Chapter 18, Problem 59P

To determine

Find the output voltage when the input voltage $v_i\left(t\right)=4e^{-t}u\left(t\right)\text{V}$

Answer to Problem 59P

The output voltage when the input voltage $v_i\left(t\right)=4e^{-t}u\left(t\right)\text{V}$ is $\underset{\_}{\left(16e^{-t}-20e^{-2t}+4e^{-4t}\right)u\left(t\right)\text{V}}$.

Explanation of Solution

Given data:

$v_i\left(t\right)=2\delta \left(t\right)\text{V}$ (1)

$v_o\left(t\right)=10e^{-2t}-6e^{-4t}\text{V}$ (2)

Formula used:

Consider the general expression for transfer function.

$H\left(\omega \right)=\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}$ (3)

Here,

$V_o\left(\omega \right)$ is the output voltage.

$V_i\left(\omega \right)$ is the input voltage.

Calculation:

Apply Fourier transform on both sides of equation (1) as follows.

$\begin{array}{l}F\left[v_i\left(t\right)\right]=F\left[2\delta \left(t\right)\right]\\ V_i\left(\omega \right)=2\end{array}$

Apply Fourier transform on both sides of equation (2) as follows.

$\begin{array}{l}F\left[v_o\left(t\right)\right]=F\left[10e^{-2t}-6e^{-4t}\right]\\ V_o\left(\omega \right)=F\left[10e^{-2t}\right]-F\left[6e^{-4t}\right]\\ V_o\left(\omega \right)=\frac{10}{2+j\omega }-\frac{6}{4+j\omega }\end{array}$

Substitute $2$ for $V_i\left(\omega \right)$ and $\frac{10}{2+j\omega }-\frac{6}{4+j\omega }$ for $V_o\left(\omega \right)$ in equation (3) as follows.

$\begin{array}{c}H\left(\omega \right)=\frac{\left(\frac{10}{2+j\omega }-\frac{6}{4+j\omega }\right)}{2}\\ =\frac{5}{2+j\omega }-\frac{3}{4+j\omega }\end{array}$

Consider $v_i\left(t\right)=4e^{-t}u\left(t\right)\text{V}$.

Apply Fourier transform on both sides of equation as follows.

$\begin{array}{l}F\left[v_i\left(t\right)\right]=F\left[4e^{-t}u\left(t\right)\right]\\ V_i\left(\omega \right)=\frac{4}{1+j\omega }\end{array}$

Rearrange equation (3) as follows.

$V_o\left(\omega \right)=H\left(\omega \right)V_i\left(\omega \right)$

Substitute $\left(\frac{5}{2+j\omega }-\frac{3}{4+j\omega }\right)$ for $H\left(\omega \right)$ and $\frac{4}{1+j\omega }$ for $V_i\left(\omega \right)$ as follows.

$\begin{array}{c}{V}_o\left(\omega \right)=\left(\frac{5}{2+j\omega }-\frac{3}{4+j\omega }\right)\left(\frac{4}{1+j\omega }\right)\\ =\frac{20}{\left(2+j\omega \right)\left(1+j\omega \right)}-\frac{12}{\left(4+j\omega \right)\left(1+j\omega \right)}\end{array}$

Consider $s=j\omega$ to reduce complex algebra.

Substitute $s$ for $j\omega$ as follows.

$V_o\left(s\right)=\frac{20}{\left(2+s\right)\left(1+s\right)}-\frac{12}{\left(4+s\right)\left(1+s\right)}$

Consider $V_o\left(s\right)=V_1\left(s\right)-V_2\left(s\right)$ (4)

Here,

$\begin{array}{l}{V}_1\left(s\right)=\frac{20}{\left(2+s\right)\left(1+s\right)}\\ V_2\left(s\right)=\frac{12}{\left(4+s\right)\left(1+s\right)}\end{array}$

Take partial fraction for $V_1\left(s\right)$.

$V_1\left(s\right)=\frac{A}{1+s}+\frac{B}{2+s}$ (5)

Where,

$A=\left(s+1\right){V_1\left(s\right)|}_{s=-1}$

Substitute $\frac{20}{\left(2+s\right)\left(1+s\right)}$ for $V_1\left(s\right)$ as follows.

