Chapter 2, Problem 2.36P

To determine

(a)

The transfer functions $\frac{X\left(s\right)}{F\left(s\right)}$ and $\frac{X\left(s\right)}{G\left(s\right)}$ for the model equations as shown:

$\dot{x}=-4x+2y+f\left(t\right)and\dot{y}=-9y-5x+g\left(t\right)$.

Answer to Problem 2.36P

The transfer functions for the model equations are as:

$\frac{X\left(s\right)}{F\left(s\right)}=\frac{\left(s+9\right)}{\left(s^2+13s+46\right)}and\frac{X\left(s\right)}{G\left(s\right)}=\frac{2}{\left(s^2+13s+46\right)}$.

Explanation of Solution

Given:

The given model equations are as:

$\dot{x}=-4x+2y+f\left(t\right)and\dot{y}=-9y-5x+g\left(t\right)$.

Concept Used:

Laplace transform is employed for obtaining the transfer functions for the model conditions keeping the initial conditions zero. In addition, for the system having multiple inputs, the superposition principle is used for finding the impact of one input at a single time by disconnecting the system from other inputs.

Calculation:

Model equations to be solved are as:

$\dot{x}=-4x+2y+f\left(t\right)and\dot{y}=-9y-5x+g\left(t\right)$

For $\frac{X\left(s\right)}{F\left(s\right)}$, set $G\left(s\right)$ temporarily equal to zero.

On taking Laplace for both the model equations, that is

$\begin{array}{l}\dot{x}=-4x+2y+f\left(t\right)\\ \Rightarrow sX\left(s\right)=-4X\left(s\right)+2Y\left(s\right)+F\left(s\right)\\ \Rightarrow \frac{X\left(s\right)}{F\left(s\right)}(s+4)=2\frac{Y\left(s\right)}{X\left(s\right)}+1\left(1\right)\end{array}$

Similarly,

$\begin{array}{l}\dot{y}=-9y-5x+g\left(t\right)\\ \Rightarrow sY\left(s\right)=-9Y\left(s\right)-5X\left(s\right)+G\left(s\right)\\ \Rightarrow sY\left(s\right)=-9Y\left(s\right)-5X\left(s\right)\\ \Rightarrow \left(s+9\right)\frac{Y\left(s\right)}{F\left(s\right)}=-5\frac{X\left(s\right)}{F\left(s\right)}\left(2\right)\end{array}$

Using equations (1) and (2), we get

$\begin{array}{l}\frac{X\left(s\right)}{F\left(s\right)}(s+4)=2\frac{Y\left(s\right)}{X\left(s\right)}+1\\ \Rightarrow \frac{X\left(s\right)}{F\left(s\right)}(s+4)=-\frac{10}{\left(s+9\right)}\frac{X\left(s\right)}{F\left(s\right)}+1\\ \Rightarrow \frac{X\left(s\right)}{F\left(s\right)}\left((s+4)+\frac{10}{\left(s+9\right)}\right)=1\\ \Rightarrow \frac{X\left(s\right)}{F\left(s\right)}=\frac{\left(s+9\right)}{10+\left(s+9\right)\left(s+4\right)}=\frac{\left(s+9\right)}{\left(s^2+13s+46\right)}\end{array}$

Similarly, for $\frac{X\left(s\right)}{G\left(s\right)}$, set F(s) temporarily to zero.

On taking Laplace for both the model equations, that is

$\begin{array}{l}\dot{x}=-4x+2y+f\left(t\right)\\ \Rightarrow sX\left(s\right)=-4X\left(s\right)+2Y\left(s\right)+F\left(s\right)\\ \Rightarrow sX\left(s\right)=-4X\left(s\right)+2Y\left(s\right)\\ \Rightarrow \frac{X\left(s\right)}{G\left(s\right)}(s+4)=2\frac{Y\left(s\right)}{G\left(s\right)}\left(3\right)\end{array}$

Similarly,

$\begin{array}{l}\dot{y}=-9y-5x+g\left(t\right)\\ \Rightarrow sY\left(s\right)=-9Y\left(s\right)-5X\left(s\right)+G\left(s\right)\\ \Rightarrow \left(s+9\right)\frac{Y\left(s\right)}{G\left(s\right)}=-5\frac{X\left(s\right)}{G\left(s\right)}+1\left(4\right)\end{array}$

Using equations (3) and (4), we get

$\begin{array}{l}\frac{Y\left(s\right)}{G\left(s\right)}(s+9)=-5\frac{X\left(s\right)}{G\left(s\right)}+1\\ \Rightarrow \frac{(s+4)(s+9)}{2}\frac{X\left(s\right)}{G\left(s\right)}+5\frac{X\left(s\right)}{G\left(s\right)}=1\\ \Rightarrow \frac{X\left(s\right)}{G\left(s\right)}\left(\frac{(s+4)\left(s+9\right)}{2}+5\right)=1\\ \Rightarrow \frac{X\left(s\right)}{G\left(s\right)}=\frac{2}{10+\left(s+9\right)\left(s+4\right)}=\frac{2}{\left(s^2+13s+46\right)}\end{array}$.

