76. 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 would include—
    

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

   Since all the coefficients in all the there polynomials are positive, we need to construct Hurwitz table for further checking.

   
   

   Correct Option: C

   Since all the coefficients in all the there polynomials are positive, we need to construct Hurwitz table for further checking.

   
   

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77. The magnitude-frequency response of a control system is shown in the figure. The value of ω₁ and ω₂ are respectively—
    
    
    
    

1. 1. 10 and 200
      

   
   
   

   2. 20 and 200
      

   
   
   

   3. 20 and 400
      

   
   
   

   4. 100 and 400

   
   
   

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1. View Hint View Answer Discuss in Forum

   From given figure
   

   	6	= 20
   	log ω1 – log10 10

   
   or
   log ω₁ – 1 = 6/20
   or
   log ω₁ = 0·3 + 1
   or
   log ω₁ = 1·3
   or
   ω₁ = 10^1·3
   or
   ω₁ = 19·95 ≈ 20
   

   and	6	= 20
   	log 10ω – logω1

   
   

   or	26	= 20
   	log ω – log20

   
   or log ω – 1·3 = 26/20 = 1·3
   or
   log ω = 2·6
   or
   ω = 102·6 = 400
   Hence alternative (C) is the correct choice.

   
   

   Correct Option: C

   From given figure
   

   	6	= 20
   	log ω1 – log10 10

   
   or
   log ω₁ – 1 = 6/20
   or
   log ω₁ = 0·3 + 1
   or
   log ω₁ = 1·3
   or
   ω₁ = 10^1·3
   or
   ω₁ = 19·95 ≈ 20
   

   and	6	= 20
   	log 10ω – logω1

   
   

   or	26	= 20
   	log ω – log20

   
   or log ω – 1·3 = 26/20 = 1·3
   or
   log ω = 2·6
   or
   ω = 102·6 = 400
   Hence alternative (C) is the correct choice.

   
   

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78. The open-loop transfer function of a unity feedback control system is: G(s) = 1/(s + 2)² The closed loop transfer function will have poles at—
    

1. 1. – 2, – 2
      

   
   
   

   2. – 2, – 1
      

   
   
   

   3. – 2, ± j1
      

   
   
   

   4. – 2, ± j2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that G(s) =		1
   		(s + 2)2

   
   

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

   
   

   ∴ M(s) =		G(s)	=	1/(s + 2)2
   		1 + G(s)		1 + (1/s + 2)2

   
   = 1/(s + 2)² + 1
   The poles are at – 2 ± j1

   Correct Option: C

   Given that G(s) =		1
   		(s + 2)2

   
   

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

   
   

   ∴ M(s) =		G(s)	=	1/(s + 2)2
   		1 + G(s)		1 + (1/s + 2)2

   
   = 1/(s + 2)² + 1
   The poles are at – 2 ± j1

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