1-|\Gamma_{out}|^2)^2 \;=\; |1-S_{11}\Gamma_s|^2\,(1-|\Gamma_{out}|^2)^2.$$Also: $|\Gamma_l| = |\Gamma_{out}^*| = |\Gamma_{out}|$, so $1-|\Gamma_l|^2 = 1-|\Gamma_{out}|^2$. Substituting everything:
$$G_T \;=\; \frac{(1-|\Gamma_s|^2)\,|S_{21}|^2\,\cancel{(1-|\Gamma_{out}|^2)}}{|1-S_{11}\Gamma_s|^2\,(1-|\Gamma_{out}|^2)^{\cancel{2}}} \;=\; \frac{|S_{21}|^2\,(1-|\Gamma_s|^2)}{|1-S_{11}\Gamma_s|^2\,(1-|\Gamma_{out}|^2)} \;=\; G_A.\;\checkmark$$The key step is $(*)$: the two terms $D\,S_{22} + S_{12}S_{21}\Gamma_s$ in the denominator bundle into $D\,\Gamma_{out}$ exactly because that is how $\Gamma_{out}$ is defined. One factor of $(1-|\Gamma_{out}|^2)$ cancels between numerator and denominator, leaving the $G_A$ formula.
Using $\Delta = S_{11}S_{22} - S_{12}S_{21}$, the same algebraic identity simplifies $1 - |\Gamma_{out}|^2$ to
$$1 - |\Gamma_{out}|^2 \;=\; \frac{|1 - S_{11}\Gamma_s|^2 - |S_{22} - \Delta\Gamma_s|^2}{|1 - S_{11}\Gamma_s|^2}.$$Substituting:
$$G_A \;=\; \frac{|S_{21}|^2\,(1-|\Gamma_s|^2)}{|1 - S_{11}\Gamma_s|^2 - |S_{22} - \Delta\Gamma_s|^2}.$$Step 2 — Normalize. Define $g_A = G_A / |S_{21}|^2$:
$$g_A \;=\; \frac{1-|\Gamma_s|^2}{|1 - S_{11}\Gamma_s|^2 - |S_{22} - \Delta\Gamma_s|^2}.$$Step 3 — Expand the denominator. The same expansion (with $1\!\leftrightarrow\! 2$) yields
$$|1-S_{11}\Gamma_s|^2 - |S_{22}-\Delta\Gamma_s|^2 \;=\; 1 - |S_{22}|^2 - (|S_{11}|^2 - |\Delta|^2)|\Gamma_s|^2 - 2\,\mathrm{Re}\{C_1\,\Gamma_s\},$$with
$$\boxed{\;C_1 \;\equiv\; S_{11} - \Delta\,S_{22}^{\,*}\;}$$the input-side coefficient. The stability identity $1 - |S_{11}|^2 - |S_{22}|^2 + |\Delta|^2 = 2k\,|S_{12}S_{21}|$ is the same.
Step 4 — Rearrange to a circle equation. Cross-multiplying and grouping powers of $\Gamma_s$:
$$\bigl[1 + g_A(|S_{11}|^2 - |\Delta|^2)\bigr]\,|\Gamma_s|^2 \;-\; 2\,g_A\,\mathrm{Re}\{C_1\,\Gamma_s\} \;-\; \bigl[g_A(1-|S_{22}|^2) - 1\bigr] \;=\; 0.$$Completing the square gives $|\,\Gamma_s - C_S\,|^2 = R_S^{\,2}$.
Step 5 — Read off the center. The linear term pairs $\Gamma_s$ with $C_1^*$:
Step 6 — Compute the radius. Using $|C_1|^2 = (|S_{11}|^2 - |\Delta|^2)(1-|S_{22}|^2) + |S_{12}S_{21}|^2$ and the stability identity, the discriminant simplifies (just as on the output side) to
$$R_S^{\,2}\!\bigl[1 + g_A(|S_{11}|^2 - |\Delta|^2)\bigr]^2 \;=\; 1 - 2k\,|S_{12}S_{21}|\,g_A + |S_{12}S_{21}|^2\,g_A^{\,2}.$$Taking the positive root:
Sanity checks.
- Unilateral limit $S_{12} \to 0$: $|S_{12}S_{21}| \to 0$, $\Delta \to S_{11}S_{22}$, $C_1 \to S_{11}(1-|S_{22}|^2)$. The circle reduces to the unilateral $G_S$ circle around $S_{11}^*$.
- Maximum gain $G_A = G_{T,\max}$: the radicand zeroes out, $R_S \to 0$ — the circle collapses to the simultaneous-conjugate-match source point $\Gamma_{Ms}$.
- $G_A = 0$: $g_A = 0$, $R_S = 1$, $C_S = 0$ — the locus is the whole unit-circle boundary.
- Mirror symmetry with output: every formula is obtained from the $C_L$, $R_L$ derivation by the swap $\{S_{11}, S_{22}, \Gamma_l, g_P, G_P, C_2\} \leftrightarrow \{S_{22}, S_{11}, \Gamma_s, g_A, G_A, C_1\}$.
