s参数相关知识

1. S matrix推导：

1.1 电流电压及反射系数推导：

其中a代表入射波，b代表反射波，$a=\frac{1}{2}(u+i)$及$b=\frac{1}{2}(u-i)$推导过程如下：
例子：无穷短导线

等效电路图：

$$\begin{gathered} U(z+dz)+dU=U(z) \\ U(z+dz)-U(z)=-dU \\ \frac{\partial U}{\partial z}dz=-dU\text{\hspace{1em}(1)} \end{gathered}$$
$$\begin{gathered} dU=R^{{}^{\prime }}\cdot dz\cdot I+L^{{}^{\prime }}\cdot dz\cdot \frac{\partial I}{\partial t} \\ -\frac{\partial U}{\partial z}=R^{{}^{\prime }}\cdot I+L^{{}^{\prime }}\frac{\partial I}{\partial t}\text{\hspace{1em}(2)} \end{gathered}$$

$$\begin{gathered} I(z)=I(z+dz)+dI \\ I(z+dz)-I(z)=-dI \\ \frac{\partial I}{\partial Z}dz=-dI\text{\hspace{1em}(3)} \end{gathered}$$

$$\begin{gathered} dI=G^{{}^{\prime }}\cdot dz(U-dU)+C^{{}^{\prime }}\cdot dz\cdot \frac{\partial (U-dU)}{\partial t} \\ dI\approx G^{{}^{\prime }}\cdot dz\cdot U+C^{{}^{\prime }}\cdot dz\cdot \frac{\partial U}{\partial t} \\ -\frac{\partial I}{\partial z}=G^{{}^{\prime }}\cdot U+C^{{}^{\prime }}\frac{\partial U}{\partial t}\text{\hspace{1em}(4)} \end{gathered}$$

将（4）带入（2）中即可得到：
$$\frac{{\partial }^{2}U}{\partial z^2}=L^{\prime }{C}^{\prime }\cdot \frac{{\partial }^{2}U}{\partial t^2}+(R^{\prime }{C}^{\prime }+G^{\prime }{L}^{\prime })\cdot \frac{\partial U}{\partial t}+R^{\prime }{G}^{\prime }\cdot U\text{\hspace{1em}(5)}$$

若上述导线为无损导线，即$R^{{}^{\prime }}=0$和$G^{{}^{\prime }}=0$：
$$\frac{{\partial }^{2}U}{\partial z^2}=L^{\prime }{C}^{\prime }\frac{{\partial }^{2}U}{\partial t^2}\text{\hspace{1em}(6)}$$
因此电压公式推导为：
$$U(z,t)=f^{+}\left(t-\frac{z}{v}\right)+f^{-}\left(t+\frac{z}{v}\right)\text{\hspace{1em}(7)}$$
证明（7）成立：
对于$f^{+}\left(t-\frac{z}{v}\right)$, 我们计算一阶导数和二阶导数（$v=\frac{1}{\sqrt{L^{{}^{\prime }}{C}^{{}^{\prime }}}}$）：

$$\frac{\partial }{\partial z}{f}^{+}\left(t-\frac{z}{v}\right)=-\frac{1}{v}{f}^{\prime +}\left(t-\frac{z}{v}\right)\text{\hspace{1em}(8)}$$
$$\frac{{\partial }^2}{\partial z^2}{f}^{+}\left(t-\frac{z}{v}\right)=\frac{1}{v^2}{f}^{\prime \prime +}\left(t-\frac{z}{v}\right)\text{\hspace{1em}(9)}$$
同样的，我们对时间$t$, 求导：
$$\frac{\partial }{\partial t}{f}^{+}\left(t-\frac{z}{v}\right)=f^{\prime +}\left(t-\frac{z}{v}\right)\text{\hspace{1em}(10)}$$
$$\frac{{\partial }^2}{\partial t^2}{f}^{+}\left(t-\frac{z}{v}\right)=f^{\prime \prime +}\left(t-\frac{z}{v}\right)\text{\hspace{1em}(11)}$$

