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

Question

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Answer 1

Textbook Question

Chapter 16, Problem 30P

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

Chapter 16, Problem 30P, Find vo(t), for all t 0, in the circuit of Fig. 16.53. Figure 16.53

Figure 16.53

Expert Solution & Answer

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$

Simplify the above equation to find $vo(t)$

$vo(t)=[3.5+8.775e−0.75t(2cos(0.6614t−101.5°))]u(t) {∵cosθ=eiθ+e−iθ2}=[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$

Conclusion:

Thus, the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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Answer 2

Textbook Question

Chapter 16, Problem 30P

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

Chapter 16, Problem 30P, Find vo(t), for all t 0, in the circuit of Fig. 16.53. Figure 16.53

Figure 16.53

Expert Solution & Answer

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$

Simplify the above equation to find $vo(t)$

$vo(t)=[3.5+8.775e−0.75t(2cos(0.6614t−101.5°))]u(t) {∵cosθ=eiθ+e−iθ2}=[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$

Conclusion:

Thus, the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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Answer 3

Textbook Question

Chapter 16, Problem 30P

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

Chapter 16, Problem 30P, Find vo(t), for all t 0, in the circuit of Fig. 16.53. Figure 16.53

Figure 16.53

Expert Solution & Answer

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$

Simplify the above equation to find $vo(t)$

$vo(t)=[3.5+8.775e−0.75t(2cos(0.6614t−101.5°))]u(t) {∵cosθ=eiθ+e−iθ2}=[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$

Conclusion:

Thus, the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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Answer 4

Textbook Question

Chapter 16, Problem 30P

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

Chapter 16, Problem 30P, Find vo(t), for all t 0, in the circuit of Fig. 16.53. Figure 16.53

Figure 16.53

Expert Solution & Answer

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$

Simplify the above equation to find $vo(t)$

$vo(t)=[3.5+8.775e−0.75t(2cos(0.6614t−101.5°))]u(t) {∵cosθ=eiθ+e−iθ2}=[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$

Conclusion:

Thus, the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$

Simplify the above equation to find $vo(t)$

$vo(t)=[3.5+8.775e−0.75t(2cos(0.6614t−101.5°))]u(t) {∵cosθ=eiθ+e−iθ2}=[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$

Conclusion:

Thus, the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

Answer 8

Expert Solution & Answer

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$

Simplify the above equation to find $vo(t)$

$vo(t)=[3.5+8.775e−0.75t(2cos(0.6614t−101.5°))]u(t) {∵cosθ=eiθ+e−iθ2}=[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$

Conclusion:

Thus, the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53.

Answer 12

Answer to Problem 30P

The expression of voltage $vo(t)$ for all $t>0$ in the circuit of Figure 16.53 is $[3.5+17.55e−0.75tcos(0.6614t−101.5°)]u(t) V$ .

Answer 13

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.

$u(t)={0 t<01 t>0$

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

$ZR=R$ (1)

Here,

$R$ is the value of resistance.

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

$ZL=sL$ (2)

Here,

$L$ is the value of inductance.

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

$ZC=1sC$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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 $is(t)$ is calculated as follows:

$is(t)=3.5(0) {∵u(t)=0 for t<0}=0 A$

The voltage source $vs(t)$ is calculated as follows:

$vs(t)=7(0) {∵u(t)=0 for t<0}=0 V$

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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 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.

$iL(0−)=0 AvC(0−)=0 V$

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

$i(0)=iL(0−)=iL(0+)=0 A$

$v(0)=vC(0−)=vC(0+)=0 V$

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

Apply Laplace transform for $vs(t)$ to find $Vs(s)$ .

$Vs(s)=7s {∵ℒ(u(t))=1s}$

Apply Laplace transform for $is(t)$ to find $Is(s)$ .

$Is(s)=3.5s {∵ℒ(u(t))=1s}$

Use equation (1) to find $ZR1$ .

$ZR1=1$

Use equation (1) to find $ZR2$ .

$ZR2=1$

Substitute $1 H$ for $L$ in equation (2) to find $ZL$ .

$ZL=s(1 H)=s$

Substitute $0.5 F$ for $C$ in equation (3) to find $ZC$ .

$ZC=1s(0.5 F)=2s$

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 16, Problem 30P , additional homework tip 3

Apply nodal analysis at node $V1(s)$ in Figure 3.

$V1(s)−7s1+V1(s)s+V1(s)−Vo(s)1=0V1(s)−7s+V1(s)s+V1(s)1−Vo(s)1=0V1(s)+V1(s)s+V1(s)−Vo(s)=7s(2+1s)V1(s)−Vo(s)=7s$

$(2s+1s)V1(s)−Vo(s)=7s$ (4)

Apply nodal analysis at node $Vo(s)$ in Figure 3.

$Vo(s)(2s)+Vo(s)−V1(s)1−3.5s=0sVo(s)2+Vo(s)−V1(s)−3.5s=0(s2+1)Vo(s)−V1(s)=3.5s(s+22)Vo(s)−V1(s)=3.5s$

Simplify the above equation to find $V1(s)$ .

$V1(s)=(s+22)Vo(s)−3.5s$

Substitute $(s+22)Vo(s)−3.5s$ for $V1(s)$ in equation (4) to find $Vo(s)$ .

