Chapter 11, Problem 11.13P

To determine

(a)

The root locus of the given characteristic equation $s(s+5)+K=0$.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+5)+KforK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as $1+G(s)H(s)=0$

$1+G(s)H(s)=s(s+5)+KforK\ge 0$

On solving we get,

$s^2+5s+K=0$

Where,

$\begin{array}{l}G(s)=\frac{K}{s^2+5s}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}{s}^2+5s\\ s(s+5)\\ s=0,-5\end{array}$

$P=2$

And zero is $Z=0$

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=2-0\\ N=2\end{array}$

Centroid is calculated as:

$\sigma =\frac{{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{poles})-{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{zeros})}}}{(P-Z)}$

Where poles are $s=0,-5$ and zero is $Z=0$

$\begin{array}{l}\sigma =\frac{(0-5)-(0)}{(2-0)}\\ \sigma =-2.5\end{array}$

Angle of asymptotes is calculated as:

Angle of asymptotes

${\theta }_k=\frac{(2k+1)180}{P-Z}$

Where P = no. of poles

Z = No. of zeros

$\begin{array}{l}P=2\\ Z=0\\ k=P-Z-1\end{array}$

$k=0,1$

On replacing the values

$\begin{array}{l}{\theta }_0=\frac{(2\times 0+1)180}{2-0}=90°\\ {\theta }_1=\frac{(2\times 1+1)180}{2-0}=270°\end{array}$

The $P-Z$ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+5)+K=0\\ s^2+5s+K=0\\ K=-s^2-5s\\ \frac{dK}{ds}=-2s-5\end{array}$

To calculate breakaway point, replace $\frac{dK}{ds}=0$

So, $\begin{array}{l}-2s-5=0\\ s=-\frac{5}{2}\\ s=-2.5\end{array}$

Breakaway point is $s=-2.5$

Calculating the value of $K$ to determine the roots of the equation

$\begin{array}{l}{s}^2+5s+K=0\\ s^{2}1K\\ s^{1}50\\ s^0\frac{5K-1\times 0}{5}=K0\end{array}$

For Stability $K>0$

Auxiliary equation is defined as:

$s^2+K=0$

Replace the value of $K$ the roots of the equation are calculated

$\begin{array}{l}{s}^2+0=0\\ s=0\end{array}$

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

Fig.1

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.1.

To determine

(b)

The root locus of the given characteristic equation $s^2+3s+3+K(s+3)=0$.

Explanation of Solution

Given:

$1+G(s)H(s)=s^2+3s+3+K(s+3)forK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as $1+G(s)H(s)=0$

$1+G(s)H(s)=s^2+3s+3+K(s+3)forK\ge 0$

Where,

$\begin{array}{l}G(s)=\frac{K(s+3)}{s^2+3s+3}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}{s}^2+3s+3=0\\ s=-1.5\pm 0.87i\end{array}$

$P=2$

And zero is $\begin{array}{l}s+3=0\\ s=-3\\ Z=1\end{array}$

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=2-1\\ N=1\end{array}$

Centroid is calculated as:

$\sigma =\frac{{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{poles})-{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{zeros})}}}{(P-Z)}$

Where poles are $s=-1.5\pm 1.66i$ and zero is $s=-3$

$\begin{array}{l}\sigma =\frac{(0+(-1.5-1.5)-(-3))}{(2-1)}\\ \sigma =0\end{array}$

Angle of asymptotes is calculated as:

Angle of asymptotes

${\theta }_k=\frac{(2k+1)180}{P-Z}$

Where P = no. of poles

Z = No. of zeros

$\begin{array}{l}P=2\\ Z=1\\ k=P-Z-1\end{array}$

$k=0$

On replacing the values

${\theta }_0=\frac{(2\times 0+1)180}{2-1}=180°$

The $P-Z$ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s^2+3s+3+K(s+3)=0\\ K(s+3)=-s^2-3s-3\\ K=\frac{-s^2-3s-3}{(s+3)}\\ \frac{dK}{ds}=-\frac{s^2+6s+6}{{(s+3)}^2}\end{array}$

