Chapter 16, Problem 13P

To determine

Design a problem to better understand the circuit analysis using Laplace transform using Figure 16.36.

Explanation of Solution

Problem design:

Find the value of voltage across resistor $\left(R_1\right)$ $v_x$  using Laplace transform if the value of resistance $\left(R_1\right)$ is $2\Omega$, the value of resistance $\left(R_2\right)$ is $4\Omega$, the value of inductance $\left(L\right)$ is $1\text{H}$, the value of capacitance $\left(C\right)$ is $\frac{1}{8}\text{F}$, and the voltage source $\left(v_s\right)$ is $4u\left(t\right)\text{V}$ in the Figure 16.36 in the textbook.

Formula used:

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

$Z_R\left(s\right)=R$ (1)

Here,

$R$ is the value of resistance.

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

$Z_L\left(s\right)=sL$ (2)

Here,

$L$ is the value of inductance.

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

$Z_C\left(s\right)=\frac{1}{sC}$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Apply Laplace transform to the voltage source $v_s$.

$\begin{array}{c}{V}_s\left(s\right)=4\left(\frac{1}{s}\right)\left\{\because \mathcal{L}\left(u\left(t\right)\right)=\frac{1}{s}\right\}\\ =\frac{4}{s}\end{array}$

Substitute $2$ for $R_1$ in equation (1) to find $Z_{R_1}\left(s\right)$.

$Z_{R_1}\left(s\right)=2$

Substitute $4$ for $R_2$ in equation (1) to find $Z_{R_2}\left(s\right)$.

$Z_{R_2}\left(s\right)=4$

Substitute $1$ for $L$ in equation (2) to find $Z_L\left(s\right)$.

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

Substitute $\frac{1}{8}$ for $C$ in equation (3) to find $Z_C\left(s\right)$.

$\begin{array}{c}{Z}_C\left(s\right)=\frac{1}{s\left(\frac{1}{8}\right)}\\ =\frac{8}{s}\end{array}$

Convert the Figure 1 into s-domain.

Apply Kirchhoff’s current law at node $V_x\left(s\right)$ in Figure 2.

$\begin{array}{l}\frac{V_x\left(s\right)-\frac{4}{s}}{s}+\frac{V_x\left(s\right)}{2}+\frac{V_x\left(s\right)}{\left(4+\frac{8}{s}\right)}=0\\ \frac{V_x\left(s\right)}{s}+\frac{\left(-\frac{4}{s}\right)}{s}+\frac{V_x\left(s\right)}{2}+\frac{V_x\left(s\right)}{\left(\frac{4s+8}{s}\right)}=0\\ \frac{V_x\left(s\right)}{s}-\frac{4}{s^2}+\frac{V_x\left(s\right)}{2}+\frac{s{V}_x\left(s\right)}{4s+8}=0\\ \frac{s{V}_x\left(s\right)\left(4s+8\right)-4\left(4s+8\right)+s^2{V}_x\left(s\right)\left(2s+4\right)+s\left(s^2\right)V_x\left(s\right)}{s^2\left(4s+8\right)}=0\end{array}$

Simplify the above equation as follows,

$\begin{array}{l}s{V}_x\left(s\right)\left(4s+8\right)-4\left(4s+8\right)+s^2{V}_x\left(s\right)\left(2s+4\right)+s\left(s^2\right)V_x\left(s\right)=0\\ s\left(V_x\left(s\right)\left(4s+8\right)-\frac{4\left(4s+8\right)}{s}+\frac{s^2{V}_x\left(s\right)\left(2s+4\right)}{s}+\frac{s\left(s^2\right)V_x\left(s\right)}{s}\right)=0\\ V_x\left(s\right)\left(4s+8\right)-\frac{4\left(4s+8\right)}{s}+s{V}_x\left(s\right)\left(2s+4\right)+\left(s^2\right)V_x\left(s\right)=0\\ V_x\left(s\right)\left(4s+8\right)-\frac{\left(16s+32\right)}{s}+V_x\left(s\right)\left(2s^2+4s\right)+s^2{V}_x\left(s\right)=0\end{array}$

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

$\begin{array}{l}\left[4s+8+2s^2+4s+s^2\right]V_x\left(s\right)=\frac{16s+32}{s}\\ \left[3s^2+8s+8\right]V_x\left(s\right)=\frac{16s+32}{s}\end{array}$

$V_x\left(s\right)=\frac{16s+32}{s\left(3s^2+8s+8\right)}$ (4)

From equation (4), the characteristic equation is written as follows,

$3s^2+8s+8=0$ (5)

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

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . (6)

Here,

$a$ is the coefficient of second order term,

$b$ is the coefficient of first order term, and

$c$ is the coefficient of constant term.

