Chapter 11, Problem 11.1P

To determine

To draw:

The Root locus plot of $3s^2+12s+k=0\text{for}k\ge 0$ and determine the value of smallest possible time constant and value of k.

Answer to Problem 11.1P

The smallest possible time constant is $\frac{1}{2}$ which occur for $k\ge 12$.

Explanation of Solution

Given information:

The characteristic equation is given as $3s^2+12s+k=0\text{for}k\ge 0$.

Concept Used:

Root locus technique is used to determine the time constant and value of k.

Calculation:

Solving the quadratic equation for finding the roots of the equation:

$3s^2+12s+k=0$

$s=-2\pm \frac{\sqrt{144-12k}}{6}$

Separate the terms of $\frac{\sqrt{144-12k}}{6}$ as $\frac{\sqrt{144}}{6}-\frac{\sqrt{12k}}{6}$

$\begin{array}{l}s=-2\pm \frac{\sqrt{144}}{6}-\frac{\sqrt{12k}}{6}\\ s=-2\pm \frac{12}{6}-\frac{\sqrt{4\times 3k}}{6}\\ s=-2\pm 2-\frac{2\sqrt{3k}}{6}\\ s=-2\pm 2-\frac{\sqrt{3k}}{\sqrt{3}\sqrt{3}}\\ s=-2\pm 2-\sqrt{\frac{k}{3}}\end{array}$

$s=-2\pm \sqrt{4-\frac{k}{3}}$

Roots of the given equation is:

$\begin{array}{l}{s}_1=-2+\sqrt{4-\frac{k}{3}}\\ s_2=-2-\sqrt{4-\frac{k}{3}}\end{array}$

Finding the value of sfor $k\ge 0$

For $k\le 12$, substitute the value of kin $s_1\&s_2$, that is, $k=12$

$s_1\&s_2=-2$

Substitute $k=0$ the

$s_1=0\&s_2=-4$

The two roots are real and equal for $k=12$ and two roots are real and unequal for $k=0$

Roots are $s=-2$ for $k=12$ and $s=0$ for $k=0$

For $k>12$ value of the roots is imaginary and varies upto infinity.

Roots for $k>12is-2\pm 0.5773j$.

Time constant is $\tau =\frac{-1}{\begin{array}{l}s\\ s=-2\end{array}}$

$\tau =\frac{1}{2}$

If $s=0$ then $\tau =\infty$.

For $0<k<12$, the value of time constant is always $\frac{1}{2}$ for $k=12$ and $\infty$ for $k=0$.

The smallest possible time constant is $\frac{1}{2}$ which occur for $k\ge 12$.

Root locus plot for the equation is:

Conclusion:

The smallest possible time constant is $\frac{1}{2}$ which occur for $k\ge 12$.