中文摘要：從可用增益 $G_A$ 出發，將 $\Gamma_{out}$ 對 $\Gamma_s$ 的依賴代入，再以 $C_1 = S_{11} - \Delta S_{22}^*$ 對分母化簡，得到 $\Gamma_s$ 平面上的圓方程。配方後讀出圓心與半徑，分子根號內含穩定因子 $k$ 與 $|S_{12}S_{21}|$。本式只需把輸出端推導中的 $1\leftrightarrow 2$ 對換即得，結構完全對稱。
Maximum gain (bilateral)
When the radius = 0:
$$G_{\max} = \frac{|S_{21}|}{|S_{12}|}(k - \sqrt{k^2-1})$$In the bilateral case, choosing $\Gamma_s$ affects $\Gamma_{out}$ (and vice versa) through:
$$\Gamma_{out} = S_{22} + \frac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s}, \qquad \Gamma_{in} = S_{11} + \frac{S_{12}S_{21}\Gamma_l}{1-S_{22}\Gamma_l}$$Input and output matching are coupled. After choosing $\Gamma_s$ from a gain/noise circle, compute $\Gamma_l = \Gamma_{out}^*$ for conjugate output match.
Why $\Gamma_l = \Gamma_{out}^*$? (click to expand)
The setup. Once you pick $\Gamma_s$ (from a gain circle, noise circle, or stability constraint), the output reflection looking into port 2 becomes
$$\Gamma_{out} = S_{22} + \frac{S_{12}\,S_{21}\,\Gamma_s}{1 - S_{11}\,\Gamma_s}.$$It depends on $\Gamma_s$ through the internal feedback $S_{12}$ — not just $S_{22}$ unless the transistor is unilateral.
Why conjugate match at the output. To deliver maximum power from the transistor's output port into the load, the load must absorb all the available power — no reflection at the port-2/load interface:
$$\Gamma_l = \Gamma_{out}^{\,*}.$$This is the standard maximum-power-transfer condition restated in reflection-coefficient form: load impedance = complex conjugate of the Thevenin impedance seen at port 2.
Why we do it in this order.
- $\Gamma_s$ is constrained first by the design goal — minimum noise ($\Gamma_s = \Gamma_{opt}$), a target available gain $G_A$, or a stability margin. You sacrifice input match to get those.
- $\Gamma_l$ is then free, so you spend it on output match to recover the gain you can still get. Conjugate-matching the output maximizes the transducer gain given the chosen $\Gamma_s$ (this is exactly $G_A$), minimizes output VSWR, and is the only remaining degree of freedom.
What you give up. Because $\Gamma_s \neq \Gamma_{in}^*$ in general, the input is mismatched. That is the price of LNA design (low noise) or constrained-gain design (specified $G_A$). The input mismatch is absorbed by the input matching network's topology, or accepted as part of the noise/gain trade.
Stability caveat. Before committing to $\Gamma_l = \Gamma_{out}^*$, check that $|\Gamma_{out}| < 1$. If the device is only conditionally stable and your $\Gamma_s$ sits in an unstable region, $\Gamma_{out}$ can exceed 1 and a passive conjugate load won't exist — you have to move $\Gamma_s$ or add loss.
3.1.5 Summary of Gain Circle Formulas
| Unilateral ($S_{12}=0$) | Bilateral ($S_{12}\neq 0$) | |
|---|---|---|
| Input circle center | $\frac{g_S S_{11}^*}{1-(1-g_S)|S_{11}|^2}$ | $\frac{g_A C_1^*}{1+g_A(|S_{11}|^2-|\Delta|^2)}$ |
| Input circle radius | $\frac{\sqrt{1-g_S}(1-|S_{11}|^2)}{1-(1-g_S)|S_{11}|^2}$ | $\frac{\sqrt{1-2k|S_{12}S_{21}|g_A+|S_{12}S_{21}|^2g_A^2}}{|1+g_A(|S_{11}|^2-|\Delta|^2)|}$ |
| Output circle center | $\frac{g_L S_{22}^*}{1-(1-g_L)|S_{22}|^2}$ | $\frac{g_P C_2^*}{1+g_P(|S_{22}|^2-|\Delta|^2)}$ |
| Output circle radius | $\frac{\sqrt{1-g_L}(1-|S_{22}|^2)}{1-(1-g_L)|S_{22}|^2}$ | $\frac{\sqrt{1-2k|S_{12}S_{21}|g_P+|S_{12}S_{21}|^2g_P^2}}{|1+g_P(|S_{22}|^2-|\Delta|^2)|}$ |
| Max gain | $G_{S,\max}\cdot|S_{21}|^2\cdot G_{L,\max}$ | $\frac{|S_{21}|}{|S_{12}|}(k-\sqrt{k^2-1})$ |
| $C_1$ | — | $S_{11} - \Delta\,S_{22}^*$ |
| $C_2$ | — | $S_{22} - \Delta\,S_{11}^*$ |
| $\Delta$ | $S_{11}S_{22} - S_{12}S_{21}$ | |
