Chapter 9, Problem 9.8P

To determine

(a)

Steady state response $y_{ss}(t)$ for the transfer function for the given input.

Answer to Problem 9.8P

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0.053$.

Explanation of Solution

Given:

$T(s)=\frac{8}{s(s^2+10s+100)}$, $f(t)=6\sin 9t$.

Concept Used:

Using the final value theorem

Calculation:

$\begin{array}{l}f(t)=6\sin 9t\\ F(s)=\frac{6\times 9}{s^2+9^2}\end{array}$

$T(s)=\frac{8}{s(s^2+10s+100)}$

According to final value theorem steady state response of the system is

$\begin{array}{l}{y}_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.\frac{8}{s(s^2+10s+100)}.\frac{6\times 9}{s^2+9^2}\end{array}$

On Solving and putting the limit $s\to 0$

We get,

$y_{ss}(t)=0.053$.

Conclusion:

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0.053$.

To determine

(b)

Steady state response $y_{ss}(t)$ for the transfer function for the given input.

Answer to Problem 9.8P

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

Explanation of Solution

Given:

$T(s)=\frac{10}{s^2(s+1)}$, $f(t)=9\sin 2t$.

Concept Used:

Using the final value theorem.

Calculation:

$\begin{array}{l}f(t)=9\sin 2t\\ F(s)=\frac{9\times 2}{s^2+2^2}\end{array}$

$T(s)=\frac{10}{s^2(s+1)}$

According to final value theorem, steady state response of the system is

$\begin{array}{l}{y}_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.\frac{10}{s^2(s+1)}.\frac{9\times 2}{s^2+2^2}\end{array}$

On solving and putting the limit $s\to 0$

We get,

$y_{ss}(t)=0$

This the required steady state response of the system.

Conclusion:

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

To determine

(b)

Steady state response $y_{ss}(t)$ for the transfer function for the given input.

Answer to Problem 9.8P

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

Explanation of Solution

Given:

$T(s)=\frac{10}{s^2(s+1)}$, $f(t)=9\sin 2t$.

Concept Used:

Using the final value theorem.

Calculation:

$\begin{array}{l}f(t)=9\sin 2t\\ F(s)=\frac{9\times 2}{s^2+2^2}\end{array}$

$T(s)=\frac{10}{s^2(s+1)}$

According to final value theorem, steady state response of the system is

$\begin{array}{l}{y}_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.\frac{10}{s^2(s+1)}.\frac{9\times 2}{s^2+2^2}\end{array}$

On solving and putting the limit $s\to 0$

We get,

$y_{ss}(t)=0$

This the required steady state response of the system.

Conclusion:

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

To determine

(c)

Steady state response $y_{ss}(t)$ for the transfer function for the given input.

Answer to Problem 9.8P

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

Explanation of Solution

Given:

$T(s)=\frac{s}{(2s+1)(5s+1)}$, $f(t)=9\sin 0.7t$.

Concept Used:

Using the final value theorem

Calculation:

$\begin{array}{l}f(t)=9\sin 0.7t\\ F(s)=\frac{9\times 0.7}{s^2+{0.7}^2}\end{array}$

$T(s)=\frac{s}{(2s+1)(5s+1)}$

According to final value theorem steady state response of the system is

$\begin{array}{l}{y}_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.\frac{s}{(2s+1)(5s+1)}.\frac{9\times 0.7}{s^2+{0.7}^2}\end{array}$

On solving and putting the limit $s\to 0$.

We get,

$y_{ss}(t)=0$.

Conclusion:

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

To determine

(d)

Steady state response $y_{ss}(t)$ for the transfer function for the given input.

Answer to Problem 9.8P

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.

Explanation of Solution

Given:

$T(s)=\frac{s^2}{(2s+1)(5s+1)}$, $f(t)=9\sin 0.7t$.

Concept Used:

Using the final value theorem

Calculation:

$\begin{array}{l}f(t)=9\sin 0.7t\\ F(s)=\frac{9\times 0.7}{s^2+{0.7}^2}\end{array}$

$T(s)=\frac{s^2}{(2s+1)(5s+1)}$

According to final value theorem steady state response of the system is

$\begin{array}{l}{y}_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.T(s).F(s)\\ y_{ss}(t)=\underset{s\to 0}{\underbrace{\lim }}s.\frac{s^2}{(2s+1)(5s+1)}.\frac{9\times 0.7}{s^2+{0.7}^2}\end{array}$

On Solving and putting the limit $s\to 0$

We get,

$y_{ss}(t)=0$.

Conclusion:

Steady state response $y_{ss}(t)$ for the transfer function for the given input is $y_{ss}(t)=0$.