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

enter image description here

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)}\$: passive RC circuit in double T connection

They write in the book the next linear system:

$$U_1S_1 = GE + GU_2$$

$$U_2S_2 = GU_1 + sCU_3$$

$$U_3S_3 = sCE + sCU_2$$


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

$$S_1 = 2G + sC$$

$$S_2 = sC + G$$

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

\$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: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

  • \$\begingroup\$ 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. \$\endgroup\$
    – lm317
    Aug 28, 2014 at 5:08
  • 1
    \$\begingroup\$ \$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\$) \$\endgroup\$
    – NumLock
    Aug 29, 2014 at 8:35
  • \$\begingroup\$ The question remains open. I want to find the transfer function \$H(s) = \frac {U_2(s)}{U_1(s)} \$ for the last circuit. \$\endgroup\$
    – NumLock
    Aug 29, 2014 at 8:45
  • \$\begingroup\$ 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. \$\endgroup\$
    – WalyKu
    Nov 10, 2014 at 13:22
  • \$\begingroup\$ u can get signal flow graph through the masons formula.. \$\endgroup\$
    – user60195
    Dec 10, 2014 at 17:33

1 Answer 1


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.

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

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:

enter image description here

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


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