因此，对于$f^{+}$部分，我们有：
$$\frac{{\partial }^2}{\partial z^2}{f}^{+}\left(t-\frac{z}{v}\right)=\frac{1}{v^2}\frac{{\partial }^2}{\partial t^2}{f}^{+}\left(t-\frac{z}{v}\right)\text{\hspace{1em}(12)}$$
同理：
$$\frac{{\partial }^2}{\partial z^2}{f}^{-}\left(t+\frac{z}{v}\right)=\frac{1}{v^2}\frac{{\partial }^2}{\partial t^2}{f}^{-}\left(t+\frac{z}{v}\right)\text{\hspace{1em}(13)}$$
因此：
$$\frac{{\partial }^2}{\partial z^2}{f}^{+}\left(t-\frac{z}{v}\right)+\frac{{\partial }^2}{\partial z^2}{f}^{-}\left(t+\frac{z}{v}\right)=\frac{1}{v^2}\frac{{\partial }^2}{\partial t^2}{f}^{-}\left(t+\frac{z}{v}\right)+\frac{1}{v^2}\frac{{\partial }^2}{\partial t^2}{f}^{+}\left(t-\frac{z}{v}\right)\text{\hspace{1em}(14)}$$
将（7）代入（14）中即可得（6）。
$U(z,t)=f^{+}\left(t-\frac{z}{v}\right)+f^{-}\left(t+\frac{z}{v}\right)$物理意义如下：
导线的实际电压为前进波$f^{+}\left(t-\frac{z}{v}\right)$与反射波$f^{-}\left(t+\frac{z}{v}\right)$的叠加

电流公式：
$$I(z,t)=v\cdot C^{\prime }\left(f^{+}\left(t-\frac{z}{v}\right)-f^{-}\left(t+\frac{z}{v}\right)\right)\text{\hspace{1em}(15)}$$
推导过程如下：
$G^{{}^{\prime }}=0$代入（4）
$$-\frac{\partial I}{\partial z}=C^{\prime }\cdot \frac{\partial U}{\partial t}\text{\hspace{1em}(16)}$$
$$\frac{\partial U}{\partial t}=\frac{\partial }{\partial t}\left[f^{+}\left(w^{+}\right)+f^{-}\left(w^{-}\right)\right]\text{\hspace{1em}(17)}$$
$$\frac{\partial f^{+}(w^{+})}{\partial t}=\frac{\partial f^{+}}{\partial w^{+}}\cdot \frac{\partial w^{+}}{\partial t}\text{\hspace{1em}(18)}$$
$$w^{+}=t-\frac{z}{v},\frac{\partial w^{+}}{\partial t}=1,\frac{\partial w^{-}}{\partial t}=1\text{\hspace{1em}(19)}$$
因此：
$$\frac{\partial f^{+}}{\partial t}=\frac{\partial f^{+}}{\partial w^{+}}及\frac{\partial f^{-}}{\partial t}=\frac{\partial f^{-}}{\partial w^{-}}\text{\hspace{1em}(20)}$$
$$\frac{\partial f^{+}(w^{+})}{\partial z}=\frac{\partial f^{+}}{\partial w^{+}}\cdot \frac{\partial w^{+}}{\partial z}\text{\hspace{1em}(21)}$$

$$\frac{\partial w^{+}}{\partial z}=-\frac{1}{v}\text{及}\frac{\partial w^{-}}{\partial z}=\frac{1}{v}\text{\hspace{1em}(22)}$$
$$\frac{\partial }{\partial z}{f}^{+}=\frac{\partial f^{+}}{\partial w^{+}}\cdot \left(-\frac{1}{v}\right)\Rightarrow \frac{\partial f^{+}}{\partial w^{+}}=-v\cdot \frac{\partial f^{+}}{\partial z}\text{\hspace{1em}(23)}$$
$$\frac{\partial }{\partial z}{f}^{-}=\frac{\partial f^{-}}{\partial w^{-}}\cdot \left(\frac{1}{v}\right)\Rightarrow \frac{\partial f^{-}}{\partial w^{-}}=v\cdot \frac{\partial f^{-}}{\partial z}\text{\hspace{1em}(24)}$$
$$\frac{\partial f^{+}}{\partial t}=-v\cdot \frac{\partial f^{+}}{\partial z},\frac{\partial f^{-}}{\partial t}=v\cdot \frac{\partial f^{-}}{\partial z}\text{\hspace{1em}(25)}$$

