Chapter 17, Problem 61E

(a)

To determine

The value of output voltage, $v_0\left(t\right)$.

Answer to Problem 61E

The value of output voltage, $v_0\left(t\right)=5.444e^{-0.625t}\left(\cos \left(1.45t+{23.32}^{\text{ο}}\right)\right)\text{V}$.

Explanation of Solution

Given data:

The input voltage is, $v_i\left(t\right)=5u\left(t\right)\text{V}$

The given diagram is shown in Figure 1.

Calculation:

Mark the loop current $i$ in the given figure.

The required diagram is shown in Figure 2.

The conversion of $\text{mH}$ into $\text{H}$ is given as,

$1\text{mH}=1\times {10}^{-3}\text{H}$

The conversion of $\text{800}\text{mH}$ into $\text{H}$ is given as,

$\begin{array}{c}800\text{mH}=800\times {10}^{-3}\text{H}\\ \text{=0}\text{.8}\text{H}\end{array}$

Hence, the value of the inductor, $L=0.8\text{H}$

The conversion of $\text{mF}$ into $\text{F}$ is given as,

$1\text{mF}=1\times {10}^{-3}\text{F}$

The conversion of $\text{500}\text{mH}$ into $\text{F}$ is given as,

$\begin{array}{c}500\text{mF}=500\times {10}^{-3}\text{F}\\ \text{=0}\text{.5}\text{F}\end{array}$

Hence, the value of the capacitor, $C=0.5\text{F}$

The expression for output voltage using KVL is given by,

$i\left(t\right)+0.8\frac{di\left(t\right)}{dt}+\frac{1}{0.5}{\displaystyle \int i\left(t\right)dt}=v_i\left(t\right)$

Substitute $5u\left(t\right)$ for $v_i\left(t\right)$ in the above expression.

$i\left(t\right)+0.8\frac{di\left(t\right)}{dt}+\frac{1}{0.5}{\displaystyle \int i\left(t\right)dt}=5u\left(t\right)$

Apply Fourier transform to the above expression.

$\begin{array}{l}I\left(j\omega \right)+0.8j\omega I\left(j\omega \right)+\frac{2}{j\omega }I\left(j\omega \right)=F\left[5u\left(t\right)\right]\\ I\left(j\omega \right)\left[\frac{j\omega +0.8{\left(j\omega \right)}^2+2}{j\omega }\right]=F\left[5u\left(t\right)\right]\end{array}$        (1)

The Fourier transform of $5u\left(t\right)$ is given by,

$F\left[5u\left(t\right)\right]=5\left(\pi \delta \left(\omega \right)+\frac{1}{j\omega }\right)$

Substitute $5\left(\pi \delta \left(\omega \right)+\frac{1}{j\omega }\right)$ for $F\left[5u\left(t\right)\right]$ in equation (1).

$\begin{array}{l}I\left(j\omega \right)\left[\frac{j\omega +0.8{\left(j\omega \right)}^2+2}{j\omega }\right]=5\left(\pi \delta \left(\omega \right)+\frac{1}{j\omega }\right)\\ I\left(j\omega \right)\left[j\omega +0.8{\left(j\omega \right)}^2+2\right]=5\left[\left(j\omega \right)\pi \delta \left(\omega \right)+1\right]\\ I\left(j\omega \right)0.8\left[{\left(j\omega \right)}^2+1.25j\omega +2.5\right]=5\left[\left(j\omega \right)\pi \delta \left(\omega \right)+1\right]\\ I\left(j\omega \right)=6.25\left[\frac{\left(j\omega \right)\pi \delta \left(\omega \right)}{{\left(j\omega \right)}^2+1.25j\omega +2.5}+\frac{1}{{\left(j\omega \right)}^2+1.25j\omega +2.5}\right]\end{array}$

