Chapter 14, Problem 1P

(a)

To determine

Find the value of the cutoff frequency in hertz for the RL low-pass filter shown in given figure.

Answer to Problem 1P

The value of the cutoff frequency $\left(f_c\right)$ in hertz for the RL low-pass filter shown in given figure is $3819.72\text{Hz}$.

Explanation of Solution

Given data:

Refer to given figure in the textbook.

Formula used:

Write the expression to calculate the angular cutoff frequency.

${\omega }_c=2\pi f_c$        (1)

Here,

$f_c$ is the value of cutoff frequency.

Write the expression to calculate the cutoff frequency of the RL low-pass filter.

${\omega }_c=\frac{R}{L}$        (2)

Here,

$R$ is the value of the resistor , and

$L$ is the value of the inductor.

Calculation:

The given filter circuit is drawn as Figure 1.

Substitute $1.2\text{k}\Omega$ for $R$ and $50\text{mH}$ for $L$ in equation (2) to find ${\omega }_c$.

$\begin{array}{c}{\omega }_c=\frac{1.2\text{k}\Omega }{50\text{mH}}\\ =\frac{1.2\times {10}^3\Omega }{50\times {10}^{-3}\text{H}}\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3}\end{array}\right\}\\ =\frac{1.2\times {10}^3\Omega }{50\times {10}^{-3}\Omega \cdot \text{s}}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =24\times {10}^3\frac{\text{rad}}{\text{s}}\end{array}$

Simplify the above equation to find ${\omega }_c$.

${\omega }_c=24\frac{\text{krad}}{\text{s}}\left\{\because 1\text{k}={10}^3\right\}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in equation (1).

$\left(24\times {10}^3\frac{\text{rad}}{\text{s}}\right)=2\pi f_c$

Rearrange the above equation to find $f_c$.

$\begin{array}{c}{f}_c=\frac{\left(24\times {10}^3\frac{\text{rad}}{\text{s}}\right)}{2\pi }\\ =\frac{24\times {10}^3}{2\pi }\frac{1}{\text{s}}\\ =3819.72\text{Hz}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\end{array}$

Conclusion:

Thus, the value of the cutoff frequency $\left(f_c\right)$ in hertz for the RL low-pass filter shown in given figure is $3819.72\text{Hz}$.

(b)

To determine

Find the value of the transfer function $H\left(j\omega \right)$ at ${\omega }_c$, $0.125{\omega }_c$ and $8{\omega }_c$.

Answer to Problem 1P

The value of the transfer function $H\left(j\omega \right)$ at ${\omega }_c$, $0.125{\omega }_c$ and $8{\omega }_c$ is $0.7071\angle -45°$, $0.9923\angle -7.125°$ and $0.124\angle -82.875°$ respectively.

Explanation of Solution

Formula used:

Write the expression to calculate the impedance of the passive elements resistor and inductor.

$Z_R=R$        (3)

$Z_L=j\omega L$        (4)

Calculation:

The impedance circuit of the Figure 1 is drawn as Figure 2 using the equations (3) and (4).

Apply voltage division rule on Figure 2 to find $V_o$.

$V_o=\frac{R}{R+j\omega L}{V}_i$

Rearrange the above equation to find $\frac{V_o}{V_i}$.

$\begin{array}{c}\frac{V_o}{V_i}=\frac{R}{R+j\omega L}\\ =\frac{R}{R\left(1+j\frac{\omega L}{R}\right)}\\ =\frac{1}{\left(1+j\frac{\omega }{\left(\frac{R}{L}\right)}\right)}\end{array}$

Substitute the equation (2) in above equation to find $\frac{V_o}{V_i}$.

$\begin{array}{c}\frac{V_o}{V_i}=\frac{1}{1+\frac{j\omega }{{\omega }_c}}\\ =\frac{1}{\left(\frac{{\omega }_c+j\omega }{{\omega }_c}\right)}\\ =\frac{{\omega }_c}{{\omega }_c+j\omega }\end{array}$

Write the expression to calculate the transfer function of the circuit in Figure 2.

$H\left(j\omega \right)=\frac{V_o}{V_i}$

Substitute $\frac{{\omega }_c}{{\omega }_c+j\omega }$ for $\frac{V_o}{V_i}$ in above equation to find $H\left(j\omega \right)$.