$$\frac{\partial U}{\partial t}=\frac{\partial }{\partial t}\left[f^{+}+f^{-}\right]\Rightarrow \frac{\partial U}{\partial t}=v\frac{\partial }{\partial z}\left[-f^{+}+f^{-}\right]\text{\hspace{1em}(26)}$$
将（26）代入（16）中：
$$-\frac{\partial I}{\partial z}=v\cdot C^{\prime }\cdot \frac{\partial }{\partial z}\left[-f^{+}+f^{-}\right]\text{\hspace{1em}(27)}$$
因此：
$$I(z,t)=v\cdot C^{\prime }\left(f^{+}\left(t-\frac{z}{v}\right)-f^{-}\left(t+\frac{z}{v}\right)\right)$$
以下进行反射系数推导：

$$U_2(t)=f^{+}\left(t-\frac{\ell }{v}\right)+f^{-}\left(t+\frac{\ell }{v}\right)\text{\hspace{1em}(28)}$$
$$Z_l{I}_2(t)=f^{+}\left(t-\frac{\ell }{v}\right)-f^{-}\left(t+\frac{\ell }{v}\right)\text{\hspace{1em}(29)}$$
入射波：
$$f^{+}\left(t-\frac{\ell }{v}\right)=\frac{1}{2}\left[U_2(t)+Z_l{I}_2(t)\right]=\frac{1}{2}{I}_2(t)\left(R_L+Z_l\right)\text{\hspace{1em}(30)}$$
反射波：
$$f^{-}\left(t+\frac{\ell }{v}\right)=\frac{1}{2}\left[U_2(t)-Z_l{I}_2(t)\right]=\frac{1}{2}{I}_2(t)\left(R_L-Z_l\right)\text{\hspace{1em}(31)}$$
$$r_L=\frac{反射波}{入射波}\text{\hspace{1em}(32)}$$
$$r_L=\frac{f^{-}\left(t+\frac{\ell }{v}\right)}{f^{+}\left(t-\frac{\ell }{v}\right)}=\frac{R_L{I}_2-Z_l{I}_2}{R_L{I}_2+Z_l{I}_2}\text{\hspace{1em}(33)}$$
所以：
$$r_i=\frac{R_i-Z_{\ell }}{R_i+Z_{\ell }}\text{\hspace{1em}(34)}$$
1.2 S matrix推导：

根据（30）及（31）即可得到入射波，反射波与电流电压关系如下：
$$a=\frac{1}{2}(u+i)\text{\hspace{1em}(35)}$$
$$b=\frac{1}{2}(u-i)\text{\hspace{1em}(36)}$$
其中$u$和$i$为归一化后的值：
$$u=\frac{U}{\sqrt{Z_{\ell }}},i=I\sqrt{Z_{\ell }}\text{\hspace{1em}(37)}$$
$$s_{11}=\frac{b_1}{a_1}{|}_{a_2=0}=\frac{\frac{1}{2}(u_1-i_1)}{\frac{1}{2}(u_1+i_1)}\text{\hspace{1em}(38)}$$
针对 $s_{11}$ 的测量，要求在端口 2无反射。为了使 $a_2=0$ 成立，端口 2 必须与标准阻抗 $Z_{\ell }$ 相匹配终端，这样可以使端口 2 的反射系数为零（端口2端接电阻大小即为归一化电阻大小）。