Further solve as,

$\begin{array}{c}I\left(j\omega \right)=6.25\left[\begin{array}{l}\frac{\pi \delta \left(\omega \right)\left[\left(j\omega +0.625-j1.45\right)-0.625+j1.45\right]}{\left(j\omega +0.625-j1.45\right)\left(j\omega +0.625+j1.45\right)}\\ +\frac{1}{\left(j\omega +0.625+j1.45\right)\left(j\omega +0.625-j1.45\right)}\end{array}\right]\\ =6.25\left[\begin{array}{l}\frac{\pi \delta \left(\omega \right)}{\left(j\omega +0.625+j1.45\right)}\\ -\left(\frac{\pi \delta \left(\omega \right)\left(0.625-j1.45\right)}{\left(j\omega +0.625-j1.45\right)\left(j\omega +0.625+j1.45\right)}\right)\\ +\frac{1}{\left(j\omega +0.625+j1.45\right)\left(j\omega +0.625-j1.45\right)}\end{array}\right]\\ =6.25\left[\begin{array}{l}\frac{\pi \delta \left(\omega \right)}{\left(j\omega +0.625+j1.45\right)}\\ +\left(\frac{\pi \delta \left(\omega \right)\left(0.5+j0.215\right)}{\left(j\omega +0.625-j1.45\right)}\right)-\frac{\pi \delta \left(\omega \right)\left(0.5+j0.215\right)}{j\omega +0.625+j1.45}\\ +\frac{6.25}{j2.9}\frac{1}{\left(j\omega +0.625+j1.45\right)}-\frac{1}{j\omega +0.625-j1.45}\end{array}\right]\\ =3.125\left[\begin{array}{l}\frac{2\pi \delta \left(\omega \right)}{\left(j\omega +0.625+j1.45\right)}\\ +\left(\frac{2\pi \delta \left(\omega \right)\left(0.5+j0.215\right)}{\left(j\omega +0.625-j1.45\right)}\right)-\frac{2\pi \delta \left(\omega \right)\left(0.5+j0.215\right)}{j\omega +0.625+j1.45}\\ +2.155j\frac{1}{\left(j\omega +0.625+j1.45\right)}-\frac{1}{j\omega +0.625-j1.45}\end{array}\right]\end{array}$

Apply shifting property for $\omega \to 0$ in $\delta \left(\omega \right)$ terms.

Further solve as,

$\begin{array}{c}I\left(j\omega \right)=\left[\begin{array}{l}3.125\left(2\pi \delta \left(\omega \right)\right)\left[\left(\frac{\left(0.5-j0.215\right)}{\left(0.625+j1.45\right)}\right)+\frac{\left(0.5+j0.215\right)}{0.625-j1.45}\right]\\ +2.155j\frac{1}{\left(j\omega +0.625+j1.45\right)}-\frac{1}{j\omega +0.625-j1.45}\end{array}\right]\\ =\left[\begin{array}{l}3.125\left(2\pi \delta \left(\omega \right)\right)\left[\left(-0.2-j0.0865\right)+\left(-0.2-j0.0865\right)\right]\\ +2.155j\frac{1}{\left(j\omega +0.625+j1.45\right)}-\frac{1}{j\omega +0.625-j1.45}\end{array}\right]\\ =\left[\begin{array}{l}3.125\left(2\pi \delta \left(\omega \right)\right)\left[-0.4\right]\\ +2.155j\frac{1}{\left(j\omega +0.625+j1.45\right)}-\frac{1}{j\omega +0.625-j1.45}\end{array}\right]\\ =-1.25\left(2\pi \delta \left(\omega \right)\right)+2.155j\frac{1}{\left(j\omega +0.625+j1.45\right)}-\frac{1}{j\omega +0.625-j1.45}\end{array}$        (2)

The inverse Fourier transform for the $2\pi \delta \left(\omega \right)$ is given by,

$F^{-1}\left[2\pi \delta \left(\omega \right)\right]=1$

The inverse Fourier transform for the $\frac{1}{j\omega +0.625+j1.45}$ is given by,

$F^{-1}\left[\frac{1}{j\omega +0.625+j1.45}\right]=e^{-\left(0.625+j1.45\right)t}u\left(t\right)$

The inverse Fourier transform for the $\frac{1}{j\omega +0.625-j1.45}$ is given by,

$F^{-1}\left[\frac{1}{j\omega +0.625-j1.45}\right]=e^{-\left(0.625-j1.45\right)t}u\left(t\right)$

Substitute $1$ for $2\pi \delta \left(\omega \right)$, $e^{-\left(0.625+j1.45\right)t}u\left(t\right)$ for $\frac{1}{j\omega +0.625+j1.45}$ and $e^{-\left(0.625-j1.45\right)t}u\left(t\right)$ for $\frac{1}{j\omega +0.625-j1.45}$ in equation (2).