To calculate breakaway point, replace $\frac{dK}{ds}=0$

So,

$\begin{array}{l}{s}^2+6s+6\\ s_1=-1.27\\ s_2=-4.73\end{array}$

Calculating the value of $K$ to determine the roots of the equation

$\begin{array}{l}{s}^2+(3+K)s+(3+3K)=0\\ s^{2}1(3+3K)\\ s^1(3+K)0\\ s^0(3+3K)0\end{array}$

For Stability $K>0$

$K=-3,-1$

Auxiliary equation is defined as:

$s^2+3+3K=0$

Replace the value of $K$ the roots of the equation are calculated

For $\begin{array}{l}K=-1\\ s=0\end{array}$

For $\begin{array}{l}K=-3\\ s=\pm 2.45\end{array}$ neglected

This is the point on the imaginary axis

Angle of departure is calculated where there are either poles or zero is imaginary.

${\Phi }_D=180+\left({\displaystyle \sum {\Phi }_z-}{\displaystyle \sum {\Phi }_p}\right)$

${\Phi }_p$ = Sum of angles from poles

${\Phi }_z$ = Sum of angles from zeros

$\begin{array}{l}\Phi ={\displaystyle \sum {\Phi }_z-}{\displaystyle \sum {\Phi }_p}\\ {\displaystyle \sum {\Phi }_p}=90\\ {\displaystyle \sum {\Phi }_z}=0\\ \Phi =-90\\ {\Phi }_D=180+\Phi \\ {\Phi }_D=90\end{array}$

Root locus plot of the characteristic equation is

Fig.2

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.2.

To determine

(c)

The root locus of the given characteristic equation $s(s^2+3s+3)+K=0$.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s^2+3s+3)+KforK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as $1+G(s)H(s)=0$

$1+G(s)H(s)=s(s^2+3s+3)+KforK\ge 0$

Where,

$\begin{array}{l}G(s)=\frac{K}{s(s^2+3s+3)}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s^2+3s+3)=0\\ s=0,-1.5\pm 0.87i\end{array}$

$P=3$

And zero is $Z=0$

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=3-0\\ N=3\end{array}$

Centroid is calculated as:

$\sigma =\frac{{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{poles})-{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{zeros})}}}{(P-Z)}$

Where poles are $s=0,-1.5\pm 1.66i$ and zero is $Z=0$

$\begin{array}{l}\sigma =\frac{(0-1.5-1.5)-(0)}{(3-0)}\\ \sigma =-1\end{array}$

Angle of asymptotes is calculated as:

Angle of asymptotes

${\theta }_k=\frac{(2k+1)180}{P-Z}$

Where P = no. of poles

Z = No. of zeros

$\begin{array}{l}P=3\\ Z=0\\ k=P-Z-1\end{array}$

$k=0,1,2$

On replacing the values

$\begin{array}{l}{\theta }_0=\frac{(2\times 0+1)180}{3-0}=60°\\ {\theta }_1=\frac{(2\times 1+1)180}{3-0}=180°\\ {\theta }_2=\frac{(2\times 2+1)180}{3-0}=300°\end{array}$

The $P-Z$ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s^2+3s+3)+K=0\\ s(s^2+3s+3)+K=0\\ K=-s(s^2+3s+3)\\ K=-s^3-3s^2-3s\\ \frac{dK}{ds}=-3s^2-6s-3\end{array}$

To calculate breakaway point, replace $\frac{dK}{ds}=0$

So,

$\begin{array}{l}-3s^2-6s-3=0\\ 3s^2+6s+3=0\\ s^2+2s+1=0\\ s=-1,-1\end{array}$

Breakaway point is $s=-1$

Calculating the value of $K$ to determine the roots of the equation:

$\begin{array}{l}{s}^3+3s^2+3s+K=0\\ s^{3}13\\ s^{2}3K\\ s^1\frac{3\times 3-K}{3}0\\ s^{0}K0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{9-K}{3}=0\\ K=9\end{array}$

Auxiliary equation is defined as:

$3s^2+K=0$

Replace the value of $K$ the roots of the equation are calculated

$\begin{array}{l}3s^2+K=0\\ K=9\\ s^2=-\frac{9}{3}=-3\\ s=\pm 1.732i\end{array}$

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

Fig.3

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.3.

