Chapter 14, Problem 66P

A “general” first-order filter is shown in Fig. 14.93.

1. (a) Show that the transfer function is

   s = jω

2. (b) What condition must be satisfied for the circuit to operate as a high-pass filter?
3. (c) What condition must be satisfied for the circuit to operate as a low-pass filter?

Figure 14.93

To determine

Prove that the transfer function $H\left(s\right)$ for the circuit shown in Figure 14.93 is $\frac{R_4}{R_3+R_4}\times \frac{s+\left(\frac{1}{R_{1}C}\right)\left[\frac{R_1}{R_2}-\frac{R_3}{R_4}\right]}{s+\frac{1}{R_{2}C}}$

Explanation of Solution

Given data:

Refer to Figure 14.93 in the textbook.

Formula used:

Write the general expression to calculate the transfer function of the circuit in Figure 1.

$H\left(s\right)=\frac{V_o}{V_s}$        (1)

Here,

$V_o$ is the output response of the system, and

$V_s$ is the input response of the system.

Calculation:

The given circuit is redrawn as shown in Figure 1 in s-domain.

Apply Kirchhoff’s current law at the node $V_{+}$ in Figure 1.

$\begin{array}{l}\frac{V_{+}-V_s}{R_3}+\frac{V_{+}}{R_4}=0\\ \frac{V_{+}}{R_3}-\frac{V_s}{R_3}+\frac{V_{+}}{R_4}=0\\ \left(\frac{1}{R_3}+\frac{1}{R_4}\right)V_{+}=\frac{V_s}{R_3}\end{array}$

Rearrange the equation to find $V_{+}$.

$V_{+}=\frac{V_s}{R_3\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}$

$V_{+}=\frac{V_s}{\left(1+\frac{R_3}{R_4}\right)}$        (2)

Apply Kirchhoff’s current law at the node $V_{-}$ in Figure 1.

$\begin{array}{l}\frac{V_{-}-V_s}{R_1}+\frac{V_{-}-V_o}{\left(\frac{1}{sC}\right)}+\frac{V_{-}-V_o}{R_2}=0\\ \frac{V_{-}}{R_1}-\frac{V_s}{R_1}+sC{V}_{-}-sC{V}_o+\frac{V_{-}}{R_2}-\frac{V_o}{R_2}=0\end{array}$

$\left(\frac{1}{R_1}+\frac{1}{R_2}+sC\right)V_{-}-\frac{V_s}{R_1}-\left(sC+\frac{1}{R_2}\right)V_o=0$        (3)

Since, for an ideal operational amplifier, $V_{+}=V_{-}$. Therefore

Substitute $V_{-}$ for $V_{+}$ in equation (2).

$V_{-}=\frac{V_s}{\left(1+\frac{R_3}{R_4}\right)}$        (4)

Substitute equation (4) in equation (3).

$\begin{array}{l}\left(\frac{1}{R_1}+\frac{1}{R_2}+sC\right)\left(\frac{V_s}{\left(1+\frac{R_3}{R_4}\right)}\right)-\frac{V_s}{R_1}-\left(sC+\frac{1}{R_2}\right)V_o=0\\ \left(\frac{V_s\left(\frac{1}{R_1}+\frac{1}{R_2}+sC\right)}{\left(1+\frac{R_3}{R_4}\right)}\right)-\frac{V_s}{R_1}=\left(sC+\frac{1}{R_2}\right)V_o\\ \left(\frac{\frac{1}{R_1}+\frac{1}{R_2}+sC}{\left(1+\frac{R_3}{R_4}\right)}-\frac{1}{R_1}\right)V_s=\left(sC+\frac{1}{R_2}\right)V_o\end{array}$

Rearrange the equation as follows,

$\begin{array}{c}\frac{V_o}{V_s}=\frac{\frac{\left(\frac{1}{R_1}+\frac{1}{R_2}+sC\right)}{\left(1+\frac{R_3}{R_4}\right)}-\left(\frac{1}{R_1}\right)}{sC+\frac{1}{R_2}}\\ =\frac{\left(\frac{1}{R_1}+\frac{1}{R_2}+sC\right)-\left(\frac{1}{R_1}\right)\left(1+\frac{R_3}{R_4}\right)}{\left(sC+\frac{1}{R_2}\right)\left(1+\frac{R_3}{R_4}\right)}\\ =\frac{\frac{1}{R_1}+\frac{1}{R_2}+sC-\frac{1}{R_1}-\frac{R_3}{R_1{R}_4}}{\left(sC+\frac{1}{R_2}\right)\left(1+\frac{R_3}{R_4}\right)}\\ =\left(\frac{R_4}{R_3+R_4}\right)\times \frac{\frac{1}{R_2}+sC-\frac{R_3}{R_1{R}_4}}{\left(sC+\frac{1}{R_2}\right)}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}\frac{V_o}{V_s}=\left(\frac{R_4}{R_3+R_4}\right)\times \frac{\frac{1}{C{R}_2}+s-\frac{R_3}{C{R}_1{R}_4}}{\left(s+\frac{1}{R_{2}C}\right)}\\ =\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s+\frac{1}{R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{\left(s+\frac{1}{R_{2}C}\right)}\end{array}$

Substitute $\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s+\frac{1}{R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{\left(s+\frac{1}{R_{2}C}\right)}$ for $\frac{V_o}{V_s}$ in equation (1) to find the transfer function $H\left(s\right)$.

