Chapter 8, Problem 8.4P

To determine

(a)

The steady-state response $x_{ss}$ and the corresponding time taken by the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98%for the model equation $2\dot{x}+x=10f\left(t\right)$.

Answer to Problem 8.4P

The steady-state response of the model equation $2\dot{x}+x=10f\left(t\right)$ is $x_{ss}=10$ and the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is $t=8\sec$.

Explanation of Solution

Given information:

The given model equation is as:

$2\dot{x}+x=10f\left(t\right)$

With initial conditions as follows:

$x(0)=0$ and $f\left(t\right)=u_s\left(t\right)$.

Concept Used:

1. Laplace transform is used for obtaining the response of the model equation.
2. The steady-state response of model is obtained by using the final value theorem:

$x_{ss}=\underset{t\to \infty }{\lim }x\left(t\right)=\underset{s\to 0}{\lim }sX\left(s\right)$.

3. For the response of the form $T\left(t\right)=A{e}^{-Bt}$, the time constant $\tau =\frac{1}{B}$ such that $B>0$

Calculation:

Equation to be solved is as:

$2\dot{x}+x=10f\left(t\right)$

By taking the Laplace of the above equation,

$\begin{array}{l}2\dot{x}+x=10f\left(t\right)\\ \Rightarrow 2\dot{x}+x=10u_s\left(t\right)\because f\left(t\right)=u_s\left(t\right)\\ \Rightarrow 2(sX(s)-x(0))+X(s)=\frac{10}{s}\\ \Rightarrow 2(sX(s)-0)+X(s)=\frac{10}{s}\because x(0)=0\\ \Rightarrow X(s).(2s+1)=\frac{10}{s}\\ \Rightarrow X(s)=\frac{5}{s(s+\frac{1}{2})}\end{array}$

Now, on taking the inverse Laplace of the above transfer function, we get

$\begin{array}{l}X(s)=\frac{5}{s(s+\frac{1}{2})}=10\left(\frac{1}{s}-\frac{1}{\left(s+\frac{1}{2}\right)}\right)\\ \Rightarrow x(t)=10\left(1-e^{-\frac{1}{2}t}\right)\end{array}$

Thus, this is the response for the given model equation. Now, for steady-stateusing final value theorem

$\begin{array}{l}{x}_{ss}=\underset{t\to \infty }{\lim }x\left(t\right)=\underset{s\to 0}{\lim }sX\left(s\right)\\ \Rightarrow x_{ss}=\underset{s\to 0}{\lim }s\frac{5}{s(s+\frac{1}{2})}=10\end{array}$

Therefore, steady-state value for the model equation is $x_{ss}=10$.

Also, from the obtained response $x(t)=10\left(1-e^{-\frac{1}{2}t}\right)$, the time constant of the model is

$\tau =2\text{seconds}$.

Thus, at $t=4\tau$

$e^{-\frac{1}{2}t}=e^{-4}<0.02$

Or say, at $t=4\tau$

$x\left(t\right)\approx 9.8$

Thus, the time taken by the response to reach 98% of its steady state value or the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is

$t=4\tau =4\times 2=8\text{seconds}$.

Conclusion:

The steady-state response of the model equation $2\dot{x}+x=10f\left(t\right)$ is $x_{ss}=10$ and the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is $t=8\text{seconds}$.

To determine

(b)

The steady-state response $x_{ss}$ and the corresponding time taken by the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% for the model equation $2\dot{x}+x=10f\left(t\right)$.

Answer to Problem 8.4P

The steady-state response of the model equation $2\dot{x}+x=10f\left(t\right)$ is $x_{ss}=10$ and the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is $t=7.824\text{seconds}$.

Explanation of Solution

Given information:

The given model equation is as:

$2\dot{x}+x=10f\left(t\right)$

With initial conditions as follows:

$x(0)=5$ and $f\left(t\right)=u_s\left(t\right)$.

Concept Used:

1. Laplace transform is used for obtaining the response of the model equation.
2. The steady-state response of model is obtained by using the final value theorem:

$x_{ss}=\underset{t\to \infty }{\lim }x\left(t\right)=\underset{s\to 0}{\lim }sX\left(s\right)$.

3. For the response of the form $T\left(t\right)=A{e}^{-Bt}$, the time constant $\tau =\frac{1}{B}$ such that $B>0$.

Calculation:

Equation to be solved is as:

$2\dot{x}+x=10f\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}2\dot{x}+x=10f\left(t\right)\\ \Rightarrow 2\dot{x}+x=10u_s\left(t\right)\because f\left(t\right)=u_s\left(t\right)\\ \Rightarrow 2(sX(s)-x(0))+X(s)=\frac{10}{s}\\ \Rightarrow 2(sX(s)-5)+X(s)=\frac{10}{s}\because x(0)=5\\ \Rightarrow X(s).(2s+1)=\frac{10\left(s+1\right)}{s}\\ \Rightarrow X(s)=\frac{10\left(s+1\right)}{s(2s+1)}\end{array}$

Now, on taking the inverse Laplace of the above transfer function, we get

$\begin{array}{l}X(s)=\frac{10\left(s+1\right)}{s(2s+1)}=10\left(\frac{1}{s}-\frac{1}{2\left(s+\frac{1}{2}\right)}\right)\\ \Rightarrow x(t)=10\left(1-\frac{1}{2}{e}^{-\frac{1}{2}t}\right)\end{array}$

Thus, this is the response for the given model equation. Now, for steady-state using final value theorem

$\begin{array}{l}{x}_{ss}=\underset{t\to \infty }{\lim }x\left(t\right)=\underset{s\to 0}{\lim }sX\left(s\right)\\ \Rightarrow x_{ss}=\underset{s\to 0}{\lim }s\frac{10\left(s+1\right)}{s(2s+1)}=10\end{array}$

Therefore, steady-state value for the model equation is $x_{ss}=10$.

