Chapter 2, Problem 2.11P

To determine

(a)

The inverse Laplace transform $x(t)$ for $X(s)=\frac{6}{s(s+4)}$.

Answer to Problem 2.11P

The inverse Laplace transform $x(t)$ of $X(s)$ $=$ $\frac{3}{2}-\frac{3}{2}{e}^{-4t}$.

Explanation of Solution

Given:

The signal in frequency domain is given as shown below.

$X(s)=\frac{6}{s(s+4)}$

Concept Used:

The denominator of $X(s)$ contains two distinct real roots thus it can be simplified using the partial fraction expansion as follows:

$X(s)=\frac{6}{s(s+4)}=\frac{A}{s}+\frac{B}{(s+4)}$

And these constants $A$ and $B$ are computed as shown below:

$A=s\frac{6}{s(s+4)}$ at $s=0$

$B=(s+4)\frac{6}{s(s+4)}$ at $s=-4$

These values are then substituted in their places and the inverse Laplace of $X(s)$ is evaluated.

Calculation:

Following the above procedure, the constants $A$ and $B$ are calculated.

At $s=0$,

$A=s\frac{6}{s(s+4)}=\frac{6}{(0+4)}=\frac{3}{2}$

At $s=-4$,

$B=(s+4)\frac{6}{s(s+4)}=\frac{6}{-4}=-\frac{3}{2}$

Now, the $X(s)$ can be expressed as follows:

$X(s)=\frac{3}{2}\cdot \frac{1}{s}-\frac{3}{2}\cdot \frac{1}{(s+4)}$

On taking the inverse Laplace of $X(s)$, $x(t)$ is obtained as shown:

$x(t)=\frac{3}{2}-\frac{3}{2}{e}^{-4t}$.

Conclusion:

The inverse Laplace transform $x(t)=\frac{3}{2}-\frac{3}{2}{e}^{-4t}$.

To determine

(b)

The inverse Laplace transform $x(t)$ for $X(s)=\frac{12s+5}{s(s+3)}$.

Answer to Problem 2.11P

The inverse Laplace transform $x(t)$ of $X(s)$

$=\frac{5}{3}+\frac{31}{3}{e}^{-3t}$.

Explanation of Solution

Given:

The signal in frequency domain is given as shown below.

$X(s)=\frac{12s+5}{s(s+3)}$

Concept Used:

The denominator of $X(s)$ contains two distinct real roots thus it can be simplified using the partial fraction expansion as follows:

$X(s)=\frac{12s+5}{s(s+3)}=\frac{A}{s}+\frac{B}{(s+3)}$

And these constants $A$ and $B$ are computed as shown below:

$A=s\frac{(12s+5)}{s(s+3)}$ at $s=0$

$B=(s+3)\frac{(12s+5)}{s(s+3)}$ at $s=-3$

These values are then substituted in their places and the inverse Laplace of $X(s)$ is evaluated.

Calculation:

Following the above procedure, the constants $A$ and $B$ are calculated.

At $s=0$,

$A=s\frac{(12s+5)}{s(s+3)}=\frac{5}{3}$

At $s=-3$,

$B=(s+3)\frac{(12s+5)}{s(s+3)}=\frac{(-36+5)}{-3}=\frac{31}{3}$

Now, the $X(s)$ can be expressed as follows:

$X(s)=\frac{5}{3}\cdot \frac{1}{s}+\frac{31}{3}\cdot \frac{1}{(s+3)}$

On taking the inverse Laplace of $X(s)$, $x(t)$ is obtained as shown:

$x(t)=\frac{5}{3}+\frac{31}{3}{e}^{-3t}$.

Conclusion:

The inverse Laplace transform $x(t)=\frac{5}{3}+\frac{31}{3}{e}^{-3t}$.

To determine

(c)

The inverse Laplace transform $x(t)$ for $X(s)=\frac{4s+7}{(s+2)(s+5)}$.

Answer to Problem 2.11P

The inverse Laplace transform $x(t)$ of $X(s)$

$=-\frac{1}{3}{e}^{-2t}+\frac{13}{3}{e}^{-5t}$.

Explanation of Solution

Given:

The signal in frequency domain is given as shown below.

$X(s)=\frac{4s+7}{(s+2)(s+5)}$

Concept Used:

The denominator of $X(s)$ contains two distinct real roots thus it can be simplified using the partial fraction expansion as follows:

$X(s)=\frac{4s+7}{(s+2)(s+5)}=\frac{A}{(s+2)}+\frac{B}{(s+5)}$

And these constants $A$ and $B$ are computed as shown below:

$A=(s+2)\frac{(4s+7)}{(s+2)(s+5)}$ at $s=-2$

$B=(s+5)\frac{(4s+7)}{(s+2)(s+5)}$ at $s=-5$

These values are then substituted in their places and the inverse Laplace of $X(s)$ is evaluated.

Calculation:

Following the above procedure, the constants $A$ and $B$ are calculated.

