Loading...

Loading, please wait...

Back to Topics

Control Systems
Frequency Domain Analysis

Practice questions from Frequency Domain Analysis.

5
Total
0
Attempted
0%
0
Correct
0%
0
Incorrect
0%
Q#1 Frequency Domain Analysis GATE EE 2025 (Set 1) MCQ +1 mark -0.33 marks

The Nyquist plot of a strictly stable  having the numerator polynomial as  encircles the critical point -1 once in the anti-clockwise direction. Which one of the following statements on the closed-loop system shown in figure, is correct?

The system stability cannot be ascertained.

The system is marginally stable.

The system is stable.

The system is unstable.

Explanation:

Given that open loop system  is stable.

 No. of open -loop poles in RH of s-plane

Also, No. of encirclements about critical point  is once,

 

Assume Nyquist Contour is clockwise

N = 1

N = P-Z

P= Open Loop Poles in RHP = 0

Z = Closed Loop Poles in RHP

Hence Z = -1 which is invalid

Now assume Nyquist Contour is Anti-Clockwise

N = -1 = P – Z

Hence Z = 1, which means one closed loop pole

lies in RHP and hence closed loop system is unstable

Option (d) is correct.

Q#2 Frequency Domain Analysis GATE EE 2024 (Set 1) NAT +2 marks -0 marks

Consider the stable closed-loop system shown in the figure. The asymptotic Bode magnitude plot of has a constant slope of  decade at least till with the gain crossover frequency being . The asymptotic Bode phase plot remains constant at  at least till . The steady-state error of the closed-loop system for a unit ramp input is ________ (rounded off to 2 decimal places).

Explanation:

Starting slope

 pole at origin

 

The value of frequency at which the initial line of slope -20dB/dec intersects 0dB line gives the value of velocity error constant.

Q#3 Frequency Domain Analysis GATE EE 2024 (Set 1) NAT +2 marks -0 marks

Consider the stable closed-loop system shown in the figure. The magnitude and phase values of the frequency response of  are given in the table. The value of the gain  for a  phase margin is ________ (rounded off to 2 decimal places).

Explanation:

 

 

 

 

Based on phase angle of transfer function G, we can determine the frequency,

From table,

and

By definition of ,

 

Q#4 Frequency Domain Analysis GATE EE 2023 (Set 1) MCQ +1 mark -0.33 marks

In the Nyquist plot of the open-loop transfer function

Corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in s-plane is mapped into a point at

Explanation:

Nyquist contour:

Infinite semicircular arc:  S =

θ:   to  –  

G (s) H (s) =

=  

Q#5 Frequency Domain Analysis GATE EE 2023 (Set 1) MCQ +2 marks -0.66 marks

The magnitude and phase plots of an LTI system are shown in the figure. The transfer function of the system is

Explanation:

Magnitude of gain = 8dB = 20 log K

Phase is linear &  at

Or rad at ω = 1

∴ Slope of phase plot

So, transfer function can be expressed as