The difference between $x\left(0\right)$ and $x_{ss}$ is:

$\left|x\left(0\right)-x_{ss}\right|=\left|5-10\right|=5$

$x\left(0\right)$

And the 98% of this difference is 4.9 units.

$x(t)$ Now, at the time difference between and reduces by 98% or becomes 0.1 units, the response reaches its value equal to 99% of steady-state value or 9.9 units.

Therefore,

$\begin{array}{l}x(t)=10\left(1-\frac{1}{2}{e}^{-\frac{1}{2}t}\right)\\ \Rightarrow 9.9=10\left(1-\frac{1}{2}{e}^{-\frac{1}{2}t}\right)\\ \Rightarrow \frac{1}{2}{e}^{-\frac{1}{2}t}=1-0.99\\ \Rightarrow \frac{1}{2}{e}^{-\frac{1}{2}t}=0.01\\ \Rightarrow e^{-\frac{1}{2}t}=0.02\end{array}$

Taking log on both sides, we get

$\begin{array}{l}\Rightarrow -\frac{1}{2}t=\ln 0.02\\ \Rightarrow -\frac{1}{2}t=-3.912\\ \Rightarrow t=7.824\text{seconds}\end{array}$

Thus, the time taken for the difference between $x_{ss}$ and $x\left(0\right)$ to get reduced by 98% is

$t=7.824\text{seconds}$.

Conclusion:

The steady-state response of the model equation $2\dot{x}+x=10f\left(t\right)$ is $x_{ss}=10$ and the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is $t=7.824\text{seconds}$.

To determine

(c)

The steady-state response $x_{ss}$ and the corresponding time taken by the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% for the model equation $2\dot{x}+x=10f\left(t\right)$.

Answer to Problem 8.4P

The steady-state response of the model equation $2\dot{x}+x=10f\left(t\right)$ is $x_{ss}=200$ and the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is $t=8\text{seconds}$.

Explanation of Solution

Given information:

The given model equation is as:

$2\dot{x}+x=10f\left(t\right)$

With initial conditions as follows:

$x(0)=0$ and $f\left(t\right)=20u_s\left(t\right)$.

Concept Used:

1. Laplace transform is used for obtaining the response of the model equation.
2. The steady-state response of model is obtained by using the final value theorem:

$x_{ss}=\underset{t\to \infty }{\lim }x\left(t\right)=\underset{s\to 0}{\lim }sX\left(s\right)$.

3. For the response of the form $T\left(t\right)=A{e}^{-Bt}$, the time constant $\tau =\frac{1}{B}$ such that $B>0$.

Calculation:

Equation to be solved is as:

$2\dot{x}+x=10f\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}2\dot{x}+x=10f\left(t\right)\\ \Rightarrow 2\dot{x}+x=200u_s\left(t\right)\because f\left(t\right)=u_s\left(t\right)\\ \Rightarrow 2(sX(s)-x(0))+X(s)=\frac{200}{s}\\ \Rightarrow 2(sX(s)-0)+X(s)=\frac{200}{s}\because x(0)=0\\ \Rightarrow X(s).(2s+1)=\frac{200}{s}\\ \Rightarrow X(s)=\frac{100}{s(s+\frac{1}{2})}\end{array}$

Now, on taking the inverse Laplace of the above transfer function, we get

$X(s)=\frac{100}{s(s+\frac{1}{2})}=200\left(\frac{1}{s}-\frac{1}{\left(s+\frac{1}{2}\right)}\right)$

$\Rightarrow x(t)=200\left(1-e^{-\frac{1}{2}t}\right)$

Thus, this is the response for the given model equation. Now, for steady-state using final value theorem

$\begin{array}{l}{x}_{ss}=\underset{t\to \infty }{\lim }x\left(t\right)=\underset{s\to 0}{\lim }sX\left(s\right)\\ \Rightarrow x_{ss}=\underset{s\to 0}{\lim }s\frac{100}{s(s+\frac{1}{2})}=200\end{array}$

Therefore, steady-state value for the model equation is.

$x(t)=200\left(1-e^{-\frac{1}{2}t}\right)$

Also, from the obtained response, the time constant of the model is

$\tau =2\text{seconds}$.

Thus, at $t=4\tau$

$e^{-\frac{1}{2}t}=e^{-4}<0.02$

Or say, at $t=4\tau$

$x\left(t\right)\approx 196$

Thus, the time taken by the response to reach 98% of its steady state value or the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is

$t=4\tau =4\times 2=8\text{seconds}$.

Conclusion:

The steady-state response of the model equation $2\dot{x}+x=10f\left(t\right)$ is $x_{ss}=200$ and the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ to get reduced by 98% is $t=8\text{seconds}$.

Thus, as in part a. and in c, we see that the time taken for the difference between $x\left(0\right)$ and $x_{ss}$ remains unchanged regardless of the changes in input response $f\left(t\right)$. However, the steady-state response gets affected with a change in input response $f\left(t\right)$.