Phase Margin MCQ Quiz in मल्याळम - Objective Question with Answer for Phase Margin - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 3, 2025

നേടുക Phase Margin ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Phase Margin MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Phase Margin MCQ Objective Questions

Top Phase Margin MCQ Objective Questions

Phase Margin Question 1:

Open loop transfer function of a unity feedback system is G(s)=Ks(s+10). The gain that will result in a phase margin of 60° is approximately

Answer (Detailed Solution Below) 66.6 - 66.7

Phase Margin Question 1 Detailed Solution

Phase marginPM=180+GH|ω=ωgc=60

GH|ω=ωgc=120=90+(tan1(ωgc10))=120tan1(ωgc10)=30ωgc=103

at ω=ωgc|G(jω)|=1

|Kωω2+100|ω=ωgc=1K103.1003+100=1K=103.10.13+1=1003×23=2003=66.67

Phase Margin Question 2:

The open loop transfer function of a unity feedback system is given by

G(s)=3e2ss(s+2).

Then the gain and phase margins of the system will be

  1. -7.09 dB and 87.5°

  2. 7.09 dB and 87.5°

  3. 7.09 dB and -87.5°

  4. -7.09 and -87.5°

Answer (Detailed Solution Below)

Option 4 :

-7.09 and -87.5°

Phase Margin Question 2 Detailed Solution

 

G(s)=3e2ss(s+2)

At gain crossover frequency, |G(jωg)|=1

            3ωgωg2+4=1

            ωg=1.26rad/s

At phase crossover frequency ∠G(s) = -180°

            180=2ωp90tan1(ωp2)

            2ωp+tan1(ωp2)=90

            ωp0.63rad

G.M=20log|G(jωp)|

|G(jωp)|=3.63(.63)2+4=2.27

G.M=20log2.27=7.12dB

G(jωg)=2ωgπ2tan1(ωg2)

=2×1.26π2tan1(1.262)

= -4.65 rad or -266.5°

P.M=180+G(jωg)=86.5

Phase Margin Question 3:

Phase margin in degree of G(s)=10(s+0.1)(s+1)(s+10) using asymptotic Bode plot is _____

  1. 45° 
  2. 135° 
  3. 180° 
  4. 0° 
  5. 90° 

Answer (Detailed Solution Below)

Option 1 : 45° 

Phase Margin Question 3 Detailed Solution

Asymptotic Bode plot for the given transfer function can be drawn as:

mob GATE LEVEL ques Images Q4

At ω=1 it cuts 0dB hence its gain crossover frequency is ω=1.

The phase of the system is

φ=tan1(ω0.1)tan1(ω1)tan1(ω10)

At ω = 1

φ=135o

Phase margin is,

PM=180+φ|ωgc

PM=180135=45

Phase Margin Question 4:

Which one of the following statements is correct?

  1. Phase margin is always positive for a stable feedback system.
  2. Phase margin is always negative for a stable feedback system.
  3. Phase margin can be negative or positive for a stable feedback system.
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Phase margin is always positive for a stable feedback system.

Phase Margin Question 4 Detailed Solution

Both gain Margin (dB) and phase margin should be positive for a stable feedback system.

Phase Margin Question 5:

The gain cross-over frequency and phase margin of the transfer function  are 

G(s)=1s(s+1)

  1. 1 rad/s and 45° 
  2. 2 rad/s and 45° 
  3. 2 rad/s and 135° 
  4. 1 rad/s and 135° 

Answer (Detailed Solution Below)

Option 1 : 1 rad/s and 45° 

Phase Margin Question 5 Detailed Solution

Concept:

Phase margin is the difference between the phase and 180°, for an output signal (relative to its input) at zero dB gain.

i.e. PM = 180° + ϕg, where

ϕg = Phase angle at the gain crossover frequency (ωg)

g = frequency at which the magnitude of G(s) H(s) becomes 1).

Calculation:

G(s)H(s)=1s(s+1)

Substituting s with jω

G(jω)H(jω)=1jω(jω+1)

Magnitude (M) =|G(jω)H(jω)|=1ω1+ω2 

At the gain cross over frequency ωg, M = 1.

1=1ωgωg2+11=1ωg2(ωg2+1)

⇒ ωg2 (ωg2 + 1) = 1

⇒ ωg4 + ωg2 – 1 = 0     …1)

Let ωg2 = x

Equation 1), becomes x2 + x – 1 = 0. Solving for ‘x’, we get:

x=1±1+42=1±52

ωg2=1+52

ωg=1+52

ωg = 0.786 = 1 rad/sec

Now, the phase at gain crossover frequency ⇒ ϕg

ϕg=π2tan1(ωg)

=π2tan1(1)

= -135°

So, PM (Phase margin) = 180 + ϕg = 180 – 135° =  45°

Phase Margin Question 6:

The open-loop transfer function of a unity feedback system is G(s)=Ks(s+5). The gain K that results in a phase margin of 45° is

  1. 35
  2. 30
  3. 25
  4. 20

Answer (Detailed Solution Below)

Option 1 : 35

Phase Margin Question 6 Detailed Solution

Concept:

The phase margin is given by

PM=180+G(jωgc)H(jωgc)

ωgc is gain crossover frequency.

