Frequency response of 3 series LPF filters

[aiq25]

Hi.

I'm working on a circuit analysis where we have to graph the frequency response of a system. The system has 3 LP filters: first an RC filter, second an op-amp with a capacitor, third is another RC filter.

I was able to figure out the frequency response (bode plot) of the three individual filters but I'm not sure how to figure out the frequency response of the entire system. Any feedback would be greatly appreciated. Attached is a block diagram.

[Kleinstein]

The bode plot is logarithmic scale. Having the filters in series gives the product of the transfer functions. In logarithmic scale this is a graphic addition on the vertical axis.

[Benta]

^^^ This

That's the beauty of Bode plots. And you can just add the phase responses (which are on a linear y-scale) as well.

[The Electrician]

The bode plot is logarithmic scale. Having the filters in series gives the product of the transfer functions. In logarithmic scale this is a graphic addition on the vertical axis.

This is not true in general.  In this particular problem, if the opamp is taken to be ideal, the R1/C1 stage is not loaded by the + input of the opamp, and because the ideal opamp output impedance is zero, the third stage doesn't load the opamp stage.

If a stage could load its previous stage, the overall transfer function would not be the product of the individual transfer functions.  The overall TF would have to be calculated as a complete circuit.  aiq25 should be aware of this because a subsequent problem might not be free of loading effects.

[Dave]

The second stage isn't a proper LPF either, the gain at high frequencies approaches 1, not 0.

Assuming ideal opamp, the frequency response functions would be:
\[G_1=\frac{1}{1+j\omega R_1 C_1}\]
\[G_2=1+\frac{R_3}{R_2(1+j\omega R_3 C_2)}\]
\[G_3=\frac{1}{1+j\omega R_4 C_3}\]
\[G=G_1\cdot G_2\cdot G_3\]

To draw a Bode plot by hand, you first draw the response of each individual filter segment (simple straight line approximations are fine) and then add the responses together for the response of the whole circuit. It is a simple matter of adding the gain responses together (loglog scale, of course) and adding the phase shifts together (linlog scale).

I could go in detail about how to draw the individual reponses, but this site does a way better job at explaining it.