126. The Bode plot of a transfer function G (s) is shown in the figure below.
     
     The gain (20 log| G(s)|) is 32 dB and – 8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all ω. Then G(s) is

1. 1. 	39.8
      	s

   
   
   

   2. 	39.8
      	s2

   
   
   

   3. 	32
      	s

   
   
   

   4. 	32
      	s2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   ⇒ 20 log k = 32
   k = 10^1.6 = 39.8
   ⇒ As slope is – 40db/decade so two pols at ons’m
   

   so , T(s) =	39.8
   	52

   
   

   Correct Option: B

   
   ⇒ 20 log k = 32
   k = 10^1.6 = 39.8
   ⇒ As slope is – 40db/decade so two pols at ons’m
   

   so , T(s) =	39.8
   	52

   
   

---

127. The signal flow graph for a system is given below. The transfer function	Y(S)
     	U(S)

     
     for this system is
     

1. 1. 	s + 1
      	5s2 + 6s + 2

   
   
   

   2. 	s + 1
      	s2 + 6s + 2

   
   
   

   3. 	s + 1
      	s2 + 4s + 2

   
   
   

   4. 	1
      	5s2 + 6s + 2

      
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   Forward path
   
   Loop L₁ = – 4S^{– 1}
   L₂ = – 2S^{– 1} S^{– 1} = – 2S^{– 2}
   L₃ = – 2S^{– 1}
   L₄ = – 4
   

   T(s) =	S– 2 + S– 1
   	1 + 4 + 2S– 1 + 2S– 2 + 4S– 1

   
   

   =	S–1 + S– 2	=	S + 1
   	S + 6S–1 + 2S– 2		5S2 + 6S + 2

   Correct Option: A

   
   Forward path
   
   Loop L₁ = – 4S^{– 1}
   L₂ = – 2S^{– 1} S^{– 1} = – 2S^{– 2}
   L₃ = – 2S^{– 1}
   L₄ = – 4
   

   T(s) =	S– 2 + S– 1
   	1 + 4 + 2S– 1 + 2S– 2 + 4S– 1

   
   

   =	S–1 + S– 2	=	S + 1
   	S + 6S–1 + 2S– 2		5S2 + 6S + 2

---

---

128. The impulse response of a continuous time system is given by h(t) = δ(t – 1) + δ(t – 3). The value of the step response at t = 2 is
     

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. 3

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Apply Laplace transform
   h(s) = e ^–s + e ^–3s
   
   for input step voltage →
   

   y(s) = h(s)	1
   	s

   
   

   = [e–s + e–3s]	1
   	s

   
   y(t) = u(t – 1) + u(t – 3)
   u(t) = 1 for t 30 = 0 prt < 0
   ∴ O/P in y(2) u(t – 1) + u(t – 3) = (4 – 1)
   = 4(z – 1) + 4 (2 – 3)
   = 4(1) + (4 – 1) = 0 + 1 = 1
   

   Correct Option: B

   Apply Laplace transform
   h(s) = e ^–s + e ^–3s
   
   for input step voltage →
   

   y(s) = h(s)	1
   	s

   
   

   = [e–s + e–3s]	1
   	s

   
   y(t) = u(t – 1) + u(t – 3)
   u(t) = 1 for t 30 = 0 prt < 0
   ∴ O/P in y(2) u(t – 1) + u(t – 3) = (4 – 1)
   = 4(z – 1) + 4 (2 – 3)
   = 4(1) + (4 – 1) = 0 + 1 = 1
   

---

129. A system with transfer function G(s) =	(s2 + 9)(s + 2)	is excited by sin (ωt).
     	(s + 1)(s + 3)(s + 4)

     
     The steady-state output of the system is zero at
     

1. 1. ω = 1 rad/s

   
   
   

   2. ω = 2 rad/s

   
   
   

   3. ω = 3 rad/s

   
   
   

   4. ω = 4 rad/s

   
   
   

---

1. View Hint View Answer Discuss in Forum

   |G(s)| =	(9 - ω2) √4 + ω²	= 0
   	√ω² + 1√ω² + 9√16 + ω²

   
   ∴ ω² = 9
   ⇒ ω = 3 rad / sec

   Correct Option: C

   |G(s)| =	(9 - ω2) √4 + ω²	= 0
   	√ω² + 1√ω² + 9√16 + ω²

   
   ∴ ω² = 9
   ⇒ ω = 3 rad / sec

---

---

130. The unilateral Laplace transform of f(t) is	1	. The unilateral Laplace
     	s2 + s + 1

     
     transform of t f(t) is

1. 1. -	s
      	(s2 + s + 1)2

   
   
   

   2. -	2s + 1
      	(s2 + s + 1)2

   
   
   

   3. 	s
      	(s2 + s + 1)2

   
   
   

   4. 	2s + 1
      	(s2 + s + 1)2

      
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   L[f(t)] = F(s) =	1
   	s2 + s + 1

   
   

   L[f(t)] = (-1)	dF(s)
   	ds

   
   

   = (-1)		-(2s + 1)		=	2s + 1
   		s2 + s + 1			s2 + s + 1

   
   

   Correct Option: D

   L[f(t)] = F(s) =	1
   	s2 + s + 1

   
   

   L[f(t)] = (-1)	dF(s)
   	ds

   
   

   = (-1)		-(2s + 1)		=	2s + 1
   		s2 + s + 1			s2 + s + 1

   
   

---