Chapter 16, Problem 8P

To determine

Find the expression of voltage $v\left(t\right)$.

Answer to Problem 8P

The expression of voltage $v\left(t\right)$ is.$\left[15-15e^{-2t}\left(\cos \left(2t\right)+\sin \left(2t\right)\right)\right]u\left(t\right)\text{V}$.

Explanation of Solution

Given data:

A branch voltage in an $RLC$ circuit is given by,

$\frac{d^{2}v}{d{t}^2}+4\frac{dv}{dt}+8v=120$ (1)

The value of initial voltage $v\left(0\right)$ is $0$.

The value of $\frac{dv\left(0\right)}{dt}$ is $0$.

Calculation:

Apply Laplace transform for equation (1) with initial conditions.

$\left[\begin{array}{l}{s}^{2}V\left(s\right)-\frac{dv\left(0^{-}\right)}{dt}-sv\left(0^{-}\right)\\ +4\left(sV\left(s\right)-v\left(0^{-}\right)\right)+8V\left(s\right)\end{array}\right]=\frac{120}{s}\left\{\begin{array}{l}\because \mathcal{L}\left(\frac{d^{2}v}{d{t}^2}\right)=s^{2}V\left(s\right)-\frac{dv\left(0^{-}\right)}{dt}-sv\left(0^{-}\right)\\ \mathcal{L}\left(\frac{dv}{dt}\right)=sV\left(s\right)-v\left(0^{-}\right),\mathcal{L}\left(u\left(t\right)\right)=\frac{1}{s}\end{array}\right\}$

Simplify the above equation as follows.

$s^{2}V\left(s\right)-\frac{dv\left(0^{-}\right)}{dt}-sv\left(0^{-}\right)+4sV\left(s\right)-4v\left(0^{-}\right)+8V\left(s\right)=\frac{120}{s}$ (2)

Substitute $0$ for $v\left(0^{-}\right)$, and $0$ for $\frac{dv\left(0^{-}\right)}{dt}$ in equation (2).

$\begin{array}{l}{s}^{2}V\left(s\right)-0-s\left(0\right)+4sV\left(s\right)-4\left(0\right)+8V\left(s\right)=\frac{120}{s}\\ s^{2}V\left(s\right)+4sV\left(s\right)+8V\left(s\right)=\frac{120}{s}\\ \left[s^2+4s+8\right]V\left(s\right)=\frac{120}{s}\end{array}$

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

$V\left(s\right)=\frac{120}{s\left(s^2+4s+8\right)}$ (3)

From equation (3), the characteristic equation is written as follows,

$s^2+4s+8=0$ (4)

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

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . (5)

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

$\begin{array}{l}a=1\\ b=4\\ c=8\end{array}$

Substitute $1$ for $a$, $4$ for $b$, and $8$ for $c$ in equation (5) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-4\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(8\right)}}{2\left(1\right)}\\ =\frac{-4\pm \sqrt{-16}}{2}\\ =\frac{-4\pm j4}{2}\\ =-2+j2,-2-j2\end{array}$

Now, the equation (3) is written as follows.

$V\left(s\right)=\frac{120}{s\left(s-\left(-2+j2\right)\right)\left(s-\left(-2-j2\right)\right)}$

$V\left(s\right)=\frac{120}{s\left(s+2-j2\right)\left(s+2+j2\right)}$ (6)

Take partial fraction for equation (6).

$V\left(s\right)=\frac{120}{s\left(s+2-j2\right)\left(s+2+j2\right)}=\frac{A}{s}+\frac{B}{\left(s+2-j2\right)}+\frac{C}{\left(s+2+j2\right)}$ (7)

The equation (7) can be written as follows.

$\frac{120}{s\left(s+2-j2\right)\left(s+2+j2\right)}=\frac{A\left(s+2-j2\right)\left(s+2+j2\right)+Bs\left(s+2+j2\right)+Cs\left(s+2-j2\right)}{s\left(s+2-j2\right)\left(s+2+j2\right)}$

Simplify the above equation.