$i\left(t\right)=-1.25+j2.155\left[e^{-\left(0.625+j1.45\right)t}-e^{-\left(0.625-j1.45\right)t}\right]u\left(t\right)$

The expression for the output voltage from the figure is given by,

$v_0\left(t\right)=0.8\frac{di\left(t\right)}{dt}$

Substitute $-1.25+j2.155\left[e^{-\left(0.625+j1.45\right)t}-e^{-\left(0.625-j1.45\right)t}\right]u\left(t\right)$ for $i\left(t\right)$ in the above expression.

$\begin{array}{c}{v}_0\left(t\right)=0.8\frac{d}{dt}\left[-1.25+j2.155\left[e^{-\left(0.625+j1.45\right)t}-e^{-\left(0.625-j1.45\right)t}\right]u\left(t\right)\right]\\ =0.8\left(j2.155\right)\left[\begin{array}{l}-\left(0.625+j1.45\right)e^{-\left(0.625+j1.45\right)t}\\ +\left(0.625-j1.45\right)e^{-\left(0.625-j1.45\right)t}\end{array}\right]\\ =2.5-j1.0775e^{-\left(0.625+j1.45\right)t}+2.5+j1.0775e^{-\left(0.625-j1.45\right)t}\\ =e^{-0.625t}\left(2.722\angle {23.32}^{\text{ο}}\right)e^{j1.45t}+e^{-0.625t}\left(2.722\angle -{23.32}^{\text{ο}}\right)e^{-j1.45t}\end{array}$

Further solve as,

$\begin{array}{c}{v}_0\left(t\right)=2\left(2.2722\right)e^{-0.625t}\left(\cos \left(1.45t+{23.32}^{\text{ο}}\right)\right)\\ =5.444e^{-0.625t}\left(\cos \left(1.45t+{23.32}^{\text{ο}}\right)\right)\end{array}$

Conclusion:

Therefore, the value of output voltage, $v_0\left(t\right)=5.444e^{-0.625t}\left(\cos \left(1.45t+{23.32}^{\text{ο}}\right)\right)\text{V}$.

(b)

To determine

The value of output voltage, $v_0\left(t\right)$.

Answer to Problem 61E

The value of output voltage, $v_0\left(t\right)=5.134e^{-0.625t}\left(\cos \left(1.45t+{136.1}^{\text{ο}}\right)\right)\text{V}$.

Explanation of Solution

Given data:

The input voltage is, $v_i\left(t\right)=3\delta \left(t\right)\text{V}$

Calculation:

The expression for output voltage using KVL is given by,

$i\left(t\right)+0.8\frac{di\left(t\right)}{dt}+\frac{1}{0.5}{\displaystyle \int i\left(t\right)dt}=v_i\left(t\right)$

Substitute $3\delta \left(t\right)$ for $v_i\left(t\right)$ in the above expression.

$i\left(t\right)+0.8\frac{di\left(t\right)}{dt}+\frac{1}{0.5}{\displaystyle \int i\left(t\right)dt}=3\delta \left(t\right)$

Apply Fourier transform to the above expression.

$\begin{array}{l}I\left(j\omega \right)+0.8j\omega I\left(j\omega \right)+\frac{2}{j\omega }I\left(j\omega \right)=F\left[3\delta \left(t\right)\right]\\ I\left(j\omega \right)\left[\frac{j\omega +0.8{\left(j\omega \right)}^2+2}{j\omega }\right]=F\left[3\delta \left(t\right)\right]\end{array}$        (3)

The Fourier transform of $3\delta \left(t\right)$ is given by,

$F\left[3\delta \left(t\right)\right]=3$

Substitute $3$ for $F\left[3\delta \left(t\right)\right]$ in equation (3).

