141. The Fourier series coefficient for given periodic signal x(t) is
     

1. 1. X[k] = 2

   
   
   

   2. X[k] = 4

   
   
   

   3. X[k] = 6

   
   
   

   4. X[k] = 8

   
   
   

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   Correct Option: A

   

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142. A periodic rectangular signal, x(t) has the wave form shown in figure below. Frequency of the fifth harmonic of its spectrum is _____ Hz
     

1. 1. 125 Hz.

   
   
   

   2. 1250 Hz.

   
   
   

   3. 1.250 Hz.

   
   
   

   4. 12.50 Hz.

   
   
   

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   Signal shown has time period,
   T = 4ms.
   

   ∴ ƒ =	1 × 10³	= 250 Hz
   	4

   
   This is the fundamental frequency. Fifth harmonic frequency,
   ƒ_s = 250 × 5 = 1250 Hz.

   Correct Option: B

   Signal shown has time period,
   T = 4ms.
   

   ∴ ƒ =	1 × 10³	= 250 Hz
   	4

   
   This is the fundamental frequency. Fifth harmonic frequency,
   ƒ_s = 250 × 5 = 1250 Hz.

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143. Sinusoidal signal x(t) = 4 cos (200t + π/6) is passed through a square law device defined by the input output relation y(t) = x² (t). The DC component in the signal is

1. 1. = 8 + 8 cos		400 t +	π
      			3

   
   
   

   2. = 8 + 16 cos		40 t +	π
      			3

   
   
   

   3. = 15 + 8 cos		100 t +	π
      			3

   
   
   

   4. = 8 + 16 sin		40 t +	π
      			3

   
   
   

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   Correct Option: A

   

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144. The laplace transform of the function
     

     I(s) =	10s + 4
     	s(s+ 1)(s² + 4s + 5)

     
     its final value will be

1. 1. 5/3

   
   
   

   2. 3/5

   
   
   

   3. 4/5

   
   
   

   4. 5/4

   
   
   

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   According to final value theorem,
   

   Correct Option: C

   According to final value theorem,
   

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145. 
     The Laplace transform of the waveform shown in the figure is
     1/s² (1 + Ae – s + Be – 4s + Ce – 6s + De – 8s).
     The value of D is ______

1. 1. -1.5

   
   
   

   2. 1.5

   
   
   

   3. 0.5

   
   
   

   4. - 0.5

   
   
   

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   ƒ(t) = t[U(t) – U(t – 1)] + 1 [U(t – 1) – U(t – 4)] + (5 – t) [U(t – 4) – U(t – 6)] + (t/2 × -4) [U(t – 6) – U(t – 8)]
   

   F(s) =	1	[1 - e-s] +	1	[e-s - e-4s] +		5	-	1		[e-4s - e-6s] +		4	-	1		[e-6s - e-8s]
   	s²		s			s		s²				2s²		s

   
   Comparing, we get D = – 0.5

   Correct Option: D

   ƒ(t) = t[U(t) – U(t – 1)] + 1 [U(t – 1) – U(t – 4)] + (5 – t) [U(t – 4) – U(t – 6)] + (t/2 × -4) [U(t – 6) – U(t – 8)]
   

   F(s) =	1	[1 - e-s] +	1	[e-s - e-4s] +		5	-	1		[e-4s - e-6s] +		4	-	1		[e-6s - e-8s]
   	s²		s			s		s²				2s²		s

   
   Comparing, we get D = – 0.5

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