To determine

(d)

The root locus of the given characteristic equation $s(s+5)(s+7)+K=0$.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+5)(s+7)+KforK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as $1+G(s)H(s)=0$

$1+G(s)H(s)=s(s+5)(s+7)+KforK\ge 0$

On solving we get,

$\begin{array}{l}s(s+5)(s+7)+K=0\\ (s^2+7s)(s+5)+K=0\\ s^3+13s^2+35s+K=0\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s+5)(s+7)=0\\ s=0,-5,-7\end{array}$

$P=3$

And zero is $Z=0$

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=3-0\\ N=3\end{array}$

Centroid is calculated as:

$\sigma =\frac{{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{poles})-{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{zeros})}}}{(P-Z)}$

Where poles are $s=0,-5,-7$ and zero is $Z=0$

$\begin{array}{l}\sigma =\frac{(0-5-7)-(0)}{(3-0)}\\ \sigma =-4\end{array}$

Angle of asymptotes is calculated as:

Angle of asymptotes

${\theta }_k=\frac{(2k+1)180}{P-Z}$

Where P = no. of poles

Z = No. of zeros

$\begin{array}{l}P=3\\ Z=0\\ k=P-Z-1\end{array}$

$k=0,1,2$

On replacing the values

$\begin{array}{l}{\theta }_0=\frac{(2\times 0+1)180}{3-0}=60°\\ {\theta }_1=\frac{(2\times 1+1)180}{3-0}=180°\\ {\theta }_2=\frac{(2\times 2+1)180}{3-0}=300°\end{array}$

The $P-Z$ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+5)(s+7)+K=0\\ 1+G(s)H(s)=s(s+5)(s+7)+K=0\\ s(s+5)(s+7)+K=0\\ (s^2+7s)(s+5)+K=0\\ s^3+12s^2+35s+K=0\\ K=-s^3-12s^2-35s\\ \frac{dK}{ds}=-3s^2-24s-35\end{array}$

To calculate breakaway point, replace $\frac{dK}{ds}=0$

So,

$\begin{array}{l}-3s^2-24s-35=0\\ 3s^2+24s+35=0\\ s_1=-1.92\\ s_2=-6.1\end{array}$

Breakaway point is $s=-1.92$

Calculating the value of $K$ to determine the roots of the equation

$\begin{array}{l}{s}^3+12s^2+35s+K=0\\ s^{3}135\\ s^{2}12K\\ s^1\frac{12\times 35-K}{12}0\\ s^{0}K0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{12\times 35-K}{12}=0\\ K=420\end{array}$

Auxiliary equation is defined as:

$12s^2+K=0$

Replace the value of $K$ the roots of the equation are calculated

$\begin{array}{l}12s^2+420=0\\ s=\pm 5.92i\end{array}$

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

Fig.4

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig4.

To determine

(e)

The root locus of the given characteristic equation $s(s+3)+K(s+4)=0$.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+3)+K(s+4)forK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as $1+G(s)H(s)=0$

$1+G(s)H(s)=s(s+3)+K(s+4)forK\ge 0$

Where,

$\begin{array}{l}G(s)=\frac{K(s+4)}{s(s+3)}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s+3)\\ s=0,-3\end{array}$

$P=2$

And zero is $\begin{array}{l}s+4=0\\ s=-4\\ Z=1\end{array}$

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=2-1\\ N=1\end{array}$

Centroid is calculated as:

$\sigma =\frac{{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{poles})-{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{zeros})}}}{(P-Z)}$

Where poles are $s=0,-3$ and zero is $s=-4$

$\begin{array}{l}\sigma =\frac{(0-3)-(-4)}{(2-1)}\\ \sigma =1\end{array}$

Angle of asymptotes is calculated as:

