Chapter 11, Problem 11.7P

To determine

(a)

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

Answer to Problem 11.7P

The root locus of the given characteristic equation $s(s+5)+K=0$ is shown in Fig 1.

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

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

$P=2$

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.

Break away 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

To determine

(b)

The root locus of the given characteristic equation $s(s+7)(s+9)+K=0$.

Answer to Problem 11.7P

The root locus of the given characteristic equation $s(s+7)(s+9)+K$ is shown in Fig 2.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+7)(s+9)+KforK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

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

$1+G(s)H(s)=s(s+7)(s+9)+KforK\ge 0$

On solving we get,

$\begin{array}{l}s(s+7)(s+9)+K=0\\ (s^2+7s)(s+9)+K=0\\ s^3+16s^2+63s+K=0\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s+7)(s+9)=0\\ s=0,-7,-9\end{array}$

$P=3$

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 $s=0,-7,-9$ and zero is $Z=0$

$\begin{array}{l}\sigma =\frac{(0-7-9)-(0)}{(3-0)}\\ \sigma =-5.33\end{array}$

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.

Break away point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+7)(s+9)+K=0\\ 1+G(s)H(s)=s(s+7)(s+9)+K=0\\ s(s+7)(s+9)+K=0\\ (s^2+7s)(s+9)+K=0\\ s^3+16s^2+63s+K=0\\ K=-s^3-16s^2-63s\\ \frac{dK}{ds}=-3s^2-32s-63\end{array}$

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

So,

$\begin{array}{l}-3s^2-32s-63=0\\ 3s^2+32s+63=0\\ s_1=-2.6\\ s=-8.06\end{array}$

Breakaway point is $s=-2.6$

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

$\begin{array}{l}{s}^3+16s^2+63+K=0\\ s^{3}163\\ s^{2}16K\\ s^1\frac{16\times 63-K}{16}0\\ s^{0}K0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{16\times 63-K}{16}>0\\ K>1008\end{array}$

Auxiliary equation is defined as:

$s^2+K=0$

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

$\begin{array}{l}{s}^2+1008=0\\ s=\pm 31.75i\end{array}$

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

Fig.2

To determine

(d)

The root locus of the given characteristic equation $s(s+4)+K(s+5)=0$.

Answer to Problem 11.7P

The root locus of the given characteristic equation $s(s+4)+K(s+5)$ is shown in Fig 4.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+4)+K(s+5)forK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

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

$1+G(s)H(s)=s(s+4)+K(s+5)forK\ge 0$

Where,

$\begin{array}{l}G(s)=\frac{K(s+5)}{s(s+4)}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s+4)\\ s=0,-4\end{array}$

$P=2$

And zero is $\begin{array}{l}s+5=0\\ s=-5\\ Z=1\end{array}$

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 $s=0,-4$ and zero is $s=-5$

$\begin{array}{l}\sigma =\frac{(0-4)-(-5)}{(2-1)}\\ \sigma =-1\end{array}$

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:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+4)+K(s+5)=0\\ s^2+4s+K(s+5)=0\\ K=\frac{-s^2-4s}{(s+5)}\\ \frac{dK}{ds}=\frac{-s^2-10s-20}{{(s+5)}^2}\end{array}$

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

So,

$\begin{array}{l}-s^2-10s-20=0\\ s=-2.76\\ s=-7.24\end{array}$

Breakaway point is $s=-2.76$

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

$\begin{array}{l}{s}^2+4s+K(s+5)=0\\ s^2+(4+K)s+5K=0\\ s^{2}15K\\ s^1(4+K)0\\ s^0\frac{(4+K)5K-1\times 0}{(4+K)}=5K0\end{array}$

For Stability $K>0$

$\begin{array}{l}(4+K)=0\\ K=-4\end{array}$ & $\begin{array}{l}5K=0\\ K=0\end{array}$

Auxiliary equation is defined as:

$s^2+5K=0$

Replace the value of $K=-4$ the roots of the equation are calculated.

$\begin{array}{l}{s}^2-20=0\\ s=\sqrt{20}\\ s=\pm 4.47\end{array}$

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

Fig4

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.4.

To determine

(e)

The root locus of the given characteristic equation $s(s^2+3s+5)+K=0$.

