Chapter 16, Problem 30P

Find v_o(t), for all t > 0, in the circuit of Fig. 16.53.

Figure 16.53

To determine

Find the expression of voltage $v_o\left(t\right)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $v_o\left(t\right)$ for all $t>0$ in the circuit of Figure 16.53 is $\left[3.5+17.55e^{-0.75t}\cos \left(0.6614t-101.5°\right)\right]u\left(t\right)\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 16.53 in the textbook.

Formula used:

Write an expression to calculate the value of step input.

$u\left(t\right)=\{\begin{array}{l}0t<0\\ 1t>0\end{array}$

Write a general expression to calculate the impedance of a resistor in s-domain.

$Z_R=R$        (1)

Here,

$R$ is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

$Z_L=sL$        (2)

Here,

$L$ is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

$Z_C=\frac{1}{sC}$        (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

For a DC circuit, at steady state condition at time $t=0^{-}$ the capacitor acts like open circuit and the inductor acts like short circuit.

For time $t<0$:

The current source $i_s\left(t\right)$ is calculated as follows:

$\begin{array}{c}{i}_s\left(t\right)=3.5\left(0\right)\left\{\because u\left(t\right)=0\text{for}t<0\right\}\\ =0\text{A}\end{array}$

The voltage source $v_s\left(t\right)$ is calculated as follows:

$\begin{array}{c}{v}_s\left(t\right)=7\left(0\right)\left\{\because u\left(t\right)=0\text{for}t<0\right\}\\ =0\text{V}\end{array}$

When the value of current source is zero, it is open circuited and when the value of voltage source is zero it is short circuited.

Now, the Figure 1 is reduced as shown in Figure 2.

Refer to Figure 2, there is no current and voltage source placed in the circuit. Therefore, the value of current through the inductor and capacitor is equal to zero.

$\begin{array}{l}{i}_L\left(0^{-}\right)=0\text{A}\\ v_C\left(0^{-}\right)=0\text{V}\end{array}$

The current through inductor and voltage across capacitor is always continuous so that,

$i\left(0\right)=i_L\left(0^{-}\right)=i_L\left(0^{+}\right)=0\text{A}$

$v\left(0\right)=v_C\left(0^{-}\right)=v_C\left(0^{+}\right)=0\text{V}$

For time $t>0$, the circuit is remains same as shown in Figure 1.

Apply Laplace transform for $v_s\left(t\right)$ to find $V_s\left(s\right)$.

$V_s\left(s\right)=\frac{7}{s}\left\{\because ℒ\left(u\left(t\right)\right)=\frac{1}{s}\right\}$

Apply Laplace transform for $i_s\left(t\right)$ to find $I_s\left(s\right)$.

$I_s\left(s\right)=\frac{3.5}{s}\left\{\because ℒ\left(u\left(t\right)\right)=\frac{1}{s}\right\}$

Use equation (1) to find $Z_{R_1}$.

$Z_{R_1}=1$

Use equation (1) to find $Z_{R_2}$.

$Z_{R_2}=1$

Substitute $1\text{H}$ for $L$ in equation (2) to find $Z_L$.

$\begin{array}{c}{Z}_L=s\left(1\text{H}\right)\\ =s\end{array}$

Substitute $0.5\text{F}$ for $C$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{s\left(0.5\text{F}\right)}\\ =\frac{2}{s}\end{array}$

Convert the Figure 1 into s-domain as shown in Figure 3.

Apply nodal analysis at node $V_1\left(s\right)$ in Figure 3.

$\begin{array}{l}\frac{V_1\left(s\right)-\frac{7}{s}}{1}+\frac{V_1\left(s\right)}{s}+\frac{V_1\left(s\right)-V_o\left(s\right)}{1}=0\\ V_1\left(s\right)-\frac{7}{s}+\frac{V_1\left(s\right)}{s}+\frac{V_1\left(s\right)}{1}-\frac{V_o\left(s\right)}{1}=0\\ V_1\left(s\right)+\frac{V_1\left(s\right)}{s}+V_1\left(s\right)-V_o\left(s\right)=\frac{7}{s}\\ \left(2+\frac{1}{s}\right)V_1\left(s\right)-V_o\left(s\right)=\frac{7}{s}\end{array}$

$\left(\frac{2s+1}{s}\right)V_1\left(s\right)-V_o\left(s\right)=\frac{7}{s}$        (4)

Apply nodal analysis at node $V_o\left(s\right)$ in Figure 3.

