[Circuit] S-Parameters (Scattering parameters)

電路知識：S-Parameters (Scattering parameters)
工具：Qucs

簡介：推導與原理

$$\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]=\left[\begin{array}{cc}{S}_{11}& S_{12}\\ S_{21}& S_{22}\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$$ $$\begin{array}{cc}{a}_1=\frac{{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{1}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}& a_2=\frac{{\mathbf{V}}_{\mathbf{2}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{2}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\\ b_1=\frac{{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}^{\mathbf{\ast }}{\mathbf{I}}_{\mathbf{1}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}& b_2=\frac{{\mathbf{V}}_{\mathbf{2}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}^{\mathbf{\ast }}{\mathbf{I}}_{\mathbf{2}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\end{array}$$ $$\begin{array}{cc}{\begin{array}{c}{S}_{11}=\frac{b_1}{a_1}\end{array}|}_{a_2=0}& {\begin{array}{c}{S}_{12}=\frac{b_1}{a_2}\end{array}|}_{a_1=0}\\ {\begin{array}{c}{S}_{21}=\frac{b_2}{a_1}\end{array}|}_{a_2=0}& {\begin{array}{c}{S}_{22}=\frac{b_2}{a_2}\end{array}|}_{a_1=0}\end{array}$$

${\mathbf{Z}}_{\mathbf{0}}$ 為任意值，通常為 50Ω
$$\text{Double subscripts: use braces to clarify}$$ 可看出 $a_i$、$b_i$ 的單位其實是 $\sqrt{power}$
然而 $|a_i{|}^2$ 就像是輸入功率，然而 $|b_i{|}^2$ 就像是反射功率
兩者相減後，得到消耗功率
若 ${\mathbf{V}}_{\mathbf{i}}$ 和 ${\mathbf{I}}_{\mathbf{i}}^{\mathbf{\ast }}$ 不是 rms 值，而是 max 值
此時需乘上 $\frac{1}{2}$ 才是真正的功率
原因可參考此篇 [Circuit] AC Sinusoids 穩態分析

傳輸線傳遞的概念，也可以看作能量傳遞
$S_{11}$：回波損耗 (return loss)，值越接近 0 越好 (越低越好 ，一般 -25 ~ -40dB)
$S_{12}$：反向傳輸增益，值越接近 0 越好
$S_{21}$：插入損耗 (inset loss)，或是頻率響應：值越接近 1 越好 (0dB)
$S_{22}$：輸出端反射係數，值越接近 1 越好 (0dB)

Network Analyzer 原理

$$\begin{gathered} {\delta }_i=\beta l_i \\ S=\left[\begin{array}{cc}{S}_{11}& S_{12}\\ S_{21}& S_{22}\end{array}\right]=\left[\begin{array}{ll}S{}_{11}^{\prime }{e}^{2j{\delta }_1}& S{}_{12}^{\prime }{e}^{j({\delta }_1+{\delta }_2)}\\ S{}_{21}^{\prime }{e}^{j({\delta }_1+{\delta }_2)}& S{}_{22}^{\prime }{e}^{2j{\delta }_2}\end{array}\right] \end{gathered}$$ 令 ${\mathbf{Z}}_{\mathbf{L}}={\mathbf{Z}}_{\mathbf{0}}$ 則 $a{}_2^{\prime }=0$，可求 $S{}_{11}^{\prime }$ & $S{}_{21}^{\prime }$
交換兩邊，可求 $S{}_{12}^{\prime }$ & $S{}_{22}^{\prime }$

