10.4: Frequency-Response Function of an RC Band-Pass Filter

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We re-visit once again Section 5.4 and the example of Section 9.10, where it is demonstrated that a typical \(RC\) band-pass filter, as depicted below, is an overdamped 2^nd order system with right-hand-side dynamics. Our goal now is to derive the FRF of this filter circuit. Rather than analyze 2^nd order ODE 9.10.9 directly, it is instructive to start with the coupled 1^st order ODEs of the two stages: Equation 5.4.1, \(\tau_{L} \dot{e}_{m}+e_{m}=e_{i}\), with \(\tau_{L}=R_{L} C_{L}\), for the upstream, low-pass filter; and Equation 5.4.2, \(\tau_{H} \dot{e}_{o}+e_{o}=\tau_{H} \dot{e}_{m}\), with \(\tau_{H}=R_{H} C_{H}\), for the downstream, high-pass filter. In this approach, we represent a system (the band-pass filter, in this case) as an assemblage of stages or lower-order sub-systems (the low-pass and high-pass filters, in this case). This section is a preview of the more general discussion beginning in Chapter 13.

Figure \(\PageIndex{1}\): \(RC\) band-pass filter

Taking the Laplace transform of the ODE for the low-pass stage, with the assumption that all ICs are zero, gives

\[\left.\left(\tau_{L} s+1\right) L\left[e_{m}\right]\right|_{I C s=0}=\left.L\left[e_{i}\right] \Rightarrow L\left[e_{m}\right]\right|_{/ C s=0}=\frac{1}{\tau_{L} s+1} L\left[e_{i}\right]\label{eqn:10.25} \]

Next, we take the Laplace transform of the ODE for the high-pass stage, still with the assumption that all ICs are zero, and then substitute Equation \(\ref{eqn:10.25}\) into the result:

\[\left(\tau_{H} s+1\right) L\left[e_{o}\right]_{/ C s=0}=\left.\tau_{H} s L\left[e_{m}\right]\right|_{I C s=0} \nonumber \]

\[\left.\Rightarrow \quad L\left[e_{o}\right]\right|_{I C s=0}=\left.\frac{\tau_{H} S}{\tau_{H} S+1} L\left[e_{m}\right]\right|_{I C s=0}=\frac{\tau_{H} s}{\tau_{H} s+1} \times \frac{1}{\tau_{L} s+1} L\left[e_{i}\right]\label{eqn:10.26} \]

So the transfer function of the entire \(RC\) band-pass filter circuit, clearly displaying right-hand-side dynamics, is

\[T F(s) \equiv \frac{\left.L\left[e_{o}\right]\right|_{I C_{s=0}}}{L\left[e_{i}\right]}=\frac{\tau_{H} s}{\left(\tau_{H} s+1\right)\left(\tau_{L} s+1\right)}\label{eqn:10.27} \]

Finally, the system complex frequency-response function is

\[F R F(\omega)=T F(j \omega)=\frac{j \omega \tau_{H}}{\left(j \omega \tau_{H}+1\right)\left(j \omega \tau_{L}+1\right)}\label{eqn:10.28} \]

See homework Problem 10.11 for a numerical evaluation of this frequency response that demonstrates the practical function for which a band-pass filter is designed.