61. The number of roots of the equation 2s⁴ + s³ + 3s² + 5s + 7 = 0 that lie in the right half of s plane is
    

1. 1. zero

   
   
   

   2. one

   
   
   

   3. two

   
   
   

   4. three

   
   
   

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1. View Hint View Answer Discuss in Forum

   Routh Hurwitz criterion
   
   Since sign changes twice, therefore 2 roots in right half plane.

   Correct Option: C

   Routh Hurwitz criterion
   
   Since sign changes twice, therefore 2 roots in right half plane.

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62. The open loop transfer function of a system is
    

    G(s) H(s) =	K(1 + s)2
    	s3

    
    The Nyquist plot for this system is

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   G(jω) H(jω) =	K(1 + jω)2
   	(jω)3

   
   

   |G(jω) H(jω)| =	K(1 + ω2)
   	ω3

   
   ∠ G(jω) H(jω) = – 270° + 2tan^-1 ω
   For ω = 0 , GH(jω) = ∞ ∠– 270°
   For ω = 1 , ∠GH(jω) = – 180°
   For ω = ∞ , GH(jω) = 0 ∠– 90°
   As ω increases from 0 to ∞, phase goes – 270° to – 90° . Due to s³ term there will be 3 infinite semicircle.
   

   Correct Option: B

   G(jω) H(jω) =	K(1 + jω)2
   	(jω)3

   
   

   |G(jω) H(jω)| =	K(1 + ω2)
   	ω3

   
   ∠ G(jω) H(jω) = – 270° + 2tan^-1 ω
   For ω = 0 , GH(jω) = ∞ ∠– 270°
   For ω = 1 , ∠GH(jω) = – 180°
   For ω = ∞ , GH(jω) = 0 ∠– 90°
   As ω increases from 0 to ∞, phase goes – 270° to – 90° . Due to s³ term there will be 3 infinite semicircle.
   

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63. A system has a complex conjugate root pair of multiplicity two or more in its characteristic equation. The impulse response of the system will be
    

1. 1. a sinusoidal oscillation which decays exponentially; the system is therefore stable.

   
   
   

   2. a sinusoidal oscillation with time multiplier; the system is therefore unstable.

   
   
   

   3. a sinusoidal oscillation which rises exponentially with time; the system is therefore unstable.

   
   
   

   4. a dc term and harmonic oscillation; the system therefore becomes limitingly stable.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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64. The open-loop transfer function of a unity feedback control system is
    

    G(s) H(s) =	30
    	s(s + 1)(s + T)

    
    where T is a variable parameter. The closed-loop system will be stable for all values of
    

1. 1. T > 0

   
   
   

   2. 0 < T < 3

   
   
   

   3. T > 5

   
   
   

   4. 3 < T < 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   G(s) H(s) =	30
   	s(s + 1)(s + T)

   
   The characteristics equation becomes
   s (s + 1) (s + T) + 30 = 0
   ⇒ s³ + s² (1 + T) + sT + 30 = 0
   Constructing Routh Tables
   
   (1 + T) > 0 or T > – 1
   

   ∴	(T2 + T - 30)	> 0
   	1 + T

   
   

   ⇒	(T + 6)(T - 5)	> 0
   	(1 + T)

   
   ⇒ T > – 6, or T > 5
   Therefore T > 5

   Correct Option: C

   G(s) H(s) =	30
   	s(s + 1)(s + T)

   
   The characteristics equation becomes
   s (s + 1) (s + T) + 30 = 0
   ⇒ s³ + s² (1 + T) + sT + 30 = 0
   Constructing Routh Tables
   
   (1 + T) > 0 or T > – 1
   

   ∴	(T2 + T - 30)	> 0
   	1 + T

   
   

   ⇒	(T + 6)(T - 5)	> 0
   	(1 + T)

   
   ⇒ T > – 6, or T > 5
   Therefore T > 5

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65. Consider the following polynomials :
    1. s⁴ + 7s³ + 17s² + 17s + 6.
    2. s⁴ + 11s³ + 41s² + 61s + 30.
    3. s⁴ + s³ + 2s² + 3s + 2.
    Among these polynomials, those which are Hurwitz are
    

1. 1. 1 and 3

   
   
   

   2. 2 and 3
      

   
   
   

   3. 1 and 2

   
   
   

   4. 1, 2 and 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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