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$\mathit{L}\mathbf{=}\mathbf{1}$and$\mathit{R}\mathbf{=}\mathbf{1}$

E(t) is the sawtooth function in Problem 55 with amplitude 1

and b = 1.

Short answer

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

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,

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

$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 55,

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=\frac{a}{s}\left(\frac{1}{bs}-\frac{1}{e^{bs}-1}\right) \\ \mathcal{L}\left\{E\right(t\left)\right\}=\frac{1}{s}\left(\frac{1}{s}-\frac{1}{e^s-1}\right) \end{gathered}$$$(a=1,b=1)$

Substitute this equation in equation (1)

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

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

Evaluate the equation using partial fraction technique:

$\frac{1}{s^2(s+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+1}$

$=\frac{As(s+1)+B(s+1)+C{s}^2}{s^2(s+1)}$

Have same numerator and denominator

$$\begin{gathered} 1=A{s}^2+As+Bs+B+C{s}^2 \\ 1=(A+C)s^2+(A+B)s+B \end{gathered}$$

Compare coefficient on both sides

localid="1664030959794" $$\begin{gathered} A+C=0 \\ A+B=0 \\ \underline{B}=1 \end{gathered}$$

localid="1664030978655" $$\begin{gathered} C=-A \\ C=-B \\ \underline{B}=1 \end{gathered}$$

localid="1664031005760" $$\begin{gathered} C=1 \\ C=-1 \\ \underline{B}=1 \end{gathered}$$

Substitute the value in equation (3)

$\frac{1}{s^2(s+1)}=-\frac{1}{s}+\frac{1}{s^2}+\frac{1}{s+1}$----------------(4)

$\frac{1}{s(s+1)}=\frac{A}{s}+\frac{B}{s+1}$--------------(5)

$=\frac{A(s+1)+Bs}{s(s+1)}$

$1=As+A+Bs$

Consider$s=0$

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

Consider$s=-1$

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

Substitute the value in equation (5)

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

width="151">1s(s+1)=1s-1s+1

Find the Laplace transform

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

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

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

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

Inverse Laplace transform of both side

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

$+{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}{e}^{-s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}{e}^{-2s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}{e}^{-3s}\right\}+\cdots$

$\left\{\begin{array}{l}If{\mathcal{L}}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)anda>0,thenf(t-a)U(t-a)={\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}\\ e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\},{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=1,{\mathcal{L}}^{-1}\left\{\frac{1}{s^2}\right\}=t\\ U(t-a)={\mathcal{L}}^{-1}\left\{\frac{e^{-as}}{s}\right\}\end{array}\right.$

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

Therefore, Laplace transform is

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

Graph it :