46. If x(t) = sin 2πt e^{– t} u(t), then its Fourier transform is—

1. 1. 1			1	–	1
      2			1 + j(ω – 2π)		1 + j(ω + 2π)

   
   
   

   2. 1			1	–	1
      2j			1 + j(ω – 2π)		1 + j(ω + 2π)

   
   
   

   3. 1			1	–	1
      2j			1 + j(ω + 2π)		1 + j(ω – 2π)

   
   
   

   4. 1			1	–	1
      2j			1 + j(ω + 2π)		1 + j(ω – 2π)

   
   
   

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1. View Hint View Answer Discuss in Forum

   x(t) = sin 2πte^–t u(t)
   

   x1(t) = e–t u(t) ←F.T.→	1	= x1(jω)
   	1 + jω

   
   

   ∴ sin 2πt =	ej2πt – e–j2πt
   	j2

   
   So,
   

   x(t) =		ej2πt – e–j2πt		e–t u(t)
   		2j

   
   By using frequency shifting property
   

   x(t) =	1ej2πt	x1(t) –	e–j2πt	x1(t) ←F.T.→
   	j2		j2

   
   

   1		1	−	1		= x(jω)
   2j		1 + j(ω – 2π)		1 + j(ω + 2π)

   
   Hence, alternative (C) is the correct choice.

   Correct Option: C

   x(t) = sin 2πte^–t u(t)
   

   x1(t) = e–t u(t) ←F.T.→	1	= x1(jω)
   	1 + jω

   
   

   ∴ sin 2πt =	ej2πt – e–j2πt
   	j2

   
   So,
   

   x(t) =		ej2πt – e–j2πt		e–t u(t)
   		2j

   
   By using frequency shifting property
   

   x(t) =	1ej2πt	x1(t) –	e–j2πt	x1(t) ←F.T.→
   	j2		j2

   
   

   1		1	−	1		= x(jω)
   2j		1 + j(ω – 2π)		1 + j(ω + 2π)

   
   Hence, alternative (C) is the correct choice.

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47. If X(jω) = 2πδ(ω) + πδ(ω – 4π) + πδ(ω + 4π) then x(t) is—

1. 1. 2π(1 – cos 4πt)

   
   
   

   2. π(1 – cos 4πt)

   
   
   

   3. 1 + cos 4πt

   
   
   

   4. 2π(1 + cos 4πt)

   
   
   

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1. View Hint View Answer Discuss in Forum

   Inverse Fourier transform of X(jω) is given as
   

   x(t) =	1		∞	X(jω).ejωtdω
   	2π		– ∞

   
   

   x(t) =	1		∞	[2πδ(ω) + πδ(ω – 4π) + πδ(ω + 4π)].ejωtdω
   	2π		– ∞

   
   

   x(t) =	1	[2π + π.ej4πt + πe–j4πt]
   	2π

   
   

   = 1 +	1ej4πt	+	1	e–j4πt
   	2		2

   
   = 1 + cos 4πt
   

   Correct Option: C

   Inverse Fourier transform of X(jω) is given as
   

   x(t) =	1		∞	X(jω).ejωtdω
   	2π		– ∞

   
   

   x(t) =	1		∞	[2πδ(ω) + πδ(ω – 4π) + πδ(ω + 4π)].ejωtdω
   	2π		– ∞

   
   

   x(t) =	1	[2π + π.ej4πt + πe–j4πt]
   	2π

   
   

   = 1 +	1ej4πt	+	1	e–j4πt
   	2		2

   
   = 1 + cos 4πt
   

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48. X(jω) = e^{– 2ω} u(ω) then x(t)—

1. 1. 1

      π(2 + jt)

   
   
   

   2. 1

      π(2 – jt)

   
   
   

   3. 1

      2π(2 + jt)

   
   
   

   4. 1

      2π(2 – jt)

   
   
   

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1. View Hint View Answer Discuss in Forum

   x(t) =	1		∞	X(jω).ejωtdω
   	2π		– ∞

   
   

   =	1		∞	e–2ω.ejωt dω
   	2π		– ∞

   
   

   =	1		e(jt – 2)ω		∞
   	2π		(jt – 2)		0

   
   

   =	1		- 1		=	- 1
   	2π		(jt – 2)			2π(2 - jt)

   
   Hence, alternative (D) is the correct choice.

