Test: Nyquist Plot - 2 - Electrical Engineering (EE) MCQ

# Test: Nyquist Plot - 2 - Electrical Engineering (EE) MCQ

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## 10 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Nyquist Plot - 2

Test: Nyquist Plot - 2 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Nyquist Plot - 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Nyquist Plot - 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Nyquist Plot - 2 below.
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Test: Nyquist Plot - 2 - Question 1

### A single-input single-output feedback system has forward transfer function G(s) and feedback transfer function H(s). It is given that |G(s)H(s)| < 1. Which of the following is true about the stability of the system?

Detailed Solution for Test: Nyquist Plot - 2 - Question 1

Concept:
D(s) = 1 + G(s)H(s)

D(s) gives the roots of characteristic equation i.e. closed-loop poles.

Nyquist stability criteria state that the number of unstable closed-loop poles is equal to the number of unstable open-loop poles plus the number of encirclements of the origin of the Nyquist plot of the complex function D(s).

It can be slightly simplified if instead of plotting the function D(s) = 1 + G(s)H(s), we plot only the function G(s)H(s) around the point and count encirclement of the Nyquist plot of around the point (-1, j0).

From the principal of argument theorem, the number of encirclements about (-1, j0) is

N = P - Z

Where

Where P = Number of open-loop poles on the right half of s plane

Z = Number of closed-loop poles on the right half of s plane

Calculation:

For the given system, we have |G(s)H(s)| < 1

So, we may easily conclude that the Nyquist-plot intersect the negative real axis between 0 and -1 point, i.e. the Nyquist plot does not enclose the point (-1, 0) or in other words number of encirclements is zero.

⇒ N = 0

For a stable closed loop system, we must have Z = 0.

To get Z = 0, from the Nyquist criteria, P must be equal to zero.

Therefore, the system is stable if all poles of G(s)H(s) are in left half of the s-plane.

Test: Nyquist Plot - 2 - Question 2

### From the below given Nyquist plot, calculate the number of open-loop poles on the right-hand side of the s-plane for the closed-loop system to be stable.

Detailed Solution for Test: Nyquist Plot - 2 - Question 2

Principle arguments

• It states that if there are “P” poles and “Z” zeroes for a closed, random selected path then the corresponding G(s)H(s) plane encircles the origin with P – Z times.
• Encirclements in s – plane and GH – plane are shown below.

• In GH plane Anti clockwise encirclements are taken as positive and clockwise encirclements are taken as negative.

It is applied to the total RH plane by selecting a closed path with r = ∞

Nyquist stability completely deals with the right half of s – plane.
N(0, 0) = P – Z
N(0, 0): Number of encirclements around critical point (- 1, 0)

P: Open loop poles

Z: Open-loop zeroes.

Note:

1) To get the Closed-loop stability we require 1 + GH plane but available is GH plane, hence the origin is shifted to “-1” to get the closed-loop stability.

2) To become the system stable there should not be any closed-loop pole in the right of s – plane.

3)  The closed-loop pole is the same as that of the zeroes of Characteristic Equation which must be zero in the right. i.e, Z = 0

N = P  is the criteria.

Calculation:

From the given Nyquist plot there is one encirclement about ( -1, 0 ) in the Anti-clockwise direction.

So, N = 1

Now to satisfy the stability criteria N should be equal to P.

N = P = 1

So the number of open-loop poles in the Right-hand side of the system is 1.

Test: Nyquist Plot - 2 - Question 3

### The most important technique used for stability and the transient response of the system is

Detailed Solution for Test: Nyquist Plot - 2 - Question 3

Definition:

The root locus plots the poles of the closed-loop transfer function in the complex s-plane as K varies from 0 to ∞

The purpose of Root locus is defined as:

• To find the nature of the system and ‘K’ value for stability.
• To find the relative stability.
• If the root locus branches move towards the right, system is less stable and if they move towards the left then the system is more relative stable.
• It is the best method to find the Relative stability and RH criteria is best to find absolute stability

The relation between the closed-loop and open-loop poles

D(s) = 0 gives open-loop poles and N(s) = 0 gives closed-loop poles

Characteristic equation is D(s) + kN(s) = 0.

Closed-loop poles are nothing but the sum of open-loop poles and zeroes.

Test: Nyquist Plot - 2 - Question 4

The number and direction of encirclements around the point −1+j0 in the complex plane by the Nyquist plot of

Detailed Solution for Test: Nyquist Plot - 2 - Question 4

Concept:

The Nyquist plot is equal to the polar plot & mirror Image of the polar plot with respect to the real axis, with opposite direction + semicircle of the infinite radius in a clockwise direction as many as the type of system.