How are these circle formulas derived? (click to expand)General principle. The gain formula is a bilinear (Möbius) function of $\Gamma$. Bilinear transforms map circles to circles, so any constant-gain contour is automatically a circle on the Smith chart. Unilateral — $G_S$ circle derivation. Start from: $$G_S = \frac{1-|\Gamma_s|^2}{|1-S_{11}\Gamma_s|^2}$$Normalize $g_S = G_S(1-|S_{11}|^2)$, set $g_S = \text{const}$: $$g_S\,|1-S_{11}\Gamma_s|^2 \;=\; (1-|\Gamma_s|^2)(1-|S_{11}|^2)$$Expand, collect $|\Gamma_s|^2$ and $\mathrm{Re}\{S_{11}\Gamma_s\}$ terms, divide through by $[1-(1-g_S)|S_{11}|^2]$, complete the square $\Rightarrow$ reads off $C_S$ and $R_S$ directly. The numerator of $R_S^2$ factors as $(1-g_S)(1-|S_{11}|^2)^2$, giving $R_S = \sqrt{1-g_S}(1-|S_{11}|^2)\,/\,[1-(1-g_S)|S_{11}|^2]$. Bilateral — $G_P$ circle derivation. Start from the operating power gain and substitute $\Gamma_{in}$. The key identity: $$1-|\Gamma_{in}|^2 = \frac{|1-S_{22}\Gamma_l|^2 - |S_{11}-\Delta\Gamma_l|^2}{|1-S_{22}\Gamma_l|^2}$$cancels $|1-S_{22}\Gamma_l|^2$, simplifying $G_P$ to: $$G_P = \frac{|S_{21}|^2(1-|\Gamma_l|^2)}{|1-S_{22}\Gamma_l|^2 - |S_{11}-\Delta\Gamma_l|^2}$$Normalize $g_P = G_P/|S_{21}|^2$, set $g_P = \text{const}$, expand and collect $\Rightarrow$ circle in $\Gamma_l$ plane with center $C_L$ and radius $R_L$. The stability factor $k$ appears in $R_L$ because the discriminant simplifies using $1-|S_{11}|^2-|S_{22}|^2+|\Delta|^2 = 2k|S_{12}S_{21}|$. The bilateral input ($G_A$) circle is identical with $1\leftrightarrow 2$. | ||
3.1.6 Interactive: Constant Gain Circles on Smith Chart
Version 1 — Unilateral ($S_{12} \approx 0$): independent $G_S$ and $G_L$ circles
Assumes the device has zero reverse gain so the two ports decouple. The source-side gain $G_S$ depends only on $S_{11}$ and the chosen $\Gamma_s$; the load-side gain $G_L$ depends only on $S_{22}$ and $\Gamma_l$. Centers sit on the ray from origin through $S_{11}^*$ (or $S_{22}^*$).
Version 2 — Bilateral, full dual-plane view ($G_A$ in $\Gamma_s$ plane & $G_P$ in $\Gamma_l$ plane)
Side-by-side Smith charts show both bilateral gain circles simultaneously. Left: constant available power gain $G_A$ in the $\Gamma_s$ plane, centered around $C_S$ with radius $R_S$. Right: constant operating power gain $G_P$ in the $\Gamma_l$ plane, centered around $C_L$ with radius $R_L$. The simultaneous-conjugate-match points $\Gamma_{Ms}$ and $\Gamma_{Ml}$ (where both circles collapse to a point at $G_{T,\max}$) are marked when $K > 1$. Stability circles are shown as gray dashed.
| Aspect | Unilateral ($S_{12}\!=\!0$) | Bilateral ($S_{12}\!\ne\!0$) |
|---|---|---|
| Port coupling | Independent — $\Gamma_{in}=S_{11}$ | Coupled — $\Gamma_{in}=S_{11}+\dfrac{S_{12}S_{21}\Gamma_l}{1-S_{22}\Gamma_l}$ |
| Design strategy | Match $\Gamma_s$ vs $S_{11}^*$ and $\Gamma_l$ vs $S_{22}^*$ separately | Solve simultaneously for $\Gamma_{Ms}$, $\Gamma_{Ml}$ |
| Gain circle (input) | Constant $G_S$ around $S_{11}^*$ | Constant available gain $G_A$ around $C_1^* / D_1$-style center |
| Gain circle (output) | Constant $G_L$ around $S_{22}^*$ | Constant operating gain $G_P$ around $C_2^*$-style center (this widget) |
| Max gain (linear) | $G_{TU,\max}=\dfrac{|S_{21}|^2}{(1-|S_{11}|^2)(1-|S_{22}|^2)}$ | $G_{T,\max}=\dfrac{|S_{21}|}{|S_{12}|}\!\bigl(K-\sqrt{K^2-1}\bigr)$ |
| When valid | $|S_{12}| \to 0$ or Unilateral Figure of Merit $U \ll 1$ | Always (must use this when $S_{12}$ is significant) |
| Stability concern | None (inherently stable if $|S_{11}|, |S_{22}| < 1$) | Requires $K>1$, $|\Delta|<1$ for simultaneous match |
3.1.7 Interactive: How to choose $\Gamma_s$ on the Smith chart
- Conjugate match — $\Gamma_s = S_{11}^*$ ⇒ maximum gain ($G_{S,\max}$). Trade-off: noise figure is whatever it lands on（落在哪就是哪）; bandwidth is narrowest.
- Noise match — $\Gamma_s = \Gamma_{opt}$ ⇒ minimum noise figure $F_{\min}$. Trade-off: gain drops below $G_{S,\max}$.