Answer to Problem 11.7P

The root locus of the given characteristic equation $s(s^2+3s+5)+K$ is shown in Fig 5.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s^2+3s+5)+KforK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

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

$1+G(s)H(s)=s(s^2+3s+5)+KforK\ge 0$

Where,

$\begin{array}{l}G(s)=\frac{K}{s(s^2+3s+5)}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s^2+3s+5)=0\\ s=0,-1.5\pm 1.66i\end{array}$

$P=3$

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:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s^2+3s+5)+K=0\\ s(s^2+3s+5)+K=0\\ K=-s(s^2+3s+5)\\ K=-s^3-3s^2-5s\\ \frac{dK}{ds}=-3s^2-6s-5\end{array}$

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

So,

$\begin{array}{l}-3s^2-6s-5=0\\ 3s^2+6s+5=0\\ s=-1\pm 0.82i\end{array}$

Breakaway point is $s=-1\pm 0.82i$

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

$\begin{array}{l}{s}^3+3s^2+5s+K=0\\ s^{3}15\\ s^{2}3K\\ s^1\frac{5\times 3-K}{3}0\\ s^{0}K0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{15-K}{3}=0\\ K=15\end{array}$

Auxiliary equation is defined as:

$3s^2+K=0$

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

$\begin{array}{l}3s^2+K=0\\ K=15\\ s^2=-\frac{15}{3}=-5\\ s=\pm 2.24\end{array}$

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 and is shown in Fig.5

To determine

(f)

The root locus of the given characteristic equation $s(s+3)(s+7)+K(s+4)=0$.

Answer to Problem 11.7P

The root locus of the given characteristic equation $s(s+3)(s+7)+K(s+4)$ is shown in Fig 6.

Explanation of Solution

Given:

$1+G(s)H(s)=s(s+3)(s+7)+K(s+4)forK\ge 0$.

Concept Used:

Root Locus technique.

Calculation:

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

$1+G(s)H(s)=s(s+3)(s+7)+K(s+4)forK\ge 0$

Where,

$\begin{array}{l}G(s)=\frac{K(s+4)}{s(s+3)(s+7)}\\ H(s)=1\end{array}$

Therefore, Poles are

$\begin{array}{l}s(s+3)(s+7)=0\\ s=0,-3,-7\end{array}$

$P=3$

And zero is

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

Total number of branches is

$\begin{array}{l}N=P-Z\\ N=3-1\\ N=2\end{array}$

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 $s=0,-3,-7$ and zero is $s=-4$

$\begin{array}{l}\sigma =\frac{(0-3-7)-(-4)}{(3-1)}\\ \sigma =-3\end{array}$

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

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

$k=1$

On replacing the values

$\begin{array}{l}{\theta }_0=\frac{(2\times 0+1)180}{3-1}=90°\\ {\theta }_1=\frac{(2\times 1+1)180}{3-1}=270°\end{array}$

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

Breakaway point is calculated as:

$\begin{array}{l}1+G(s)H(s)=0\\ 1+G(s)H(s)=s(s+3)(s+7)+K(s+4)=0\\ K(s+4)=-s^3-7s^2-3s^2-21s\\ K=\frac{-s^3-7s^2-3s^2-21s}{(s+4)}\\ \frac{dK}{ds}=\frac{-2s^3-22s^2-80s-84}{{(s+4)}^2}\end{array}$

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

So,

$\begin{array}{l}\frac{-2s^3-22s^2-80s-84}{{(s+4)}^2}=0\\ -2s^3-22s^2-80s-84=0\\ 2s^3+22s^2+80s+84=0\\ s=-1.78,-4.61\pm 1.53i\end{array}$

Breakaway point is $s=-1.78$

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

$\begin{array}{l}s(s+3)(s+7)+K(s+4)=0\\ s^3+10s^2+(21+K)s+4K=0\\ s^{3}121+K\\ s^{2}104K\\ s^1\frac{10(21+K)-4K}{10}0\\ s^{0}4K0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{10(21+K)-4K}{10}=0\\ K=-35\\ K+21=0\\ K=-21\end{array}$

Auxiliary equation is defined as:

$10s^2+4K=0$

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

$\begin{array}{l}K=-35\\ s=\pm 3.74\end{array}$ neglected for

$\begin{array}{l}K=-21\\ s=\pm 2.89\end{array}$ neglected

Root locus plot of the characteristic equation is

Fig. 6

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.6.