At $s=-2$,

$A=(s+2)\frac{(4s+7)}{(s+2)(s+5)}=\frac{(-8+7)}{(-2+5)}=-\frac{1}{3}$

At $s=-5$,

$B=(s+5)\frac{(4s+7)}{(s+2)(s+5)}=\frac{(-20+7)}{(-5+2)}=\frac{13}{3}$

Now, the $X(s)$ can be expressed as follows:

$X(s)=-\frac{1}{3}\cdot \frac{1}{(s+2)}+\frac{13}{3}\cdot \frac{1}{(s+5)}$

On taking the inverse Laplace of $X(s)$, $x(t)$ is obtained as shown:

$x(t)=-\frac{1}{3}{e}^{-2t}+\frac{13}{3}{e}^{-5t}$.

Conclusion:

The inverse Laplace transform $x(t)=-\frac{1}{3}{e}^{-2t}+\frac{13}{3}{e}^{-5t}$.

To determine

(d)

The inverse Laplace transform $x(t)$ for $X(s)=\frac{5}{s^2(2s+8)}$.

Answer to Problem 2.11P

The inverse Laplace transform $x(t)$ of $X(s)$

$=\frac{5}{8}t-\frac{5}{32}+\frac{5}{32}{e}^{-4t}$.

Explanation of Solution

Given:

The signal in frequency domain is given as shown below.

$X(s)=\frac{5}{s^2(2s+8)}$

Concept Used:

The denominator of $X(s)$ contains three roots out of which two are repeated thus it can be simplified using the partial fraction expansion as follows:

$X(s)=\frac{5}{s^2(2s+8)}=\frac{A}{s^2}+\frac{B}{s}+\frac{C}{(2s+8)}$

And these constants $A$, $B$ and $C$ are computed as shown below:

$A=s^2\frac{5}{s^2(2s+8)}$ at $s=0$

$C=(2s+8)\frac{5}{s^2(2s+8)}$ at $s=-4$

Now, in order to determine the value of constant $B$ the lowest common denominator method is employed as follows:

$X(s)=\frac{5}{s^2(2s+8)}=\frac{A}{s^2}+\frac{B}{s}+\frac{C}{(2s+8)}$

$\frac{5}{s^2(2s+8)}=\frac{A(2s+8)+Bs(2s+8)+C{s}^2}{s^2(2s+8)}$

On comparing the numerators on both sides that is

$5=(2B+C)s^2+(2A+8B)s+8A$

Here, on further comparing above equation, the coefficients of $s^2$ and $s$ are found to be zero. Therefore,

$B=-\frac{C}{2}$ and $B=-\frac{A}{4}$

These values are then substituted in their places and the inverse Laplace of $X(s)$ is evaluated.

Calculation:

Following the above procedure, the constants $A$ and $B$ are calculated.

At $s=0$,

$A=s^2\frac{5}{s^2(2s+8)}=\frac{5}{(0+8)}=\frac{5}{8}$

At $s=-4$,

$C=(2s+8)\frac{5}{s^2(2s+8)}=\frac{5}{{(-4)}^2}=\frac{5}{16}$

Thus,

$B=-\frac{C}{2}=-\frac{(\frac{5}{16})}{2}=-\frac{5}{32}$

Now, the $X(s)$ can be expressed as follows:

$\begin{array}{l}X(s)=\frac{5}{8}\cdot \frac{1}{s^2}-\frac{5}{32}\cdot \frac{1}{s}+\frac{5}{16}\cdot \frac{1}{(2s+8)}\\ \Rightarrow X(s)=\frac{5}{8}\cdot \frac{1}{s^2}-\frac{5}{32}\cdot \frac{1}{s}+\frac{5}{32}\cdot \frac{1}{(s+4)}\end{array}$

On taking the inverse Laplace of $X(s)$, $x(t)$ is obtained as shown:

$x(t)=\frac{5}{8}t-\frac{5}{32}+\frac{5}{32}{e}^{-4t}$.

Conclusion:

The inverse Laplace transform $x(t)=\frac{5}{8}t-\frac{5}{32}+\frac{5}{32}{e}^{-4t}$.

To determine

(e)

The inverse Laplace transform $x(t)$ for $X(s)=\frac{(3s+2)}{s^2(s+5)}$.

Answer to Problem 2.11P

The inverse Laplace transform $x(t)$ of $X(s)$

$=\frac{2}{5}t+\frac{13}{25}-\frac{13}{25}{e}^{-5t}$.

Explanation of Solution

Given:

The signal in frequency domain is given as shown below.

$X(s)=\frac{(3s+2)}{s^2(s+5)}$

Concept Used:

The denominator of $X(s)$ contains three roots out of which two are repeated thus it can be simplified using the partial fraction expansion as follows:

$X(s)=\frac{(3s+2)}{s^2(s+5)}=\frac{A}{s^2}+\frac{B}{s}+\frac{C}{(s+5)}$

And these constants $A$, $B$ and $C$ are computed as shown below:

$A=s^2\frac{(3s+2)}{s^2(s+5)}$ at $s=0$

$C=(s+5)\frac{(3s+2)}{s^2(s+5)}$ at $s=-5$

Now, in order to determine the value of constant $B$ the lowest common denominator method is employed as follows:

$X(s)=\frac{(3s+2)}{s^2(s+5)}=\frac{A}{s^2}+\frac{B}{s}+\frac{C}{(s+5)}$

$\frac{(3s+2)}{s^2(s+5)}=\frac{A(s+5)+Bs(s+5)+C{s}^2}{s^2(s+5)}$

On comparing the numerators on both sides that is

$3s+2=(B+C)s^2+(A+5B)s+5A$

Here, on further comparing above equation, the coefficients of $s^2$ is found to be zero. Therefore,

$B=-C$ and $B=\frac{3-A}{5}$

These values are then substituted in their places and the inverse Laplace of $X(s)$ is evaluated.