$\begin{array}{c}A=\left(s+1\right){\left[\frac{20}{\left(2+s\right)\left(1+s\right)}\right]}_{s=-1}\\ ={\left[\frac{20}{\left(2+s\right)}\right]}_{s=-1}\\ =\left[\frac{20}{1}\right]\\ =20\end{array}$

Similarly,

$B=\left(s+2\right){V_1\left(s\right)|}_{s=-2}$

Substitute $\frac{20}{\left(2+s\right)\left(1+s\right)}$ for $V_1\left(s\right)$ as follows.

$\begin{array}{c}B=\left(s+2\right){\left[\frac{20}{\left(2+s\right)\left(1+s\right)}\right]}_{s=-2}\\ ={\left[\frac{20}{1+s}\right]}_{s=-2}\\ =\left[\frac{20}{-1}\right]\\ =-20\end{array}$

Substitute $20$ for $A$ and $-20$ for $B$ in equation (5) as follows.

$V_1\left(s\right)=\frac{20}{1+s}-\frac{20}{2+s}$

Take partial fraction for $V_2\left(s\right)$.

$V_2\left(s\right)=\frac{C}{1+s}+\frac{D}{4+s}$ (6)

Where,

$C=\left(s+1\right){V_2\left(s\right)|}_{s=-1}$

Substitute $\frac{12}{\left(4+s\right)\left(1+s\right)}$ for $V_2\left(s\right)$ as follows.

$\begin{array}{c}C=\left(s+1\right){\left[\frac{12}{\left(4+s\right)\left(1+s\right)}\right]}_{s=-1}\\ ={\left[\frac{12}{4+s}\right]}_{s=-1}\\ =\left[\frac{12}{3}\right]\\ =4\end{array}$

Similarly,

$D=\left(s+4\right){V_2\left(s\right)|}_{s=-4}$

Substitute $\frac{12}{\left(4+s\right)\left(1+s\right)}$ for $V_2\left(s\right)$ as follows.

$\begin{array}{c}D=\left(s+4\right){\left[\frac{12}{\left(4+s\right)\left(1+s\right)}\right]}_{s=-4}\\ ={\left[\frac{12}{1+s}\right]}_{s=-4}\\ =\frac{12}{-3}\\ =-4\end{array}$

Substitute $4$ for $C$ and $-4$ for $D$ in equation (6) as follows.

$V_2\left(s\right)=\frac{4}{1+s}-\frac{4}{4+s}$

Substitute $\frac{20}{1+s}-\frac{20}{2+s}$ for $V_1\left(s\right)$ and $\frac{4}{1+s}-\frac{4}{4+s}$ for $V_2\left(s\right)$ in equation (4) as follows.

$\begin{array}{c}{V}_o\left(s\right)=\left[\frac{20}{1+s}-\frac{20}{2+s}\right]-\left[\frac{4}{1+s}-\frac{4}{4+s}\right]\\ =\frac{20}{1+s}-\frac{20}{2+s}-\frac{4}{1+s}+\frac{4}{4+s}\\ =\frac{16}{1+s}-\frac{20}{2+s}+\frac{4}{4+s}\end{array}$

Substitute $j\omega$ for $s$ as follows.

$V_o\left(\omega \right)=\frac{16}{j\omega +1}-\frac{20}{j\omega +2}+\frac{4}{j\omega +4}$

Apply inverse Fourier transform on both sides of equation.

$\begin{array}{l}{F}^{-1}\left[V_o\left(\omega \right)\right]=F^{-1}\left[\frac{16}{j\omega +1}-\frac{20}{j\omega +2}+\frac{4}{j\omega +4}\right]\\ v_o\left(t\right)=F^{-1}\left[\frac{16}{j\omega +1}\right]-F^{-1}\left[\frac{20}{j\omega +2}\right]+F^{-1}\left[\frac{4}{j\omega +4}\right]\\ v_o\left(t\right)=16e^{-t}u\left(t\right)-20e^{-2t}u\left(t\right)+4e^{-4t}u\left(t\right)\\ v_o\left(t\right)=\left(16e^{-t}-20e^{-2t}+4e^{-4t}\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the output voltage when the input voltage $v_i\left(t\right)=4e^{-t}u\left(t\right)\text{V}$ is $\underset{\_}{\left(16e^{-t}-20e^{-2t}+4e^{-4t}\right)u\left(t\right)\text{V}}$.