181. Let the Laplace transform of a function f(t) which exists for t > 0 be F₁ (s) and the Laplace transform of its delayed version f(t – π) be F₂ * (s). Let F₁ * (s) be the complex conjugate of F₁ (s) with the Laplace variable set as s = σ + jω.
     

     G(s) =	F2(s).F1*(s)	, then the inverse Laplace transform of G(s) is
     	|F1(s)|2

1. 1. an ideal impulse δ(t)
      

   
   
   

   2. an ideal delayed impulse δ(t – τ)

   
   
   

   3. an ideal step function u(t)

   
   
   

   4. an ideal delayed step function impulse u(t – τ)

   
   
   

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1. View Hint View Answer Discuss in Forum

   F₂ (t) = L{f(t – τ)} = e ^{– Sτ} F₁ (S)
   

   G(s) =	e– sτ F1 (s) . F1* (s)	= e– sτ
   	|F1 (s)|2

   
   G(t) = L^{– 1} {G(S)} = δ(t – τ)

   Correct Option: B

   F₂ (t) = L{f(t – τ)} = e ^{– Sτ} F₁ (S)
   

   G(s) =	e– sτ F1 (s) . F1* (s)	= e– sτ
   	|F1 (s)|2

   
   G(t) = L^{– 1} {G(S)} = δ(t – τ)

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182. The open loop transfer function G(s) of a unity feedback control system is given as,
     

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

     
     From the root locus, it can be inferred that when k tends to positive infinity,
     

1. 1. three roots with nearly equal real parts exist on the left half of the s-plane

   
   
   

   2. one real root is found on the right half of s-plane

   
   
   

   3. the root loci cross the jω axis for a finite value of k; k ≠ 0

   
   
   

   4. three real roots are found on the right half of the s-plane

   
   
   

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1. View Hint View Answer Discuss in Forum

   Centroid , σ =	-2 -		-	2		=	-6 + 2	=	-4	= -	2
   				3
   		3 - 1					6		6		3

   
   
   

   Asymptotes =	(29 ± 1) 180
   	p - z

   
   

   θ1 =	180	= 90°
   	2

   
   

   and θ2 =	180 × 3	= 270°
   	2

   
   

   Correct Option: A

   Centroid , σ =	-2 -		-	2		=	-6 + 2	=	-4	= -	2
   				3
   		3 - 1					6		6		3

   
   
   

   Asymptotes =	(29 ± 1) 180
   	p - z

   
   

   θ1 =	180	= 90°
   	2

   
   

   and θ2 =	180 × 3	= 270°
   	2

   
   

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183. A two-loop position control system is shown below.
     
     The gain k of the Tacho-generator influences mainly
     

1. 1. peak overshoot

   
   
   

   2. natural frequency of oscillation

   
   
   

   3. phase shift of the closed loop transfer function at very low frequencies (ω → 0)

   
   
   

   4. phase shift of the closed loop transfer function at very high frequencies (ω → ∞)

   
   
   

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1. View Hint View Answer Discuss in Forum

   	Y(s)	=	1
   	R(s)		s2 + (k + 1)s + 1

   
   2ω_n = k + 1
   

   ⇒ ξ =		k + 1		;
   		2

   
   Peak over shoot = e^{– πξ} / (1 – ξ²)
   

   Correct Option: A

   	Y(s)	=	1
   	R(s)		s2 + (k + 1)s + 1

   
   2ω_n = k + 1
   

   ⇒ ξ =		k + 1		;
   		2

   
   Peak over shoot = e^{– πξ} / (1 – ξ²)
   

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184. As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10% tolerance in
     

     the forward path, and an attenuator of value	9	in the feedback path. The overall system
     	100

     
     gain is approximately
     
     

1. 1. 10 ± 1%
      

   
   
   

   2. 10 ± 2%

   
   
   

   3. 10 ± 5%

   
   
   

   4. 10 ± 10%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Gain, G(s) = 100 ± 10%
   

   H(s) =	9
   	100

   
   
   

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

   
   When G(s) = 100 + 10 = 110,
   

   C(s)	=	90				=	900
   R(s)		1 + 90 ×	9				91
   			100

   
   = 9.89 ≈ 9.9
   = 10 – 0.1 ≠ 10 – 1% of 10
   Hence overall system gain = 10 ± 1%

   Correct Option: A

   Gain, G(s) = 100 ± 10%
   

   H(s) =	9
   	100

   
   
   

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

   
   When G(s) = 100 + 10 = 110,
   

   C(s)	=	90				=	900
   R(s)		1 + 90 ×	9				91
   			100

   
   = 9.89 ≈ 9.9
   = 10 – 0.1 ≠ 10 – 1% of 10
   Hence overall system gain = 10 ± 1%

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185. For the system	2	the approximate time taken for a step response to
     	(s + 1)

     
     reach 98% of its final value is
     

1. 1. 1 s

   
   
   

   2. 2 s

   
   
   

   3. 4 s

   
   
   

   4. 8 s

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : G(s) =	2	=	A
   	s + 1		Ts + 1

   
   ∴ Time constant, T = 1
   For 98%, time required = 4T = 4 sec
   

   ∴	C(s)	=	2
   	R(s)		s + 1

   
   

   ∴ C(s) =	2	.	1
   	s + 1		s

   
   

   ⇒ C(s) = 2		1	-	1
   		s		s + 1

   
   ⇒ C(t) = 2 [1 - e^-t ] = 2 × 0.98
   ⇒ 1 – e^-t = 0.98
   ⇒ e ^–t = 0.02
   ⇒ t = - ln 0.2 = 3.91 sec

   Correct Option: C

   Given : G(s) =	2	=	A
   	s + 1		Ts + 1

   
   ∴ Time constant, T = 1
   For 98%, time required = 4T = 4 sec
   

   ∴	C(s)	=	2
   	R(s)		s + 1

   
   

   ∴ C(s) =	2	.	1
   	s + 1		s

   
   

   ⇒ C(s) = 2		1	-	1
   		s		s + 1

   
   ⇒ C(t) = 2 [1 - e^-t ] = 2 × 0.98
   ⇒ 1 – e^-t = 0.98
   ⇒ e ^–t = 0.02
   ⇒ t = - ln 0.2 = 3.91 sec

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