Chapter 2, Problem 2.38P

To determine

(a)

The response $x\left(t\right)$ for the model equation $7\dot{x}+5x=4\dot{g}\left(t\right)$.Also, compare the values of $x\left(0^{-}\right)andx\left(0^{+}\right)$.

Answer to Problem 2.38P

The response $x\left(t\right)$ for the model equations is as:

$x\left(t\right)=\frac{25}{7}\cdot e^{-\frac{5}{7}t}$.

The response $x\left(t\right)$ changes from $t=0^{-}tot=0^{+}$ such that

$x\left(0^{-}\right)=3andx\left(0^{+}\right)=\frac{25}{7}$.

Explanation of Solution

Given:

The given model equation is as:

$7\dot{x}+5x=4\dot{g}\left(t\right)$

Where, $g\left(t\right)=u_s\left(t\right)$ and the initial conditions are

$x\left(0^{-}\right)=3andg\left(0^{-}\right)=0$.

Concept Used:

First, evaluate the response $x\left(t\right)$ from the given model equation using the Laplace transform. Then, find the right-hand limit of $x\left(t\right)$ at $t=0$.

Calculation:

The model equation to be solved is as:

$7\dot{x}+5x=4\dot{g}\left(t\right)$

Where, $g\left(t\right)=u_s\left(t\right)$

Therefore, $\dot{g}\left(t\right)=\delta \left(t\right)$

And $7\dot{x}+5x=4\delta \left(t\right)$

On taking Laplace transform, that is

$\begin{array}{l}7\dot{x}+5x=4\delta \left(t\right)\\ \Rightarrow 7\left(sX\left(s\right)-x\left(0^{-}\right)\right)+5X\left(s\right)=4\\ \Rightarrow 7\left(sX\left(s\right)-3\right)+5X\left(s\right)=4\\ \Rightarrow X\left(s\right)=\frac{25}{\left(7s+5\right)}\end{array}$

On taking the inverse Laplace of this,

$x\left(t\right)=\frac{25}{7}\cdot e^{-\frac{5}{7}t}$

Therefore,

$x\left(0^{+}\right)=\underset{t\to 0^{+}}{\lim }\frac{25}{7}\cdot e^{-\frac{5}{7}t}=\frac{25}{7}$

Thus the response $x\left(t\right)$ changes from 3 to $\frac{25}{7}$ as time varies from $t=0^{-}tot=0^{+}$.

Conclusion:

The obtained response $x\left(t\right)$ is as:

$x\left(t\right)=\frac{25}{7}\cdot e^{-\frac{5}{7}t}$

And $x\left(t\right)$ changes to $\frac{25}{7}$ from 3 as time varies from $t=0^{-}tot=0^{+}$.

To determine

(b)

The response $x\left(t\right)$ for the model equation $7\dot{x}+5x=4\dot{g}\left(t\right)+6g\left(t\right)$. Compare the values of $x\left(0^{-}\right)andx\left(0^{+}\right)$.

Answer to Problem 2.38P

The response $x\left(t\right)$ for the model equation is as:

$x\left(t\right)=\frac{6}{5}+\frac{83}{35}\cdot e^{-\frac{5}{7}t}$.

The response $x\left(t\right)$ changes from $t=0^{-}tot=0^{+}$ such that

$x\left(0^{-}\right)=3andx\left(0^{+}\right)=\frac{25}{7}$.

Explanation of Solution

Given:

The given model equation is as:

$7\dot{x}+5x=4\dot{g}\left(t\right)+6g\left(t\right)$

Where, $g\left(t\right)=u_s\left(t\right)$ and the initial conditions are

$x\left(0^{-}\right)=3andg\left(0^{-}\right)=0$.

Concept Used:

First, evaluate the response $x\left(t\right)$ from the given model equation using the Laplace transform. Then, find the right-hand limit of $x\left(t\right)$ at $t=0$.

Calculation:

The model equation to be solved is as:

$7\dot{x}+5x=4\dot{g}\left(t\right)+6g\left(t\right)$

Where, $g\left(t\right)=u_s\left(t\right)$

Therefore, $\dot{g}\left(t\right)=\delta \left(t\right)$

And $7\dot{x}+5x=4\delta \left(t\right)+6u_s\left(t\right)$

On taking Laplace transform, that is

$\begin{array}{l}7\dot{x}+5x=4\delta \left(t\right)+6u_s\left(t\right)\\ \Rightarrow 7\left(sX\left(s\right)-x\left(0^{-}\right)\right)+5X\left(s\right)=4+\frac{6}{s}\\ \Rightarrow 7\left(sX\left(s\right)-3\right)+5X\left(s\right)=4+\frac{6}{s}\\ \Rightarrow X\left(s\right)\cdot \left(7s+5\right)=25+\frac{6}{s}\\ \Rightarrow X\left(s\right)=\frac{83}{35}\cdot \frac{1}{\left(s+\frac{5}{7}\right)}+\frac{6}{5}\cdot \frac{1}{s}\end{array}$

On taking the inverse Laplace of this,

$x\left(t\right)=\frac{6}{5}+\frac{83}{35}\cdot e^{-\frac{5}{7}t}$

Therefore,

$x\left(0^{+}\right)=\underset{t\to 0^{+}}{\lim }\left(\frac{6}{5}+\frac{83}{35}\cdot e^{-\frac{5}{7}t}\right)=\left(\frac{6}{5}+\frac{83}{35}\right)=\frac{25}{7}$

Thus the response $x\left(t\right)$ changes from 3 to $\frac{25}{7}$ as time varies from $t=0^{-}tot=0^{+}$.

