Chapter 16, Problem 7P

To determine

Find the expression of current $i\left(t\right)$.

Answer to Problem 7P

The expression of current $i\left(t\right)$ is $\left[6+12e^{-t}\left(\cos \left(2t\right)+2\sin \left(2t\right)\right)\right]u\left(t\right)\text{A}$.

Explanation of Solution

Given data:

The step response of an $RLC$ circuit is given as,

$\frac{d^{2}i}{d{t}^2}+2\frac{di}{dt}+5i=30$ (1)

The value of initial current $i\left(0\right)$ is $18\text{A}$.

The value of $\frac{di\left(0\right)}{dt}$ is $36\frac{\text{A}}{\text{s}}$.

Calculation:

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

$\left[\begin{array}{l}{s}^{2}I\left(s\right)-\frac{di\left(0^{-}\right)}{dt}-si\left(0^{-}\right)\\ +2\left(sI\left(s\right)-i\left(0^{-}\right)\right)+5I\left(s\right)\end{array}\right]=\frac{30}{s}\left\{\begin{array}{l}\because \mathcal{L}\left(\frac{d^{2}i}{d{t}^2}\right)=s^{2}I\left(s\right)-\frac{di\left(0^{-}\right)}{dt}-si\left(0^{-}\right)\\ \mathcal{L}\left(\frac{di}{dt}\right)=sI\left(s\right)-i\left(0^{-}\right),\mathcal{L}\left(u\left(t\right)\right)=\frac{1}{s}\end{array}\right\}$

Simplify the above equation as follows.

$s^{2}I\left(s\right)-\frac{di\left(0^{-}\right)}{dt}-si\left(0^{-}\right)+2sI\left(s\right)-2i\left(0^{-}\right)+5I\left(s\right)=\frac{30}{s}$ (2)

Substitute $18$ for $i\left(0^{-}\right)$, and $36$ for $\frac{di\left(0^{-}\right)}{dt}$ in equation (2).

$\begin{array}{l}{s}^{2}I\left(s\right)-36-s\left(18\right)+2sI\left(s\right)-2\left(18\right)+5I\left(s\right)=\frac{30}{s}\\ s^{2}I\left(s\right)-36-18s+2sI\left(s\right)-36+5I\left(s\right)=\frac{30}{s}\\ \left[s^2+2s+5\right]I\left(s\right)=\frac{30}{s}+36+36+18s\\ \left[s^2+2s+5\right]I\left(s\right)=\frac{30+72s+18s^2}{s}\end{array}$

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

$I\left(s\right)=\frac{18\left(s^2+4s\right)+30}{s\left(s^2+2s+5\right)}$ (3)

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

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

$\begin{array}{l}a=1\\ b=2\\ c=5\end{array}$

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

$\begin{array}{c}{s}_{1,2}=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ =\frac{-2\pm \sqrt{-16}}{2}\\ =\frac{-2\pm j4}{2}\\ =-1+j2,-1-j2\end{array}$

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

$I\left(s\right)=\frac{18\left(s^2+4s\right)+30}{s\left(s-\left(-1+j2\right)\right)\left(s-\left(-1-j2\right)\right)}$

$I\left(s\right)=\frac{18\left(s^2+4s\right)+30}{s\left(s+1-j2\right)\left(s+1+j2\right)}$ (6)

Take partial fraction for equation (6).

$I\left(s\right)=\frac{18\left(s^2+4s\right)+30}{s\left(s+1-j2\right)\left(s+1+j2\right)}=\frac{A}{s}+\frac{B}{\left(s+1-j2\right)}+\frac{C}{\left(s+1+j2\right)}$ (7)

The equation (7) can be written as follows.

$\frac{18\left(s^2+4s\right)+30}{s\left(s+1-j2\right)\left(s+1+j2\right)}=\frac{A\left(s+1-j2\right)\left(s+1+j2\right)+Bs\left(s+1+j2\right)+Cs\left(s+1-j2\right)}{s\left(s+1-j2\right)\left(s+1+j2\right)}$

Simplify the above equation.

$18\left(s^2+4s\right)+30=A\left(s+1-j2\right)\left(s+1+j2\right)+Bs\left(s+1+j2\right)+Cs\left(s+1-j2\right)$ (8)

Substitute $0$ for $s$ in equation (8).

$\begin{array}{l}18\left({\left(0\right)}^2+4\left(0\right)\right)+30=A\left(0+1-j2\right)\left(0+1+j2\right)+B\left(0\right)\left(0+1+j2\right)+C\left(0\right)\left(0+1-j2\right)\\ 30=A\left(1-j2\right)\left(1+j2\right)+0+0\\ A\left(1-j2\right)\left(1+j2\right)=30\\ A\left(1^2-{\left(j2\right)}^2\right)=30\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\end{array}$

Simplify the above equation.

