16. 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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17. The feedback system with characteristic equation
    s⁴ + 20 Ks³ + 5s² + 10s + 15 = 0 is
    

1. 1. stable for all value of K

   
   
   

   2. stable for positive value of K

   
   
   

   3. stable for ∞ > K > 7.0

   
   
   

   4. unstable for all values of K

   
   
   

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1. View Hint View Answer Discuss in Forum

   By Routh criterion
   
   For this, any value at K the system is unstable.

   Correct Option: D

   By Routh criterion
   
   For this, any value at K the system is unstable.

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18. Match List-I with List-II and select the correct answer using the codes given below the Lists :
    
    

1. 1. A - 4 , B - 3 , C - 2 , D - 1

   
   
   

   2. A - 3 , B - 4 , C - 1 , D - 2

   
   
   

   3. A - 3 , B - 4 , C - 2 , D - 1

   
   
   

   4. A - 4 , B - 2 , C - 3 , D - 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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19. 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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20. A linear discrete-time system has t he characteristic equation, z³ – 0.81z = 0
    The system
    

1. 1. is stable

   
   
   

   2. is marginally stable

   
   
   

   3. is unstable

   
   
   

   4. stability cannot be assessed from the given information

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here linear discrete time system has characteristic equation
   z³ – 0. 81 z = 0
   ⇒ z (z² – 0.92) = 0
   ⇒ z (z – 0.9) (z + 0.9) = 0
   Hence z = 0, 0.9, – 0.9
   All roots lie inside unit circle. Hence the system is stable .

   Correct Option: A

   Here linear discrete time system has characteristic equation
   z³ – 0. 81 z = 0
   ⇒ z (z² – 0.92) = 0
   ⇒ z (z – 0.9) (z + 0.9) = 0
   Hence z = 0, 0.9, – 0.9
   All roots lie inside unit circle. Hence the system is stable .

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