# Signal Flow Graph for electrical circuits

I am a student and my question is about finding the signal flow graph for a simple circuit.

I found the above formula for a $k$ node having $U_k$ potential. In the book it is said that this is a base for building the signal flow graph using nodes potentials.

$k$ 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 $H(s)= \frac{U_2(s)}{E(s)}$:

They write in the book the next linear system:

$$U_1S_1 = GE + GU_2$$

$$U_2S_2 = GU_1 + sCU_3$$

$$U_3S_3 = sCE + sCU_2$$

where:

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

$$S_1 = 2G + sC$$

$$S_2 = sC + G$$

$$S_3 = 2(sC + G)$$

$G$ 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:

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

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

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

The resulting signal flow graph is:

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:

Hope someone can help me! Thanks in advance!

Best regards, Daniel

• 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 Aug 28 '14 at 5:08
• $G (Conductance)$ is the real part of the admittance $Y = G+jX$ which is the inverse of impedance $Z$. The impedance of a resistor is $Z=R$, so $Y=\frac {1}{R}$ or $G = \frac {1}{R}$ (because $\Im{Y} = 0$) – NumLock Aug 29 '14 at 8:35
• The question remains open. I want to find the transfer function $H(s) = \frac {U_2(s)}{U_1(s)}$ for the last circuit. – NumLock Aug 29 '14 at 8:45
• Nicely formulated question. For the conductance, I would rather formulate it as $Y=G-jB$, and not use $X$ there. In this case $B$ is the susceptance, you can look it up on the internet. The minus sign is optional depending on your conventions. – WalyKu Nov 10 '14 at 13:22
• u can get signal flow graph through the masons formula.. – user60195 Dec 10 '14 at 17:33

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

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

\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{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:

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

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

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

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