171. Let x(t) = sin2 t be represented as the complex Fourier series representation i.e.
     

     	∞
     		= Ck.ejkω0t
     	k = – ∞

     
     

     where, ω0 =	2π	← Fundamental period
     	T0

     
     The value of complex Fourier coefficient C₁ for x(t) is—

1. 1. 1

      2

   
   
   

   2. –	1
      	4

   
   
   

   3. –	1
      	2

   
   
   

   4. 1

      4

   
   
   

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1. View Hint View Answer Discuss in Forum

   To solve this problem first calculate the fundamental period of the given signal i.e.,
   x(t) = sin² t
   or
   

   x(t) =	1 – cos2x	= x(t) =	1 – cos2x
   	2		2

   
   (∴ cos² x = 1 – 2 sin² x)
   

   ∴ sin2x =	1 – cos2x
   	2

   
   FundamentaI period of x(t)
   

   x(t) = T0 =	2π
   	ω

   
   

   =	2π	= π
   	2

   
   

   Now, x(t) = sin2t =		ejt – e–jf		2
   		2j

   
   

   = –	1	[e2jt – 2 + e– 2jt] ....…(A)
   	4

   
   Also given that Fourier series representation of x(t), i.e.,
   

   	∞		∞
   x(t) =		Ck ejkω0t =		Ck ejk2t …(B)
   	– ∞		– ∞

   
   In order to find the Fourier coefficients compare equation (A) and (B), we get
   

   C1 = –	1
   	4

   
   

   C0 =	1
   	2

   
   

   C-1 = –	1
   	4

   
   and all other C_k = 0
   Hence, alternative (B) is the correct choice.

   Correct Option: B

   To solve this problem first calculate the fundamental period of the given signal i.e.,
   x(t) = sin² t
   or
   

   x(t) =	1 – cos2x	= x(t) =	1 – cos2x
   	2		2

   
   (∴ cos² x = 1 – 2 sin² x)
   

   ∴ sin2x =	1 – cos2x
   	2

   
   FundamentaI period of x(t)
   

   x(t) = T0 =	2π
   	ω

   
   

   =	2π	= π
   	2

   
   

   Now, x(t) = sin2t =		ejt – e–jf		2
   		2j

   
   

   = –	1	[e2jt – 2 + e– 2jt] ....…(A)
   	4

   
   Also given that Fourier series representation of x(t), i.e.,
   

   	∞		∞
   x(t) =		Ck ejkω0t =		Ck ejk2t …(B)
   	– ∞		– ∞

   
   In order to find the Fourier coefficients compare equation (A) and (B), we get
   

   C1 = –	1
   	4

   
   

   C0 =	1
   	2

   
   

   C-1 = –	1
   	4

   
   and all other C_k = 0
   Hence, alternative (B) is the correct choice.

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172. The function f(t) has the Fourier transform g(ω). The Fourier transform of
     

     =		∞	g(t) e– jωt dt is—
     		– ∞

1. 1. 	1	f(ω)
      	2π

   
   
   

   2. 	1	f(– ω)
      	2π

   
   
   

   3. 2πf(– ω)

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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173. A periodic signal x(t) of period T₀ is given by
     

     x(t) =		1, |t| < T,
     		0, T1 < |t| < T0/2

     
     The d.c. component of x(t) is—

1. 1. T1

      T0

   
   
   

   2. T1

      2T0

   
   
   

   3. 2T1

      T0

   
   
   

   4. T0

      T1

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given waveform like
   
   The d.c. component of x(t) is
   

   =	1		T0/2	x(t).dt
   	T0		– T0/2

   
   

   =	1		T1	1.dt =	2T1
   	T0		– T1		T0

   
   Hence, alternative (C) is correct choice.

   Correct Option: C

   The given waveform like
   
   The d.c. component of x(t) is
   

   =	1		T0/2	x(t).dt
   	T0		– T0/2

   
   

   =	1		T1	1.dt =	2T1
   	T0		– T1		T0

   
   Hence, alternative (C) is correct choice.

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174. 	∞
     The z-transform of time function		δ(n – k) is—
     	k = 0

1. 1. z – 1

      z

   
   
   

   2. z

      (z – 1)2

   
   
   

   3. z

      z – 1

   
   
   

   4. (z – 1)2

      z

   
   
   

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1. View Hint View Answer Discuss in Forum

   	∞
   Since, the value of		δ(n – k) = 1
   	k = 0

   
   

   Therefore, 1.u(n) ←z→	z
   	z – 1

   Correct Option: C

   	∞
   Since, the value of		δ(n – k) = 1
   	k = 0

   
   

   Therefore, 1.u(n) ←z→	z
   	z – 1

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175. The solution of integral		2	(t3 + 3) δ(t + 2)dt
     		– 1

1. 1. – 5

   
   
   

   2. – 2

   
   
   

   3. Zero

   
   
   

   4. 11

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given		2	(t3 + 3) δ(t + 2)dt
   		– 1

   
   

   Since,		2	x(t) δ(t + t0)dt = x(– t0)
   		– 1

   
   

   Therefore,		2	(t3 + 3) δ(t + 2)dt = t3 + 3|at t = –2
   		– 1

   
   = (– 2)³ + 3
   = – 8 + 3 = – 5
   Hence, alternative (A) is the correct choice.

   Correct Option: A

   Given		2	(t3 + 3) δ(t + 2)dt
   		– 1

   
   

   Since,		2	x(t) δ(t + t0)dt = x(– t0)
   		– 1

   
   

   Therefore,		2	(t3 + 3) δ(t + 2)dt = t3 + 3|at t = –2
   		– 1

   
   = (– 2)³ + 3
   = – 8 + 3 = – 5
   Hence, alternative (A) is the correct choice.

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