$(2s+1s)[(s+22)Vo(s)−3.5s]−Vo(s)=7s(2s+1s)(s+22)Vo(s)−(2s+1s)3.5s−Vo(s)=7s(2s2+4s+s+22s)Vo(s)−(7s+3.5s2)−Vo(s)=7s(2s2+5s+22s)Vo(s)−Vo(s)=7s+(7s+3.5s2)$

Simplify the above equation as follows:

$[(2s2+5s+22s)−1]Vo(s)=7s+7s+3.5s2(2s2+5s+2−2s2s)Vo(s)=14s+3.5s2(2s2+3s+22s)Vo(s)=14s+3.5s2(2(s2+1.5s+1)2s)Vo(s)=14s+3.5s2$

Simplify the above equation to find $Vo(s)$ .

$Vo(s)=(14s+3.5s2)(ss2+1.5s+1)$

$Vo(s)=14s+3.5s(s2+1.5s+1)$ (5)

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

$s2+1.5s+1=0$ (6)

Write a general expression to calculate the roots of quadratic equation $(as2+bs+c=0)$ .

$s1,2=−b±b2−4ac2a$ (7)

Comparing the equation (6) with the equation $(as2+bs+c=0)$ .

$a=1b=1.5c=1$

Substitute $1$ for $a$ , $1.5$ for $b$ , and $1$ for $c$ in equation (7) to find $s1,2$ .

$s1,2=−1.5±(1.5)2−4(1)(1)2(1)=−1.5±2.25−42(1)=−1.5±−1.752=−1.5±j1.32282$

Simplify the above equation to find $s1,2$ .

$s1,2=−1.5+j1.32282,−1.5−j1.32282=−0.75+j0.6614,−0.75−j0.6614$

Substitute the roots of characteristic equation in equation (5) to find $Vo(s)$ .

$Vo(s)=14s+3.5s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Take partial fraction for above equation.

$Vo(s)=14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[As+B(s−(−0.75+j0.6614))+C(s−(−0.75−j0.6614))]$ (8)

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

$14s+3.5[s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))]=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]s(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))$

Simplify the above equation as follows:

$14s+3.5=[A(s−(−0.75+j0.6614))(s−(−0.75−j0.6614))+Bs(s−(−0.75−j0.6614))+Cs(s−(−0.75+j0.6614))]$ (9)

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

$14(0)+3.5=[A(0−(−0.75+j0.6614))(0−(−0.75−j0.6614))+B(0)((0−(−0.75−j0.6614)))+C(0)(0−(−0.75+j0.6614))]3.5=A(0.75−j0.6614)(0.75+j0.6614)+0+03.5=A(0.752−(j0.6614)2) {∵a2−b2=(a+b)(a−b)}3.5=A(0.5625−j20.4374)$

Simplify the above equation to find $A$ .

$A=3.5(0.5625−j20.4374)=3.50.5625−(−1)0.4374 {∵j2=−1}=3.51=3.5$

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

$14(−0.75+j0.6614)+3.5=[A((−0.75+j0.6614)−(−0.75+j0.6614))((−0.75+j0.6614)−(−0.75−j0.6614))+B(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75−j0.6614))+C(−0.75+j0.6614)((−0.75+j0.6614)−(−0.75+j0.6614))]−10.5+j9.2596+3.5=A(0)+B(−0.75+j0.6614)(j1.3228)+C(0)−7+j9.2596=B(−0.875−j0.9921)$

Simplify the above equation to find $B$ .

$B=−7+j9.2596(−0.875−j0.9921)=8.775∠−101.5°$

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

$14(−0.75−j0.6614)+3.5=[A((−0.75−j0.6614)−(−0.75+j0.6614))((−0.75−j0.6614)−(−0.75−j0.6614))+B(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75−j0.6614))+C(−0.75−j0.6614)((−0.75−j0.6614)−(−0.75+j0.6614))]−10.5−j9.2596+3.5=A(0)+B(0)+C(−0.75−j0.6614)(−j1.3228)−7−j9.2596=C(−0.875+j0.9921)$

Simplify the above equation to find $C$ .

$C=−7−j9.2596(−0.875+j0.9921)=8.775∠101.5°$

Substitute $3.5$ for $A$ , $8.775∠−101.5°$ for $B$ , and $8.775∠101.5°$ for $C$ in equation (8) to find $Vo(s)$ .

$Vo(s)=[3.5s+8.775∠−101.5°(s−(−0.75+j0.6614))+8.775∠101.5°(s−(−0.75−j0.6614))]=[3.5s+8.775e−j101.5°(s−(−0.75+j0.6614))+8.775ej101.5°(s−(−0.75−j0.6614))] {∵r∠ϕ=rejϕ}$

Take inverse Laplace transform for above equation to find $vo(t)$ .

$vo(t)=[3.5u(t)+8.775e−j101.5°e(−0.75+j0.6614)tu(t)+8.775ej101.5°e(−0.75−j0.6614)tu(t)] {∵ℒ−1(1s)=u(t)ℒ−1(1s+a)=e−atu(t)}=[3.5+8.775e(−0.75t+j0.6614t−j101.5°)+8.775e(−0.75t−j0.6614t+j101.5°)]u(t) {∵ea⋅eb=ea+b}=[3.5+8.775e−0.75tej(0.6614t−101.5°)+8.775e−0.75te−j(0.6614t−101.5°)]u(t)=[3.5+8.775e−0.75t[ej(0.6614t−101.5°)+e−j(0.6614t−101.5°)]]u(t)$