16. In the unity feedback system shown below, the value of derivative feedback constant a, which will increase the damping ratio of the system to 0.7 is—
    
    
    
    

1. 1. 0.152
      

   
   
   

   2. 0.245
      

   
   
   

   3. 0.353
      

   
   
   

   4. 0.558

   
   
   

---

1. View Hint View Answer Discuss in Forum

   On solving block 2
   

   C(s)	=	8/s(s+2)
   R′(s)		1+8/s(s+2).as

   
   

   C(s)	=	8
   R′(s)		s2 + s (8a + 2)

   
   Now, again simplifying block 1
   

   C(s)	=	8/s2 + s(8a +2)
   R′(s)		8/s2 + s (8a + 2).1 + 1

   
   

   =	8
   	s2 + (8a + 2)s + 8

   
   C.E. is s² + (2 + 8a) s + 8 = 0
   on comparing with the standard equation
   s² + 2 ξ ω_ns + ω_n² = 0
   ω_n² = 8 ⇒ ω_n = 2√2
   2 ξ ω_n = 2 + 8a
   or
   

   ξ =	2 + 8a	⇒ 0·7 =	2 + 8a
   	2.ωn		2.2√2

   
   on solving we get a = 0.245

   
   

   Correct Option: B

   On solving block 2
   

   C(s)	=	8/s(s+2)
   R′(s)		1+8/s(s+2).as

   
   

   C(s)	=	8
   R′(s)		s2 + s (8a + 2)

   
   Now, again simplifying block 1
   

   C(s)	=	8/s2 + s(8a +2)
   R′(s)		8/s2 + s (8a + 2).1 + 1

   
   

   =	8
   	s2 + (8a + 2)s + 8

   
   C.E. is s² + (2 + 8a) s + 8 = 0
   on comparing with the standard equation
   s² + 2 ξ ω_ns + ω_n² = 0
   ω_n² = 8 ⇒ ω_n = 2√2
   2 ξ ω_n = 2 + 8a
   or
   

   ξ =	2 + 8a	⇒ 0·7 =	2 + 8a
   	2.ωn		2.2√2

   
   on solving we get a = 0.245

   
   

---

17. 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 + 4^{e– 1} – 1
      

   
   
   

   2. – 3e^–2t – 4e^–t + 1
      

   
   
   

   3. zero
      

   
   
   

   4. infinity

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given r (t) = 1 then R (s) = 1/s
   

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

   
   

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

   
   

   =	0.2 (s - 1)
   	s(s + 2)(s + 1)

   
   

   C(s) =	A	+	B	+	C
   	s		s + 1		s + 2

   
   

   A =	2 (s + 1)	|s=0 = -1
   	(s + 1)(s + 2)

   
   

   B =	2 (s - 1)	|s=1 = 4
   	s(s + 2)

   
   

   C =	2 (s - 1)	|s=2 = -3
   	s(s + 1)

   
   on putting the value of A, B, C we get
   

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

   
   Taking inverse Laplace transform, we get
   C (t) = – 1 + 4e^{– t} – 3e^{– 2t}

   Correct Option: A

   Given r (t) = 1 then R (s) = 1/s
   

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

   
   

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

   
   

   =	0.2 (s - 1)
   	s(s + 2)(s + 1)

   
   

   C(s) =	A	+	B	+	C
   	s		s + 1		s + 2

   
   

   A =	2 (s + 1)	|s=0 = -1
   	(s + 1)(s + 2)

   
   

   B =	2 (s - 1)	|s=1 = 4
   	s(s + 2)

   
   

   C =	2 (s - 1)	|s=2 = -3
   	s(s + 1)

   
   on putting the value of A, B, C we get
   

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

   
   Taking inverse Laplace transform, we get
   C (t) = – 1 + 4e^{– t} – 3e^{– 2t}

---

---

18. The overshoot of the system
    

    16K

    s(s2 + 2s + 16

    
    for a step input applied would be—
    

1. 1. 60%
      

   
   
   

   2. 40%
      

   
   
   

   3. 20%
      

   
   
   

   4. 10%

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given C.E. = s² + 2s + 16 = 0
   NOTE : Here reader not confused with
   C.E. = s (s² + 2s + 16).
   Because here s is the input term, on comparing this equation with the standard equation
   s² + 2ξ ω_ns + ω_n² = 0
   we get
   2 ξ ω_n = 2, ω_n ² = 16 so ω_n = 4.
   

   ξ =	2	=	2	= 0.25
   	2ωn		2 × 4

   
   so % M_p = e^{–π ξ}√1 – ξ² × 100
   = e ^{– 3·14 × 0·25} √1 – (.25)²
   = 40%
   

   Correct Option: B

   Given C.E. = s² + 2s + 16 = 0
   NOTE : Here reader not confused with
   C.E. = s (s² + 2s + 16).
   Because here s is the input term, on comparing this equation with the standard equation
   s² + 2ξ ω_ns + ω_n² = 0
   we get
   2 ξ ω_n = 2, ω_n ² = 16 so ω_n = 4.
   

   ξ =	2	=	2	= 0.25
   	2ωn		2 × 4

   
   so % M_p = e^{–π ξ}√1 – ξ² × 100
   = e ^{– 3·14 × 0·25} √1 – (.25)²
   = 40%
   

---

19. The position and acceleration error coefficients for the open-loop transfer function
    

    G(s) =	K
    	s2(s + 10) (s + 100)

    
    respectively are—
    

1. 1. zero and infinity
      

   
   
   

   2. infinity and zero
      

   
   
   

   3. K/100 and zero
      

   
   
   

   4. infinity and K/100

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Kp = _{s → 0}Lim
   

   G (s) = s → 0Lim	K	= ∞
   	s2(s + 10)(s + 100)

   
   Ka = _{s → 0}Lim s² G(s)
   

   = s → 0Lim	s2·K	= 0.001 K,
   	s2(s + 10)(s + 100)

   
   or
   K/1000

   Correct Option: D

   Kp = _{s → 0}Lim
   

   G (s) = s → 0Lim	K	= ∞
   	s2(s + 10)(s + 100)

   
   Ka = _{s → 0}Lim s² G(s)
   

   = s → 0Lim	s2·K	= 0.001 K,
   	s2(s + 10)(s + 100)

   
   or
   K/1000

---

---

20. 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

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Kp = s → 0Lim G (s) = s → 0Lim	50	= 50
   	(1 + 0·1s)(1 + 2s)

   
   Kv = _{s → 0}Lim sG (s)
   

   = s → 0Lim	s.50	= 0
   	(1 + 0·1s)(1 + 2s)

   
   

   Correct Option: C

   Kp = s → 0Lim G (s) = s → 0Lim	50	= 50
   	(1 + 0·1s)(1 + 2s)

   
   Kv = _{s → 0}Lim sG (s)
   

   = s → 0Lim	s.50	= 0
   	(1 + 0·1s)(1 + 2s)

   
   

---