根據 [Circuit] Transmission line
$$\begin{array}{rl}\mathbf{V}(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}\\ \mathbf{I}(z)& ={\mathbf{I}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{I}}_{\mathbf{r}}{e}^{\gamma z}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{Z}}_{\mathbf{0}}}{e}^{-\gamma z}-\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{Z}}_{\mathbf{0}}}{e}^{\gamma z}\\ {\mathbf{Z}}_{\mathbf{0}}\mathbf{I}(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}\\ {\mathbf{V}}_{\mathbf{f}}& =\frac{\mathbf{V}(z)+{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}(z)}{2}{e}^{\gamma z}\\ {\mathbf{V}}_{\mathbf{r}}& =\frac{\mathbf{V}(z)-{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}(z)}{2}{e}^{-\gamma z}\end{array}$$ 當 $z=0$
$$\begin{array}{cc}{a}_1=\frac{{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{1}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}=\frac{{\mathbf{V}}_{\mathbf{1}\mathbf{f}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}& a_2=\frac{{\mathbf{V}}_{\mathbf{2}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{2}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}=\frac{{\mathbf{V}}_{\mathbf{2}\mathbf{f}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\\ b_1=\frac{{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}^{\mathbf{\ast }}{\mathbf{I}}_{\mathbf{1}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}=\frac{{\mathbf{V}}_{\mathbf{1}\mathbf{r}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}& b_2=\frac{{\mathbf{V}}_{\mathbf{2}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}^{\mathbf{\ast }}{\mathbf{I}}_{\mathbf{2}}}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}=\frac{{\mathbf{V}}_{\mathbf{2}\mathbf{r}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\end{array}$$ 假設為無損耗傳輸線
$$\begin{array}{rl}\gamma & =j\beta \\ \\ \left[\begin{array}{c}a{}_1^{\prime }\\ b{}_1^{\prime }\end{array}\right]& =\left[\begin{array}{c}\frac{\mathbf{V}{}_{\mathbf{1}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{1}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\\ \frac{\mathbf{V}{}_{\mathbf{1}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{1}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\end{array}\right]\\ & =\left[\begin{array}{c}\frac{\mathbf{V}{}_{\mathbf{1}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{1}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{\gamma (-l_1)}{e}^{-\gamma (-l_1)}\\ \frac{\mathbf{V}{}_{\mathbf{1}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{1}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{-\gamma (-l_1)}{e}^{\gamma (-l_1)}\end{array}\right]\\ & =\left[\begin{array}{c}\frac{{\mathbf{V}}_{\mathbf{1}\mathbf{f}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{-\gamma (-l_1)}\\ \frac{{\mathbf{V}}_{\mathbf{1}\mathbf{r}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{\gamma (-l_1)}\end{array}\right]\\ & =\left[\begin{array}{c}{a}_1{e}^{j\beta l_1}\\ b_1{e}^{-j\beta l_1}\end{array}\right]\\ & =\left[\begin{array}{cc}{e}^{j\beta l_1}& 0\\ 0& e^{-j\beta l_1}\end{array}\right]\left[\begin{array}{c}{a}_1\\ b_1\end{array}\right]\\ \left[\begin{array}{c}a{}_2^{\prime }\\ b{}_2^{\prime }\end{array}\right]& =\left[\begin{array}{c}\frac{\mathbf{V}{}_{\mathbf{2}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{2}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\\ \frac{\mathbf{V}{}_{\mathbf{2}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{2}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\end{array}\right]\\ & =\left[\begin{array}{c}\frac{\mathbf{V}{}_{\mathbf{2}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{2}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{\gamma (-l_2)}{e}^{-\gamma (-l_2)}\\ \frac{\mathbf{V}{}_{\mathbf{2}}^{\prime }\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}\mathbf{I}{}_{\mathbf{2}}^{\prime }}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{-\gamma (-l_2)}{e}^{\gamma (-l_2)}\end{array}\right]\\ & =\left[\begin{array}{c}\frac{{\mathbf{V}}_{\mathbf{2}\mathbf{f}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{-\gamma (-l_2)}\\ \frac{{\mathbf{V}}_{\mathbf{2}\mathbf{r}}}{\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}{e}^{\gamma (-l_2)}\end{array}\right]\\ & =\left[\begin{array}{c}{a}_2{e}^{j\beta l_2}\\ b_2{e}^{-j\beta l_2}\end{array}\right]\\ & =\left[\begin{array}{cc}{e}^{j\beta l_2}& 0\\ 0& e^{-j\beta l_2}\end{array}\right]\left[\begin{array}{c}{a}_2\\ b_2\end{array}\right]\\ \left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]& =D\left[\begin{array}{c}b{}_1^{\prime }\\ b{}_2^{\prime }\end{array}\right]\\ \left[\begin{array}{c}a{}_1^{\prime }\\ a{}_2^{\prime }\end{array}\right]& =D\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\\ D& =\left[\begin{array}{cc}{e}^{j\beta l_1}& 0\\ 0& e^{j\beta l_2}\end{array}\right]\\ & =\left[\begin{array}{cc}{e}^{j{\delta }_1}& 0\\ 0& e^{j{\delta }_2}\end{array}\right]\end{array}$$ $$\begin{array}{rl}\left[\begin{array}{c}b{}_1^{\prime }\\ b{}_2^{\prime }\end{array}\right]& =\left[\begin{array}{cc}S{}_{11}^{\prime }& S{}_{12}^{\prime }\\ S{}_{21}^{\prime }& S{}_{22}^{\prime }\end{array}\right]\left[\begin{array}{c}a{}_1^{\prime }\\ a{}_2^{\prime }\end{array}\right]\\ \left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]& =D\left[\begin{array}{c}b{}_1^{\prime }\\ b{}_2^{\prime }\end{array}\right]\\ & =D\left[\begin{array}{cc}S{}_{11}^{\prime }& S{}_{12}^{\prime }\\ S{}_{21}^{\prime }& S{}_{22}^{\prime }\end{array}\right]\left[\begin{array}{c}a{}_1^{\prime }\\ a{}_2^{\prime }\end{array}\right]\\ & =D\left[\begin{array}{cc}S{}_{11}^{\prime }& S{}_{12}^{\prime }\\ S{}_{21}^{\prime }& S{}_{22}^{\prime }\end{array}\right]D\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\\ & =DS{}^{\prime }D\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\\ & =S\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\end{array}$$ $$\begin{array}{rl}S& =DS{}^{\prime }D\\ & =\left[\begin{array}{cc}{e}^{j{\delta }_1}& 0\\ 0& e^{j{\delta }_2}\end{array}\right]\left[\begin{array}{cc}S{}_{11}^{\prime }& S{}_{12}^{\prime }\\ S{}_{21}^{\prime }& S{}_{22}^{\prime }\end{array}\right]\left[\begin{array}{cc}{e}^{j{\delta }_1}& 0\\ 0& e^{j{\delta }_2}\end{array}\right]\\ & =\left[\begin{array}{cc}{e}^{j{\delta }_1}S{}_{11}^{\prime }& e^{j{\delta }_1}S{}_{12}^{\prime }\\ e^{j{\delta }_2}S{}_{21}^{\prime }& e^{j{\delta }_2}S{}_{22}^{\prime }\end{array}\right]\left[\begin{array}{cc}{e}^{j{\delta }_1}& 0\\ 0& e^{j{\delta }_2}\end{array}\right]\\ & =\left[\begin{array}{ll}S{}_{11}^{\prime }{e}^{2j{\delta }_1}& S{}_{12}^{\prime }{e}^{j({\delta }_1+{\delta }_2)}\\ S{}_{21}^{\prime }{e}^{j({\delta }_1+{\delta }_2)}& S{}_{22}^{\prime }{e}^{2j{\delta }_2}\end{array}\right]\end{array}$$