$H\left(s\right)=\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s+\frac{1}{R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{\left(s+\frac{1}{R_{2}C}\right)}$

Conclusion:

Thus, the transfer function $H\left(s\right)$ for the given circuit shown in Figure 14.92 is $\left(1+\frac{R_f}{R_i}\right)\left(\frac{j\omega RC}{1+j\omega RC}\right)$ and is proved.

To determine

Examine the condition to be satisfied to operate as a high-pass filter of the given circuit.

Explanation of Solution

Given data:

Refer to Part (a),

$H\left(s\right)=\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s+\frac{1}{R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{\left(s+\frac{1}{R_{2}C}\right)}$        (5)

Calculation:

For high-pass filter, apply the limits $s\to 0$ in equation (5). That is,

$\begin{array}{l}\left(\frac{R_4}{R_3+R_4}\right)\times \frac{0+\frac{1}{R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{\left(0+\frac{1}{R_{2}C}\right)}=0\\ \frac{1}{R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)=0\\ \frac{R_1}{R_2}-\frac{R_3}{R_4}=0\end{array}$

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$        (6)

Substitute equation (6) in equation (5) to find the transfer function for high-pass filter $H\left(s\right)$.

$\begin{array}{c}H\left(s\right)=\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s+\frac{1}{R_{1}C}\left(\frac{R_3}{R_4}-\frac{R_3}{R_4}\right)}{\left(s+\frac{1}{R_{2}C}\right)}\\ =\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s}{\left(s+\frac{1}{R_{2}C}\right)}\end{array}$

Conclusion:

Thus, the condition to be satisfied to operate as a high-pass filter of the given circuit is at $\frac{R_1}{R_2}=\frac{R_3}{R_4}$.

To determine

Examine the condition to be satisfied to operate as a low-pass filter of the given circuit.

Explanation of Solution

Reduce the equation (5) as follows,

$H\left(s\right)=\left(\frac{R_4}{R_3+R_4}\right)\times \frac{s\left[1+\frac{1}{s{R}_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)\right]}{s\left(1+\frac{1}{s{R}_{2}C}\right)}$

$H\left(s\right)=\left(\frac{R_4}{R_3+R_4}\right)\times \frac{1+\frac{1}{s{R}_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{1+\frac{1}{s{R}_{2}C}}$        (7)

For low-pass filter, apply the limits $s\to \infty$ in equation (7). That is,

$\begin{array}{l}\left(\frac{R_4}{R_3+R_4}\right)\times \frac{1+\frac{1}{\left(\infty \right)R_{1}C}\left(\frac{R_1}{R_2}-\frac{R_3}{R_4}\right)}{1+\frac{1}{\left(\infty \right)R_{2}C}}=0\\ \frac{R_4}{R_3+R_4}=0\\ \frac{R_4}{R_3\left(1+\frac{R_4}{R_3}\right)}=0\end{array}$

$\frac{R_4}{R_3}=0$        (8)

Therefore, if the resistance $R_3$ tends to infinity only the equation (8) satisfies the condition to exist for low-pass filter. That is, $R_3\to \infty$.

Rearrange the equation (5) as follows,

$H\left(s\right)=\left(\frac{R_4}{R_3\left(1+\frac{R_4}{R_3}\right)}\right)\times \frac{R_3\left[\frac{s}{R_3}+\frac{1}{R_{1}C}\left(\frac{R_1}{R_3{R}_2}-\frac{1}{R_4}\right)\right]}{\left(s+\frac{1}{R_{2}C}\right)}$

$H\left(s\right)=\left(\frac{R_4}{\left(1+\frac{R_4}{R_3}\right)}\right)\times \frac{\left[\frac{s}{R_3}+\frac{1}{R_{1}C}\left(\frac{R_1}{R_3{R}_2}-\frac{1}{R_4}\right)\right]}{\left(s+\frac{1}{R_{2}C}\right)}$        (9)

Substitute $\infty$ for $R_3$ to find the transfer function for low-pass filter $H\left(s\right)$.

$\begin{array}{c}H\left(s\right)=\left(\frac{R_4}{\left(1+\frac{R_4}{\infty }\right)}\right)\times \frac{\left[\frac{s}{\infty }+\frac{1}{R_{1}C}\left(\frac{R_1}{\left(\infty \right)R_2}-\frac{1}{R_4}\right)\right]}{\left(s+\frac{1}{R_{2}C}\right)}\\ =\left(R_4\right)\times \frac{\left[\frac{1}{R_{1}C}\left(-\frac{1}{R_4}\right)\right]}{\left(s+\frac{1}{R_{2}C}\right)}\\ =-\frac{\left(\frac{1}{R_{1}C}\right)}{\left(s+\frac{1}{R_{2}C}\right)}\end{array}$

Conclusion:

Thus, the condition to be satisfied to operate as a low-pass filter of the given circuit is at $R_3\to \infty$.