$$s_{11}=\frac{\frac{1}{2}(u_1-i_1)}{\frac{1}{2}(u_1+i_1)}=\frac{\frac{Z_1+Z_{\ell }}{Z_{\ell }}{i}_1-i_1}{\frac{Z_1+Z_{\ell }}{Z_{\ell }}{i}_1+i_1}=\frac{Z_1}{Z_1+2Z_{\ell }}\text{\hspace{1em}(39)}$$
或者：
$$s_{11}=r_E=\frac{Z_E-Z_{\ell }}{Z_E+Z_{\ell }}\text{其中}{Z}_E=Z_1+Z_{\ell }\Rightarrow s_{11}=\frac{Z_1}{Z_1+2Z_{\ell }}\text{\hspace{1em}(40)}$$
同理：
$$s_{21}=\frac{\frac{1}{2}(u_2-i_2)}{\frac{1}{2}(u_1+i_1)}=\frac{i_1-(-i_1)}{\frac{Z_1+Z_{\ell }}{Z_{\ell }}{i}_1+i_1}=\frac{2Z_{\ell }}{Z_1+2Z_{\ell }}\text{\hspace{1em}(41)}$$
根据无源器件对称性及互易性：$s_{11}=s_{22}$，$s_{12}=s_{21}$。
所以S matrix为：
$$S=\left(\begin{array}{cc}\frac{Z_1}{Z_1+2Z_{\ell }}& \frac{2Z_{\ell }}{Z_1+2Z_{\ell }}\\ \frac{2Z_{\ell }}{Z_1+2Z_{\ell }}& \frac{Z_1}{Z_1+2Z_{\ell }}\end{array}\right)\text{\hspace{1em}(42)}$$

2. 如何使用$m$端口VNA测量$n$端口器件（$m\le n$）：

由（42）可知S matrix与归一化电阻$Z_{\ell }$有关。
此结论为论文核心：
A Rigorous Technique for Measuring the Scattering Matrix of a Multiport Device with a 2-Port Network Analyzer
问题：如何通过2端口网分仪器测量4端口器件？
核心思想：
首先通过2端口网分测量：1-2,1-3,1-4,2-3,2-4和3-4端口，即将4端口4x4 的S-matrix通过6次测量填满

测量1-2端口时，1,2端口接网分（默认接50欧姆电阻），3,4端口端接端口对应电阻$Z_3$,$Z_4$
记此时测得的S matrix为$S$，后将S matrix根据$Z_{\ell 1}=50\to Z_{\ell 1}^{{}^{\prime }}=Z_1$, $Z_{\ell 2}=50\to Z_{\ell 2}^{{}^{\prime }}=Z_2$进行重新归一化，得到S matrix $S^{{}^{\prime }}$
$$S^{\prime }=(I-S{)}^{-1}(S-\Gamma )(I-S\cdot \Gamma {)}^{-1}(I-S)\text{\hspace{1em}(43)}$$
其中，$S$ 是归一化到一组端口阻抗 ${\zeta }_i(i=1\text{到}n)$ 的原始散射矩阵，$S^{\prime }$ 是归一化到一组新的端口阻抗 $Z_i(i=1\text{到}n)$ 的变换后的散射矩阵。$I$ 是 $n\times n$ 的单位矩阵，$\Gamma$ 是包含了反射系数 ${\Gamma }_i(i=1\text{到}n)$ 的 $n\times n$ 对角矩阵。
其他思路：
Multiport S-parameter calculation from two-port network analyzer measurements with or without switch matrix
例子：使用2端口VNA测量三端口器件
三次测量：1-2，1-3，2-3

$$\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\text{\hspace{1em}(44)}$$
两端口校准的矢量网络分析仪（VNA）提供了测量的散射参数$S_{ii}^{mi}$，其中 ( i = 1, 2, 3 ) 表示反射参数，而 $S_{ij}^m$（( i = 1, 2, 3 ) ， ( j = 1, 2, 3 )）表示测试设备（DUT）的传输参数。第一次校准测得的散射参数$S_{11}^{m1}$与第二次校准测得的散射参数$S_{11}^{m2}$并不相同。两者都是 DUT 的反射值$S_{11}$和匹配端口2或3的终端匹配的函数。
三端口器件的反射系数分别为：
$$S_a=\frac{a_1}{b_1},S_b=\frac{a_2}{b_2},S_c=\frac{a_3}{b_3}\text{\hspace{1em}(45)}$$
将（45）代入（44）得：
$$\frac{a_3}{S_c}=S_{31}{a}_1+S_{32}{a}_2+S_{33}{a}_3\text{\hspace{1em}(46)}$$
$$a_3=\frac{S_{31}}{\frac{1}{S_c}-S_{33}}{a}_1+\frac{S_{32}}{\frac{1}{S_c}-S_{33}}{a}_2\text{\hspace{1em}(47)}$$