Conclusion:

The obtained transfer functions are as:

$\frac{X\left(s\right)}{F\left(s\right)}=\frac{\left(s+9\right)}{\left(s^2+13s+46\right)}and\frac{X\left(s\right)}{G\left(s\right)}=\frac{2}{\left(s^2+13s+46\right)}$.

To determine

(b)

The values for $\tau ,\zeta ,{\omega }_{d}and{\omega }_n$ of the model equations as shown:

$\dot{x}=-4x+2y+f\left(t\right)and\dot{y}=-9y-5x+g\left(t\right)$.

Answer to Problem 2.36P

The parameter values are:

$\begin{array}{l}\tau =\frac{2}{13},\\ \zeta =\frac{13}{2\sqrt{46}},\\ {\omega }_n=\sqrt{46}\\ and{\omega }_d=\frac{15}{2\sqrt{15}}\end{array}$.

Explanation of Solution

Given:

The given system model equations are as:

$\dot{x}=-4x+2y+f\left(t\right)and\dot{y}=-9y-5x+g\left(t\right)$.

Concept Used:

Characteristic equation is obtained from transfer function obtained in part a, that is

Since,

$\frac{X\left(s\right)}{F\left(s\right)}=\frac{\left(s+9\right)}{\left(s^2+13s+46\right)}and\frac{X\left(s\right)}{G\left(s\right)}=\frac{2}{\left(s^2+13s+46\right)}$

Therefore, the characteristic equation is as

$s^2+13s+46=0$

This equation is compared with $m{s}^2+cs+k=0$ and the following formulae for the parameters are used:

$\tau =\frac{2m}{c}$,

$\zeta =\frac{c}{2\sqrt{mk}}$,

${\omega }_n=\sqrt{\frac{k}{m}}$

And ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$.

Calculation:

Characteristic equation of the system is as:

$s^2+13s+46=0$

On comparing this with $m{s}^2+cs+k=0$, we get

$\begin{array}{l}\tau =\frac{2m}{c}=\frac{2}{13},\\ \zeta =\frac{c}{2\sqrt{mk}}=\frac{13}{2\sqrt{46}},\\ {\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{46}\\ and{\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=\sqrt{46}\sqrt{1-{\left(\frac{13}{2\sqrt{46}}\right)}^2}=\sqrt{46}\sqrt{1-\frac{169}{184}}=\frac{\sqrt{15}}{2}=\frac{15}{2\sqrt{15}}\end{array}$.

Conclusion:

The obtained values for the parameters are as shown:

$\begin{array}{l}\tau =\frac{2}{13},\\ \zeta =\frac{13}{2\sqrt{46}},\\ {\omega }_n=\sqrt{46}\\ and{\omega }_d=\frac{15}{2\sqrt{15}}\end{array}$.

To determine

(c)

The time taken for the responses $x\left(t\right)andy\left(t\right)$ to disappear when response $f\left(t\right)=g\left(t\right)=0$.

Answer to Problem 2.36P

At $t=4\tau =\frac{8}{13}$ seconds, the responses $x\left(t\right)andy\left(t\right)$ disappear.

Explanation of Solution

Given:

The obtained transfer functions in the first sub-part ‘a’ are:

$\frac{X\left(s\right)}{F\left(s\right)}=\frac{\left(s+9\right)}{\left(s^2+13s+46\right)}and\frac{X\left(s\right)}{G\left(s\right)}=\frac{2}{\left(s^2+13s+46\right)}$

Also, $f\left(t\right)=g\left(t\right)=0$.

Concept Used:

Since

$f\left(t\right)=g\left(t\right)=0$

Therefore,

$\dot{x}=-4x+2yand\dot{y}=-9y-5x$

Using these modified model equations, expressions for $X(s)andY(s)$ are evaluated using Laplace transform consequently, the responses for $x\left(t\right)andy\left(t\right)$ are computed.

Using these responses, the time taken by oscillations to decay is observed.