$\begin{array}{l}\frac{V_o\left(s\right)}{\left(\frac{2}{s}\right)}+\frac{V_o\left(s\right)-V_1\left(s\right)}{1}-\frac{3.5}{s}=0\\ \frac{s{V}_o\left(s\right)}{2}+V_o\left(s\right)-V_1\left(s\right)-\frac{3.5}{s}=0\\ \left(\frac{s}{2}+1\right)V_o\left(s\right)-V_1\left(s\right)=\frac{3.5}{s}\\ \left(\frac{s+2}{2}\right)V_o\left(s\right)-V_1\left(s\right)=\frac{3.5}{s}\end{array}$

Simplify the above equation to find $V_1\left(s\right)$.

$V_1\left(s\right)=\left(\frac{s+2}{2}\right)V_o\left(s\right)-\frac{3.5}{s}$

Substitute $\left(\frac{s+2}{2}\right)V_o\left(s\right)-\frac{3.5}{s}$ for $V_1\left(s\right)$ in equation (4) to find $V_o\left(s\right)$.

$\begin{array}{l}\left(\frac{2s+1}{s}\right)\left[\left(\frac{s+2}{2}\right)V_o\left(s\right)-\frac{3.5}{s}\right]-V_o\left(s\right)=\frac{7}{s}\\ \left(\frac{2s+1}{s}\right)\left(\frac{s+2}{2}\right)V_o\left(s\right)-\left(\frac{2s+1}{s}\right)\frac{3.5}{s}-V_o\left(s\right)=\frac{7}{s}\\ \left(\frac{2s^2+4s+s+2}{2s}\right)V_o\left(s\right)-\left(\frac{7s+3.5}{s^2}\right)-V_o\left(s\right)=\frac{7}{s}\\ \left(\frac{2s^2+5s+2}{2s}\right)V_o\left(s\right)-V_o\left(s\right)=\frac{7}{s}+\left(\frac{7s+3.5}{s^2}\right)\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}\left[\left(\frac{2s^2+5s+2}{2s}\right)-1\right]V_o\left(s\right)=\frac{7}{s}+\frac{7}{s}+\frac{3.5}{s^2}\\ \left(\frac{2s^2+5s+2-2s}{2s}\right)V_o\left(s\right)=\frac{14}{s}+\frac{3.5}{s^2}\\ \left(\frac{2s^2+3s+2}{2s}\right)V_o\left(s\right)=\frac{14s+3.5}{s^2}\\ \left(\frac{2\left(s^2+1.5s+1\right)}{2s}\right)V_o\left(s\right)=\frac{14s+3.5}{s^2}\end{array}$

Simplify the above equation to find $V_o\left(s\right)$.

$V_o\left(s\right)=\left(\frac{14s+3.5}{s^2}\right)\left(\frac{s}{s^2+1.5s+1}\right)$

$V_o\left(s\right)=\frac{14s+3.5}{s\left(s^2+1.5s+1\right)}$        (5)

From equation (5), the characteristic equation of denominator is written as follows:

$s^2+1.5s+1=0$        (6)

Write a general expression to calculate the roots of quadratic equation $\left(a{s}^2+bs+c=0\right)$.

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$        (7)

Comparing the equation (6) with the equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=1.5\\ c=1\end{array}$

Substitute $1$ for $a$, $1.5$ for $b$, and $1$ for $c$ in equation (7) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-1.5\pm \sqrt{{\left(1.5\right)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ =\frac{-1.5\pm \sqrt{2.25-4}}{2\left(1\right)}\\ =\frac{-1.5\pm \sqrt{-1.75}}{2}\\ =\frac{-1.5\pm j1.3228}{2}\end{array}$

Simplify the above equation to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-1.5+j1.3228}{2},\frac{-1.5-j1.3228}{2}\\ =-0.75+j0.6614,-0.75-j0.6614\end{array}$

Substitute the roots of characteristic equation in equation (5) to find $V_o\left(s\right)$.

