1. The Nyquist plot of a loop transfer funct ion G(jω) H(jω) of a system encloses the (– 1, j0) point. The gain margin of the system is
   

1. 1. less than zero

   
   
   

   2. zero

   
   
   

   3. greater than zero

   
   
   

   4. infinity

   
   
   

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   Gain margin of the system for which Nyquist plot of a loop tr ansfer function G(jω) H(jω) encloses (– 1, + j0) point is less than zero.

   Correct Option: A

   Gain margin of the system for which Nyquist plot of a loop tr ansfer function G(jω) H(jω) encloses (– 1, + j0) point is less than zero.

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2. The position and velocity error coefficients for the system of transfer function
   

   G(s) =	50
   	(1 + 0.1s)(1 + 2s)

   
   respectively are

1. 1. zero and zero

   
   
   

   2. zero and infinity

   
   
   

   3. 50 and zero

   
   
   

   4. 50 and infinity

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: C

   

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3. Consider the following statements regarding timedomain analysis of a control system :
   1. Derivative control improves system’s transient performance
   2. I ntegral control does not improve system’s steady state performance
   3. I ntegral control can convert a second order system into a third order system
   Of these statements
   

1. 1. 1 and 2 are correct

   
   
   

   2. 1 and 3 are correct

   
   
   

   3. 2 and 3 are correct

   
   
   

   4. 1, 2 and 3 are correct

   
   
   

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1. View Hint View Answer Discuss in Forum

   For same K_V, the natural frequency ω_n remains unchanged in derivative control while damping is increased.
   Thus it improves the transient response.
   Integral compensator introduces a term 1/s, thus changes second order to third order.

   Correct Option: B

   For same K_V, the natural frequency ω_n remains unchanged in derivative control while damping is increased.
   Thus it improves the transient response.
   Integral compensator introduces a term 1/s, thus changes second order to third order.

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4. For the system shown in the figure, with a damping ratio ξ of 0.7. and undamped natural frequency ω_n of 4 rad / sec, the values of K and a are
   
   

1. 1. K = 4, a = 0.35

   
   
   

   2. K = 8, a = 0.455

   
   
   

   3. K = 16, a = 0.225

   
   
   

   4. K = 64, a = 0.9

   
   
   

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1. View Hint View Answer Discuss in Forum

   M(s) =	G(s)	=	K / s ( + 2)
   	1 + G(s) H(s)		1 + (1 + as) K / s (s + 2)

   
   The characteristic equation is
   s (s + 2) + K (1 + as) = 0
   ⇒ s² + s (2 + aK) + K = 0
   Compare with equation s ² + 2 ξ ω_n s + ω_n² = 0, we get
   K = ω_n ² = 4² = 16
   ∴ 2 ξ ω_n = (2 + aK)
   

   ⇒ a =	2 × 0.7 × 4 - 2	=	3.6	= 0.225
   	16		16

   
   

   Correct Option: C

   M(s) =	G(s)	=	K / s ( + 2)
   	1 + G(s) H(s)		1 + (1 + as) K / s (s + 2)

   
   The characteristic equation is
   s (s + 2) + K (1 + as) = 0
   ⇒ s² + s (2 + aK) + K = 0
   Compare with equation s ² + 2 ξ ω_n s + ω_n² = 0, we get
   K = ω_n ² = 4² = 16
   ∴ 2 ξ ω_n = (2 + aK)
   

   ⇒ a =	2 × 0.7 × 4 - 2	=	3.6	= 0.225
   	16		16

   
   

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5. The closed-loop transfer function of a control system is given by
   

   	C(s)	=	2(s - 1)
   	R(s)		(s + 2)(s + 1)

   
   For a unit step input the output is
   

1. 1. – 3e ^{– 2t} + 4e ^{– t} – 1

   
   
   

   2. – 3e ^{– 2t} – 4e ^{– t} + 1

   
   
   

   3. zero

   
   
   

   4. infinity

   
   
   

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1. View Hint View Answer Discuss in Forum

   	C(s)	=	2(s - 1)	, R(s) =	1
   	R(s)		(s + 2)(s + 1)		s

   
   

   Hence , C(s) =	2(s - 1)
   	s(s + 1)(s + 2)

   
   

   =	K1	+	K2	+	K3
   	s		s + 1		s + 2

   
   
   

   Hence , C(s) =	-1	+	4	-	3
   	s		s + 1		s + 2

   
   ⇒ c(t) = [– 1 + 4e^{– t} – 3e^{– 2t}]

   Correct Option: A

   	C(s)	=	2(s - 1)	, R(s) =	1
   	R(s)		(s + 2)(s + 1)		s

   
   

   Hence , C(s) =	2(s - 1)
   	s(s + 1)(s + 2)

   
   

   =	K1	+	K2	+	K3
   	s		s + 1		s + 2

   
   
   

   Hence , C(s) =	-1	+	4	-	3
   	s		s + 1		s + 2

   
   ⇒ c(t) = [– 1 + 4e^{– t} – 3e^{– 2t}]

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