System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 11, Problem 11.1P
To determine

To draw:

The Root locus plot of 3s2+12s+k=0fork0 and determine the value of smallest possible time constant and value of k.

Expert Solution & Answer
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Answer to Problem 11.1P

The smallest possible time constant is 12 which occur for k12.

Explanation of Solution

Given information:

The characteristic equation is given as 3s2+12s+k=0fork0.

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:

3s2+12s+k=0

s=2±14412k6

Separate the terms of 14412k6 as 144612k6

s=2±144612k6s=2±1264×3k6s=2±223k6s=2±23k33s=2±2k3

s=2±4k3

Roots of the given equation is:

s1=2+4k3s2=24k3

Finding the value of sfor k0

For k12, substitute the value of kin s1&s2, that is, k=12

s1&s2=2

Substitute k=0 the

s1=0&s2=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>12is2±0.5773j.

Time constant is τ=1ss=2

τ=12

If s=0 then τ=.

For 0<k<12, the value of time constant is always 12 for k=12 and for k=0.

The smallest possible time constant is 12 which occur for k12.

Root locus plot for the equation is:

System Dynamics, Chapter 11, Problem 11.1P

Conclusion:

The smallest possible time constant is 12 which occur for k12.

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