$V_o\left(s\right)=\frac{14s+3.5}{s\left(s-\left(-0.75+j0.6614\right)\right)\left(s-\left(-0.75-j0.6614\right)\right)}$

Take partial fraction for above equation.

$V_o\left(s\right)=\frac{14s+3.5}{\left[\begin{array}{l}s\left(s-\left(-0.75+j0.6614\right)\right)\\ \left(s-\left(-0.75-j0.6614\right)\right)\end{array}\right]}=\left[\begin{array}{l}\frac{A}{s}+\frac{B}{\left(s-\left(-0.75+j0.6614\right)\right)}\\ +\frac{C}{\left(s-\left(-0.75-j0.6614\right)\right)}\end{array}\right]$        (8)

The equation (8) can also be written as follows:

$\frac{14s+3.5}{\left[\begin{array}{l}s\left(s-\left(-0.75+j0.6614\right)\right)\\ \left(s-\left(-0.75-j0.6614\right)\right)\end{array}\right]}=\frac{\left[\begin{array}{l}A\left(s-\left(-0.75+j0.6614\right)\right)\left(s-\left(-0.75-j0.6614\right)\right)\\ +Bs\left(s-\left(-0.75-j0.6614\right)\right)\\ +Cs\left(s-\left(-0.75+j0.6614\right)\right)\end{array}\right]}{s\left(s-\left(-0.75+j0.6614\right)\right)\left(s-\left(-0.75-j0.6614\right)\right)}$

Simplify the above equation as follows:

$14s+3.5=\left[\begin{array}{l}A\left(s-\left(-0.75+j0.6614\right)\right)\left(s-\left(-0.75-j0.6614\right)\right)\\ +Bs\left(s-\left(-0.75-j0.6614\right)\right)+Cs\left(s-\left(-0.75+j0.6614\right)\right)\end{array}\right]$        (9)

Substitute $0$ for $s$ in equation (9) to find $A$.

$\begin{array}{l}14\left(0\right)+3.5=\left[\begin{array}{l}A\left(0-\left(-0.75+j0.6614\right)\right)\left(0-\left(-0.75-j0.6614\right)\right)\\ +B\left(0\right)\left(\left(0-\left(-0.75-j0.6614\right)\right)\right)\\ +C\left(0\right)\left(0-\left(-0.75+j0.6614\right)\right)\end{array}\right]\\ 3.5=A\left(0.75-j0.6614\right)\left(0.75+j0.6614\right)+0+0\\ 3.5=A\left({0.75}^2-{\left(j0.6614\right)}^2\right)\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ 3.5=A\left(0.5625-j^{2}0.4374\right)\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{3.5}{\left(0.5625-j^{2}0.4374\right)}\\ =\frac{3.5}{0.5625-\left(-1\right)0.4374}\left\{\because j^2=-1\right\}\\ =\frac{3.5}{1}\\ =3.5\end{array}$

Substitute $-0.75+j0.6614$ for $s$ in equation (9) to find $B$.

$\begin{array}{l}14\left(-0.75+j0.6614\right)+3.5=\left[\begin{array}{l}A\left(\left(-0.75+j0.6614\right)-\left(-0.75+j0.6614\right)\right)\\ \left(\left(-0.75+j0.6614\right)-\left(-0.75-j0.6614\right)\right)\\ +B\left(-0.75+j0.6614\right)\left(\left(-0.75+j0.6614\right)-\left(-0.75-j0.6614\right)\right)\\ +C\left(-0.75+j0.6614\right)\left(\left(-0.75+j0.6614\right)-\left(-0.75+j0.6614\right)\right)\end{array}\right]\\ -10.5+j9.2596+3.5=A\left(0\right)+B\left(-0.75+j0.6614\right)\left(j1.3228\right)+C\left(0\right)\\ -7+j9.2596=B\left(-0.875-j0.9921\right)\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{-7+j9.2596}{\left(-0.875-j0.9921\right)}\\ =8.775\angle -101.5°\end{array}$

Substitute $-0.75-j0.6614$ for $s$ in equation (9) to find $B$.