Angle of asymptotes

${\theta }_k=\frac{(2k+1)180}{P-Z}$

Where P = no. of poles

Z = No. of zeros

$\begin{array}{l}P=2\\ Z=1\\ k=P-Z-1\end{array}$

$k=0$

On replacing the values

${\theta }_0=\frac{(2\times 0+1)180}{2-1}=180°$

The $P-Z$ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+3)+K(s+4)=0\\ s^2+3s+K(s+4)=0\\ K=\frac{-s^2-3s}{(s+4)}\\ \frac{dK}{ds}=\frac{-s^2-8s-12}{{(s+4)}^2}\end{array}$

To calculate breakaway point, replace $\frac{dK}{ds}=0$

So,

$\begin{array}{l}-s^2-8s-12=0\\ s^2+8s+12=0\\ s=-2\\ s=-6\end{array}$

Breakaway point is $s=-2$

Calculating the value of $K$ to determine the roots of the equation

$\begin{array}{l}{s}^2+(3+K)s+4K=0\\ s^{2}14K\\ s^1(3+K)0\\ s^0\frac{(3+K)4K-1\times 0}{(3+K)}=4K0\end{array}$

For Stability $K>0$

$\begin{array}{l}(3+K)=0\\ K=-3\end{array}$ & $\begin{array}{l}4K=0\\ K=0\end{array}$

Auxiliary equation is defined as:

$s^2+4K=0$

Replace the value of $K=-3$ the roots of the equation are calculated

$\begin{array}{l}{s}^2-4\times 3=0\\ s=\sqrt{12}\\ s=\pm 3.46\end{array}$

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

Fig.5

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.5.

To determine

(f)

The root locus of the given characteristic equation $s(s+6)+K(s-4)=0$.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+6)+K(s-4)=0$.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as $1+G(s)H(s)=0$

$1+G(s)H(s)=s(s+6)+K(s-4)=0$

Where,

$\begin{array}{l}G(s)=\frac{K(s-4)}{s(s+6)}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s+6)=0\\ s=0,-6\end{array}$

$P=2$

And zero is

$\begin{array}{l}s-4=0\\ s=4\\ Z=1\end{array}$

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=2-1\\ N=1\end{array}$

Centroid is calculated as:

$\sigma =\frac{{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{poles})-{\displaystyle \sum (\text{Real}\text{part}\text{of}\text{open}\text{loop}\text{zeros})}}}{(P-Z)}$

Where poles are $s=0,-6$ and zero is $s=4$

$\begin{array}{l}\sigma =\frac{(0-6)-(4)}{(2-1)}\\ \sigma =-10\end{array}$

Angle of asymptotes is calculated as:

Angle of asymptotes

${\theta }_k=\frac{(2k+1)180}{P-Z}$

Where P = no. of poles

Z = No. of zeros

$\begin{array}{l}P=2\\ Z=1\\ k=P-Z-1\end{array}$

$k=0$

On replacing the values

${\theta }_0=\frac{(2\times 0+1)180}{2-1}=180°$

The $P-Z$ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+6)+K(s-4)=0\\ K(s-4)=-s^2-6s\\ K=\frac{-s^2-6s}{(s-4)}\\ \frac{dK}{ds}=\frac{-s^2+8s+24}{{(s-4)}^2}\end{array}$

To calculate breakaway point, replace $\frac{dK}{ds}=0$

So,

$\begin{array}{l}\frac{-s^2+8s+24}{{(s-4)}^2}=0\\ -s^2+8s+24=0\\ s^2-8s-24=0\\ s=10.32,-2.33\end{array}$

Breakaway point is $s=-2.33$

Calculating the value of $K$ to determine the roots of the equation

$\begin{array}{l}s(s+6)+K(s-4)=0\\ s^2+(6+K)s-4K=0\\ s^{2}1-4K\\ s^1(6+K)0\\ s^0-4K0\end{array}$

For Stability $K>0$

$\begin{array}{l}6+K=0\\ K=-6\\ -4K=0\\ K=0\end{array}$

Auxiliary equation is defined as:

$s^2-4K=0$

Replace the value of $K$ the roots of the equation are calculated

For

$\begin{array}{l}K=-6\\ s=\pm 4.89i\end{array}$

For

$\begin{array}{l}K=0\\ s=0\end{array}$ neglected

Root locus plot of the characteristic equation is

Fig.6

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig6.