$\begin{array}{l}A\left(1-j^{2}4\right)=30\\ A\left(1-\left(-1\right)4\right)=30\left\{\because j^2=-1\right\}\\ A\left(1+4\right)=30\\ 5A=30\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{30}{5}\\ =6\end{array}$

Substitute $-1+j2$ for $s$ in equation (8) to find $B$.

$\begin{array}{l}18\left({\left(-1+j2\right)}^2+4\left(-1+j2\right)\right)+30=\left[\begin{array}{l}A\left(-1+j2+1-j2\right)\left(-1+j2+1+j2\right)\\ +B\left(-1+j2\right)\left(-1+j2+1+j2\right)\\ +C\left(-1+j2\right)\left(-1+j2+1-j2\right)\end{array}\right]\\ \left[\begin{array}{l}18\left(\begin{array}{l}{\left(-1\right)}^2+2\left(-1\right)\left(j2\right)\\ +{\left(j2\right)}^2-4+j8\end{array}\right)\\ +30\end{array}\right]=\left[\begin{array}{l}A\left(0\right)\left(j4\right)\\ +B\left(-1+j2\right)\left(j4\right)\\ +C\left(-1+j2\right)\left(0\right)\end{array}\right]\left\{\because {\left(a+b\right)}^2=a^2+2ab+b^2\right\}\\ 18\left(1-j4+j^{2}4-4+j8\right)+30=0+B\left(-1+j2\right)\left(j4\right)+0\\ 18\left(1-j4+\left(-1\right)4-4+j8\right)+30=B\left(-1+j2\right)\left(j4\right)\left\{\because j^2=-1\right\}\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}18\left(1-j4-4-4+j8\right)+30=B\left(-1+j2\right)\left(j4\right)\\ 18\left(j4-7\right)+30=B\left(-1+j2\right)\left(j4\right)\\ j72-126+30=B\left(-j4+j^{2}8\right)\\ j72-96=B\left(-j4+\left(-1\right)8\right)\\ j72-96=B\left(-j4-8\right)\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{-96+j72}{-8-j4}\\ =\frac{24\left(-4+j3\right)}{4\left(-2-j\right)}\\ =\frac{6\left(-4+j3\right)}{\left(-2-j\right)}\end{array}$

Multiply and divide by $-2+j$ on right hand side of above equation to find $B$.

$\begin{array}{c}B=\frac{6\left(-4+j3\right)\left(-2+j\right)}{\left(-2-j\right)\left(-2+j\right)}\\ =\frac{6\left(8-j4-j6+j^{2}3\right)}{{\left(-2\right)}^2-{\left(j\right)}^2}\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ =\frac{6\left(8-j10+\left(-1\right)3\right)}{{\left(-2\right)}^2-\left(-1\right)}\left\{\because j^2=-1\right\}\\ =\frac{6\left(5-j10\right)}{4+1}\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{30\left(1-j2\right)}{5}\\ =6\left(1-j2\right)\end{array}$

Substitute $-1-j2$ for $s$ in equation (8) to find $C$.

$\begin{array}{l}18\left({\left(-1-j2\right)}^2+4\left(-1-j2\right)\right)+30=\left[\begin{array}{l}A\left(-1-j2+1-j2\right)\left(-1-j2+1+j2\right)\\ +B\left(-1-j2\right)\left(-1-j2+1+j2\right)\\ +C\left(-1-j2\right)\left(-1-j2+1-j2\right)\end{array}\right]\\ \left[\begin{array}{l}18\left(\begin{array}{l}{\left(-1\right)}^2+2\left(-1\right)\left(-j2\right)\\ +{\left(-j2\right)}^2-4-j8\end{array}\right)\\ +30\end{array}\right]=\left[\begin{array}{l}A\left(-j4\right)\left(0\right)\\ +B\left(-1-j2\right)\left(0\right)\\ +C\left(-1-j2\right)\left(-j4\right)\end{array}\right]\left\{\because {\left(a+b\right)}^2=a^2+2ab+b^2\right\}\\ 18\left(1+j4+j^{2}4-4-j8\right)+30=0+0+C\left(j4+j^{2}8\right)\\ 18\left(1+j4+\left(-1\right)4-4-j8\right)+30=C\left(j4+\left(-1\right)8\right)\left\{\because j^2=-1\right\}\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}18\left(1+j4-4-4-j8\right)+30=C\left(j4-8\right)\\ 18\left(-7-j4\right)+30=C\left(j4-8\right)\\ -126-j72+30=C\left(j4-8\right)\\ -96-j72=C\left(j4-8\right)\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{-96-j72}{j4-8}\\ =\frac{24\left(-4-j3\right)}{4\left(-2+j\right)}\\ =\frac{6\left(-4-j3\right)}{\left(-2+j\right)}\end{array}$

Multiply and divide by $-2-j$ on right hand side of above equation.

