Problem

Determine io(t) in the network shown in Fig. 16.70.Figure 16.70:

Determine io(t) in the network shown in Fig. 16.70.

Figure 16.70:

Step-by-Step Solution

Solution 1

Refer to circuit diagram in Figure 16.70 in the textbook.

It is known that, the step function is,

For , the value of voltage source is 20 V, inductor acts as short circuit and capacitor acts as open circuit.

Picture 4

Calculate the initial current through the inductor.

$$ \begin{aligned} i_{o}(0) &=\frac{20 \mathrm{~V}}{1 \Omega} \\ &=20 \mathrm{~A} \end{aligned} $$

Calculate the initial voltage across the capacitor.

$$ v(0)=0 \mathrm{~V} $$

Calculate the voltage source value for \(t>0\).

$$ \begin{aligned} V &=20+40 u(t) \\ &=20+40(1) \quad \text { (since } u(t)=1 \text { for } t>0) \\ &=60 \mathrm{~V} \end{aligned} $$

The frequency domain equivalent circuit for is as shown in Figure 2.

Picture 6

Apply Kirchhoff's current law at node \(V_{1}\).

$$ \begin{aligned} &\frac{V_{1}-\frac{60}{s}}{1}+\frac{V_{1}}{2 s}+\frac{20}{s}+\frac{V_{1}}{4+\frac{4}{s}}=0 \\ &\left(1+\frac{1}{2 s}+\frac{s}{4 s+4}\right) V_{1}=\frac{60}{s}-\frac{20}{s} \\ &\left(\frac{2 s(4 s+4)+4 s+4+2 s^{2}}{2 s(4 s+4)}\right) V_{1}=\frac{40}{s} \\ &\left(\frac{10 s^{2}+12 s+4}{2 s(4 s+4)}\right) V_{1}=\frac{40}{s} \\ &V_{1}=\frac{40(4 s+4)}{5 s^{2}+6 s+2} \ldots \ldots(1) \end{aligned} $$

Calcualte the current \(I_{o}(s)\)

$$ I_{o}(s)=\frac{V_{1}}{2 s}+\frac{20}{s} $$

Substiute \(\frac{40(4 s+4)}{5 s^{2}+6 s+2}\) for \(V_{1}\).

$$ \begin{aligned} I_{o}(s) &=\frac{\left(\frac{40(4 s+4)}{5 s^{2}+6 s+2}\right)}{2 s}+\frac{20}{s} \\ &=\frac{20(4 s+4)+20\left(5 s^{2}+6 s+2\right)}{s\left(5 s^{2}+6 s+2\right)} \\ &=\frac{20\left(5 s^{2}+10 s+6\right)}{s\left(5 s^{2}+6 s+2\right)} \end{aligned} $$

Apply partial transfom.

$$ \begin{aligned} &I_{o}(s)=\frac{A}{s}+\frac{B s+C}{5 s^{2}+6 s+2} \ldots \ldots(2) \\ &\frac{20\left(5 s^{2}+10 s+6\right)}{s\left(5 s^{2}+6 s+2\right)}=\frac{A\left(5 s^{2}+6 s+2\right)+B s^{2}+C s}{s\left(5 s^{2}+6 s+2\right)} \\ &100 s^{2}+200 s+120=A\left(5 s^{2}+6 s+2\right)+B s^{2}+C s \ldots \ldots (3) \end{aligned} $$

Equate constant terms from equation (3).

$$ \begin{aligned} &120=2 A \\ &A=60 \end{aligned} $$

Equate \(s\) terms from equation (3).

$$ \begin{aligned} &200=6 A+C \\ &200=(6)(60)+C \\ &C=-160 \end{aligned} $$

Equate \(s^{2}\) terms from equation (3).

$$ \begin{aligned} &100=5 A+B \\ &100=(5)(60)+B \\ &B=-200 \end{aligned} $$

Recall equation (2).

$$ I_{o}(s)=\frac{A}{s}+\frac{B s+C}{5 s^{2}+6 s+2} $$

Substiute 60 for \(s,-200\) for \(B\) and \(-160\) for \(C\).

$$ \begin{aligned} I_{o}(s) &=\frac{60}{s}+\frac{-200 s-160}{5 s^{2}+6 s+2} \\ &=\frac{60}{s}-\frac{200 s+160}{5\left(s^{2}+1.2 s+0.4\right)} \\ &=\frac{60}{s}-\frac{40 s+32}{s^{2}+1.2 s+0.36+0.04} \\ &=\frac{60}{s}-\frac{40(s+0.6)}{(s+0.6)^{2}+(0.2)^{2}}-\frac{40(0.2)}{(s+0.6)^{2}+(0.2)^{2}} \end{aligned} $$

Apply inverse Laplace transform.

$$ \begin{aligned} i_{o}(t) &=60 u(t)-40 e^{-0.6 t} \cos (0.2 t) u(t)-40 e^{-0.6 t} \sin (0.2 t) u(t) \mathrm{A} \\ &=\left[60-40 e^{-0.6 t}\{\cos (0.2 t)+\sin (0.2 t)\}\right] u(t) \mathrm{A} \end{aligned} $$

Therefore, the current \(i_{o}(t)\) is \(\left[60-40 e^{-0.6 t}\{\cos (0.2 t)+\sin (0.2 t)\}\right] u(t) \mathrm{A}\).

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