将这个$a_3$的表达式代入$b_1$和$b_2$的方程中，我们会得到如下形式：

$$b_1=S_{11}{a}_1+S_{12}{a}_2+S_{13}\left(\frac{S_{31}}{1/S_c-S_{33}}{a}_1+\frac{S_{32}}{1/S_c-S_{33}}{a}_2\right)\text{\hspace{1em}(48)}$$

$$b_2=S_{21}{a}_1+S_{22}{a}_2+S_{23}\left(\frac{S_{31}}{1/S_c-S_{33}}{a}_1+\frac{S_{32}}{1/S_c-S_{33}}{a}_2\right)\text{\hspace{1em}(49)}$$

$$r_c=\frac{S_c}{1-S_{33}{S}_c}\approx \frac{S_c}{1-S_{33}^m{S}_c}\text{\hspace{1em}(50)}$$

$$b_1=(S_{11}+S_{13}{S}_{31}{r}_c)a_1+(S_{12}+S_{13}{S}_{32}{r}_c)a_2\text{\hspace{1em}(51)}$$
$$b_2=(S_{21}+S_{23}{S}_{31}{r}_c)a_1+(S_{22}+S_{23}{S}_{32}{r}_c)a_2\text{\hspace{1em}(52)}$$
$$\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)=\left(\begin{array}{cc}{S}_{11}+S_{13}{S}_{31}{r}_c& S_{12}+S_{13}{S}_{32}{r}_c\\ S_{21}+S_{23}{S}_{31}{r}_c& S_{22}+S_{23}{S}_{32}{r}_c\end{array}\right)\cdot \left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)\text{\hspace{1em}(53)}$$
因此：
$$\left(\begin{array}{cc}{S}_{11}^{m1}& S_{12}^m\\ S_{21}^m& S_{22}^{m1}\end{array}\right)=\left(\begin{array}{cc}{S}_{11}+S_{13}{S}_{31}{r}_c& S_{12}+S_{13}{S}_{32}{r}_c\\ S_{21}+S_{23}{S}_{31}{r}_c& S_{22}+S_{23}{S}_{32}{r}_c\end{array}\right)\text{\hspace{1em}(54)}$$
同理：
$$r_b=\frac{S_b}{1-S_{22}{S}_b}\approx \frac{S_b}{1-S_{22}^m{S}_b}\text{\hspace{1em}(55)}$$

$$r_a=\frac{S_a}{1-S_{11}{S}_a}\approx \frac{S_a}{1-S_{11}^m{S}_a}\text{\hspace{1em}(56)}$$
$$\left(\begin{array}{cc}{S}_{11}^{m2}& S_{13}^m\\ S_{31}^m& S_{33}^{m2}\end{array}\right)=\left(\begin{array}{cc}{S}_{11}+S_{12}{S}_{21}{r}_b& S_{13}+S_{12}{S}_{23}{r}_b\\ S_{31}+S_{32}{S}_{21}{r}_b& S_{33}+S_{23}{S}_{32}{r}_b\end{array}\right)\text{\hspace{1em}(57)}$$
$$\left(\begin{array}{cc}{S}_{22}^{m3}& S_{23}^m\\ S_{32}^m& S_{33}^{m3}\end{array}\right)=\left(\begin{array}{cc}{S}_{22}+S_{21}{S}_{12}{r}_c& S_{23}+S_{21}{S}_{13}{r}_c\\ S_{32}+S_{31}{S}_{12}{r}_c& S_{33}+S_{31}{S}_{13}{r}_c\end{array}\right)\text{\hspace{1em}(58)}$$

$$S_{11}=\frac{1}{2}\left(S_{11}^{m1}+S_{11}^{m2}-S_{13}^m{S}_{31}^m{r}_c-S_{12}^m{S}_{21}^m{r}_b\right),\text{\hspace{1em}(59)}$$

$$S_{22}=\frac{1}{2}\left(S_{22}^{m1}+S_{22}^{m3}-S_{12}^m{S}_{21}^m{r}_a-S_{32}^m{S}_{23}^m{r}_c\right),\text{\hspace{1em}(60)}$$