$H\left(j\omega \right)=\frac{{\omega }_c}{{\omega }_c+j\omega }$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $H\left(j\omega \right)$.

$H\left(j\omega \right)=\frac{\left(24\times {10}^3\right)}{\left(24\times {10}^3\right)+j\omega }$

$H\left(j\omega \right)=\frac{24000}{24000+j\omega }$        (5)

Substitute ${\omega }_c$ for $\omega$ in equation (5) to find $H\left(j{\omega }_c\right)$.

$H\left(j{\omega }_c\right)=\frac{24000}{24000+j{\omega }_c}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $H\left(j{\omega }_c\right)$.

$\begin{array}{c}H\left(j{\omega }_c\right)=\frac{24000}{24000+j\left(24\times {10}^3\right)}\\ =\frac{24000}{24000+j24000}\\ =0.7071\angle -45°\end{array}$

Substitute $0.125{\omega }_c$ for $\omega$ in equation (5) to find $H\left(j0.125{\omega }_c\right)$.

$H\left(j0.125{\omega }_c\right)=\frac{24000}{24000+j\left(0.125{\omega }_c\right)}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $H\left(j0.125{\omega }_c\right)$.

$\begin{array}{c}H\left(j0.125{\omega }_c\right)=\frac{24000}{24000+j\left(0.125\left(24\times {10}^3\right)\right)}\\ =\frac{24000}{24000+j3000}\\ =0.9923\angle -7.125°\end{array}$

Substitute $8{\omega }_c$ for $\omega$ in equation (5) to find $H\left(j8{\omega }_c\right)$.

$H\left(j8{\omega }_c\right)=\frac{24000}{24000+j\left(8{\omega }_c\right)}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $H\left(j8{\omega }_c\right)$.

$\begin{array}{c}H\left(j8{\omega }_c\right)=\frac{24000}{24000+j\left(8\left(24\times {10}^3\right)\right)}\\ =\frac{24000}{24000+j192000}\\ =0.124\angle -82.875°\end{array}$

Conclusion:

Thus, the value of the transfer function $H\left(j\omega \right)$ at ${\omega }_c$, $0.125{\omega }_c$ and $8{\omega }_c$ is $0.7071\angle -45°$, $0.9923\angle -7.125°$ and $0.124\angle -82.875°$ respectively.

(c)

To determine

Find the steady state expression for the output voltage $v_o\left(t\right)$ when $\omega ={\omega }_c$, $\omega =0.125{\omega }_c$ and $\omega =8{\omega }_c$.

Answer to Problem 1P

The steady state expression for the output voltage $v_o\left(t\right)$ when $\omega ={\omega }_c$, $\omega =0.125{\omega }_c$ and $\omega =8{\omega }_c$ is $14.142\cos \left(24000t-45°\right)\text{V}$, $19.846\cos \left(3000t-7.125°\right)\text{V}$ and $2.48\cos \left(192000t-82.875°\right)\text{V}$ respectively.

Explanation of Solution

Given data:

The input voltage is,

$v_i\left(t\right)=20\cos \omega t\text{V}$

Calculation:

From part (b),

$\frac{V_o}{V_i}=\frac{{\omega }_c}{{\omega }_c+j\omega }$

Rearrange the above equation to find $V_o$.

$V_o=\left(\frac{{\omega }_c}{{\omega }_c+j\omega }\right)V_i$

The time domain expression for the above equation is written as,

$v_o\left(t\right)=\left(\frac{{\omega }_c}{{\omega }_c+j\omega }\right)v_i\left(t\right)$

Substitute $20\cos \omega t\text{V}$ for $v_i$ and $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $v_o\left(t\right)$.

$v_o\left(t\right)=\left(\frac{24\times {10}^3}{\left(24\times {10}^3\right)+j\omega }\right)\left(20\cos \omega t\text{V}\right)$

$v_o\left(t\right)=\left(\frac{24000}{24000+j\omega }\right)\left(20\cos \omega t\right)\text{V}$        (6)

Substitute ${\omega }_c$ for $\omega$ in equation (6) to find $v_o\left(t\right)$.