$\begin{array}{l}14\left(-0.75-j0.6614\right)+3.5=\left[\begin{array}{l}A\left(\left(-0.75-j0.6614\right)-\left(-0.75+j0.6614\right)\right)\\ \left(\left(-0.75-j0.6614\right)-\left(-0.75-j0.6614\right)\right)\\ +B\left(-0.75-j0.6614\right)\left(\left(-0.75-j0.6614\right)-\left(-0.75-j0.6614\right)\right)\\ +C\left(-0.75-j0.6614\right)\left(\left(-0.75-j0.6614\right)-\left(-0.75+j0.6614\right)\right)\end{array}\right]\\ -10.5-j9.2596+3.5=A\left(0\right)+B\left(0\right)+C\left(-0.75-j0.6614\right)\left(-j1.3228\right)\\ -7-j9.2596=C\left(-0.875+j0.9921\right)\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{-7-j9.2596}{\left(-0.875+j0.9921\right)}\\ =8.775\angle 101.5°\end{array}$

Substitute $3.5$ for $A$, $8.775\angle -101.5°$ for $B$, and $8.775\angle 101.5°$ for $C$ in equation (8) to find $V_o\left(s\right)$.

$\begin{array}{c}{V}_o\left(s\right)=\left[\frac{3.5}{s}+\frac{8.775\angle -101.5°}{\left(s-\left(-0.75+j0.6614\right)\right)}+\frac{8.775\angle 101.5°}{\left(s-\left(-0.75-j0.6614\right)\right)}\right]\\ =\left[\frac{3.5}{s}+\frac{8.775e^{-j101.5°}}{\left(s-\left(-0.75+j0.6614\right)\right)}+\frac{8.775e^{j101.5°}}{\left(s-\left(-0.75-j0.6614\right)\right)}\right]\left\{\because r\angle \varphi =r{e}^{j\varphi }\right\}\end{array}$

Take inverse Laplace transform for above equation to find $v_o\left(t\right)$.

$\begin{array}{c}{v}_o\left(t\right)=\left[\begin{array}{l}3.5u\left(t\right)+8.775e^{-j101.5°}{e}^{\left(-0.75+j0.6614\right)t}u\left(t\right)\\ +8.775e^{j101.5°}{e}^{\left(-0.75-j0.6614\right)t}u\left(t\right)\end{array}\right]\left\{\begin{array}{l}\because {ℒ}^{-1}\left(\frac{1}{s}\right)=u\left(t\right)\\ {ℒ}^{-1}\left(\frac{1}{s+a}\right)=e^{-at}u\left(t\right)\end{array}\right\}\\ =\left[3.5+8.775e^{\left(-0.75t+j0.6614t-j101.5°\right)}+8.775e^{\left(-0.75t-j0.6614t+j101.5°\right)}\right]u\left(t\right)\left\{\because e^a\cdot e^b=e^{a+b}\right\}\\ =\left[3.5+8.775e^{-0.75t}{e}^{j\left(0.6614t-101.5°\right)}+8.775e^{-0.75t}{e}^{-j\left(0.6614t-101.5°\right)}\right]u\left(t\right)\\ =\left[3.5+8.775e^{-0.75t}\left[e^{j\left(0.6614t-101.5°\right)}+e^{-j\left(0.6614t-101.5°\right)}\right]\right]u\left(t\right)\end{array}$

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

$\begin{array}{c}{v}_o\left(t\right)=\left[3.5+8.775e^{-0.75t}\left(2\cos \left(0.6614t-101.5°\right)\right)\right]u\left(t\right)\left\{\because \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}\right\}\\ =\left[3.5+17.55e^{-0.75t}\cos \left(0.6614t-101.5°\right)\right]u\left(t\right)\text{V}\end{array}$

Conclusion:

Thus, the expression of voltage $v_o\left(t\right)$ for all $t>0$ in the circuit of Figure 16.53 is $\left[3.5+17.55e^{-0.75t}\cos \left(0.6614t-101.5°\right)\right]u\left(t\right)\text{V}$.