   Correct Option: D

   x(t) =	1		∞	X(jω).ejωtdω
   	2π		– ∞

   
   

   =	1		∞	e–2ω.ejωt dω
   	2π		– ∞

   
   

   =	1		e(jt – 2)ω		∞
   	2π		(jt – 2)		0

   
   

   =	1		- 1		=	- 1
   	2π		(jt – 2)			2π(2 - jt)

   
   Hence, alternative (D) is the correct choice.

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49. Given X(jω) =	jω + 3	then x(t)—
    	(jω + 1)2

1. 1. e^{– t} (1 + 2t) u(t)

   
   
   

   2. e^{– t} [1 + 2δ(t)] u(t)

   
   
   

   3. e^{– t} (1 + cos ωt + sin ωt) u(t)

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given
   

   X(jω) =	jω + 3
   	(jω + 1)2

   
   

   =	1	+	2
   	1 + jω		(1 + jω)2

   
   then
   x(t) = (e^–t + 2te^–t) u(t)

   Correct Option: A

   Given
   

   X(jω) =	jω + 3
   	(jω + 1)2

   
   

   =	1	+	2
   	1 + jω		(1 + jω)2

   
   then
   x(t) = (e^–t + 2te^–t) u(t)

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50. Calculate y(t) = x(1 – t) + x(– 1 – t)

1. 1. – 2ω3 sin ω

      1 + ω2

   
   
   

   2. j2ω3 cos ω

      1 + ω2

   
   
   

   3. – j2ω3 cos ω

      1 + ω2

   
   
   

   4. 2ω3 sin ω

      1 + ω2

   
   
   

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1. View Hint View Answer Discuss in Forum

   If x(t) ←F.T.→ X(jω)
   then
   x(– t) ←F.T.→ X(– jω)
   then
   x(– t + 1) ←F.T.→ e^jω X(– jω)
   then
   x(– t – 1) ←F.T.→ e^{- jω} X(– jω)
   Therefore,
   x(1 – t) + x(– t – 1) ←F.T.→ e^{- jω} X(– jω) + e^jω X(– jω)
   or
   

   Y(jω) = e–jω	(– jω)3	+	ejω (– jω)3
   	1 + (- ω)2		1 + (- ω)2

   
   

   = – e–jω	jω3	-	ejω jω3
   	1 + ω2		1 + ω2

   
   

   =	– jω2	2		ejω + e–jω
   	1 + ω2			2

   
   

   =	– jω2 2 cos ω
   	1 + ω2

   
   

   =	– 2jω2 cos ω
   	1 + ω2

   
   Hence, alternative (C) is the correct choice.

   Correct Option: C

   If x(t) ←F.T.→ X(jω)
   then
   x(– t) ←F.T.→ X(– jω)
   then
   x(– t + 1) ←F.T.→ e^jω X(– jω)
   then
   x(– t – 1) ←F.T.→ e^{- jω} X(– jω)
   Therefore,
   x(1 – t) + x(– t – 1) ←F.T.→ e^{- jω} X(– jω) + e^jω X(– jω)
   or
   

   Y(jω) = e–jω	(– jω)3	+	ejω (– jω)3
   	1 + (- ω)2		1 + (- ω)2

   
   

   = – e–jω	jω3	-	ejω jω3
   	1 + ω2		1 + ω2

   
   

   =	– jω2	2		ejω + e–jω
   	1 + ω2			2

   
   

   =	– jω2 2 cos ω
   	1 + ω2

   
   

   =	– 2jω2 cos ω
   	1 + ω2

   
   Hence, alternative (C) is the correct choice.

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