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So, I'm trying to derive the transfer function of the following circuit: enter image description here

with $$R_L=3R$$

So my attempt was to treat the circuit as 3 cascaded blocks, obtaining the following transfer functions:

$$\frac{V_{o1}(s)}{V_i(s)}=\frac{1}{1+sCR}$$ $$\frac{V_{o2}(s)}{V_{o1}(s)}=\frac{1}{1+sCR}$$ $$\frac{V_{o}(s)}{V_{o2}(s)}=\frac{3R}{4R+s3CR^2}$$

Then to obtain the transfer function I multiply the 3, obtaining:

$$\frac{V_{o1}(s)}{V_i(s)}=\frac{3R}{3C^3R^4s^3+10C^2R^3s^2+11CR^2s+4R}$$

And putting in canonical form:

$$\frac{V_{o}(s)}{V_i(s)}=\frac{\frac{1}{R^3C^3}}{s^3+\frac{10}{3RC}s^2+\frac{11}{3R^2C^2}s+\frac{4}{3R^3C^3}}$$

However my book obtains this answer instead

$$\frac{V_{o}(s)}{V_i(s)}=\frac{\frac{1}{R^3C^3}}{s^3+\frac{16}{3RC}s^2+\frac{22}{3R^2C^2}s+\frac{2}{R^3C^3}}$$

So I might be making some sort of mistake with the coefficients in the original transfer function. I've already redone this multiple times and can't find my mistake. Can someone help me please?

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    \$\begingroup\$ Read here for something that is almost exactly what you want. It's only missing the trivial addition of your final R. And I'm certain you'll have no problem adding it in. \$\endgroup\$ – jonk Aug 17 '20 at 3:30
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    \$\begingroup\$ @jonk Thank you, I realized my mistake was considering the circuit was unidirectional. By using two-port network theory, using transmission matrices, I was able to reach the book answer! \$\endgroup\$ – Granger Obliviate Aug 17 '20 at 4:15
  • \$\begingroup\$ First you can't cascade individual transfer functions considering the loading of each section. Second, the transfer function you gave is not expressed in a low-entropy format which should be in your case \$H(s)=H_0\frac{1}{D(s)}\$ with \$H_0=\frac{R_L}{3R+R_L}\$. Using the fast analytical circuits techniques or FACTs is truly the fastest way to go without resorting to matrices. The link suggested by jonk is a typical example of the path to follow. \$\endgroup\$ – Verbal Kint Aug 17 '20 at 15:02
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You seem to be considering that each of the low-pass filters will be independent and their cascading will be as having

$$ \frac{V_{o}(s)}{V_i(s)} = \frac{V_{o1}(s)}{V_i(s)} \frac{V_{o2}(s)}{V_{o1}(s)} \frac{V_{o}(s)}{V_{o2}(s)} $$

But, unless you have buffers between each stage you do not have those individual $$ \frac{V_{o1}(s)}{V_i(s)}, \frac{V_{o2}(s)}{V_{o1}(s)}, \frac{V_{o}(s)}{V_{o2}(s)}. $$

Take these high-pass filter and a resistor as example, individually they have some TF. But when combined they do not have the product of those two TF. One high-pass filter, one resistor (all-pass filter). And the fact that the combined circuit with them in cascade is not the same as having their transfer functions multiplied

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  • \$\begingroup\$ Thank you! Yes, I was assuming that the circuit was unidirectional. By using two-port network theory, using transmission matrices, I was able to reach the book answer! \$\endgroup\$ – Granger Obliviate Aug 17 '20 at 4:14
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    \$\begingroup\$ @GrangerObliviate good. If this answer was helpful, please upvote it. And if you think this answer worked to solve your problem, do consider marking it as "accepted". \$\endgroup\$ – jDAQ Aug 17 '20 at 4:23

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