$\begin{array}{l}I\left(j\omega \right)\left[\frac{j\omega +0.8{\left(j\omega \right)}^2+2}{j\omega }\right]=3\\ I\left(j\omega \right)\left[j\omega +0.8{\left(j\omega \right)}^2+2\right]=3j\omega \\ I\left(j\omega \right)0.8\left[{\left(j\omega \right)}^2+1.25j\omega +2.5\right]=3j\omega \\ I\left(j\omega \right)=3.75\left[\frac{j\omega }{{\left(j\omega \right)}^2+1.25j\omega +2.5}\right]\end{array}$

Further solve as,

$\begin{array}{c}I\left(j\omega \right)=3.75\left[\frac{j\omega }{\left(j\omega +0.625+j1.45\right)\left(j\omega +0.625-j1.45\right)}\right]\\ =3.75\left[\frac{0.5+j0.215}{\left(j\omega +0.625-j1.45\right)}+\frac{0.5-j0.215}{j\omega +0.625+j1.45}\right]\\ =\frac{1.875+j0.7875}{\left(j\omega +0.625-j1.45\right)}+\frac{1.875-j0.7875}{j\omega +0.625+j1.45}\end{array}$        (4)

The inverse Fourier transform for the $\frac{1}{j\omega +0.625+j1.45}$ is given by,

$F^{-1}\left[\frac{1}{j\omega +0.625+j1.45}\right]=e^{-\left(0.625+j1.45\right)t}u\left(t\right)$

The inverse Fourier transform for the $\frac{1}{j\omega +0.625-j1.45}$ is given by,

$F^{-1}\left[\frac{1}{j\omega +0.625-j1.45}\right]=e^{-\left(0.625-j1.45\right)t}u\left(t\right)$

Substitute $e^{-\left(0.625+j1.45\right)t}u\left(t\right)$ for $\frac{1}{j\omega +0.625+j1.45}$ and $e^{-\left(0.625-j1.45\right)t}u\left(t\right)$ for $\frac{1}{j\omega +0.625-j1.45}$ in equation (4).

$i\left(t\right)=\left[\left(1.875+j0.7875\right)e^{-\left(0.625-j1.45\right)t}+\left(1.875-j0.7875\right)e^{-\left(0.625+j1.45\right)t}\right]u\left(t\right)$

The expression for the output voltage from the figure is given by,

$v_0\left(t\right)=0.8\frac{di\left(t\right)}{dt}$

Substitute $\left[\left(1.875+j0.7875\right)e^{-\left(0.625-j1.45\right)t}+\left(1.875-j0.7875\right)e^{-\left(0.625+j1.45\right)t}\right]u\left(t\right)$ for $i\left(t\right)$ in the above expression.

$\begin{array}{c}{v}_0\left(t\right)=0.8\frac{d}{dt}\left[\left(1.875+j0.7875\right)e^{-\left(0.625-j1.45\right)t}+\left(1.875-j0.7875\right)e^{-\left(0.625-j1.45\right)t}\right]u\left(t\right)\\ =\left[\begin{array}{l}\left(-1.5+j0.63\right)\left(0.625-j1.45\right)e^{-\left(0.625-j1.45\right)t}\\ -\left(1.5-j0.63\right)\left(0.625+j1.45\right)e^{-\left(0.625+j1.45\right)t}\end{array}\right]\\ =-\left(1.85-j1.78\right)e^{-\left(0.625-j1.45\right)t}-\left(1.85+j1.78\right)e^{-\left(0.625+j1.45\right)t}\\ =e^{-0.625t}\left(2.567\angle {136.1}^{\text{ο}}\right)e^{j1.45t}+e^{-0.625t}\left(2.567\angle -{136.1}^{\text{ο}}\right)e^{-j1.45t}\end{array}$

Further solve as,

$\begin{array}{c}{v}_0\left(t\right)=2\left(2.567\right)e^{-0.625t}\left(\cos \left(1.45t+{136.1}^{\text{ο}}\right)\right)\\ =5.134e^{-0.625t}\left(\cos \left(1.45t+{136.1}^{\text{ο}}\right)\right)\end{array}$

Conclusion:

Therefore, the value of output voltage, $v_0\left(t\right)=5.134e^{-0.625t}\left(\cos \left(1.45t+{136.1}^{\text{ο}}\right)\right)\text{V}$.