- Compromise on a circle — pick a $\Gamma_s$ that sits on both an acceptable constant-$G_S$ circle and an acceptable constant-$F$ circle. The closer to the chart center, the more bandwidth and the easier the matching network.
Use the widget below: drag the orange $\Gamma_s$ marker around the chart, or click the preset buttons to jump to a strategy. The readouts show $G_S$, $F$, the normalized impedance $z_s$, and the un-normalized $Z_s$ at $Z_0=50\,\Omega$.
$\Gamma_s = (Z_s - Z_0)/(Z_s + Z_0)$ — but watch which $Z$
Same formula shape, different $Z$ — that's the only thing that distinguishes the two reflection coefficients you'll see on every amplifier-design page:
| Symbol | Formula | What $Z$ is |
|---|---|---|
| $\Gamma_s$ — source-side | $\dfrac{Z_s - Z_0}{Z_s + Z_0}$ | Source impedance, looking back into the input matching network from the transistor input plane |
| $\Gamma_{in}$ — into the device | $\dfrac{Z_{in} - Z_0}{Z_{in} + Z_0}$ | Input impedance of the transistor itself |
Where each lives. They sit on opposite sides of the same plane — the transistor's input port:
Source [Input matching Transistor [Output matching Load
network] (S-params) network]
←── Γ_s here Γ_L here ──→
(looking back (looking forward
into source side) into load side)
Γ_in here ──→ ←── Γ_out here
(looking into (looking into
transistor input) transistor output)
For a non-unilateral transistor, $\Gamma_{in}$ depends on what you put on the output:
$$\Gamma_{in} = S_{11} + \frac{S_{12}\,S_{21}\,\Gamma_L}{1 - S_{22}\,\Gamma_L}$$For unilateral devices ($S_{12}\approx 0$): $\Gamma_{in} \approx S_{11}$.
Conjugate-match condition at the input plane (max power transfer):
$$\boxed{\;\Gamma_{in} = \Gamma_s^{*}\;}$$so once you know $\Gamma_{in}$ from the device, you set $\Gamma_s$ to its complex conjugate. Same logic on the output: $\Gamma_{out} = \Gamma_L^{*}$.
What each parameter means
$|S_{11}|$ — magnitude of the input reflection coefficient measured with the output terminated in $Z_0$.
- Range $0$ – $1$. $0$ = device input is perfectly matched ($Z_{in}=Z_0$); $1$ = total reflection.
- Typical LNA at its design frequency: $0.4$ – $0.8$.
- Example: GaAs MESFET at 4 GHz, $|S_{11}| = 0.60$.
$\angle S_{11}$ (°) — phase of the input reflection.
- Range $-180°$ to $+180°$. Together with $|S_{11}|$ it locates $S_{11}$ on the Smith chart.
- The conjugate-match point $S_{11}^*$ has the same magnitude but the opposite-sign phase.
- Example: $S_{11} = 0.60\angle{-60°}$ ⇒ $S_{11}^* = 0.60\angle{+60°}$ (mirror across the real axis).
$|\Gamma_{opt}|$ — magnitude of the source reflection that minimizes the noise figure.
- Datasheet value. Generally $\Gamma_{opt} \neq S_{11}^*$, so the noise-match point and the gain-match point are different — that is the whole reason gain–NF trade-offs exist.
- Example: The same MESFET has $|\Gamma_{opt}| = 0.62$ — close in magnitude to $|S_{11}|$ here, but different in phase.
$\angle \Gamma_{opt}$ (°) — phase of $\Gamma_{opt}$.
- Locates the green dot ($\Gamma_{opt}$) on the Smith chart.
- Example: $\Gamma_{opt} = 0.62\angle{+100°}$ — sits in the upper-left of the chart, well separated from $S_{11}^* = 0.60\angle{+60°}$.
$F_{\min}$ (dB) — the minimum noise figure the device can achieve, attained when $\Gamma_s = \Gamma_{opt}$.
- Datasheet value. Sets the floor of the blue $F$ readout in this widget.
- Example: $F_{\min} = 1.6$ dB means the best possible NF of this device is 1.6 dB; any other $\Gamma_s$ gives $F > 1.6$ dB.
$R_n / Z_0$ — normalized noise resistance, the "stiffness" of the noise-figure vs. $\Gamma_s$ relationship.
- Recall $F = F_{\min} + \dfrac{4\,r_n\,|\Gamma_s - \Gamma_{opt}|^2}{(1-|\Gamma_s|^2)\,|1+\Gamma_{opt}|^2}$ with $r_n = R_n/Z_0$.
- Large $r_n$ (≳ 0.5) → noise figure rises quickly as $\Gamma_s$ moves away from $\Gamma_{opt}$. The constant-$F$ circles in the widget shrink rapidly. Device is "fussy" about noise matching.
- Small $r_n$ (≲ 0.1) → noise figure is insensitive to mismatch. Constant-$F$ circles balloon out. Easy to get near $F_{\min}$ with a sloppy match.
- Example: $R_n/Z_0 = 0.40$ means $R_n = 0.40 \times 50 = 20\,\Omega$. Try sliding it to $0.05$ vs $1.0$ in the widget — watch the blue circle expand or shrink dramatically.
