Question 14.2: For the series RC circuit in Fig.14.7:a) Find the transfer f...

a) To derive an expression for the transfer function, we first construct the s-domain equivalent of the circuit in Fig.14.7,as shown in Fig.14.8.

Using s-domain voltage division on the equivalent circuit, we find

H\left(s\right) =\frac{\frac{1}{RC} }{s+\frac{1}{RC} }.

Now, substitute s=j\omega  and compute the magnitude of the resulting complex expression:

\mid H\left(j\omega \right)\mid =\frac{\frac{1}{RC}}{\sqrt{\omega ^{2}+\left(\frac{1}{RC}\right)^{2}} }.

b) At the cutoff frequency \omega _{c},  \mid H\left(j\omega \right)\mid is equal to   \left(1/ \sqrt{2} \right) H_{max}.   For   a   low-pass   filter,

H_{max}=H\left(j0\right), and for the circuit in Fig. 14.8, H\left(j0\right)=1. We can then describe the relationship among the quantities R,C, and \omega _{c}:

\mid H\left(j\omega \right)\mid =\frac{1}{\sqrt{2} } \left(1\right)=\frac{\frac{1}{RC}}{\sqrt{\omega _{c}^{2}+\left(\frac{1}{RC}\right)^{2}} }.

Solving this equation for \omega _{c} we get

\omega _{c}=\frac{1}{RC}.

c) From the results in (b), we see that the cutoff frequency is determined by the values of R and C. Because R and C cannot be computed independently, let’s choose C=1\mu F. Given a choice, we will usually specify a value for C first, rather than for R or L, because the number of available capacitor values is much smaller than the number of resistor or inductor values. Remember that we have to convert the specified cutoff frequency from 3  kHz to \left(2\pi \right) \left(3\right)Krad/ s:

=53.05  \Omega.