Multiplying the transfer functions of multiple cascaded circuit fragments to get the frequency response of the system doesn’t work because of loading between stages. However, I came across the following equation from here:

On page 7-14 it says that you can do this if you introduce load factors into the equation:

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Say I break up their simple example circuit into three sections and use their method as follows:

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However, the transfer function I get is different than the transfer function I get with simple circuit analysis of the entire combined circuit using the following:

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Plotting the responses in Matlab they are similar, but not identical, and the transfer function in the second case is much simpler compared to the first. Is there something wrong with my calculations? Or can someone verify if/when this method works?

I used the above 2 methods and compared them to a more complex circuit built on a breadboard, the second method bode plot is very close to measured, but the cascaded transfer function method is way off.


1 Answer 1


I'm sorry for I cannot answer your particular question, I can however give you a solution to your underlying question which I assume is:

How can I perform cascade-by-cascade calculations and end up with H(s)?

Well, I've used the ABCD matrix for this, I am not 100% sure if I've used it correctly, but I've come to the same answer as when I use KCL, so it does work.

I will continue assuming you've clicked the link and seen how every impedance is translated into a 2x2 matrix for series/parallel connections.

For clarity I will show you how I would apply the ABCD-parameters for your particular case.

$$ Z_{s} = \begin{pmatrix} 1 & R_s \\ 0 & 1 \\ \end{pmatrix} ~~~ Z_f = \begin{pmatrix} 1 & R_f \\ 0 & 1 \\ \end{pmatrix} ~~~ Z_c = \begin{pmatrix} 1 & 0 \\ sC & 1 \\ \end{pmatrix} ~~~ Z_l = \begin{pmatrix} 1 & 0 \\ \frac{1}{R_l} & 1 \\ \end{pmatrix} $$

These are your impedances, then we place the currents and voltages at the appropriate places with appropriate signs and we get the following:

$$ \begin{pmatrix} V_1(s) \\ I_1(s) \\ \end{pmatrix} =Z_sZ_fZ_cZ_l \begin{pmatrix} V_2(s) \\ -I_2(s)\\ \end{pmatrix} $$

Then we get something that looks like this:

$$ \begin{pmatrix} V_1(s) \\ I_1(s) \\ \end{pmatrix} = \begin{pmatrix} 1+(R_s+R_f)(sC+\frac{1}{R_l}) & R_s+R_f \\ sC+\frac{1}{R_l} & 1 \\ \end{pmatrix} \begin{pmatrix} V_2(s) \\ -I_2(s)\\ \end{pmatrix} $$

The equation above shows how the current and voltages are related between the input and output. But we don’t care about the input current. For simplicity, let's assume that you are driving a high-impedance load meaning that \$I_2\$ ≃ 0. With output current equal to 0 the left elements remains. Only caring about \$V_1\$ leaves the top elements, this equation is simplified to what we only care about, namely the voltages. The top left element.

In other words, this is the important equation we can dissect out of the matrix equation above: $$ \begin{align} V_1(s)&=(1+(R_s+R_f)(sC+\frac{1}{R_l}))V_2(s)\\ \\ \frac{V_2(s)}{V_1(s)}=H(s)&=\frac{1}{1+(R_s+R_f)(sC+\frac{1}{R_l})} \end{align} $$

I have not verified if the equation I came up with is the same as the one you have, but I will let you verify that.

I'm sorry for not answering your specific question, but I believe this answers your underlying question/problem.


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