Concrete worked example. The Pozar/Misra LNA-design example uses a single GaAs MESFET at $f=4$ GHz with:
Three landings to compare in this widget:
- Click "Conjugate match" → $\Gamma_s = 0.60\angle{+60°}$, $G_S = G_{S,\max} = 1.94$ dB, but $F$ is whatever it lands on (often $\sim 2.5$ dB or worse here).
- Click "Noise match" → $\Gamma_s = 0.62\angle{+100°}$, $F = F_{\min} = 1.6$ dB, but $G_S$ drops below $G_{S,\max}$.
- Click "$Z_0$ match" → $\Gamma_s = 0$, no input matching network. Cheapest physically; both $G_S$ and $F$ are sub-optimal.
3.2 Low-Noise Amplifier Design
The first stage of a receiver dominates the overall noise performance (Friis formula). Generally, minimum noise figure and maximum gain cannot be achieved simultaneously — a trade-off is required.
3.2.1 Noise Figure of a Two-Port Amplifier
where
| Parameter | Meaning |
|---|---|
| $F_{\min}$ | Minimum noise figure (achieved when $Y_s = Y_{opt}$) |
| $Y_{opt} = G_{opt} + jB_{opt}$ | Optimum source admittance for minimum NF |
| $R_n$ | Equivalent noise resistance (sensitivity of $F$ to source mismatch) |
| $Y_s = G_s + jB_s$ | Actual source admittance presented to the amplifier |
$F_{\min}$ is the best possible noise figure. $R_n$ controls how fast $F$ degrades as the source deviates from $Y_{opt}$. Large $R_n$ → very sensitive to mismatch. Small $R_n$ → forgiving.
3.2.2 Constant Noise Figure Circles
Convert to reflection coefficient form. Using $Y_s = \frac{1}{Z_0}\frac{1-\Gamma_s}{1+\Gamma_s}$ and $Y_{opt} = \frac{1}{Z_0}\frac{1-\Gamma_{opt}}{1+\Gamma_{opt}}$:
$$F(\Gamma_s) = F_{\min} + \frac{4R_n/Z_0 \cdot |\Gamma_s - \Gamma_{opt}|^2}{(1-|\Gamma_s|^2)|1+\Gamma_{opt}|^2}$$Define the noise figure parameter:
$$N = \frac{|\Gamma_s - \Gamma_{opt}|^2}{1-|\Gamma_s|^2} = \frac{F - F_{\min}}{4R_n/Z_0}|1+\Gamma_{opt}|^2$$Setting $F = \text{const}$ → $N = \text{const}$ → circle on $\Gamma_s$ plane:
- When $F = F_{\min}$: $N = 0$ → $C_F = \Gamma_{opt}$, $R_F = 0$ (point at $\Gamma_{opt}$)
- As $F$ increases, circles grow larger around $\Gamma_{opt}$
- All circles are centered along the line from origin to $\Gamma_{opt}$
Show derivation
From $|\Gamma_s - \Gamma_{opt}|^2 = N(1-|\Gamma_s|^2)$, expand:
$$|\Gamma_s|^2 - \Gamma_s\Gamma_{opt}^* - \Gamma_s^*\Gamma_{opt} + |\Gamma_{opt}|^2 = N - N|\Gamma_s|^2$$ $$|\Gamma_s|^2(N+1) - \Gamma_s\Gamma_{opt}^* - \Gamma_s^*\Gamma_{opt} = N - |\Gamma_{opt}|^2$$Divide by $(N+1)$ and complete the square:
$$\left|\Gamma_s - \frac{\Gamma_{opt}}{N+1}\right|^2 = \frac{N(N+1-|\Gamma_{opt}|^2)}{(N+1)^2}$$3.2.3 Gain vs. Noise Figure Trade-off
Maximum gain needs $\Gamma_s = S_{11}^*$ (conjugate match). Minimum noise needs $\Gamma_s = \Gamma_{opt}$. In general, $S_{11}^* \neq \Gamma_{opt}$, so you can't have both. The design strategy: plot NF circles and gain circles together, find the tangent point that gives acceptable NF with maximum gain.
