136. Given H (s) =		X	=	s + 1	. x (t) = cos t 0º. The phasor Y is given by—
     		Y		s2 + s + 1

     
     

1. 1. 2 45°
      

   
   
   

   2. 1	45°
      2

   
   
   

   3. 1 0°

   
   
   

   4. 2	tan– 1 1– tan– 1 2
      5

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given H (S) =	X	=	s + 1	.....(1)
   	Y		s2 + s + 1

   
   X (t) = cost = 1∠ 0º.
   From input x (t) = cos t it is clear that ω = 1. By putting S = jω in equation (1), with ω = 1, we get
   

   	X	=	jω + 1	=	j + 1
   	Y		j 2 ω2 + jω + 1		–1 + jω + 1

   
   

   	j + 1	= √2	∠ 45°	= j2 ∠ – 45°
   	j		∠ 90°

   
   

   ⇒	1 ∠ 0°	= √2∠ – 45°
   	y

   
   

   or y =	1	– 45°
   	√2

   Correct Option: B

   Given H (S) =	X	=	s + 1	.....(1)
   	Y		s2 + s + 1

   
   X (t) = cost = 1∠ 0º.
   From input x (t) = cos t it is clear that ω = 1. By putting S = jω in equation (1), with ω = 1, we get
   

   	X	=	jω + 1	=	j + 1
   	Y		j 2 ω2 + jω + 1		–1 + jω + 1

   
   

   	j + 1	= √2	∠ 45°	= j2 ∠ – 45°
   	j		∠ 90°

   
   

   ⇒	1 ∠ 0°	= √2∠ – 45°
   	y

   
   

   or y =	1	– 45°
   	√2

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137. S is in position (a) for a long time. S moved to position (b) at t = 0—, At t = 0⁺, the values of I_C and V_L are—
     
     
     

1. 1. – 5, 0

   
   
   

   2. 5, 0

   
   
   

   3. 0, 4

   
   
   

   4. – 2, 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Where switch S is in position (a) for a long time in steady state in opened L is short. The circuit looks like
   
   

   l L = 5 ×	4	= 4A
   	4 + 1

   
   V_C (0^–) =V_C (0⁺) = 4 × 1 = 4 V
   As the switch is moved to position (b) current through 4 Ω resistor
   

   =4	4	= 1 A
   	4

   
   ∴ Current through capacitor = – (1 + 4) = – 5 A
   V_L = 4 × 1 = 4
   ∴ as the two voltage are in opposite sign
   so V_L = 0 V

   Correct Option: A

   Where switch S is in position (a) for a long time in steady state in opened L is short. The circuit looks like
   
   

   l L = 5 ×	4	= 4A
   	4 + 1

   
   V_C (0^–) =V_C (0⁺) = 4 × 1 = 4 V
   As the switch is moved to position (b) current through 4 Ω resistor
   

   =4	4	= 1 A
   	4

   
   ∴ Current through capacitor = – (1 + 4) = – 5 A
   V_L = 4 × 1 = 4
   ∴ as the two voltage are in opposite sign
   so V_L = 0 V

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138. S is open, V₀ = 6 V, i = 0. S is closed at t = 0, the values of i and at t = 0⁺ are given by—
     
     
     

1. 1. 0, 6

   
   
   

   2. 2, 0

   
   
   

   3. 1, 3

   
   
   

   4. 0, 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   V = L	di
   	dt

   
   
   

   6 = 2	di
   	dt

   
   

   or	di	= 3
   	dt

   
   when the switch is closed at t = 0⁺ there will be only transient current.
   i.e. i _L (0⁺)=0

   Correct Option: D

   V = L	di
   	dt

   
   
   

   6 = 2	di
   	dt

   
   

   or	di	= 3
   	dt

   
   when the switch is closed at t = 0⁺ there will be only transient current.
   i.e. i _L (0⁺)=0

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139. If the unit step response of a network is (1 – e^–αt), then its unit impulse response will be—
     

1. 1. α e^{– αt}

   
   
   

   2. 1	e–t/α
      α

   
   
   

   3. 1
      αe–t/α

   
   
   

   4. (1 – α) e^{– αt}

   
   
   

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1. View Hint View Answer Discuss in Forum

   In order to calculate the unit impulse response first we will obtain transfer function i.e. H (S)
   given y (t) = 1 – e^{– α t}
   

   	Y (s) =	1	-	1	=	α
   		s		s2 (s + α)		s(s + α)

   
   we know that
   

   H (s) =	Y (s)
   	X (s)

   
   X (t) = u (t)
   

   X (s) =	1
   	s

   
   

   	so, H (s) =	α	=	α
   		S(s + α) 1 / S		s + α

   
   Now, according to the question
   x (t) = δ (t)
   so, X (s)=1
   Y (t)=?
   Now, Y (s) = H (S) X (S)
   

   or Y (S) =	α	. 1
   	s + α

   
   

   or Y (S) =	α
   	s + α

   
   or Y (t) = α. e^{– αt}
   

   Correct Option: A

   In order to calculate the unit impulse response first we will obtain transfer function i.e. H (S)
   given y (t) = 1 – e^{– α t}
   

   	Y (s) =	1	-	1	=	α
   		s		s2 (s + α)		s(s + α)

   
   we know that
   

   H (s) =	Y (s)
   	X (s)

   
   X (t) = u (t)
   

   X (s) =	1
   	s

   
   

   	so, H (s) =	α	=	α
   		S(s + α) 1 / S		s + α

   
   Now, according to the question
   x (t) = δ (t)
   so, X (s)=1
   Y (t)=?
   Now, Y (s) = H (S) X (S)
   

   or Y (S) =	α	. 1
   	s + α

   
   

   or Y (S) =	α
   	s + α

   
   or Y (t) = α. e^{– αt}
   

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140. The impulse response of an R-L circuit is a—

1. 1. rising exponential function

   
   
   

   2. decaying exponential function

   
   
   

   3. step function

   
   
   

   4. parabolic function

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since I (s) =	V (s)
   	Z (s)

   
   
   V (t) = δ (t)
   V (s)= 1
   

   so, I (s) =	1
   	R + Ls

   
   

   L (t) =		1		e– (R / L) x t
   		L

   
   (which in a decaying exponential function)

   Correct Option: B

   Since I (s) =	V (s)
   	Z (s)

   
   
   V (t) = δ (t)
   V (s)= 1
   

   so, I (s) =	1
   	R + Ls

   
   

   L (t) =		1		e– (R / L) x t
   		L

   
   (which in a decaying exponential function)

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