1. The impulse response of an initially relaxed linear system is e^{– 2t} u(t). To produce a response of t e^–2t u (t), the input must be equal to—
   

1. 1. 2^e– t u(t)
      

   
   
   

   2. 1/2^e–2t u(t)
      

   
   
   

   3. e^–2t u(t)
      

   
   
   

   4. e^–t u(t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given
   y (t) = e^{– 2t} u(t)
   then
   

   Y(s) =	1
   	s + 2

   
   and R (s) = 1 (for impulse input)
   so,
   

   H(s) =	Y(s)	=	1	=	1
   	R(s)		S + 2/1		s + 2

   
   Now, for
   y(t) = te ^{– 2t} u(t)
   then
   

   Y(s) =	1
   	(s + 2)2

   
   so,
   

   X(s) =	Y(s)	=	1/(s + 2)2	=	1
   	H(s)		1/S + 2		s + 2

   
   or
   x (t) = e^–2tu(t)

   Correct Option: D

   Given
   y (t) = e^{– 2t} u(t)
   then
   

   Y(s) =	1
   	s + 2

   
   and R (s) = 1 (for impulse input)
   so,
   

   H(s) =	Y(s)	=	1	=	1
   	R(s)		S + 2/1		s + 2

   
   Now, for
   y(t) = te ^{– 2t} u(t)
   then
   

   Y(s) =	1
   	(s + 2)2

   
   so,
   

   X(s) =	Y(s)	=	1/(s + 2)2	=	1
   	H(s)		1/S + 2		s + 2

   
   or
   x (t) = e^–2tu(t)

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2. For a unit step input, a system with forward path transfer function G(s) = 20/s² and feedback path transfer function H(s) = (s + 5) has a steady state output of—
   

1. 1. 20
      

   
   
   

   2. 5
      

   
   
   

   3. 0.2
      

   
   
   

   4. zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   Steady state error, e_ss = _{s →∞}Lim s E (s)
   Given
   

   G(s) =	20
   	s2

   
   H(s) = s + 5
   r(t) = 1, then R (s) = 1/s
   we know that
   

   E(s)	=	1
   R(s)		1 + G(s) H(s)

   
   

   E(s)	=	1	=	1
   R(s)		1 + 20/s2(s + 5)		1/s2 + 20(s + 5)

   
   or
   

   E(s) = R(s)	s2
   	s2 + 20(s + 5)

   
   

   E(s) =	1	.	s2
   	s		s2 + 20(s + 5)

   
   so,
   

   s →∞Lim s·	s
   	s2 + 20(s + 5)

   
   

   = s →∞Lim s·	1
   	1 + 20/s + 5/s2

   
   e_ss = 0 (apply the limit)

   Correct Option: D

   Steady state error, e_ss = _{s →∞}Lim s E (s)
   Given
   

   G(s) =	20
   	s2

   
   H(s) = s + 5
   r(t) = 1, then R (s) = 1/s
   we know that
   

   E(s)	=	1
   R(s)		1 + G(s) H(s)

   
   

   E(s)	=	1	=	1
   R(s)		1 + 20/s2(s + 5)		1/s2 + 20(s + 5)

   
   or
   

   E(s) = R(s)	s2
   	s2 + 20(s + 5)

   
   

   E(s) =	1	.	s2
   	s		s2 + 20(s + 5)

   
   so,
   

   s →∞Lim s·	s
   	s2 + 20(s + 5)

   
   

   = s →∞Lim s·	1
   	1 + 20/s + 5/s2

   
   e_ss = 0 (apply the limit)

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3. The close-loop transfer function of a control system is given by

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

   For the input r (t) = sin t, the steady state value of C(t) is equal to—
   

1. 1. 1	cos t
      √2

      
      

   
   
   

   2. 1
      

   
   
   

   3. 1	sin t
      √2

      
      

   
   
   

   4. 1	sin t (1-π/4)
      √2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given
   

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

   
   r (t) = sin t
   In order to calculate the C(t), we will calculate the magnitude and phase by the term 1/(1 + s)
   

   Mag. of		1		=	1	=	1
   		1 + s			√1 + 1		√2

   
   

   Phase of		1		= - tan-1		1		=	–π
   		1 + s				1			4

   
   so,
   

   C(t) =	1	sin		t -	π
   	√2				4

   Correct Option: D

   Given
   

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

   
   r (t) = sin t
   In order to calculate the C(t), we will calculate the magnitude and phase by the term 1/(1 + s)
   