3.2.4 LNA Design Procedure
- Plot constant NF circles for desired $F$ on the $\Gamma_s$ Smith chart
- Plot constant gain circles ($G_S$ or $G_A$) on the same chart
- Find the tangent point where the desired NF circle just touches a gain circle — this gives maximum $G_S$ for that NF
- Read $\Gamma_s$ from the tangent point
- Compute $\Gamma_l = \Gamma_{out}^*$ for conjugate output match: $$\Gamma_l = \left(S_{22} + \frac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s}\right)^*$$
- Compute the transducer gain $G_T$
3.2.5 Interactive: NF Circles & Gain Trade-off
GaAs MESFET at 4 GHz: $S_{11}=0.60\angle{-60°}$, $S_{21}=1.9\angle{81°}$, $S_{12}=0.05\angle{26°}$, $S_{22}=0.5\angle{-60°}$
Noise parameters: $F_{\min}=1.6$ dB, $\Gamma_{opt}=0.62\angle{100°}$, $R_n=20\,\Omega$
▶ Click to show solution
(a) Minimum noise figure design: $\Gamma_s = \Gamma_{opt} = 0.62\angle{100°}$
$$\Gamma_l = \Gamma_{out}^* = \left(S_{22}+\frac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s}\right)^* = 0.526\angle{68°}$$ $$G_T = 5.4 = 7.3\text{ dB}, \quad F = F_{\min} = 1.6\text{ dB}$$▸ How $G_T = 5.4 = 7.3$ dB is computed — step by step
Formula used. Output is conjugate matched ($\Gamma_l = \Gamma_{out}^*$), so transducer gain equals available gain (bilateral $G_A$, eq. (3.1.4-1)):
$$G_T \;\stackrel{\text{(out conj matched)}}{=}\; G_A \;=\; \frac{|S_{21}|^2\,(1-|\Gamma_s|^2)}{|1 - S_{11}\Gamma_s|^2\,(1-|\Gamma_{out}|^2)}$$The full bilateral $G_T$ (eq. (3.1.3-1)) has an output factor $\dfrac{1-|\Gamma_l|^2}{|1-\Gamma_{out}\Gamma_l|^2}$; when $\Gamma_l = \Gamma_{out}^*$ it collapses to $\dfrac{1}{1-|\Gamma_{out}|^2}$. The $\Gamma_{out}$ formula used in Step 3 is the coupling equation (3.1-1). The unilateral $G_{TU,\max}$ used in the sanity check below is eq. (3.1.2-2) (with $G_{S,\max}$, $G_{L,\max}$ from eq. (3.1.2-1)).
S-parameters in rectangular form (for reference):
| Quantity | Polar | Rectangular |
|---|---|---|
| $S_{11}$ | $0.60\angle{-60°}$ | $0.300 - j0.520$ |
| $S_{12}$ | $0.05\angle{26°}$ | $0.0449 + j0.0219$ |
| $S_{21}$ | $1.90\angle{81°}$ | $0.297 + j1.876$ |
| $S_{22}$ | $0.50\angle{-60°}$ | $0.250 - j0.433$ |
| $\Gamma_s$ | $0.62\angle{100°}$ | $-0.108 + j0.611$ |
Step 1 — numerator pieces.
$$|S_{21}|^2 = 1.9^2 = 3.61, \qquad 1-|\Gamma_s|^2 = 1 - 0.62^2 = 0.6156$$Numerator $= 3.61 \times 0.6156 = 2.222$.
Step 2 — denominator piece 1: $|1 - S_{11}\Gamma_s|^2$.
$$S_{11}\Gamma_s = (0.60)(0.62)\angle(-60°+100°) = 0.372\angle{40°} = 0.285 + j0.239$$ $$1 - S_{11}\Gamma_s = 0.715 - j0.239 = 0.754\angle{-18.5°}$$ $$|1 - S_{11}\Gamma_s|^2 = 0.715^2 + 0.239^2 = 0.5683$$Step 3 — find $\Gamma_{out}$, then $1-|\Gamma_{out}|^2$.
$$S_{12}S_{21}\Gamma_s = (0.05)(1.9)(0.62)\angle(26°+81°+100°) = 0.0589\angle{207°}$$ $$\frac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s} = \frac{0.0589}{0.754}\angle(207°+18.5°) = 0.0781\angle{225.5°} = -0.055 - j0.056$$ $$\Gamma_{out} = S_{22} + (-0.055 - j0.056) = 0.195 - j0.489 = 0.526\angle{-68.2°}$$So $\Gamma_l = \Gamma_{out}^* = 0.526\angle{+68°}$ ✓ matches the printed value.
$$1-|\Gamma_{out}|^2 = 1 - 0.277 = 0.723$$Step 4 — assemble.
$$G_T = \frac{2.222}{0.5683 \times 0.723} = \frac{2.222}{0.411} = 5.41$$Step 5 — to dB.
$$G_T(\text{dB}) = 10\log_{10}(5.41) = 7.33\;\text{dB}$$Rounded: $G_T = 5.4 = 7.3$ dB. ✓
Sanity check（合理性驗證）— why this is below max. Unilateral max-gain estimate (eq. (3.1.2-2)):
$$G_{TU,\max} = \frac{|S_{21}|^2}{(1-|S_{11}|^2)(1-|S_{22}|^2)} = \frac{3.61}{0.64 \times 0.75} = 7.52 \approx 8.76\;\text{dB}$$Our $7.3$ dB is about $1.5$ dB below the gain-optimal point — exactly the price paid by setting $\Gamma_s = \Gamma_{opt}$ (noise match) instead of $\Gamma_s = S_{11}^*$ (gain match). That sacrifice bought $F = F_{\min} = 1.6$ dB.