$v_o\left(t\right)=\left(\frac{24000}{24000+j{\omega }_c}\right)\left(20\cos {\omega }_{c}t\right)\text{V}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=\left(\frac{24000}{24000+j\left(24\times {10}^3\right)}\right)\left(20\cos \left(24\times {10}^3\right)t\right)\text{V}\\ =\left(\frac{24000}{24000+j24000}\right)\left(20\cos \left(24000t+0°\right)\right)\text{V}\\ =\left(0.7071\angle -45°\right)\left(20\angle 0°\right)\text{V}\left\{\because A\cos \left(\omega t+\varphi \right)=A\angle \varphi \right\}\\ =\left(14.142\angle -45°\right)\text{V}\end{array}$

Simplify the above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=14.142\cos \left(\omega t-45°\right)\text{V}\left\{\because A\angle \varphi =A\cos \left(\omega t+\varphi \right)\right\}\\ =14.142\cos \left({\omega }_{c}t-45°\right)\text{V}\left\{\because \omega ={\omega }_c\right\}\\ =14.142\cos \left(24000t-45°\right)\text{V}\end{array}$

Substitute $0.125{\omega }_c$ for $\omega$ in equation (6) to find $v_o\left(t\right)$.

$v_o\left(t\right)=\left(\frac{24000}{24000+j\left(0.125{\omega }_c\right)}\right)\left(20\cos \left(0.125{\omega }_c\right)t\right)\text{V}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=\left(\frac{24000}{24000+j\left(0.125\left(24\times {10}^3\right)\right)}\right)\left(20\cos \left(0.125\left(24\times {10}^3\right)\right)t\right)\text{V}\\ =\left(\frac{24000}{24000+j3000}\right)\left(20\cos \left(3000t+0°\right)\right)\text{V}\\ =\left(0.9923\angle -7.125°\right)\left(20\angle 0°\right)\text{V}\left\{\because A\cos \left(\omega t+\varphi \right)=A\angle \varphi \right\}\\ =\left(19.846\angle -7.125°\right)\text{V}\end{array}$

Simplify the above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=19.846\cos \left(\omega t-7.125°\right)\text{V}\left\{\because A\angle \varphi =A\cos \left(\omega t+\varphi \right)\right\}\\ =19.846\cos \left(0.125{\omega }_{c}t-7.125°\right)\text{V}\left\{\because \omega =0.125{\omega }_c\right\}\\ =19.846\cos \left(0.125\left(24\times {10}^3\right)t-7.125°\right)\text{V}\\ =19.846\cos \left(3000t-7.125°\right)\text{V}\end{array}$

Substitute $8{\omega }_c$ for $\omega$ in equation (6) to find $v_o\left(t\right)$.

$v_o\left(t\right)=\left(\frac{24000}{24000+j\left(8{\omega }_c\right)}\right)\left(20\cos \left(8{\omega }_c\right)t\right)\text{V}$

Substitute $24\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=\left(\frac{24000}{24000+j\left(8\left(24\times {10}^3\right)\right)}\right)\left(20\cos \left(8\left(24\times {10}^3\right)\right)t\right)\text{V}\\ =\left(\frac{24000}{24000+j192000}\right)\left(20\cos \left(192000t+0°\right)\right)\text{V}\\ =\left(0.124\angle -82.875°\right)\left(20\angle 0°\right)\text{V}\left\{\because A\cos \left(\omega t+\varphi \right)=A\angle \varphi \right\}\\ =\left(2.48\angle -82.875°\right)\text{V}\end{array}$

Simplify the above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=2.48\cos \left(\omega t-82.875°\right)\text{V}\left\{\because A\angle \varphi =A\cos \left(\omega t+\varphi \right)\right\}\\ =2.48\cos \left(8{\omega }_{c}t-82.875°\right)\text{V}\left\{\because \omega =8{\omega }_c\right\}\\ =2.48\cos \left(8\left(24\times {10}^3\right)t-82.875°\right)\text{V}\\ =2.48\cos \left(192000t-82.875°\right)\text{V}\end{array}$

Conclusion:

Thus, the steady state expression for the output voltage $v_o\left(t\right)$ when $\omega ={\omega }_c$, $\omega =0.125{\omega }_c$ and $\omega =8{\omega }_c$ is $14.142\cos \left(24000t-45°\right)\text{V}$, $19.846\cos \left(3000t-7.125°\right)\text{V}$ and $2.48\cos \left(192000t-82.875°\right)\text{V}$ respectively.