$$S_{33}=\frac{1}{2}\left(S_{33}^{m2}+S_{33}^{m3}-S_{31}^m{S}_{13}^m{r}_a-S_{23}^m{S}_{32}^m{r}_b\right),\text{\hspace{1em}(61)}$$
此处我们认为$S_{ij}^m$的测量误差要远小于$S_{ii}^m$的测量误差，因此（59）~（61）用$S_{ij}^m$代替$S_{ij}$。
$$S_{12}=S_{12}^m-S_{13}^m{S}_{32}^m{r}_c,\text{\hspace{1em}(62)}$$

$$S_{13}=S_{13}^m-S_{12}^m{S}_{23}^m{r}_b,\text{\hspace{1em}(63)}$$

$$S_{21}=S_{21}^m-S_{23}^m{S}_{31}^m{r}_c,\text{\hspace{1em}(64)}$$

$$S_{31}=S_{31}^m-S_{32}^m{S}_{12}^m{r}_a,\text{\hspace{1em}(65)}$$

$$S_{23}=S_{23}^m-S_{21}^m{S}_{13}^m{r}_b,\text{\hspace{1em}(66)}$$

$$S_{32}=S_{32}^m-S_{31}^m{S}_{12}^m{r}_a.\text{\hspace{1em}(67)}$$
因此，此方法利用多次测量的$S_{ii}^{mi}$所得值以及各端口间的相互关系推导出最终反射系数$S_{ii}$，充分利用了$m$端口VNA测量$n$端口器件时，因VNA端口不足而重复测量的数据。
同理：当使用4端口VNA测量7端口器件时：
$$S_{11}=\frac{1}{6}\left(S_{11}^{m1}+S_{11}^{m2}+S_{11}^{m3}+S_{11}^{m4}+S_{11}^{m5}+S_{11}^{m6}-S_{17}^m{S}_{71}^m{r}_g-S_{16}^m{S}_{61}^m{r}_f-S_{15}^m{S}_{51}^m{r}_e-S_{14}^m{S}_{41}^m{r}_d-S_{13}^m{S}_{31}^m{r}_c-S_{12}^m{S}_{21}^m{r}_b\right),\text{\hspace{1em}(68)}$$

$$S_{12}=S_{12}^m-S_{13}^m{S}_{32}^m{r}_c-S_{14}^m{S}_{42}^m{r}_d-S_{15}^m{S}_{52}^m{r}_e-S_{16}^m{S}_{62}^m{r}_f-S_{17}^m{S}_{72}^m{r}_g\text{\hspace{1em}(69)}$$
所以使用4端口VNA测量7端口器件时，需要确保选取测量的对角线元素$S_{ii}^m$能够重复测量6次：
$$\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\\ b_4\\ b_5\\ b_6\\ b_7\end{array}\right)=\left(\begin{array}{ccccccc}{S}_{11}& S_{12}& S_{13}& S_{14}& S_{15}& S_{16}& S_{17}\\ S_{21}& S_{22}& S_{23}& S_{24}& S_{25}& S_{26}& S_{27}\\ S_{31}& S_{32}& S_{33}& S_{34}& S_{35}& S_{36}& S_{37}\\ S_{41}& S_{42}& S_{43}& S_{44}& S_{45}& S_{46}& S_{47}\\ S_{51}& S_{52}& S_{53}& S_{54}& S_{55}& S_{56}& S_{57}\\ S_{61}& S_{62}& S_{63}& S_{64}& S_{65}& S_{66}& S_{67}\\ S_{71}& S_{72}& S_{73}& S_{74}& S_{75}& S_{76}& S_{77}\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\\ a_5\\ a_6\\ a_7\end{array}\right)\text{\hspace{1em}(70)}$$
补充：信号流图
计算方法：

例子：

根据图示可知此器件信号只能从2->5，5->3, 所以：
$$\left(\begin{array}{c}{b}_2\\ b_3\\ b_5\end{array}\right)=\left(\begin{array}{ccc}{S}_{22}& S_{23}& S_{25}\\ S_{32}& S_{33}& S_{35}\\ S_{52}& S_{53}& S_{55}\end{array}\right)\left(\begin{array}{c}{a}_2\\ a_3\\ a_5\end{array}\right)\text{\hspace{1em}(1)}$$
$S_{23}$，$S_{32}$，$S_{35}$，$S_{53}=0$
根据图示及所提供S matrix可画出如下信号流图：