為何 sanity check 可以用單邊（unilateral）公式？
兩個理由——這裡用的是上界，不是精確答案。
1. $G_{TU,\max}$ 是增益的上界。
單邊公式把 $S_{12}$ 設為零，忽略了輸出→輸入的回授效應，因此高估了可用增益。真實的雙邊最大增益 $G_{MAG}$ 一定 $\le G_{TU,\max}$。用它來檢查「我的答案有沒有超過物理上限」是合理的——如果算出的 $G_T$ 比 $G_{TU,\max}$ 還大，代表哪裡算錯了。
2. 本題 $|S_{12}| = 0.05$ 很小，誤差有限。
單邊誤差由 unilateral figure of merit $U$ 控制：
誤差範圍為
$$\frac{1}{(1+U)^2} \;<\; \frac{G_T}{G_{TU}} \;<\; \frac{1}{(1-U)^2}$$本題 $U \approx 0.05$，對應誤差約 $\pm 0.4$ dB——用作 sanity check 已足夠準確。
精確的雙邊最大增益應用 $G_{MAG} = \dfrac{|S_{21}|}{|S_{12}|}(k - \sqrt{k^2-1})$，但需要先算 $k$ 和 $\Delta$，計算量較大。$G_{TU,\max}$ 一行就能估出上界，作為方向性確認（我的答案確實在合理範圍內）已足夠。
(b) 2 dB noise figure design:
$F = 2$ dB $= 1.585$ (linear). Calculate:
$$N = \frac{1.585-1.445}{4\times20/50}|1+0.62\angle{100°}|^2 = 0.0986$$ $$C_F = \frac{0.62\angle{100°}}{1.0986} = 0.564\angle{100°}, \qquad R_F = 0.242$$$R_F = 0.242$ 怎麼來的？
代入 NF 圓半徑公式（$N=0.0986$，$|\Gamma_{opt}|=0.62$）：
$$R_F = \frac{\sqrt{N\,(N+1-|\Gamma_{opt}|^2)}}{N+1}$$| 步驟 | 數值 |
|---|---|
| $N+1$ | $1.0986$ |
| $|\Gamma_{opt}|^2 = 0.62^2$ | $0.3844$ |
| $N+1-|\Gamma_{opt}|^2$ | $1.0986 - 0.3844 = 0.7142$ |
| $N \times 0.7142$ | $0.0986 \times 0.7142 = 0.07042$ |
| $\sqrt{0.07042}$ | $0.2654$ |
| $\div\,(N+1)$ | $0.2654\,/\,1.0986 = \mathbf{0.242}$ |
當 $F = F_{\min}$ 時 $N=0$，$R_F=0$（圓退化為點 $\Gamma_{opt}$）。此處 $F=2$ dB 略高於 $F_{\min}=1.6$ dB，所以圓有小但非零的半徑 0.242。
Tangent point with $G_S = 1.7$ dB gain circle: $\Gamma_s = 0.53\angle{75°}$
$$\Gamma_l = 0.479\angle{68°}, \qquad G_T = 6.9 = 8.4\text{ dB}$$▸ How $G_T = 6.9 = 8.4$ dB is computed — step by step
Formula. Output conjugate matched ($\Gamma_l = \Gamma_{out}^*$), same as part (a) (eq. (3.1.4-1)), now with $\Gamma_s = 0.53\angle{75°}$ (noise-gain compromise):
$$G_T = G_A = \frac{|S_{21}|^2\,(1-|\Gamma_s|^2)}{|1-S_{11}\Gamma_s|^2\,(1-|\Gamma_{out}|^2)}$$Step 1 — numerator.
$$|S_{21}|^2(1-|\Gamma_s|^2) = 3.61\times(1-0.53^2) = 3.61\times0.7191 = 2.596$$Step 2 — $|1-S_{11}\Gamma_s|^2$.
$\Gamma_s = 0.53\angle{75°} = 0.1372+j0.5119$, $\;S_{11} = 0.300-j0.5196$
$$S_{11}\Gamma_s = 0.3071+j0.0823 \;\Rightarrow\; 1-S_{11}\Gamma_s = 0.6929-j0.0823$$ $$|1-S_{11}\Gamma_s|^2 = 0.6929^2+0.0823^2 = 0.4868$$Step 3 — $\Gamma_{out}$.
$$\frac{S_{12}S_{21}\Gamma_s}{1-S_{11}\Gamma_s} = -0.0713-j0.0110$$ $$\Gamma_{out} = S_{22}+(-0.0713-j0.0110) = 0.1787-j0.4440$$ $$|\Gamma_{out}|^2 = 0.1787^2+0.4440^2 = 0.2291 \;\Rightarrow\; 1-|\Gamma_{out}|^2 = 0.7709$$Step 4 — assemble.
$$G_T = \frac{2.596}{0.4868\times0.7709} = \frac{2.596}{0.3753} = 6.92 \approx 6.9$$Step 5 — dB.
$$G_T(\text{dB}) = 10\log_{10}(6.9) = 8.4\text{ dB}$$Gain is higher than part (a)'s 7.3 dB because $\Gamma_s = 0.53\angle{75°}$ is closer to $S_{11}^*$ than $\Gamma_{opt} = 0.62\angle{100°}$ — +1.1 dB gain at the cost of +0.4 dB worse noise figure.
Comparison:
| Design | $F$ (dB) | $G_T$ (dB) | Trade-off |
|---|---|---|---|
| (a) Min NF | 1.6 | 7.3 | Best noise, less gain |
| (b) 2 dB NF | 2.0 | 8.4 | +0.4 dB NF buys +1.1 dB gain |
3.3 Power Amplifier Design
Power amplifiers are used in the final stages of wireless transmitters. Typical output: 0.5–2 W (handset), 10–100 W (base station). Key considerations: output power, efficiency, gain, and intermodulation.