參數轉換

$$S=(\mathbf{Z}-{\mathbf{Z}}_{\mathbf{0}}I)(\mathbf{Z}+{\mathbf{Z}}_{\mathbf{0}}I{)}^{-1}$$

$$\begin{array}{rl}\mathbf{V}& =\left[\begin{array}{c}{\mathbf{V}}_{\mathbf{1}}\\ {\mathbf{V}}_{\mathbf{2}}\end{array}\right]\mathbf{I}=\left[\begin{array}{c}{\mathbf{I}}_{\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{2}}\end{array}\right]\\ \mathbf{a}& =\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\mathbf{b}=\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]\\ \mathbf{V}& =\mathbf{Z}\mathbf{I}\\ \mathbf{a}& =\frac{1}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}(\mathbf{V}+{\mathbf{Z}}_{\mathbf{0}}\mathbf{I})=\frac{1}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}(\mathbf{Z}+{\mathbf{Z}}_{\mathbf{0}}I)\mathbf{I}\\ \mathbf{b}& =\frac{1}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}(\mathbf{V}-{\mathbf{Z}}_{\mathbf{0}}\mathbf{I})=\frac{1}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}(\mathbf{Z}-{\mathbf{Z}}_{\mathbf{0}}I)\mathbf{I}\\ \frac{1}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}\mathbf{I}& =(\mathbf{Z}+{\mathbf{Z}}_{\mathbf{0}}I{)}^{-1}\mathbf{a}\\ \mathbf{b}& =\frac{1}{2\sqrt{{\mathbf{Z}}_{\mathbf{0}}}}(\mathbf{Z}-{\mathbf{Z}}_{\mathbf{0}}I)\mathbf{I}\\ & =(\mathbf{Z}-{\mathbf{Z}}_{\mathbf{0}}I)(\mathbf{Z}+{\mathbf{Z}}_{\mathbf{0}}I{)}^{-1}\mathbf{a}\\ & =S\mathbf{a}\end{array}$$

