Test: Nyquist Plot - 3 - Electrical Engineering (EE) MCQ

Nyquist plot:

1. Nyquist plots are an extension of polar plots for finding the stability of the closed-loop control systems. This is done by varying ω from −∞ to ∞, i.e. Nyquist plots are used to draw the complete frequency response of the open-loop transfer function.
2. In practice, it is not enough that a system is stable. There must also be some margins of stability that describe how stable the system is and its robustness. 
3. There are many ways to express this, but one of the most common is the use of gain and phase margins, inspired by Nyquist’s stability criterion. 
4. An increase in controller gain simply expands the Nyquist plot radially and an increase in the phase of the controller twists the Nyquist plot.
5. Hence from the Nyquist plot, we can easily pick off the amount of gain or phase that can be added without causing the system to become unstable. Hence option (c) is the correct answer.

Method of drawing Nyquist plot:

• Locate the poles and zeros of the open-loop transfer function G(s)H(s) in the ‘s’ plane.
• Draw the polar plot by varying ω from zero to infinity.
• Draw the mirror image of the above polar plot for values of ω ranging from −∞ to zero.
• The number of infinite radii half circles will be equal to the number of poles at the origin.
• The infinite radius half-circle will start at the point where the mirror image of the polar plot ends. And this infinite radius half-circle will end at the point where the polar plot starts.

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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

Calculation:

Number of open loops RHP pole (P) = 1
Characteristic equation: 1 + G(s) H(s) = 0

⇒ s² – 2s + 1 = 0

⇒ s = 1, 1

Number of closed loop RHP poles (Z) = 2

From Nyquist stability condition,

Number of encirclements N = P – Z = 1 – 2 = -1

Therefore, the Nyquist plot encircles (–1, 0) once in the clockwise direction.

In G(s) plane, the Nyquist plot of G(S) passes through the negative real axis at the point (-a, j0)

a = magnitude of G(s) at ω = ω_pc

ω_pc is phase cross over frequency.

at ω = ω_pc,∠G(jω) = −180^∘

⇒ -0.25 ω_pc – 90 = -180

⇒ ω_pc = 2π

(-a, j0) = (0.5, j0)

Concept:

For standard transfer function:

Nyquist plot = [Polar Plot] + [mirror image of polar plot w.r.t. x-axis but in opposite Direction] + [No. of semicircle of infinite radius in clockwise direction as many as type of system]

For ω(frequency) changing from 0 to ∞

Analysis:

Note: Above concept is only Valid for std. transfer function for Non-std. transfer function use the conventional method

Concept:

Cauchy principles argument states that the closed contour Γ is mapped into the G(s)-plane will encircle the origin as many times as the difference between the number of poles (P) and zeros (Z) of the open-loop transfer function G(s) that are encircled by the S – plane locus Γ, i.e.

No. of encirclement is given by:

N = P – Z

Calculation:

The closed contour Γ of a pole-zero map of a rational function G(s) contains 2 poles and 3 zeros.

So, the number of encirclement will be:

N = P – Z

N = 2 – 3 = -1

Hence,

It encircles the origin once in the clockwise direction.

Another method to solve:

The closed contour Γ of a pole-zero map of a rational function G(s) is encircling 2 poles and 3 zeros in a clockwise direction, hence the corresponding G(s) plane contour encircles origin 2 times in anti-clockwise direction and 3 times in clockwise direction.

Hence, Effectively it encircles origin once in the clockwise direction.

Special note:

• If we discuss the stability of the open-loop transfer function then we take encirclement around the origin.
• If we discuss the stability of closed-loop transfer function then we take encirclement around

 -1 + j0. (∴ Option 3 and 4 are incorrect)

G (jω) = 5 + jω

 ω = 0  

 G (jω) = 5 + j 0 

 ω = 10  

 G (jω) = 5 + j 10 

 ω = ∞ 

 G (jω) = 5 + j ∞ 

 ∴ G (jω) is a straight line parallel to jω axis

• Both Nyquist and RH criteria can be used to find the range of gain (k) for the system to be stable
• Bode plot can be used to find the transfer function which further can be used to determine the Nyquist Plot.
• RH criteria do not give accurate terms in case of exponential terms, an approximation can be obtained using exponential expansion up to 1st degree
• Nyquist plots are the continuation of polar plots for finding the stability of the closed-loop control systems by varying ω from −∞ to ∞.

