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 $\underset{\_}{179\Omega }$ and $\underset{\_}{318.31\Omega }$ respectively.

Explanation of Solution

Given data:

The value of capacitor $C$ is $50\text{nF}$.

The value of passband gain is $5\text{dB}$.

Cutoff frequency is $10\text{kHz}$.

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,

$R_1$ and $R_2$ is the 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 $10\text{kHz}$ for $f_c$ and $50\text{nF}$ for C to find the value of $R_2$.

$\begin{array}{c}{R}_2=\frac{1}{2\pi \left(10\text{kHz}\right)\left(50\text{nF}\right)}\\ =\frac{1}{2\pi \left(10\times {10}^3\text{Hz}\right)\left(50\times {10}^{-9}\text{F}\right)}\left\{\begin{array}{l}\because 1\text{kHz}=1\times {10}^3\text{Hz}\\ \because 1\text{nF}=1\times {10}^{-9}\text{F}\end{array}\right\}\\ =318.309\Omega \\ \cong 318.31\Omega \end{array}$

Convert dB value of passband gain into normal value.

$\begin{array}{l}5\text{dB}={\text{20log}}_{10}K\\ {\text{log}}_{10}K=\frac{5}{20}\\ K={10}^{\left(\frac{5}{20}\right)}\\ K=1.778\end{array}$

Substitute $1.778$ for $K$ and $318.31\Omega$ for $R_2$ in equation (1) to find the value of $R_1$.

$\begin{array}{l}1.788=\frac{318.31\Omega }{R_1}\\ R_1=\frac{318.31\Omega }{1.788}\\ R_1=179\Omega \end{array}$

Conclusion:

Thus, the obtained value of resistor $R_1$ and $R_2$ of first order low-pass filter is $\underset{\_}{179\Omega }$ and $\underset{\_}{318.31\Omega }$ is 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.