Conclusion:

The obtained response $x\left(t\right)$ is as:

$x\left(t\right)=\frac{6}{5}+\frac{83}{35}\cdot e^{-\frac{5}{7}t}$

And $x\left(t\right)$ changes to $\frac{25}{7}$ from 3 as time varies from $t=0^{-}tot=0^{+}$.

To determine

(c)

The response $x\left(t\right)$ for the model equation $3\ddot{x}+30\dot{x}+63x=4\dot{g}\left(t\right)$. Compare the values of $x\left(0^{-}\right)andx\left(0^{+}\right)$.Also, do the same for $\dot{x}\left(0^{-}\right)and\dot{x}\left(0^{+}\right)$.

Answer to Problem 2.38P

The response $x\left(t\right)$ for the model equation is as:

$x\left(t\right)=\frac{55}{12}\cdot e^{-3t}-\frac{31}{12}\cdot e^{-7t}$.

The response $x\left(t\right)$ doesn’t change from $t=0^{-}tot=0^{+}$ such that

$x\left(0^{-}\right)=x\left(0^{+}\right)=2$

While the response $\dot{x}\left(t\right)$ changes from $t=0^{-}tot=0^{+}$ such that

$\dot{x}\left(0^{-}\right)=3and\dot{x}\left(0^{+}\right)=\frac{13}{3}$.

Explanation of Solution

Given:

The given model equations are as:

$3\ddot{x}+30\dot{x}+63x=4\dot{g}\left(t\right)$

Where, $g\left(t\right)=u_s\left(t\right)$ and the initial conditions are

$x\left(0^{-}\right)=2,\dot{x}\left(0^{-}\right)=3andg\left(0^{-}\right)=0$.

Concept Used:

First, evaluate the response $x\left(t\right)$ from the given model equation using the Laplace transform. Then, find the right-hand limit of $x\left(t\right)$ and $\dot{x}\left(t\right)$ at $t=0$.

Calculation:

The model equation to be solved is as:

$3\ddot{x}+30\dot{x}+63x=4\dot{g}\left(t\right)$

On taking Laplace transform, that is

$\begin{array}{l}3\ddot{x}+30\dot{x}+63x=4\delta \left(t\right)\\ \Rightarrow 3\left(s^{2}X\left(s\right)-sx\left(0^{-}\right)-\dot{x}\left(0^{-}\right)\right)+30\left(sX\left(s\right)-x\left(0^{-}\right)\right)+63X\left(s\right)=4\\ \Rightarrow 3\left(s^{2}X\left(s\right)-2s-3\right)+30\left(sX\left(s\right)-2\right)+63X\left(s\right)=4\\ \Rightarrow X\left(s\right)\cdot \left(3s^2+30s+63\right)=6s+73\\ \Rightarrow X\left(s\right)=\frac{\left(6s+73\right)}{\left(3s^2+30s+63\right)}=\frac{55}{12}\cdot \frac{1}{\left(s+3\right)}-\frac{31}{12}\cdot \frac{1}{\left(s+7\right)}\end{array}$

On taking the inverse Laplace of this,

$x\left(t\right)=\frac{55}{12}\cdot e^{-3t}-\frac{31}{12}\cdot e^{-7t}$

On taking the derivative,

$\dot{x}\left(t\right)=-\frac{55}{4}\cdot e^{-3t}+\frac{217}{12}\cdot e^{-7t}$

Therefore,

$x\left(0^{+}\right)=\underset{t\to 0^{+}}{\lim }\left(\frac{55}{12}\cdot e^{-3t}-\frac{31}{12}\cdot e^{-7t}\right)=\left(\frac{55}{12}-\frac{31}{12}\right)=\frac{24}{12}=2$

Similarly,

$\dot{x}\left(0^{+}\right)=\underset{t\to 0^{+}}{\lim }\left(-\frac{55}{4}\cdot e^{-3t}+\frac{217}{12}\cdot e^{-7t}\right)=\left(-\frac{55}{4}+\frac{217}{12}\right)=\frac{52}{12}=\frac{13}{3}$

Thus the response $x\left(t\right)$ doesn’t change as time varies from $t=0^{-}tot=0^{+}$.

Whereas, the response $\dot{x}\left(t\right)$ changes from 3 to $\frac{13}{3}$ as time varies from $t=0^{-}tot=0^{+}$.

Conclusion:

The obtained response $x\left(t\right)$ is as:

$x\left(t\right)=\frac{55}{12}\cdot e^{-3t}-\frac{31}{12}\cdot e^{-7t}$

And $x\left(t\right)$ remains unchanged as time varies from $t=0^{-}tot=0^{+}$.

