Question

In Problems 59 solve equation (15) subject to $\mathbf{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$with E(t) as given. Use a graphing utility to graph the solution for$\mathbf{0}⩽\mathbf{t}⩽\mathbf{4}$in the case when$\mathbf{L}\mathbf{=}\mathbf{1}$and$\mathit{R}\mathbf{=}\mathbf{1}$

E(t) is the meander function in Problem 53 with amplitude 1and a=1.

Short answer

Laplace transform$i\left(t\right)=1-e^{-t}+2(-1{)}^n\sum _{n=1}^{\infty }\left(1-e^{-(t-n)}\right)U(t-n)$

Step-by-step solution

Define Laplace transform:

The conversion of the function f (x) to the function $\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\int \infty \mathbf{0}\mathbf{e}\mathbf{-}\mathbf{xtf}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}$is particularly useful in reducing the solution of a standard dividing line equation with constant coefficients in the polynomial equation solution.

Find Is:

$\text{L}\frac{\text{d}i}{\text{d}t}+Ri=E\left(t\right)$

Substitute $L=1$and $R=1$

$\frac{\text{d}i}{\text{d}t}+i=E\left(t\right)$

By taking Laplace transform on both side,

width="168">Ldidt+L{i}=L{E(t)}

$s\mathcal{L}\left\{i\right\}+\mathcal{L}\left\{i\right\}=\mathcal{L}\left\{E\right(t\left)\right\}$$\left(\mathcal{L}\left\{\frac{\text{d}i}{\text{d}t}\right\}=s\mathcal{L}\left\{i\right\}-i\left(0\right)i\left(0\right)=0\right)$

Consider $\mathcal{L}\left\{i\right\}=I\left(s\right)$replace in above equation

$sI\left(s\right)+I\left(s\right)=\mathcal{L}\left\{E\right(t\left)\right\}$ ----------------------------- (1)

From solution 53,

$\mathcal{L}\left\{E\right(t\left)\right\}=\frac{1-e^{-as}}{s\left(1+e^{-as}\right)}$

Consider$(a=1)$

$\mathcal{L}\left\{E\right(t\left)\right\}=\frac{1-e^{-s}}{s\left(1+e^{-s}\right)}$

Substitute this equation in equation (1)

localid="1664033013078" $$\begin{gathered} sI\left(s\right)+I\left(s\right)=\frac{1-e^{-s}}{s\left(1+e^{-s}\right)} \\ (s+1)I\left(s\right)=\frac{1-e^{-s}}{s\left(1+e^{-s}\right)} \end{gathered}$$

$I\left(s\right)=\frac{1-e^{-s}}{s(s+1)\left(1+e^{-s}\right)}------\left(2\right)$

Evaluate the equation using partial fraction technique:

$$\begin{gathered} \frac{1}{s(s+1)}=\frac{A}{s}+\frac{B}{s+1} \\ =\frac{A(s+1)+Bs}{s(s+1)} \end{gathered}$$-------------------(3)

Have same numerator and denominator

$1=As+A+Bs$

Consider$s=0$

Then

$$\begin{gathered} 1=0+A+0 \\ A=1 \end{gathered}$$

Consider $s=-1$

Then$$\begin{gathered} 1=-A+A-B \\ B=-1 \end{gathered}$$

Substitute the value in equation (3)

$\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}$

Find the Laplace transform:

$I\left(s\right)=\left(1-e^{-s}\right)\left[\frac{1}{s}-\frac{1}{s+1}\right]\frac{1}{\left(1+e^{-s}\right)}$

$$\begin{gathered} =\left(1-e^{-s}\right)\left[\frac{1}{s}-\frac{1}{s+1}\right]\left(1-e^{-s}+e^{-2s}-e^{-3s}+\cdots \right) \\ =\left[\frac{1}{s}-\frac{1}{s+1}\right]\left(1-2e^{-s}+2e^{-2s}-2e^{-3s}+\cdots \right) \end{gathered}$$

$$\begin{gathered} =\frac{1}{s}-\frac{2}{s}{e}^{-s}+\frac{2}{s}{e}^{-2s}-\frac{2}{s}{e}^{-3s}+\cdots \\ -\frac{1}{s+1}+\frac{2}{s+1}{e}^{-s}-\frac{2}{s+1}{e}^{-2s}+\frac{2}{s+1}{e}^{-3s}+\cdots \end{gathered}$$

Inverse Laplace transform of both side

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{I\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{2}{s}{e}^{-s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2}{s}{e}^{-2s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{2}{s}{e}^{-3s}\right\}+\cdots \\ -{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2}{s+1}{e}^{-s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{2}{s+1}{e}^{-2s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2}{s+1}{e}^{-3s}\right\}+\cdots \end{gathered}$$

$\left\{\begin{array}{l}If{\mathcal{L}}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)anda>0,thenf(t-a)U(t-a)={\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}\\ e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\},{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=1\\ U(t-a)={\mathcal{L}}^{-1}\left\{\frac{e^{-as}}{s}\right\}\end{array}\right.$

$$\begin{gathered} =1-2U(t-1)+2U(t-2)-2U(t-3)+\cdots \\ -e^{-t}+2e^{-(t-1)}U(t-1)-2e^{-(t-2)}U(t-2)+2e^{-(t-3)}U(t-3)-2e^{-(t-4)}+\cdots \end{gathered}$$

Therefore, Laplace transform is

$i\left(t\right)={\mathcal{L}}^{-1}\left\{I\right(s\left)\right\}=1-e^{-t}+2(-1{)}^n\sum _{n=1}^{\infty }\left(1-e^{-(t-n)}\right)U(t-n)$

$i\left(t\right)={\mathcal{L}}^{-1}\left\{I\right(s\left)\right\}=1-e^{-t}+2(-1{)}^n\sum _{n=1}^{\infty }\left(1-e^{-(t-n)}\right)U(t-n)$

$i\left(t\right)={\mathcal{L}}^{-1}\left\{I\right(s\left)\right\}=1-e^{-t}+2(-1{)}^n\sum _{n=1}^{\infty }\left(1-e^{-(t-n)}\right)U(t-n)$

Graph of it :