即然可以從 $\mathbf{Z}$ 轉換為 $S$，那麼 $S$ 運用反矩陣也可以轉為 $\mathbf{Z}$
所以其他的參數也就可以轉換，可利用此篇的轉換表 [Circuit] Two-Port Networks 分析

範例

Qucs 自動

若 $Z0\ne 50\mathrm{\Omega }$，stoz(S, Zref=50) 需更改 Zref 參數，預設為 50Ω

算式推導

$$\begin{array}{rl}S& =\left[\begin{array}{cc}{\mathbf{z}}_{\mathbf{11}}-{\mathbf{Z}}_{\mathbf{0}}& {\mathbf{z}}_{\mathbf{12}}\\ {\mathbf{z}}_{\mathbf{21}}& {\mathbf{z}}_{\mathbf{22}}-{\mathbf{Z}}_{\mathbf{0}}\end{array}\right]{\left[\begin{array}{cc}{\mathbf{z}}_{\mathbf{11}}+{\mathbf{Z}}_{\mathbf{0}}& {\mathbf{z}}_{\mathbf{12}}\\ {\mathbf{z}}_{\mathbf{21}}& {\mathbf{z}}_{\mathbf{22}}+{\mathbf{Z}}_{\mathbf{0}}\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}50& 50\\ 50& 50\end{array}\right]{\left[\begin{array}{cc}150& 50\\ 50& 150\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}50& 50\\ 50& 50\end{array}\right]\left[\begin{array}{cc}\frac{3}{400}& \frac{-1}{400}\\ \frac{-1}{400}& \frac{3}{400}\end{array}\right]\\ & =\left[\begin{array}{cc}0.25& 0.25\\ 0.25& 0.25\end{array}\right]\end{array}$$

Qucs 手動
根據 Network Analyzer 原理，因 $l_1=0$ 跟 $l_2=0$，故 $S{}^{\prime }=S$
將 V2 Short，並將 RL_R 設為跟 Z0 一致，此時 a2 才會為 0，也才能求 S11 & S21
且此時 RL_L 的值無任何影響，沒有一定需為 Z0

將 V1 Short，並將 RL_L 設為跟 Z0 一致，此時 a1 才會為 0，也才能求 S12 & S22
且此時 RL_R 的值無任何影響，沒有一定需為 Z0

算式推導

$$\begin{array}{rl}S& =\left[\begin{array}{cc}{\mathbf{z}}_{\mathbf{11}}-{\mathbf{Z}}_{\mathbf{0}}& {\mathbf{z}}_{\mathbf{12}}\\ {\mathbf{z}}_{\mathbf{21}}& {\mathbf{z}}_{\mathbf{22}}-{\mathbf{Z}}_{\mathbf{0}}\end{array}\right]{\left[\begin{array}{cc}{\mathbf{z}}_{\mathbf{11}}+{\mathbf{Z}}_{\mathbf{0}}& {\mathbf{z}}_{\mathbf{12}}\\ {\mathbf{z}}_{\mathbf{21}}& {\mathbf{z}}_{\mathbf{22}}+{\mathbf{Z}}_{\mathbf{0}}\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}25& 50\\ 50& 25\end{array}\right]{\left[\begin{array}{cc}175& 50\\ 50& 175\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}25& 50\\ 50& 25\end{array}\right]\left[\begin{array}{cc}\frac{7}{1125}& \frac{-2}{1125}\\ \frac{-2}{1125}& \frac{7}{1125}\end{array}\right]\\ & =\left[\begin{array}{cc}0.0667& 0.267\\ 0.267& 0.0667\end{array}\right]\end{array}$$

參考

Scattering parameters
S-Parameters
Reflection and Transmission
Transmission Lines
Multilayer Structures
RF Engineering Basic Concepts: S-Parameters S-parameter -- 基礎篇