Compare equation (5) with quadratic equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=3\\ b=8\\ c=8\end{array}$

Substitute $1$ for $a$, $2$ for $b$, and $1$ for $c$ in equation (6) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-8\pm \sqrt{{\left(8\right)}^2-4\left(3\right)\left(8\right)}}{2\left(3\right)}\\ =\frac{-8\pm \sqrt{64-96}}{6}\\ =\frac{-8\pm j2\sqrt{8}}{6}\\ =\frac{-4}{3}+\frac{j\sqrt{8}}{3},\frac{-4}{3}-\frac{j\sqrt{8}}{3}\end{array}$

Now, the equation (4) is written as follows,

$V_x\left(s\right)=\frac{16s+32}{s\left(s-\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)\right)\left(s-\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\right)}$

$V_x\left(s\right)=\frac{16s+32}{s\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)}$ (7)

Take partial fraction for equation (7).

$V_x\left(s\right)=\frac{16s+32}{s\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)}=\frac{A}{s}+\frac{B}{\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)}+\frac{C}{\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)}$ (8)

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

$\frac{16s+32}{s\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)}=\frac{\left[\begin{array}{l}A\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)+Bs\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)\\ +Cs\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\end{array}\right]}{s\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)}$

Simplify the above equation as follows,

$16s+32=\left\{\begin{array}{l}A\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)+\\ Bs\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)+Cs\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\end{array}\right\}$ (9)

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

$\begin{array}{l}16\left(0\right)+32=A\left(0+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(0+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)+B\left(0\right)\left(0+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)+C\left(0\right)\left(0+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\\ 32=A\left(\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)+0+0\\ A\left({\left(\frac{4}{3}\right)}^2-{\left(\frac{j\sqrt{8}}{3}\right)}^2\right)=32\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ A\left(\frac{16}{9}-j^2\left(\frac{8}{9}\right)\right)=32\end{array}$

Simplify the above equation as follows,

$\begin{array}{l}A\left(\frac{16}{9}-\left(-1\right)\left(\frac{8}{9}\right)\right)=32\left\{\because j^2=-1\right\}\\ A\left(\frac{16}{9}+\frac{8}{9}\right)=32\\ A\left(\frac{24}{9}\right)=32\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=32\left(\frac{9}{24}\right)\\ =12\end{array}$

Substitute $\frac{-4}{3}+\frac{j\sqrt{8}}{3}$ for $s$ in equation (9) to find $B$.

$\begin{array}{l}16\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)+32=\left[\begin{array}{l}A\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)\\ +B\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)\\ +C\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\end{array}\right]\\ 16\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)+32=A\left(0\right)\left(\frac{j2\sqrt{8}}{3}\right)+B\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)\left(\frac{j2\sqrt{8}}{3}\right)+C\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)\left(0\right)\\ B\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)\left(\frac{j2\sqrt{8}}{3}\right)=16\left(\frac{-4}{3}+\frac{j\sqrt{8}}{3}\right)+32\\ B\left(\frac{-4+j\sqrt{8}}{3}\right)\left(\frac{j2\sqrt{8}}{3}\right)=\left(\frac{-64+j16\sqrt{8}+96}{3}\right)\end{array}$

Simplify the above equation as follows,

$\begin{array}{l}B\left(\frac{-8j\sqrt{8}+j^{2}16}{3}\right)\left(\frac{1}{3}\right)=16\left(\frac{2+j\sqrt{8}}{3}\right)\\ B\left(\frac{-8j\sqrt{8}-16}{3}\right)\left(\frac{1}{3}\right)=16\left(\frac{2+j\sqrt{8}}{3}\right)\\ B\left(-8\right)\left(\frac{j\sqrt{8}+2}{3}\right)\left(\frac{1}{3}\right)=16\left(\frac{2+j\sqrt{8}}{3}\right)\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=16\left(\frac{2+j\sqrt{8}}{3}\right)\left(\frac{3}{2+j\sqrt{8}}\right)\left(\frac{3}{-8}\right)\\ =-6\end{array}$

Substitute $\frac{-4}{3}-\frac{j\sqrt{8}}{3}$ for $s$ in equation (9) to find $C$.