Derivation:

Let us consider O.L.T.F of a unity feedback system as:

 

This represents a type = 1 and order – 2 system for which the Nyquist plot will be as shown below:

 

The closed-loop transfer function for the system is given as:

We can draw its Nyquist plot by simply substituting s = jω and obtain polar coordinates of T(jω) for different values of ω as:


So, the statement given in option (D) is false. 

Concept:

For Nyquist stability

N = P

P = Number of open-loop poles in RHS of s-plane

N = Number of encirclement about the point (-1 + j0)

N = +ve for anticlockwise

N = -ve for clockwise enduement

Calculation:

The polar plot is drawn as:

 

Let two points A and B representing -1 + j0 for different values of K,

The Nyquist plot is not encircling -1 + j 0 at A, i.e, K = 10

at B, the Nyquist plot is encircling -1 + j 0 i. e, K = 100

Let us now find the stability for K = 10 and K = 100 using Nyquist

Case: 1

K = 10

N = P - Z

Z = P - N

The Routh table can be drawn as follows:

No sign change, hence P = 0 and also N = 0

∴ Z = 0 → CLTP is stable

Case: 2

Z = P - N

N = -2 (two clockwise encirclement)

∴ Z = 2.

hence, CLTP is unstable.

We will observe the encirclement around the origin: (for open loop transfer function → G(s)H(s)= 0 )

N = P - Z

where,

P = the no. of open-loop poles lie in the right half of plane

Z = the no. of closed-loop poles or open-loop zeros lie in the right half of plane

N = no. of encirclement about (-1 + j0) in G(S) H(S) plane

OR,

no. of encirclement about the origin in 1 + G(S) H(S) plane.

Given;

Z = 1 (for open loop system)

N = 2 (at the origin in ACW direction)

then

2 = P - 1

∴ P = 3 

For the analysis of closed loop system:

observe the encirclement around the (-1 , j0) [∵ 1 + GH = 0 → GH = -1]

N = 0

P = 3

Then,

N = P - Z

Z= P - N

∴ Z= 3 - 0 = 3

There are 3 poles in the right half-plane. 

Nyquist plot:

• Nyquist plots are the continuation of polar plots for finding the stability of the closed-loop control systems by varying ω from −∞ to ∞.
• It provides information regarding stability as well as the relative stability of a system without the need to evaluate the roots of the characteristic equation.

Nyquist stability criteria:

Encirclement:

• A point (or) a region is encircled if it inside the closed path.
• Encirclements(N) are considered as positive in the counter-clockwise direction and negative in a clockwise direction.

Nyquist stability criteria:

The Nyquist plot will encircle critical point (-1, j0) as many number of times as the difference between, the number of right side poles and zeros of the characteristic equation.

N = P - Z

Here

N = Number of encirclements of (-1, j0) by the Nyquist plot.

P = Number of right side poles of G(s) H(s)

Z = Number of right side zeros or roots of the characteristic equation (or) right side poles of the closed-loop transfer function.

For stability, 'Z' must be zero.

N = P -Z

N = P - 0

⇒ N = P

Hence, for closed-loop stability, NSC states that critical point (-1, j0) should be encircled in the counterclockwise direction as many numbers of times as the number of right side poles of G(s) H(s) by the Nyquist plot if the Nyquist contour is defined in the clockwise direction.
Note: For open-loop stable system P = 0, that means for closed-loop stability N = 0 - Z, for stability Z should be zero, it is possible if there are no encirclements made by Nyquist plot.