电路的信号流图

我是一名学生，我的问题是找到简单电路的信号流程图。

在此处输入图片说明

我找到了具有势的节点的上述公式。书中说这是使用节点电位建立信号流图的基础。 $k$ $U_k$

$k$ is the number of the node,

$U_k$ it’s potential,

$S_k$ the sum of the admittances from node $k$

$Y_{jk}$ is the admittance between $j$ node having $U_j$ potential and $k$ node

$I_{gk}$ is the algebraic sum of currents in the $k$ node (positive sign if he current enters in the node, negative sign if the current exits from the node)

Next , an example for this circuit for which we need to find the transfer function $H(s)= \frac{U_2(s)}{E(s)}$ : passive RC circuit in double T connection

They write in the book the next linear system:

U_{1} S_{1} = G E + G U_{2}

$U_1S_1 = GE + GU_2$

U_{2} S_{2} = G U_{1} + s C U_{3}

$U_2S_2 = GU_1 + sCU_3$

U_{3} S_{3} = s C E + s C U_{2}

$U_3S_3 = sCE + sCU_2$

where:

S_{1} = 2 (s C + G)

$\require {cancel} \cancel{S_1 = 2(sC + G)}$

S_{1} = 2 G + s C

$S_1 = 2G + sC$

S_{2} = s C + G

$S_2 = sC + G$

S_{3} = 2 (s C + G)

$S_3 = 2(sC + G)$

$G$ is the real part of the admittance $Y_{jk}$ or $G = \frac {1}{R}$ .

From the above equations they find the equation of the potential in each node as:

U_{1} = \frac{G}{S_{1}} E + \frac{G}{S_{1}} U_{2}

$U_1 = \frac{G}{S_1}E + \frac{G}{S_1}U_2$

U_{2} = \frac{G}{S_{2}} U_{1} + \frac{s C}{S_{2}} U_{3}

$U_2 = \frac {G}{S_2}U_1 + \frac {sC}{S_2}U_3$

U_{3} = \frac{s C}{S_{3}} E + \frac{s C}{S_{3}} U_{2}

$U_3 = \frac{sC}{S_3}E + \frac{sC}{S_3}U_2$

The resulting signal flow graph is: enter image description here

If the $S_k$ is the sum of the admittances from $k$ node, how they calculated $S_1 = 2(sC + G)$

I understand for node 2: $S_2 = sC + G$ (because I have one resistor from node 1 to node 2 and one capacitor from node 3 to node 2).

Why for the node 1: $S_1$ expression is not $S_1 = 2G + sC$ ? It is wrong in the book?

Later edit: the correct expression for $S_1$ is indeed $S_1 = 2G + sC$ .

Where are the currents from the first formula?

Later edit: that term is equal to zero.

I need to understand because I have to find the signal flow graph for this circuit and based on the graph to find the transfer function using Mason rule: enter image description here

Hope someone can help me! Thanks in advance!

Best regards, Daniel

transfer-function signal-theory

— NumLock
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As an aside, in North America we call this type of analysis as "Nodal Analysis". Hopefully you can find more homework tutorials under this name. We tend to use different variable letters. Instead of U for voltage we use V. ex. V1 is voltage at node 1. Personally I find it more convenient to keep resistances as resistances and not convert them to admittance. Could you please clarify what G is? I think you mean it is the current.

— lm317

G (C o n d u c t a n c e)

$G (Conductance)$ is the real part of the admittance

Y = G + j X

$Y = G+jX$ which is the inverse of impedance

Z

$Z$ . The impedance of a resistor is

Z = R

$Z=R$ , so

Y = \frac{1}{R}

$Y=\frac {1}{R}$ or

G = \frac{1}{R}

$G = \frac {1}{R}$ (because

ℑ Y = 0

$\Im{Y} = 0$ )

— NumLock

The question remains open. I want to find the transfer function

H (s) = \frac{U_{2} (s)}{U_{1} (s)}

$H(s) = \frac {U_2(s)}{U_1(s)}$ for the last circuit.

— NumLock

Nicely formulated question. For the conductance, I would rather formulate it as

Y = G - j B

$Y=G-jB$ , and not use

X

$X$ there. In this case

B

$B$ is the susceptance, you can look it up on the internet. The minus sign is optional depending on your conventions.

— WalyKu

u can get signal flow graph through the masons formula..

Let me label the intermediate nodes in the circuit using the letters A, B and C as shown below.

enter image description here

The nodal equations at the nodes A,B and C can be re-arranged to get the three equations given below.

\begin{aligned} (1) & U_{A} S_{A} & = C s U_{1} + C s U_{B} \\ (2) & U_{B} S_{B} & = \frac{G}{2} U_{1} + G U_{2} + C s U_{A} \\ (3) & U_{C} S_{C} & = G U_{1} + G^{'} U_{2} \end{aligned}

$\begin{align}U_AS_A &= CsU_1 + CsU_B\tag1\\ U_BS_B &= \frac{G}{2}U_1 + GU_2+CsU_A\tag2\\ U_CS_C &= GU_1 + G'U_2\tag3\end{align}$

Where, $G=\frac{1}{R}$ , $G'=\frac{1}{K}$ . $U_A, U_B$ and $U_C$ are the potential at the nodes A, B and C respectively. And $S_A, S_B$ and $S_C$ are defined as follows:

\begin{aligned} S_{A} & = G + 2 C s \\ S_{B} & = \frac{3}{2} G + C s \\ S_{C} & = G + G^{'} \end{aligned}

$\begin{align}S_A &= G+2Cs\\ S_B &= \frac{3}{2}G + Cs\\ S_C &=G+G'\end{align}$

Let the gain of the operational amplifier be $A_{op}$ and $A_{op}\rightarrow \infty$ . Now the output voltage of op-amp can be written as:

\begin{matrix} (4) & U_{2} = A_{o p} (U_{B} - U_{C}) |_{A_{o p} \to \infty} \end{matrix}

$U_2 = A_{op}(U_B-U_C)|_{A_{op}\rightarrow\infty}\tag4$

From the equations (1) to (3) potential at nodes can be written as:

\begin{aligned} (5) & U_{A} & = \frac{C s}{S_{A}} U_{1} + \frac{C s}{S_{A}} U_{B} \\ (6) & U_{B} & = \frac{G}{2 S_{B}} U_{1} + \frac{G}{S_{B}} U_{2} + \frac{C s}{S_{B}} U_{A} \\ (7) & U_{C} & = \frac{G}{S_{C}} U_{1} + \frac{G^{'}}{S_{C}} U_{2} \end{aligned}

$\begin{align}U_A &= \frac{Cs}{S_A}U_1+\frac{Cs}{S_A}U_B\tag5\\ U_B &= \frac{G}{2S_B}U_1+\frac{G}{S_B}U_2+\frac{Cs}{S_B}U_A\tag6\\ U_C &= \frac{G}{S_C}U_1+\frac{G'}{S_C}U_2\tag7\end{align}$

The signal flow graph can be drawn using the equations from (4) to (7) as given below:

enter image description here

You can apply the limit $A_{op}\rightarrow \infty$ while simplifying the calculations.

— nidhin
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