$\begin{array}{c}C=\frac{6\left(-4-j3\right)\left(-2-j\right)}{\left(-2+j\right)\left(-2-j\right)}\\ =\frac{6\left(8+j4+j6+j^{2}3\right)}{{\left(-2\right)}^2-{\left(j\right)}^2}\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ =\frac{6\left(8+j10+\left(-1\right)3\right)}{{\left(-2\right)}^2-\left(-1\right)}\left\{\because j^2=-1\right\}\\ =\frac{6\left(5+j10\right)}{4+1}\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{30\left(1+j2\right)}{5}\\ =6\left(1+j2\right)\end{array}$

Substitute $6$ for $A$, $6\left(1-j2\right)$ for $B$, and $6\left(1+j2\right)$ for $C$ in equation (7) to find $I\left(s\right)$.

$I\left(s\right)=\frac{6}{s}+\frac{6\left(1-j2\right)}{\left(s+1-j2\right)}+\frac{6\left(1+j2\right)}{\left(s+1+j2\right)}$

$I\left(s\right)=\frac{6}{s}+\frac{6-j12}{\left(s+1-j2\right)}+\frac{6+j12}{\left(s+1+j2\right)}$ (9)

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

$\begin{array}{c}i\left(t\right)=\left[\begin{array}{l}6u\left(t\right)+\left(6-j12\right)e^{\left(-1+j2\right)t}u\left(t\right)\\ +\left(6+j12\right)e^{\left(-1-j2\right)t}u\left(t\right)\end{array}\right]\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s+a}\right)=e^{-at}u\left(t\right),{\mathcal{L}}^{-1}\left(\frac{1}{s}\right)=u\left(t\right)\right\}\\ =\left[6+\left(6-j12\right)e^{\left(-1+j2\right)t}+\left(6+j12\right)e^{\left(-1-j2\right)t}\right]u\left(t\right)\text{A}\\ \text{=}\left[6+\left(6-j12\right)e^{\left(-1\right)t}{e}^{\left(j2\right)t}+\left(6+j12\right)e^{\left(-1\right)t}{e}^{\left(-j2\right)t}\right]u\left(t\right)\text{A}\\ \text{=}\left[6+\left(6-j12\right)e^{\left(-1\right)t}{e}^{\left(j2\right)t}+\left(6+j12\right)e^{\left(-1\right)t}{e}^{\left(-j2\right)t}\right]u\left(t\right)\text{A}\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}i\left(t\right)\text{=}\left[6+6e^{\left(-1\right)t}{e}^{\left(j2\right)t}-j12e^{\left(-1\right)t}{e}^{\left(j2\right)t}+6e^{\left(-1\right)t}{e}^{\left(-j2\right)t}+j12e^{\left(-1\right)t}{e}^{\left(-j2\right)t}\right]u\left(t\right)\text{A}\\ \text{=}\left[6+6e^{-t}\left(e^{\left(j2\right)t}+e^{\left(-j2\right)t}\right)-j12e^{-t}\left(e^{\left(j2\right)t}-e^{\left(-j2\right)t}\right)\right]u\left(t\right)\text{A}\\ =\left[\begin{array}{l}6+6e^{-t}\left(2\cos \left(2t\right)\right)\\ -j12e^{-t}\left(2j\sin \left(2t\right)\right)\end{array}\right]u\left(t\right)\text{A}\left\{\because \cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2},\sin \theta =\frac{e^{j\theta }-e^{-j\theta }}{2j}\right\}\\ =\left[6+12e^{-t}\cos \left(2t\right)-j^{2}24e^{-t}\sin \left(2t\right)\right]u\left(t\right)\text{A}\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}i\left(t\right)\text{=}\left[6+12e^{-t}\cos \left(2t\right)-\left(-1\right)24e^{-t}\sin \left(2t\right)\right]u\left(t\right)\text{A}\left\{\because j^2=-1\right\}\\ =\left[6+12e^{-t}\cos \left(2t\right)+24e^{-t}\sin \left(2t\right)\right]u\left(t\right)\text{A}\\ \text{=}\left[6+12e^{-t}\left(\cos \left(2t\right)+2\sin \left(2t\right)\right)\right]u\left(t\right)\text{A}\end{array}$

Conclusion:

Thus, the expression of current $i\left(t\right)$ is $\left[6+12e^{-t}\left(\cos \left(2t\right)+2\sin \left(2t\right)\right)\right]u\left(t\right)\text{A}$.