Chapter 9, Problem 9.1P

To determine

(a)

The steady state response.

Answer to Problem 9.1P

The steady state response is,

$y_{\text{ss}}=13.5582\sin \left(1.5t-0.8622\right)$.

Explanation of Solution

Given Information:

Transfer function is given by,

$T(s)=\frac{Y(s)}{F(s)}=\frac{25}{14s+18}$.

Input signal is, $f(t)=15\sin 1.5t$.

Calculation:

From the input given, $f(t)=15\sin 1.5t$, the angular frequency is,

$\omega =1.5\text{rad/s}$.

Transfer function is,

$\begin{array}{l}T(s)=\frac{25}{14s+18}\\ \Rightarrow T(j\omega )=\frac{25}{14\left(j\omega \right)+18}\\ \Rightarrow T(j\omega )=\frac{25}{14\left(j1.5\right)+18}\\ \Rightarrow T(j\omega )=\frac{25}{18+j21}\end{array}$

So, the magnitude of transfer function is,

$\begin{array}{l}M=\frac{25}{\sqrt{{18}^2+{21}^2}}\\ \Rightarrow M=0.90388\end{array}$

Phase angle,

$\begin{array}{l}\varphi =0-{\tan }^{-1}\left(\frac{21}{18}\right)\\ \Rightarrow \varphi =0.8622\text{rad}\end{array}$

Thus, the steady state response,

$\begin{array}{l}{y}_{\text{ss}}=15M\sin \left(1.5t+\varphi \right)\\ \Rightarrow y_{\text{ss}}=13.5582\sin \left(1.5t-0.8622\right)\end{array}$.

Conclusion:

The steady state response is,

$y_{\text{ss}}=13.5582\sin \left(1.5t-0.8622\right)$.

To determine

(b)

The steady state response.

Answer to Problem 9.1P

The steady state response is,

$y_{\text{ss}}=20.8013\sin \left(1.5t+0.588\right)$.

Explanation of Solution

Given Information:

Transfer function is given by,

$T(s)=\frac{Y(s)}{F(s)}=\frac{15s}{3s+4}$.

Input signal is, $f(t)=5\sin 2t$.

Calculation:

From the input given, $f(t)=5\sin 2t$, the angular frequency is,

$\omega =2\text{rad/s}$.

Transfer function is,

$\begin{array}{l}T(s)=\frac{15s}{3s+4}\\ \Rightarrow T(j\omega )=\frac{15\left(j\omega \right)}{3\left(j\omega \right)+4}\\ \Rightarrow T(j\omega )=\frac{15\left(j2\right)}{3\left(j2\right)+4}\\ \Rightarrow T(j\omega )=\frac{j30}{4+j6}\end{array}$

So, the magnitude of transfer function is,

$\begin{array}{l}M=\frac{30}{\sqrt{4^2+6^2}}\\ \Rightarrow M=4.16025\end{array}$

Phase angle,

$\begin{array}{l}\varphi =\frac{\pi }{2}-{\tan }^{-1}\left(\frac{6}{4}\right)\\ \Rightarrow \varphi =0.588\text{rad}\end{array}$

Thus, the steady state response,

$\begin{array}{l}{y}_{\text{ss}}=5M\sin \left(2t+\varphi \right)\\ \Rightarrow y_{\text{ss}}=20.8013\sin \left(1.5t+0.588\right)\end{array}$.

Conclusion:

The steady state response is,

$y_{\text{ss}}=20.8013\sin \left(1.5t+0.588\right)$.

To determine

(c)

The steady state response.

Answer to Problem 9.1P

The steady state response is,

$y_{\text{ss}}=1.8605\sin \left(100t+0.5191\right)$.

Explanation of Solution

Given Information:

Transfer function is given by,

$T(s)=\frac{Y(s)}{F(s)}=\frac{s+50}{s+150}$.

Input signal is, $f(t)=3\sin 100t$.

Calculation:

From the input given,, the angular frequency is,

$\omega =100\text{rad/s}$.

Transfer function is,

$\begin{array}{l}T(s)=\frac{s+50}{s+150}\\ \Rightarrow T(j\omega )=\frac{\left(j\omega \right)+50}{\left(j\omega \right)+150}\\ \Rightarrow T(j\omega )=\frac{50+j100}{150+j100}\end{array}$

So, the magnitude of transfer function is,

$\begin{array}{l}M=\frac{\sqrt{{50}^2+{100}^2}}{\sqrt{{150}^2+{100}^2}}\\ \Rightarrow M=0.62017\end{array}$

Phase angle,

$\begin{array}{l}\varphi ={\tan }^{-1}\left(\frac{100}{50}\right)-{\tan }^{-1}\left(\frac{100}{150}\right)\\ \Rightarrow \varphi =0.5191\text{rad}\end{array}$

Thus, the steady state response,

$\begin{array}{l}{y}_{\text{ss}}=3M\sin \left(100t+\varphi \right)\\ \Rightarrow y_{\text{ss}}=1.8605\sin \left(100t+0.5191\right)\end{array}$.

Conclusion:

The steady state response is,

$y_{\text{ss}}=1.8605\sin \left(100t+0.5191\right)$.

To determine

(d)

The steady state response.

Answer to Problem 9.1P

The steady state response is,

$y_{\text{ss}}=2.4634\sin \left(50t-0.5238\right)$.

Explanation of Solution

Given Information:

Transfer function is given by,

$T(s)=\frac{Y(s)}{F(s)}=\frac{33}{200}\frac{s+100}{s+33}$.

Input signal is, $f(t)=8\sin 50t$.

Calculation:

From the input given,, the angular frequency is,

$\omega =50\text{rad/s}$.

Transfer function is,

$\begin{array}{l}T(s)=\frac{33}{200}\left(\frac{s+100}{s+33}\right)\\ \Rightarrow T(j\omega )=\frac{33}{200}\left(\frac{\left(j\omega \right)+100}{\left(j\omega \right)+33}\right)\\ \Rightarrow T(j\omega )=\frac{33}{200}\left(\frac{100+j50}{33+j50}\right)\end{array}$

So, the magnitude of transfer function is,

$\begin{array}{l}M=\frac{33}{200}\times \frac{\sqrt{{100}^2+{50}^2}}{\sqrt{{33}^2+{50}^2}}\\ \Rightarrow M=0.30793\end{array}$

Phase angle,

$\begin{array}{l}\varphi ={\tan }^{-1}\left(\frac{50}{100}\right)-{\tan }^{-1}\left(\frac{50}{33}\right)\\ \Rightarrow \varphi =-0.5238\text{rad}\end{array}$

Thus, the steady state response,

$\begin{array}{l}{y}_{\text{ss}}=8M\sin \left(50t+\varphi \right)\\ \Rightarrow y_{\text{ss}}=2.4634\sin \left(50t-0.5238\right)\end{array}$.

Conclusion:

The steady state response is,

$y_{\text{ss}}=2.4634\sin \left(50t-0.5238\right)$.