Calculation:

Given:

So,

Polar plot of Given G(s) is

Now,
Nyquist plot of Given G(s) is:

Hence, the number of Encirclements of (-1 + j0) = 0

Test: Nyquist Plot - 2 - Question 5

In Nyquist plot of a system on adding a pole at s = 0, then plot will -

Detailed Solution for Test: Nyquist Plot - 2 - Question 5

Due to adding a pole at s = 0, the angle of shifting in the Nyquist plot of the system will be a shift of 90° clockwise.

Due to adding a zero at s = 0, the angle of shifting in the Nyquist plot of the system will be a shift of 90° anti-clockwise.

The shapes of the Nyquist plot for different transfer functions are given below:

Test: Nyquist Plot - 2 - Question 6

The frequency at which the Nyquist plot of a unity feedback system with the open loop transfer function crosses the negative real axis is

Detailed Solution for Test: Nyquist Plot - 2 - Question 6

Concept:

• Nyquist plot crosses the negative real axis at phase cross over frequency (ωpc)
• The phase crossover frequency is the frequency at which the phase angle first reaches −180°
• i.e., ∠(G(ωpc)) = -1800

Calculation:

Given G(s) =
∠(G(ωpc)) = 900 – 0.5 tan-1pc) – 4 tan-1pc) = -1800
By solving we get ωpc = √3 rad/sec

Test: Nyquist Plot - 2 - Question 7

If the Nyquist plot cuts the negative real axis at a distance of 0.8, then the gain margin of the system is

Detailed Solution for Test: Nyquist Plot - 2 - Question 7

Concept:
The gain margin is defined as the amount of change in open-loop gain needed to make a closed-loop system unstable.

where, |G| = Gain of the system
ωpc = Phase crossover frequency
The frequency at which the gain of the system is zero is known as the phase crossover frequency.
Calculation:

GM = 1/0.8
GM = 1.25

Test: Nyquist Plot - 2 - Question 8

The figure shows the Nyquist plot of the open-loop transfer function G(s)H(s) of a system. If G(s)H(s) has one right-hand pole, the closed-loop system is

Detailed Solution for Test: Nyquist Plot - 2 - Question 8

Concept:

The stability from the Nyquist plot is given by:

N = P - Z

where N = No. of encirclement of the critical point -1+j0 in an anticlockwise direction

P = No. of open-loop poles

Z = No. of zeroes of the characteristic equation

For a system to be stable, the value of Z = 0.

Calculation:

Given, P = 1

From the figure, N =1

N = P - Z

1 = 1 - Z

Z = 0

Hence, the system is stable.

Test: Nyquist Plot - 2 - Question 9

Consider a closed-loop control system with unity negative feedback and KG(s) in the forward path, where the gain K = 2. The complete Nyquist plot of the transfer function G(s) is shown in the figure. Note that the Nyquist contour has been chosen to have the clockwise sense. Assume G(s) has no poles on the closed right-half of the complex plane. The number of poles of the closed-loop transfer function in the closed right-half of the complex plane is ___________.

Detailed Solution for Test: Nyquist Plot - 2 - Question 9

Concept:

Nyquist stability criterion:

N = P – Z

N is the number of encirclements of (-1+j0) point by the Nyquist contour in an anticlockwise direction.

P is the open-loop RHP poles

Z is the closed-loop RHP poles

Analysis:

For K = 1,

For K = 2, the plot will be

N = No. of encirclement about (-1, 0) in anticlockwise.

P = Total number of open loop poles, in R.H.S.

Z = P - N

N = -2, P = 0

Z = 0 - (-2) = 2

Z = 2

Two poles in right side.

Test: Nyquist Plot - 2 - Question 10

The open loop transfer function of a unity gain negative feedback system is given as

The Nyquist contour in the ��-plane encloses the entire right half plane and a small neighbourhood around the origin in the left half plane, as shown in the figure below. The number of encirclements of the point (−1 + j0) by the Nyquist plot of G(s), corresponding to the Nyquist contour, is denoted as N. Then N equals to

Detailed Solution for Test: Nyquist Plot - 2 - Question 10

Concept

N = P - 2

N = no. of encirclements of )-1, 0) critical point by the Nyquist plot.

P = no. of right half of s-plane of G(s) H(s) as F(s)

z = no. of lright half of s-plane of CLTF as zero of F(s)

For stability z = 0

N - P = 0

N = P

for Nyquist stability criteria

Calculation

Open loop function:

Close loop transfer function =

no pole in right hand side

z = 0, P ⇒ 1

N = 1 - 0

N = 1

no. of oncirclements N = 1

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