$\begin{array}{l}16\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)+32=\left[\begin{array}{l}A\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)\\ +B\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)\\ +C\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\end{array}\right]\\ 16\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)+32=A\left(-\frac{j2\sqrt{8}}{3}\right)\left(0\right)+B\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\left(0\right)+C\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\left(-\frac{j2\sqrt{8}}{3}\right)\\ \left(\frac{-64}{3}-\frac{j16\sqrt{8}}{3}\right)+32=0+0+C\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\left(-\frac{j2\sqrt{8}}{3}\right)\\ \left(\frac{-64-j16\sqrt{8}+96}{3}\right)=0+0+C\left(\frac{-4}{3}-\frac{j\sqrt{8}}{3}\right)\left(-\frac{j2\sqrt{8}}{3}\right)\end{array}$

Simplify the above equation as follows,

$\begin{array}{l}C\left(\frac{8j\sqrt{8}+j^{2}16}{3}\right)\left(\frac{1}{3}\right)=\left(\frac{32-j16\sqrt{8}}{3}\right)\\ C\left(\frac{8j\sqrt{8}+\left(-1\right)16}{3}\right)\left(\frac{1}{3}\right)=16\left(\frac{2-j\sqrt{8}}{3}\right)\\ C\left(-8\right)\left(\frac{2-j\sqrt{8}}{3}\right)\left(\frac{1}{3}\right)=16\left(\frac{2-j\sqrt{8}}{3}\right)\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=16\left(\frac{2-j\sqrt{8}}{3}\right)\left(\frac{3}{2-j\sqrt{8}}\right)\left(\frac{3}{-8}\right)\\ =-6\end{array}$

Substitute $12$ for $A$, $-6$ for $B$, and $-6$ for $C$ in equation (8) to find $V_x\left(s\right)$.

$V_x\left(s\right)=\frac{12}{s}+\frac{\left(-6\right)}{\left(s+\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)}+\frac{\left(-6\right)}{\left(s+\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)}$

$V_x\left(s\right)=\frac{12}{s}-\frac{6}{\left(s+\left(\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)\right)}-\frac{6}{\left(s+\left(\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)\right)}$ (10)

Take inverse Laplace transform for equation (10) to find $v_x\left(t\right)$.

$\begin{array}{c}{v}_x\left(t\right)=12u\left(t\right)-6e^{\left(-\frac{4}{3}+\frac{j\sqrt{8}}{3}\right)t}u\left(t\right)-6e^{\left(-\frac{4}{3}-\frac{j\sqrt{8}}{3}\right)t}u\left(t\right)\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s+a}\right)=e^{-at}u\left(t\right),{\mathcal{L}}^{-1}\left(\frac{1}{s}\right)=u\left(t\right)\right\}\\ =\left[12-6e^{\left(-\frac{4}{3}\right)t}{e}^{\frac{j\sqrt{8}}{3}t}-6e^{\left(-\frac{4}{3}\right)t}{e}^{\left(-\frac{j\sqrt{8}}{3}\right)t}\right]u\left(t\right)\text{V}\\ \text{=}\left[12-6e^{\left(-\frac{4}{3}\right)t}\left[e^{\frac{j\sqrt{8}}{3}t}+e^{\left(-\frac{j\sqrt{8}}{3}\right)t}\right]\right]u\left(t\right)\text{V}\\ =\left[12-6e^{\left(-\frac{4}{3}\right)t}\left[2\cos \left(\frac{\sqrt{8}}{3}t\right)\right]\right]u\left(t\right)\text{V}\left\{\because \cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2}\right\}\end{array}$

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

$v_x\left(t\right)=\left[12-12e^{\left(-\frac{4}{3}\right)t}\left[\cos \left(\frac{2\sqrt{2}}{3}t\right)\right]\right]u\left(t\right)\text{V}$

Conclusion:

Thus, the problem to better understand the circuit analysis using Laplace transform is designed.