$120=A\left(s+2-j2\right)\left(s+2+j2\right)+Bs\left(s+2+j2\right)+Cs\left(s+2-j2\right)$ (8)

Substitute $0$ for $s$ in equation (8) to find $A$.

$\begin{array}{l}120=A\left(0+2-j2\right)\left(0+2+j2\right)+B\left(0\right)\left(0+2+j2\right)+C\left(0\right)\left(0+2-j2\right)\\ 120=A\left(2-j2\right)\left(2+j2\right)+0+0\\ A\left(2-j2\right)\left(2+j2\right)=120\\ A\left(2^2-{\left(j2\right)}^2\right)=120\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}A\left(4-j^{2}4\right)=120\\ A\left(4-\left(-1\right)4\right)=120\left\{\because j^2=-1\right\}\\ A\left(4+4\right)=120\\ 8A=120\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{120}{8}\\ =15\end{array}$

Substitute $-2+j2$ for $s$ in equation (8) to find $B$.

$\begin{array}{l}120=\left[\begin{array}{l}A\left(-2+j2+2-j2\right)\left(-2+j2+2+j2\right)+B\left(-2+j2\right)\left(-2+j2+2+j2\right)\\ +C\left(-2+j2\right)\left(-2+j2+2-j2\right)\end{array}\right]\\ 120=\left[A\left(0\right)\left(j4\right)+B\left(-2+j2\right)\left(j4\right)+C\left(-2+j2\right)\left(0\right)\right]\\ 120=0+B\left(-2+j2\right)\left(j4\right)+0\\ 120=B\left(-2+j2\right)\left(j4\right)\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}120=B\left(-2+j2\right)\left(j4\right)\\ 120=B\left(-j8+j^{2}8\right)\\ 120=B\left(-j8+\left(-1\right)8\right)\left\{\because j^2=-1\right\}\\ 120=B\left(-j8-8\right)\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{120}{\left(-j8-8\right)}\\ =\frac{120}{8\left(-1-j\right)}\\ =\frac{15}{\left(-1-j\right)}\end{array}$

Multiply and divide by $-1+j$ on right hand side of above equation.

$\begin{array}{c}B=\frac{15\left(-1+j\right)}{\left(-1-j\right)\left(-1+j\right)}\\ =\frac{15\left(-1+j\right)}{{\left(-1\right)}^2-{\left(j\right)}^2}\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ =\frac{15\left(-1+j\right)}{{\left(-1\right)}^2-\left(-1\right)}\left\{\because j^2=-1\right\}\\ =\frac{15\left(-1+j\right)}{1+1}\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{15\left(-1+j\right)}{2}\\ =7.5\left(-1+j\right)\end{array}$

Substitute $-2-j2$ for $s$ in equation (8) to find $C$.

$\begin{array}{l}120=\left[\begin{array}{l}A\left(-2-j2+2-j2\right)\left(-2-j2+2+j2\right)\\ +B\left(-2-j2\right)\left(-2-j2+2+j2\right)\\ +C\left(-2-j2\right)\left(-2-j2+2-j2\right)\end{array}\right]\\ 120=\left[A\left(-j4\right)\left(0\right)+B\left(-2-j2\right)\left(0\right)+C\left(-2-j2\right)\left(-j4\right)\right]\\ 120=0+0+C\left(j8+j^{2}8\right)\\ 120=C\left(j8+\left(-1\right)8\right)\left\{\because j^2=-1\right\}\end{array}$

Simplify the above equation as follows.

$120=C\left(j8-8\right)$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{120}{\left(j8-8\right)}\\ =\frac{120}{8\left(-1+j\right)}\\ =\frac{15}{\left(-1+j\right)}\end{array}$

Multiply and divide by $-1-j$ on right hand side of above equation.