While $\dot{x}\left(t\right)$ attains a value of $\frac{13}{3}$ from 3 as time increases from $t=0^{-}tot=0^{+}$.

To determine

(d)

The response $x\left(t\right)$ for the model equation $3\ddot{x}+30\dot{x}+63x=4\dot{g}\left(t\right)+6g\left(t\right)$. Compare the values of $x\left(0^{-}\right)andx\left(0^{+}\right)$.Also, do the same for $\dot{x}\left(0^{-}\right)and\dot{x}\left(0^{+}\right)$.

Answer to Problem 2.38P

The response $x\left(t\right)$ for the model equation is as:

$x\left(t\right)=\frac{2}{21}-\frac{211}{84}\cdot e^{-7t}+\frac{53}{12}\cdot e^{-3t}$.

The response $\dot{x}\left(t\right)$ decreases from $t=0^{-}tot=0^{+}$ such that

$x\left(0^{-}\right)=4andx\left(0^{+}\right)=2$

While the response $\dot{x}\left(t\right)$ also decreases from $t=0^{-}tot=0^{+}$ such that

$\dot{x}\left(0^{-}\right)=7and\dot{x}\left(0^{+}\right)=\frac{13}{3}$.

Explanation of Solution

Given:

The given model equation is as:

$3\ddot{x}+30\dot{x}+63x=4\dot{g}\left(t\right)+6g\left(t\right)$

Where, $g\left(t\right)=u_s\left(t\right)$ and the initial conditions are

$x\left(0^{-}\right)=4,\dot{x}\left(0^{-}\right)=7andg\left(0^{-}\right)=0$.

Concept Used:

First, evaluate the response $x\left(t\right)$ from the given model equation using the Laplace transform. Then, find the right-hand limit of $x\left(t\right)$ and $\dot{x}\left(t\right)$ at $t=0$.

Calculation:

The model equation to be solved is as:

$3\ddot{x}+30\dot{x}+63x=4\dot{g}\left(t\right)+6g\left(t\right)$

On taking Laplace transform, that is

$\begin{array}{l}3\ddot{x}+30\dot{x}+63x=4\delta \left(t\right)+6u_s\left(t\right)\\ \Rightarrow 3\left(s^{2}X\left(s\right)-sx\left(0^{-}\right)-\dot{x}\left(0^{-}\right)\right)+30\left(sX\left(s\right)-x\left(0^{-}\right)\right)+63X\left(s\right)=4+\frac{6}{s}\\ \Rightarrow 3\left(s^{2}X\left(s\right)-2s-3\right)+30\left(sX\left(s\right)-2\right)+63X\left(s\right)=4+\frac{6}{s}\\ \Rightarrow X\left(s\right)\cdot \left(3s^2+30s+63\right)=\frac{6s^2+73s+6}{s}\\ \Rightarrow X\left(s\right)=\frac{\left(6s^2+73s+6\right)}{s\left(3s^2+30s+63\right)}=\frac{\left(6s^2+73s+6\right)}{3s\left(s+3\right)\left(s+7\right)}\\ \Rightarrow X\left(s\right)=\frac{2}{21}\cdot \frac{1}{s}-\frac{211}{84}\cdot \frac{1}{\left(s+7\right)}+\frac{53}{12}\cdot \frac{1}{\left(s+3\right)}\end{array}$

On taking the inverse Laplace of this,

$x\left(t\right)=\frac{2}{21}-\frac{211}{84}\cdot e^{-7t}+\frac{53}{12}\cdot e^{-3t}$

On taking the derivative,

$\dot{x}\left(t\right)=\frac{211}{12}\cdot e^{-7t}-\frac{53}{4}\cdot e^{-3t}$

Therefore,

$x\left(0^{+}\right)=\underset{t\to 0^{+}}{\lim }\left(\frac{2}{21}-\frac{211}{84}\cdot e^{-7t}+\frac{53}{12}\cdot e^{-3t}\right)=\left(\frac{2}{21}-\frac{211}{84}+\frac{53}{12}\right)=\frac{168}{24}=2$

Similarly,

$\dot{x}\left(0^{+}\right)=\underset{t\to 0^{+}}{\lim }\left(\frac{211}{12}\cdot e^{-7t}-\frac{53}{4}\cdot e^{-3t}\right)=\left(\frac{211}{12}-\frac{53}{4}\right)=\frac{52}{12}=\frac{13}{3}$

Thus the response $x\left(t\right)$ decreases from 4 to2 as time varies from $t=0^{-}tot=0^{+}$.

In addition, the response $\dot{x}\left(t\right)$ also decreases from 7 to $\frac{13}{3}$ as time varies from $t=0^{-}tot=0^{+}$.

Conclusion:

The obtained response $x\left(t\right)$ is as:

$x\left(t\right)=\frac{2}{21}-\frac{211}{84}\cdot e^{-7t}+\frac{53}{12}\cdot e^{-3t}$

And $x\left(t\right)$ decreases from 4 to2 as time varies from $t=0^{-}tot=0^{+}$.

While $\dot{x}\left(t\right)$ attains a value of $\frac{13}{3}$ from 7 as time increases from $t=0^{-}tot=0^{+}$.