   Mag. of		1		=	1	=	1
   		1 + s			√1 + 1		√2

   
   

   Phase of		1		= - tan-1		1		=	–π
   		1 + s				1			4

   
   so,
   

   C(t) =	1	sin		t -	π
   	√2				4

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4. The error response of a second order system to a step input is obtained as
   E (t) = 1.66 e^–8t sin (6t + 37°).
   The damping ratio is—
   

1. 1. 0.4
      

   
   
   

   2. 0.5
      

   
   
   

   3. 0.8
      

   
   
   

   4. 1.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given E(t) = 1.66 e^–8t sin (6t + 37°) …(A)
   on comparing above equation (A) with the standard equation
   

   E(t) =	ωn	e–ωnt sin (ωn√1 – ξ2 t ± φ)
   	√1 - ξ2

   …(B)
   we get,
   ω_n √1 – ξ₂ = 6 …(i)
   ξω_n = 8 …(ii)
   From (i) and (ii)
   √1 – ξ² /ξ
   

   =	6	=	3
   	8		4

   
   

   or,	1 - ξ2	=	9	⇒ 16 - 16ξ2 = 9ξ2
   	ξ2		16

   
   25ξ² = 16
   

   ξ =	4	= 0.8
   	5

   Correct Option: C

   Given E(t) = 1.66 e^–8t sin (6t + 37°) …(A)
   on comparing above equation (A) with the standard equation
   

   E(t) =	ωn	e–ωnt sin (ωn√1 – ξ2 t ± φ)
   	√1 - ξ2

   …(B)
   we get,
   ω_n √1 – ξ₂ = 6 …(i)
   ξω_n = 8 …(ii)
   From (i) and (ii)
   √1 – ξ² /ξ
   

   =	6	=	3
   	8		4

   
   

   or,	1 - ξ2	=	9	⇒ 16 - 16ξ2 = 9ξ2
   	ξ2		16

   
   25ξ² = 16
   

   ξ =	4	= 0.8
   	5

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5. Given a system represented by equations
   

   x = -		0	1		x +		0		u and y = [1 0]x
   		-2	-3				1

   
   The equivalent transfer function representation G(s) of the system is—
   

1. 1. G(s) =	1
      	s2 + 5s + 2

      
      

   
   
   

   2. G(s) =	1
      	s2 + 3s + 2

      
      

   
   
   

   3. G(s) =	3
      	s2 + 3s + 2

      
      

   
   
   

   4. G(s) =	2
      	s3 + 3s2 + 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given equation
   

   x =		0	1		x +		0
   		-2	-3				1

   
   Y = [1 , 0]x
   Here,
   

   x =		x1		and Y =		y1
   		x2				y2

   
   However from the output expression it is seen that only y1 exists i.e.,
   y₁ = x_{2 …(i)
   Also,
   x1 = x2 …(ii)
   x2 = – 2x1 – 3x2 + u …(iii)
   from equations (ii) and (iii)
   x1 = – 2x1 – 3x1 + u …(iv)
   By taking Laplace transform of eqn. (iv), we get
   s² x1 (s) = – 2x11}or
   (s² + 3s + 2) x₁ (x) = u (s)
   or
   

   x1(s)	=	1
   u(s)		s2 + 3s + 2

   
   or
   

   G(s) =	1
   	s2 + 3s + 2

   Correct Option: B

   Given equation
   

   x =		0	1		x +		0
   		-2	-3				1

   
   Y = [1 , 0]x
   Here,
   

   x =		x1		and Y =		y1
   		x2				y2

   
   However from the output expression it is seen that only y1 exists i.e.,
   y₁ = x_{2 …(i)
   Also,
   x1 = x2 …(ii)
   x2 = – 2x1 – 3x2 + u …(iii)
   from equations (ii) and (iii)
   x1 = – 2x1 – 3x1 + u …(iv)
   By taking Laplace transform of eqn. (iv), we get
   s² x1 (s) = – 2x11}or
   (s² + 3s + 2) x₁ (x) = u (s)
   or
   

   x1(s)	=	1
   u(s)		s2 + 3s + 2

   
   or
   

   G(s) =	1
   	s2 + 3s + 2

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