Transfer function of a filter: where is my mistake?

So, I'm trying to derive the transfer function of the following circuit:

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?

• 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.
– jonk
Aug 17, 2020 at 3:30
• @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! Aug 17, 2020 at 4:15
• 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. Aug 17, 2020 at 15:02

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