Calculation:

Since,

$\dot{x}=-4x+2y+f\left(t\right)and\dot{y}=-9y-5x+g\left(t\right)$

On taking $f\left(t\right)=g\left(t\right)=0$, we get

$\dot{x}=-4x+2yand\dot{y}=-9y-5x$

On taking Laplace transform, we have

$\begin{array}{l}\dot{x}=-4x+2y\\ \Rightarrow sX\left(s\right)-x\left(0\right)=-4X\left(s\right)+2Y\left(s\right)\\ \Rightarrow X\left(s\right)\left(s+4\right)=2Y\left(s\right)+x\left(0\right)\left(1\right)\\ and\\ \dot{y}=-9y-5x\\ \Rightarrow sY\left(s\right)-y\left(0\right)=-9Y\left(s\right)-5X\left(s\right)\\ \Rightarrow Y\left(s\right)\left(s+9\right)=-5X\left(s\right)+y\left(0\right)\left(2\right)\end{array}$

Using equations (1) and (2), we have

$X\left(s\right)\left(s+4\right)=2Y\left(s\right)+x\left(0\right)$

$\Rightarrow X\left(s\right)\left(s+4\right)=2\left(\frac{-5X\left(s\right)+y\left(0\right)}{\left(s+9\right)}\right)+x\left(0\right)$

$\Rightarrow X\left(s\right)\left(\left(s+4\right)+\frac{10}{\left(s+9\right)}\right)=\frac{2y\left(0\right)}{\left(s+9\right)}+x\left(0\right)$

$\Rightarrow X\left(s\right)\left(\frac{\left(s+4\right)\left(s+9\right)+10}{\left(s+9\right)}\right)=\frac{2y\left(0\right)}{\left(s+9\right)}+x\left(0\right)$

$\Rightarrow X\left(s\right)\left(\frac{\left(s^2+13s+46\right)}{\left(s+9\right)}\right)=\frac{2y\left(0\right)}{\left(s+9\right)}+x\left(0\right)$

$\Rightarrow X\left(s\right)=\frac{2y\left(0\right)}{\left(s^2+13s+46\right)}+x\left(0\right)\frac{\left(s+9\right)}{\left(s^2+13s+46\right)}$

$\Rightarrow X\left(s\right)=\frac{2y\left(0\right)}{\left({\left(s+\frac{13}{2}\right)}^2+{\left(\frac{\sqrt{15}}{2}\right)}^2\right)}+x\left(0\right)\frac{\left(s+9\right)}{\left({\left(s+\frac{13}{2}\right)}^2+{\left(\frac{\sqrt{15}}{2}\right)}^2\right)}$

$\Rightarrow X\left(s\right)=\frac{2}{\sqrt{15}}\left(2y\left(0\right)+\frac{5}{2}x\left(0\right)\right)\frac{\frac{\sqrt{15}}{2}}{\left({\left(s+\frac{13}{2}\right)}^2+{\left(\frac{\sqrt{15}}{2}\right)}^2\right)}+x\left(0\right)\frac{\left(s+\frac{13}{2}\right)}{\left({\left(s+\frac{13}{2}\right)}^2+{\left(\frac{\sqrt{15}}{2}\right)}^2\right)}$

On taking inverse Laplace of this, we get

$x\left(t\right)=x\left(0\right)e^{-\frac{13}{2}t}\cos \frac{\sqrt{15}}{2}t+\frac{2}{\sqrt{15}}\left(2y\left(0\right)+\frac{5}{2}x\left(0\right)\right)e^{-\frac{13}{2}t}\sin \frac{\sqrt{15}}{2}t$

On observing the response $x\left(t\right)$, we get

Here, at $t=4\tau =\frac{8}{13}$, $e^{-\frac{13}{2}t}<0.02$ that is the oscillations of response $x\left(t\right)$ disappear at $t=\frac{8}{13}$ seconds.

Similarly, we have

$\begin{array}{l}\dot{x}=-4x+2y\\ \Rightarrow y=\frac{\dot{x}+4x}{2}\end{array}$

That is the response $y\left(t\right)$ will be of the form as shown

$\begin{array}{l}\dot{x}=-4x+2y\\ \Rightarrow y=\frac{\dot{x}+4x}{2}=-\frac{13}{4}{e}^{-\frac{13}{2}t}A\left(t\right)+\frac{e^{-\frac{13}{2}t}}{2}\frac{dA\left(t\right)}{dt}+2e^{-\frac{13}{2}t}A\left(t\right)\\ where,A\left(t\right)=\left(x\left(0\right)\cos \frac{\sqrt{15}}{2}t+\frac{2}{\sqrt{15}}\left(2y\left(0\right)+\frac{5}{2}x\left(0\right)\right)\sin \frac{\sqrt{15}}{2}t\right)\\ \Rightarrow y=\frac{e^{-\frac{13}{2}t}}{2}\frac{dA\left(t\right)}{dt}-\frac{5}{4}{e}^{-\frac{13}{2}t}A\left(t\right)\\ \Rightarrow y=e^{-\frac{13}{2}t}\left(-\frac{5}{4}A\left(t\right)+\frac{1}{2}\frac{dA\left(t\right)}{dt}\right)\end{array}$

Therefore, at $t=4\tau =\frac{8}{13}$, $e^{-\frac{13}{2}t}<0.02$ that is the response $y\left(t\right)$ will also disappear at $t=\frac{8}{13}$ seconds.

Conclusion:

The obtained responses $x\left(t\right)andy\left(t\right)$ are disappearing at $t=\frac{8}{13}$ seconds.