Chapter 15, Problem 1P

a.

To determine

Design a low pass filter for a given specification.

Answer to Problem 1P

The obtained value of resistor $R_1$ and $R_2$ of first order low-pass filter is $\underset{\_}{67.16\Omega }$ and $\underset{\_}{212.21\Omega }$ respectively.

Explanation of Solution

Given data:

The value of capacitor $C$ is $750\text{nF}$.

The value of passband gain is $10\text{dB}$.

Cutoff frequency is $1\text{kHz}$.

Formula used:

Refer to Figure 15.1 in the textbook for a first order low-pass filter.

Write the expression for passband gain.

$K=\frac{R_2}{R_1}$        (1)

Here,

$R_1$ is the forward path resistor, and

$R_2$ is the feedback resistors.

Write the expression for cutoff frequency.

${\omega }_c=\frac{1}{R_{2}C}$        (2)

Calculation:

Write the expression for ${\omega }_c$.

${\omega }_c=2\pi f_c$

Re-arrange equation (2) as follows.

$R_2=\frac{1}{{\omega }_{c}C}$

Substitute $2\pi f_c$ for ${\omega }_c$.

$R_2=\frac{1}{2\pi f_{c}C}$

Substitute $1\text{kHz}$ for $f_c$ and $750\text{nF}$ for C to find the value of $R_2$.

$\begin{array}{c}{R}_2=\frac{1}{2\pi \left(1\text{kHz}\right)\left(750\text{nF}\right)}\\ =\frac{1}{2\pi \left(1\times {10}^3\text{Hz}\right)\left(750\times {10}^{-9}\text{F}\right)}\left\{\begin{array}{l}\because 1\text{kHz}=1\times {10}^3\text{Hz}\\ \because 1\text{nF}=1\times {10}^{-9}\text{F}\end{array}\right\}\\ =212.206\Omega \\ \cong 212.21\Omega \end{array}$

Convert dB value of passband gain into normal value.

$\begin{array}{l}10\text{dB}={\text{20log}}_{10}K\\ {\text{log}}_{10}K=\frac{10}{20}\\ K={10}^{\left(\frac{10}{20}\right)}\\ K=3.16\end{array}$

Substitute $3.16$ for $K$ and $212.21\Omega$ for $R_2$ in equation (1) to find the value of $R_1$.

$\begin{array}{l}3.16=\frac{212.21\Omega }{R_1}\\ R_1=\frac{212.21\Omega }{3.16}\\ R_1=67.155\Omega \\ R_1\cong 67.16\Omega \end{array}$

Conclusion:

Thus, the obtained value of resistor $R_1$ and $R_2$ of first order low-pass filter is $\underset{\_}{67.16\Omega }$ and $\underset{\_}{212.21\Omega }$ respectively.

b.

To determine

Draw the circuit diagram of designed low pass filter.

Explanation of Solution

Calculation:

Modify the Figure 15.1 for designed value as shown in Figure 1.

Conclusion:

Thus, the circuit diagram of designed low pass filter is shown in Figure 1.