At gain cross over frequency, the magnitude is unity.

|G(jωgc)H(jωgc)|=1

Calculation:

G(s)H(s)=Ks(s+5)

G(jωgc)H(jωgc)=90tan1(ωgc5)

Given that phase margin = 45°

18090tan1(ωgc5)=45

⇒ ωgc = 5 rad/sec

G(jω)H(jω)=Kjω(jω+5)

|G(jω)H(jω)|=Kωω2+25

|G(jωgc)H(jωgc)|=1

Kωgcωgc2+25=1

K=525+25=35.35

Phase Margin Question 7:

Consider a unity gain feedback control system whose open loop transfer function is G(S)=as+1S2. This system has a phase margin equal to π/4. Then the value of unit impulse response of the open loop system at t = 2 second is equal to ________.

Answer (Detailed Solution Below) 2.7 - 3

Phase Margin Question 7 Detailed Solution

G(S)=as+1S2

Given, Phase margin = π/4

180+GHjω|ω=ωgc=π4180+tan1(aω)180=π4

⇒ ωgc = 1/a rad/sec

At ωgc, |G(S)| = 1

|a2ω2+1ω2|=1a2(1a2)+1=1a2a2=12a=0.84

Now, G(S)=0.84S+1S2

H(S) = 1, R(S) = 1

⇒ C(S) = G(S) R(S)

=0.84s+1S2

Apply inverse Laplace on both sides

C(t)=L1[1+0.84sS2]

C(t) = (t + 0.84) u(t)

at = 2, C(t) = 2.84

Phase Margin Question 8:

A transfer function of a system is given by G(s)H(s)=10s+20 find P.M of the transfer function.

  1. 1
  2. 17.32
  3. 0
  4. 1.732

Answer (Detailed Solution Below)

Option 1 : ∞

Phase Margin Question 8 Detailed Solution

G(s)H(s)=10s+20

We know that P.M = 180 + (ϕ at ωgc)

|G(s)H(s)|=0.51+ωg2400=1

400400+ωgc2=4

ωgc=Imaginaryvalue

∴ ωgc does not exit

∴ P.M of the system will be ∞

Phase Margin Question 9:

The phase margin (PM) and the damping ration (ζ ) are related by

  1. PM=90tan1{2ζ2+1+4ζ42}
  2. PM=tan1{2ζ2ζ2+1+4ζ4}
  3. PM=90+tan1{2ζ2+1+4ζ22ζ}
  4. PM=180tan1{2ζ21+4ζ22}
  5. None of these

Answer (Detailed Solution Below)

Option 2 : PM=tan1{2ζ2ζ2+1+4ζ4}

Phase Margin Question 9 Detailed Solution

The OLTF of unity feedback second order prototype system is

G(s)=ωn2s(s+2ζωn)

Phase margin

= 180 + (∠GH)ω = ωgc

=180+[90tan1(ωgc2ζωn)]

=90tan1(ωgc2ζωn)      ----(1)

To calculate ωgc

G(jω)H(jω)=ωn2jω(jω+2εωn)

At, ω = ωgc

|G(jω)H(jω)|=1

ωn2ωgcωgc2+4ζ2ωn2=1

ωgc4+4ζ2ω2nωgc2ωn4=0

(ωgcωn)2=4ζ4+12ζ2

substitute value of ωgcωin (1)

=90tan1[12ζ[[4ζ4+1]122ζ2]1/2]

90tan1θ=cot1θ

=cot1[[(4ζ4+1)2ζ2]1/22ζ]

=tan1[2ζ[(4ζ4+1)122ζ2]1/2]

Phase Margin Question 10:

Phase margin of a system is used to specify

  1. Relative stability
  2. Absolute stability
  3. Time response
  4. Frequency response
  5. All of the above

Answer (Detailed Solution Below)

Option 1 : Relative stability

Phase Margin Question 10 Detailed Solution

Gain margin (GM): The gain margin of the system defines by how much the system gain can be increased so that the system moves on the edge of stability.

It is determined from the gain at the phase cross over frequency.

GM=1|G(jω)H(jω)|ω=ωpc

Phase crossover frequency (ωpc): It is the frequency at which phase angle of G(s) H(s) is -180°.

G(jω)H(jω)|ω=ωpc=180

Phase margin (PM): The phase margin of the system defines by how much the phase of the system can increase to make the system unstable.

PM=180+G(jω)H(jω)|ω=ωgc=180

It is determined from the phase at the gain cross over frequency.

Gain crossover frequency (ωgc): It is the frequency at which the magnitude of G(s) H(s) is unity.

|G(jω)H(jω)|ω=ωgc=1

Both gain margin and phase margin specify the relative stability of the system

Important Points:

  • If both GM and PM are positive, the system is stable (ωgc < ωpc)
  • If both GM and PM are negative, the system is unstable (ωgc > ωpc)
  • If both GM and PM are zero, the system is just stable (ωgc = ωpc)
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