81. The performance specifications for a unity feedback control system having an open-loop transfer function
    

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

    
    are
    (i) Velocity error coefficient K_v > 10 sec^{– 1}
    (ii) Stable closed-loop operation.
    The value of K, satisfying the above specifications, is
    

1. 1. K > 6

   
   
   

   2. 6 < K < 10

   
   
   

   3. K > 10

   
   
   

   4. none of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Given : K_v > 10, then K > 20 sec^{– 1} .

   Correct Option: D

   
   Given : K_v > 10, then K > 20 sec^{– 1} .

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82. Given : KK_t = 99; s = j1 rad/s
    The sensitivity of the closed-loop system (shown in the figure) to variation in parameter K is approximately
    
    

1. 1. 0.01

   
   
   

   2. 0.1
      

   
   
   

   3. 1.0

   
   
   

   4. 10

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : KK_t = 99; s = j1 rad/sec.
   
   

   SKM =		dM
   		M		=	K	dM
   	1 +	10			M	dK
   		K

   
   

   Now M =	G	=		K		=	K
   				(10s + 1)
   	1 + GH		1 +	KKt			10s + 1 + KKt
   				(10s + 1)

   
   

   ∴	dM	=	1 × (10s + 1 + KKt) - KKt
   	dK		(10s + 1 + KKt)2

   
   

   =	10s + 1
   	(10s + 1 + KKt)2

   
   

   ⇒	K	dM	=	K
   	M	dK		K
   				(10s + 1 + KKt)

   
   

   =	(10s + 1)
   	(10s + 1 + KKt)2

   
   

   =	(10s + 1)
   	(10s + 1 + KKt)

   
   

   =	(10s + 1)
   	10(10 + s)

   
   

   With s = j1,	K	dM	=	√100 + 1	= 0.1
   	M	dK		10 √100 + 1

   
   

   Correct Option: B

   Given : KK_t = 99; s = j1 rad/sec.
   
   

   SKM =		dM
   		M		=	K	dM
   	1 +	10			M	dK
   		K

   
   

   Now M =	G	=		K		=	K
   				(10s + 1)
   	1 + GH		1 +	KKt			10s + 1 + KKt
   				(10s + 1)

   
   

   ∴	dM	=	1 × (10s + 1 + KKt) - KKt
   	dK		(10s + 1 + KKt)2

   
   

   =	10s + 1
   	(10s + 1 + KKt)2

   
   

   ⇒	K	dM	=	K
   	M	dK		K
   				(10s + 1 + KKt)

   
   

   =	(10s + 1)
   	(10s + 1 + KKt)2

   
   

   =	(10s + 1)
   	(10s + 1 + KKt)

   
   

   =	(10s + 1)
   	10(10 + s)

   
   

   With s = j1,	K	dM	=	√100 + 1	= 0.1
   	M	dK		10 √100 + 1

   
   

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83. If the unit step response of a network is (1 – e ^{– α t}), then its unit impulse response will be
    

1. 1. α e ^{– α t}
      

   
   
   

   2. α e ^{– t / α}
      

   
   
   

   3. 	1
      	αe – α t

      
      

   
   
   

   4. (1 - α ) e ^{– α t}

   
   
   

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1. View Hint View Answer Discuss in Forum

   Unit impulse response of a linear, time invariant network is the derivative of unit step function response of the network.
   Unit step response is (1 – e ^{– α t})
   ∴ Unit impulse response = – (– α) e^{– α t} = α e ^{– α t}

   Correct Option: A

   Unit impulse response of a linear, time invariant network is the derivative of unit step function response of the network.
   Unit step response is (1 – e ^{– α t})
   ∴ Unit impulse response = – (– α) e^{– α t} = α e ^{– α t}

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84. A system is represented by
    

    	dy	+ 2y = 4t u(t)
    	dt

    
    The ramp component in the forced response will be

1. 1. t u(t)
      

   
   
   

   2. 2t u (t)

   
   
   

   3. 3t u (t)

   
   
   

   4. 4t u (t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given :	dy	+ 2y = 4t u(t)
   	dt

   
   Laplace transofrming is
   

   sY(s) + 2Y(s) =	4
   	s2

   
   

   Y(s)[s + 2] =	4
   	s2

   
   

   ⇒ Y(s) =	4	= -	1	+	2	+	1
   	s2(s + 2)		s		s2		(s + 2)

   
   ⇒ y(t) = – u(t) + 2t u(t) + e ^{– 2t}
   The ramp component in the forced response is 2t u(t)
   

   Correct Option: B

   Given :	dy	+ 2y = 4t u(t)
   	dt

   
   Laplace transofrming is
   

   sY(s) + 2Y(s) =	4
   	s2

   
   

   Y(s)[s + 2] =	4
   	s2

   
   

   ⇒ Y(s) =	4	= -	1	+	2	+	1
   	s2(s + 2)		s		s2		(s + 2)

   
   ⇒ y(t) = – u(t) + 2t u(t) + e ^{– 2t}
   The ramp component in the forced response is 2t u(t)
   

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85. What will be the closed loop transfer function of a unity feedback control system whose step response is given by
    c(t) = K [1 – 1.66e^{– 8 t} sin (6t + 37°)]
    

1. 1. 	100 K
      	s2 + 16s + 100

      
      

   
   
   

   2. 	10 K
      	s2 + 16s + 100

      
      

   
   
   

   3. 	K
      	s2 + 16s + 100

      
      

   
   
   

   4. 	10 K
      	s2 + 8s + 10

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   M(s) =	C(s)
   	R(s)

   
   

   But R(s) =	1	=
   	c

   
   ∴ M (s) = s C (s)
   
   (ω_n √1 - δ² t - φ)
   Thus, 8 = δω_n
   and 6 = ω_n √1 - δ²
   ⇒ δ = 0.8
   and since ω_n = 10 satisfy these equations,
   therefore
   

   ≤ K [1 + 1.66 e– 8t sin (6t – 143°)] =	100 K
   	s(s2 + 16s + 100)

   
   

   ∴ M(s) = s C(s) =	100 K
   	s2 + 16s + 100

   Correct Option: A

   M(s) =	C(s)
   	R(s)

   
   

   But R(s) =	1	=
   	c

   
   ∴ M (s) = s C (s)
   
   (ω_n √1 - δ² t - φ)
   Thus, 8 = δω_n
   and 6 = ω_n √1 - δ²
   ⇒ δ = 0.8
   and since ω_n = 10 satisfy these equations,
   therefore
   

   ≤ K [1 + 1.66 e– 8t sin (6t – 143°)] =	100 K
   	s(s2 + 16s + 100)

   
   

   ∴ M(s) = s C(s) =	100 K
   	s2 + 16s + 100

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