Calculation:

Following the above procedure, the constants $A$ and $B$ are calculated.

At $s=0$

$A=s^2\frac{(3s+2)}{s^2(s+5)}=\frac{(0+2)}{(0+5)}=\frac{2}{5}$

At $s=-5$

$C=(s+5)\frac{(3s+2)}{s^2(s+5)}=\frac{(-15+2)}{{(-5)}^2}=-\frac{13}{25}$

Thus,

$B=-C=-\left(-\frac{13}{25}\right)=\frac{13}{25}$

Now, the $X(s)$ can be expressed as follows:

$X(s)=\frac{2}{5}\cdot \frac{1}{s^2}+\frac{13}{25}\cdot \frac{1}{s}-\frac{13}{25}\cdot \frac{1}{(s+5)}$

On taking the inverse Laplace of $X(s)$, $x(t)$ is obtained as shown:

$x(t)=\frac{2}{5}t+\frac{13}{25}-\frac{13}{25}{e}^{-5t}$.

Conclusion:

The inverse Laplace transform $x(t)=\frac{2}{5}t+\frac{13}{25}-\frac{13}{25}{e}^{-5t}$.

To determine

(f)

The inverse Laplace transform $x(t)$ of $X(s)=\frac{(12s+5)}{{(s+3)}^2(s+7)}$.

Answer to Problem 2.11P

The inverse Laplace transform

$$

x(t) of $=-\frac{31}{4}t{e}^{-3t}+\frac{79}{16}{e}^{-3t}-\frac{79}{16}{e}^{-7t}$.

Explanation of Solution

Given:

The signal in frequency domain is given as shown below.

$X(s)=\frac{(12s+5)}{{(s+3)}^2(s+7)}$

Concept Used:

The denominator of $X(s)$ contains three roots out of which two are repeated thus it can be simplified using the partial fraction expansion as follows:

$X(s)=\frac{(12s+5)}{{(s+3)}^2(s+7)}=\frac{A}{{(s+3)}^2}+\frac{B}{(s+3)}+\frac{C}{(s+7)}$

And these constants $A$, $B$ and $C$ are computed as shown below:

$A={(s+3)}^2\frac{(12s+5)}{{(s+3)}^2(s+7)}$ at $s=-3$

$C=(s+7)\frac{(12s+5)}{{(s+3)}^2(s+7)}$ at $s=-7$

Now, in order to determine the value of constant $B$ the lowest common denominator method is employed as follows:

$X(s)=\frac{(12s+5)}{{(s+3)}^2(s+7)}=\frac{A}{{(s+3)}^2}+\frac{B}{(s+3)}+\frac{C}{(s+7)}$

$\frac{(12s+5)}{{(s+3)}^2(s+7)}=\frac{A(s+7)+B(s+3)(s+7)+C{(s+3)}^2}{{(s+3)}^2(s+7)}$

On comparing the numerators on both sides that is

$(12s+5)=(B+C)s^2+(A+10B+6C)s+(7A+21B+9C)$

Here, on further comparing above equation, the coefficients of $s^2$ is found to be zero. Therefore,

$B=-C$ and $B=\frac{5-9C-7A}{21}$

These values are then substituted in their places and the inverse Laplace of $X(s)$ is evaluated.

Calculation:

Following the above procedure, the constants $A$ and $B$ are calculated.

At $s=-3$

$A={(s+3)}^2\frac{(12s+5)}{{(s+3)}^2(s+7)}=\frac{(-36+5)}{(-3+7)}=-\frac{31}{4}$

At $s=-7$

$C=(s+7)\frac{(12s+5)}{{(s+3)}^2(s+7)}=\frac{(-84+5)}{{(-4)}^2}=-\frac{79}{16}$

Thus,

$B=-C=-\left(-\frac{79}{16}\right)=\frac{79}{16}$

Now, the $X(s)$ can be expressed as follows:

$X(s)=-\frac{31}{4}\cdot \frac{1}{{(s+3)}^2}+\frac{79}{16}\cdot \frac{1}{(s+3)}-\frac{79}{16}\cdot \frac{1}{(s+7)}$

On taking the inverse Laplace of $X(s)$, $x(t)$ is obtained as shown:

$x(t)=-\frac{31}{4}t{e}^{-3t}+\frac{79}{16}{e}^{-3t}-\frac{79}{16}{e}^{-7t}$.

Conclusion:

The inverse Laplace transform $x(t)=-\frac{31}{4}t{e}^{-3t}+\frac{79}{16}{e}^{-3t}-\frac{79}{16}{e}^{-7t}$.