Chapter 7, Problem 7.1EP

(a) For the circuit shown in Figure 7.2, the parameters are $R_S=2\text{kΩ}$ and $R_P=8\text{kΩ}$ . (i) If the corner frequency is $f_L=50\text{Hz}$ , determine the value of $C_S$ . (ii) Find the magnitude of the transfer function at $f=20\text{Hz,}50\text{Hz},$ and 100Hz. (Ans. (i) $C_S=0.318\text{μF}$ ; (ii) 0.297, 0.566, and 0.716)
(b) Consider the circuit shown in Figure 7.3 with parameters $R_S=4.7\text{kΩ}$ , $R_P=25\text{kΩ}$ , and $C_P=120\text{pF}$ . (i) Determine the corner frequency $f_H$ . (ii) Determine the magnitude of the transfer function at $f=0.2f_H$ , $f=f_H$ , and $f=8f_H$ . (Ans. (i) $f_H=335\text{kHz}$ ; (ii) 0.825, 0.595, 0.104)

To determine

The value of the capacitor $C_S$ and the magnitude of transfer function.

Answer to Problem 7.1EP

  $\begin{array}{l}\left(\text{i}\right).C_S=0.318\mu F\\ \left(\text{ii}\right).\left|H\left(jf\right)\right|=0.297,0.566,0.716\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{R}_S=2\text{kΩ},R_P=8\text{kΩ}\\ \left(\text{i}\right).f_L=50Hz\\ \left(\text{ii}\right).f=20\text{Hz},50\text{Hz},100\text{Hz}\end{array}$

Calculation:

(i).

The value of coupling capacitor $\left(C_S\right)$ is determined as follows:

The expression of lower cutoff frequency is:

  $\begin{array}{l}{f}_L=\frac{1}{2\pi {\tau }_s}\\ f_L=\frac{1}{2\pi \left(R_S+R_P\right)C_S}\left[{\tau }_S=\left(R_S+R_P\right)C_S\right]\\ 50=\frac{1}{2\pi \left(2\times {10}^3+8\times {10}^3\right)C_S}\\ C_S=\frac{1}{100\pi \times {10}^4}\\ C_S=0.318\text{μF}\end{array}$

(ii).

The transfer function of the circuit is determined as follows:

  $\begin{array}{l}f=50\text{Hz,}\\ \left|H\left(jf\right)\right|=\left(\frac{8}{2+8}\right)\left(\frac{\frac{50}{50}}{\sqrt{1+{\left(\frac{50}{50}\right)}^2}}\right)\\ \left|H\left(jf\right)\right|=\frac{8}{10}\times \frac{1}{\sqrt{2}}\\ \left|H\left(jf\right)\right|=0.566\end{array}$

  $\begin{array}{l}f=100\text{Hz,}\\ \left|H\left(jf\right)\right|=\left(\frac{8}{2+8}\right)\left(\frac{\frac{100}{50}}{\sqrt{1+{\left(\frac{100}{50}\right)}^2}}\right)\\ \left|H\left(jf\right)\right|=\frac{8}{10}\times \frac{2}{\sqrt{5}}\\ \left|H\left(jf\right)\right|=0.716\end{array}$

To determine

The value of the corner frequency $f_H$ and the magnitude of transfer function.

Answer to Problem 7.1EP

  $\begin{array}{l}\left(\text{i}\right).f_H=335\text{kHz}\\ \left(\text{ii}\right).\left|H\left(jf\right)\right|=0.825,0.595,0.104\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{R}_S=4.7\text{kΩ},R_P=25\text{kΩ}\\ \left(\text{i}\right).C_P=120\text{pF}\\ \left(\text{ii}\right).f=0.2f_H,f_H,8f_H\end{array}$

Calculation:

The value of time constant of the circuit is:

  $\begin{array}{l}{\tau }_P=\left(R_S\left|\right|R_P\right)C_P\\ {\tau }_P=\frac{R_S{R}_P}{R_S+R_P}\times C_P\\ {\tau }_P=\frac{4.7\times {10}^3\times 25\times {10}^3}{\left(4.7+25\right)\times {10}^3}\times 120\times {10}^{-12}\\ {\tau }_P=474.747\text{ns}\end{array}$

The value of corner frequency is determined as follows:

  $\begin{array}{l}{f}_H=\frac{1}{2\pi {\tau }_P}\\ f_H=\frac{1}{2\pi \times 474.747\times {10}^{-9}}\\ f_H=335\text{kHz}\end{array}$

The transfer function of the circuit is determined as follows:

  $\begin{array}{l}f=0.2f_H\\ \left|H\left(jf\right)\right|=\frac{25}{4.7+25}\left[\frac{1}{\sqrt{1+{\left(\frac{0.2f_H}{f_H}\right)}^2}}\right]\\ \left|H\left(jf\right)\right|=\frac{25}{29.7}\left[\frac{1}{\sqrt{1.04}}\right]\\ \left|H\left(jf\right)\right|=0.825\end{array}$

  $\begin{array}{l}f=f_H\\ \left|H\left(jf\right)\right|=\frac{25}{4.7+25}\left[\frac{1}{\sqrt{1+{\left(\frac{f_H}{f_H}\right)}^2}}\right]\\ \left|H\left(jf\right)\right|=\frac{25}{29.7}\left[\frac{1}{\sqrt{2}}\right]\\ \left|H\left(jf\right)\right|=0.595\end{array}$

  $\begin{array}{l}f=8f_H\\ \left|H\left(jf\right)\right|=\frac{25}{4.7+25}\left[\frac{1}{\sqrt{1+{\left(\frac{8f_H}{f_H}\right)}^2}}\right]\\ \left|H\left(jf\right)\right|=\frac{25}{29.7}\left[\frac{1}{\sqrt{65}}\right]\\ \left|H\left(jf\right)\right|=0.104\end{array}$