$\begin{array}{c}C=\frac{15\left(-1-j\right)}{\left(-1+j\right)\left(-1-j\right)}\\ =\frac{15\left(-1-j\right)}{{\left(-1\right)}^2-{\left(j\right)}^2}\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ =\frac{15\left(-1-j\right)}{{\left(-1\right)}^2-\left(-1\right)}\left\{\because j^2=-1\right\}\\ =\frac{15\left(-1-j\right)}{1+1}\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{15\left(-1-j\right)}{2}\\ =7.5\left(-1-j\right)\end{array}$

Substitute $15$ for $A$, $7.5\left(-1+j\right)$ for $B$, and $7.5\left(-1-j\right)$ for $C$ in equation (7) to find $V\left(s\right)$.

$V\left(s\right)=\frac{15}{s}+\frac{7.5\left(-1+j\right)}{\left(s+2-j2\right)}+\frac{7.5\left(-1-j\right)}{\left(s+2+j2\right)}$

$V\left(s\right)=\frac{15}{s}+\frac{\left(-7.5+j7.5\right)}{\left(s+2-j2\right)}+\frac{\left(-7.5-j7.5\right)}{\left(s+2+j2\right)}$ (9)

Apply inverse Laplace transform for equation (9) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left[\begin{array}{l}15u\left(t\right)+\left(-7.5+j7.5\right)e^{\left(-2+j2\right)t}u\left(t\right)\\ +\left(-7.5-j7.5\right)e^{\left(-2-j2\right)t}u\left(t\right)\end{array}\right]\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s+a}\right)=e^{-at}u\left(t\right),{\mathcal{L}}^{-1}\left(\frac{1}{s}\right)=u\left(t\right)\right\}\\ =\left[15+\left(-7.5+j7.5\right)e^{\left(-2+j2\right)t}+\left(-7.5-j7.5\right)e^{\left(-2-j2\right)t}\right]u\left(t\right)\text{V}\\ \text{=}\left[15+\left(-7.5+j7.5\right)e^{\left(-2\right)t}{e}^{\left(j2\right)t}+\left(-7.5-j7.5\right)e^{\left(-2\right)t}{e}^{\left(-j2\right)t}\right]u\left(t\right)\text{V}\\ \text{=}\left[15-7.5e^{\left(-2\right)t}{e}^{\left(j2\right)t}+j7.5e^{\left(-2\right)t}{e}^{\left(j2\right)t}-7.5e^{\left(-2\right)t}{e}^{\left(-j2\right)t}-j7.5e^{\left(-2\right)t}{e}^{\left(-j2\right)t}\right]u\left(t\right)\text{V}\end{array}$

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

$\begin{array}{c}v\left(t\right)\text{=}\left[6-7.5e^{-2t}\left(e^{\left(j2\right)t}+e^{\left(-j2\right)t}\right)+j7.5e^{-t}\left(e^{\left(j2\right)t}-e^{\left(-j2\right)t}\right)\right]u\left(t\right)\text{V}\\ =\left[\begin{array}{l}15-7.5e^{-2t}\left(2\cos \left(2t\right)\right)\\ +j7.5e^{-2t}\left(2j\sin \left(2t\right)\right)\end{array}\right]u\left(t\right)\text{V}\left\{\because \cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2},\sin \theta =\frac{e^{j\theta }-e^{-j\theta }}{2j}\right\}\\ =\left[15-15e^{-2t}\cos \left(2t\right)+j^{2}15e^{-2t}\sin \left(2t\right)\right]u\left(t\right)\text{V}\\ =\left[15-15e^{-2t}\cos \left(2t\right)+\left(-1\right)15e^{-2t}\sin \left(2t\right)\right]u\left(t\right)\text{V}\left\{\because j^2=-1\right\}\end{array}$

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

$\begin{array}{c}v\left(t\right)=\left[15-15e^{-2t}\cos \left(2t\right)-15e^{-2t}\sin \left(2t\right)\right]u\left(t\right)\text{V}\\ \text{=}\left[15-15e^{-2t}\left(\cos \left(2t\right)+\sin \left(2t\right)\right)\right]u\left(t\right)\text{V}\end{array}$

Conclusion:

Thus, the expression of voltage $v\left(t\right)$ is $\left[15-15e^{-2t}\left(\cos \left(2t\right)+\sin \left(2t\right)\right)\right]u\left(t\right)\text{V}$.