3.3.1 Efficiency: Conversion & PAE
This doesn't account for input RF power. A better metric:
Conversion efficiency ignores input power. For a low-gain amplifier ($G \approx 3$ dB), one-third of the "output" power came from the input, not from DC conversion. PAE correctly penalizes low gain. $G$ is typically evaluated at the 1 dB compression point.
3.3.2 Power Amplifier Classes
| Class | Conduction Angle $\theta_0$ | Linearity | Max Efficiency |
|---|---|---|---|
| A | $2\pi$ (full cycle) | Best | 50% |
| AB | $\pi < \theta_0 < 2\pi$ | Good | 50–78.5% |
| B | $\pi$ (half cycle) | Moderate | 78.5% |
| C | $< \pi$ | Poor | Up to 100% |
General efficiency formula
The conversion efficiency as a function of conduction angle $\theta_0$:
This uses the relation $I_Q = -I_0\cos(\theta_0/2)$ (quiescent current depends on conduction angle).
Smaller conduction angle → higher efficiency but worse linearity. Class-A is the most linear (active for full cycle) but at most 50% efficient. Class-C can approach 100% efficiency but is highly nonlinear — suitable for constant-envelope signals (e.g., FM, radar).
3.3.3 Class-A Design & Optimum AC Load Line
For Class-A, the transistor is biased at a quiescent point ($V_{CC}$, $I_Q$) and operates in its active region for the entire cycle.
Optimum AC load impedance
where $V_k$ is the knee voltage. This maximizes voltage and current swing simultaneously.
Maximum linear output power
$$P_{opt} = \frac{1}{2}(V_{CC}-V_k)\cdot I_Q = \frac{(V_{CC}-V_k)^2}{2R_{opt}}$$- $R_L < R_{opt}$: load line steeper — voltage swing limited before current clips → less output power
- $R_L > R_{opt}$: load line shallower — current swing limited before voltage clips → less output power
Only $R_L = R_{opt}$ achieves simultaneous maximum voltage and current swing → maximum linear output power.
Design procedure for Class-A PA
- Compute $R_{opt} = (V_{CC}-V_k)/I_Q$
- $\Gamma_l = (R_{opt}-Z_0)/(R_{opt}+Z_0)$
- Compute $\Gamma_{in}$ from $\Gamma_l$ and S-parameters
- $\Gamma_s = \Gamma_{in}^*$ (conjugate input match for max gain)
3.3.4 Load-Pull Contours
At large signal levels, transistors are nonlinear — S-parameters depend on power level. Load-pull measurement sweeps $\Gamma_L$ and measures output power (or $P_{1dB}$) at each point, producing contours of constant power on the Smith chart.
Small-signal S-parameters are inaccurate for power amplifier design. Load-pull contours give the actual large-signal impedance that maximizes output power (or efficiency). The equipment physically varies the load impedance presented to the transistor and measures the result.
3.3.5 Interactive: Efficiency vs. Conduction Angle & Class-A Design
Power BJT at 2 GHz, $V_{CC}=5$ V, $I_C=200$ mA, $V_k=0.7$ V.
$S_{11}=0.60\angle{36°}$, $S_{21}=2.3\angle{-80°}$, $S_{12}=0.14\angle{-85°}$, $S_{22}=0.15\angle{45°}$
▶ Click to show solution
(a) Matching:
$$R_{opt} = \frac{5-0.7}{0.2} = 21.5\,\Omega$$ $$\Gamma_l = \frac{21.5-50}{21.5+50} = -0.4 = 0.4\angle{180°}$$ $$\Gamma_{in} = S_{11}+\frac{S_{12}S_{21}\Gamma_l}{1-S_{22}\Gamma_l} = 0.72\angle{32°}$$ $$\Gamma_s = \Gamma_{in}^* = 0.72\angle{-32°}$$(b) Output power:
$$P_{opt} = \frac{1}{2}(5-0.7)(0.2) = 0.43\text{ W} = 26.3\text{ dBm}$$(c) Operating power gain:
$$G_P = \frac{|S_{21}|^2(1-|\Gamma_l|^2)}{(1-|\Gamma_{in}|^2)|1-S_{22}\Gamma_l|^2} = 8.4 = 9.2\text{ dB}$$(d) PAE:
$$\text{PAE} = \frac{P_{out}-P_{in}}{P_{DC}} = \frac{0.43 - 0.43/8.4}{5\times 0.2} = 38\%$$Chapter Summary
Specified Gain Design
Constant gain circles on the Smith chart allow trade-off between gain, NF, bandwidth, and power. Unilateral case: $G_{TU} = G_S \cdot |S_{21}|^2 \cdot G_L$. Check validity with unilateral figure of merit $U$.
Low-Noise Amplifier
NF circles centered near $\Gamma_{opt}$. Plot NF and gain circles together, find tangent for best gain at acceptable NF. Output: conjugate match $\Gamma_l = \Gamma_{out}^*$.
Power Amplifier
Class-A: $R_{opt}=(V_{CC}-V_k)/I_Q$, max $\eta = 50\%$. Higher classes trade linearity for efficiency. Large-signal design uses load-pull contours.
Key Trade-offs
Gain vs. NF (LNA), gain vs. output power (PA), efficiency vs. linearity (PA class). No single design optimizes all — matching networks are the lever.