The magnitude plot for the open-loop transfer function is shown b
The magnitude plot for the open-loop transfer function is shown below :
Its open-loop transfer function, G(s)H(s), is
A. 10(s+1)
B. <span class="math-tex">\(\frac{1}{s+1}\)</span>
C. <span class="math-tex">\(\frac{10}{s+1}\)</span>
D. 20(s+1)
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Concept:
Purpose of Bode plot
- To draw the frequency response of the system
- To find the closed-loop system stability.
Initial magnitude is 20logK always.
Gain margin and Phase margin both are calculated to find the closed-loop stability.
Calculation:
From the figure given starting magnitude is 20 dB and slope = 0.
Slope = 0 states that there is no pole or zero at the origin.
20logK = 20
logK = 1
K = 101 = 10
At ω = 1 rad/s slope is -20 dB it implies that there is a pole at that frequency.
So the transfer function will be:
\(G\left( s \right)H\left( s \right) = \frac{k}{{s + 1}}\)
Substituting the k value we get the final transfer function as:
\(G\left( s \right)H\left( s \right) = \frac{{10}}{{s + 1}}\)
Important points
NOTE: This is valid for Minimum phase system only.
Minimum phase system: All poles and zeroes are present at the left side of s-plane.
|
Condition |
Stability of the closed-loop system |
Gain margin in dB |
Gain margin in linear |
Phase margin |
|
ωpc > ωgc |
Stable |
Positive |
> 1 |
Positive |
|
ωpc = ωgc |
Marginal stable |
0 |
= 1 |
o° |
|
ωpc < ωgc |
Unstable |
Negative |
< 1 |
Negative |