Nyquist Path or Contour MCQ Quiz - Objective Question with Answer for Nyquist Path or Contour - Download Free PDF

Concept:

The Nyquist plot is used to assess the stability of a closed-loop control system using the open-loop transfer function .

Given transfer function:

Poles of G(s): and ⇒ Both in Left Half Plane (LHP), so open-loop system is stable.

According to the Nyquist criterion:

, where:

• N = number of encirclements of point (clockwise = negative, anticlockwise = positive)
• Z = number of right-half plane poles of closed-loop transfer function
• P = number of right-half plane poles of open-loop transfer function (here, P = 0)

Since the open-loop poles are in LHP, the Nyquist plot of does not encircle .

Hence, the number of encirclements is: zero

Final Answer:  zero

Nyquist Contour:

Given:

Put s = Re^jθ

G(s)H(s) = 3

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.

It can be seen from the plot that (-1, j0) lies below one of the curves i.e. when the plot will be completed (-1, j0) will be encircled.

P = 0, N ≠ 0 therefore Z ≠ 0

(-1, j0) is enclosed by the Nyquist plot, the closed-loop system is unstable.

Note: Until unless mentioned it must always be assumed that all open-loop poles lie in the left-hand side of the s plane

Concept:

From the principal of Argument theorem,

The number of encirclements about (-1, 0) is  

Where P = Number of open-loop poles on Right – Half of s – plane

Z = Number of Closed Loop Poles on Right Half of s – plane

Given that the closed-loop system is stable means:

∴ So the Nyquist encircles the s – plane point (-1 +j0) in the counter-clockwise direction as many times as the number of right half s – plane poles

Note:

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).

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 

⇒ -0.25 ω_pc – 90 = -180

⇒ ω_pc = 2π

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

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:

Given the open-loop transfer function, 

Clockwise encirclements = 1 i.e. N = -1

Open loop poles on right half of s plane, P = 1

The closed loop poles lying in the open right half of the s-plane,

Z = P – N = 1 – (-1) = 2

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.

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)

Nyquist Contour:

Given:

Put s = Re^jθ

G(s)H(s) = 3

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

Concept:

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 = ∞

Calculation:

Given transfer function,

Substitute s = jω in the above equation,

​

In the plot G(s)H(s) encircle -1 + j0 once in clockwise direction.

Given the open-loop transfer function,

⇒ s3 - 3s2 + s + 3 = 0

By routh hurwitz criterion 

There are two sign changes 

Therefore it is unstable with two right half of s-plane poles

Z = 2, P = 1

N = P - Z = 1 - 2 = -1

Once in clockwise direction

From the Nyquist plot, we have

So, the product of  and  is obtained as

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

From the principal of argument theorem the number of encirclements about (-1, 0) is  

Where P = Number of open loop poles on Right – Half of s – plane

Z = Number of Closed Loop Poles on Right Half of s – plane

Given that closed loop system is stable means 

∴ So the Nyquist encircles the s – plane point (-1 +j0) in the counter clockwise direction as many times as the number of right half s – plane poles