Signals and systems electrical engineering miscellaneous
-  A sinusoid x(t) of unknown frequency is sampled by an impulse train of period 20 ms. The resulting sample train is next applied to an ideal lowpass filter with a cutoff at 25 Hz. The filter output is seen to be a sinusoid of frequency 20 Hz. This means that x(t) has a frequency of
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                        View Hint View Answer Discuss in Forum Sampling rate 50 Hz 
 Let x(t) has a frequency of fx Hz, 
 After sampling with 50 Hz spectrum will be LPF of cut off = 25 Hz
 output = 20 Hz
 ⇒ 50 – fx = 20
 ⇒ fx = 30 Hz Sampling rate 50 Hz
 Let x(t) has a frequency of fx Hz, 
 After sampling with 50 Hz spectrum will be LPF of cut off = 25 Hz
 output = 20 Hz
 ⇒ 50 – fx = 20
 ⇒ fx = 30 HzCorrect Option: CSampling rate 50 Hz 
 Let x(t) has a frequency of fx Hz, 
 After sampling with 50 Hz spectrum will be LPF of cut off = 25 Hz
 output = 20 Hz
 ⇒ 50 – fx = 20
 ⇒ fx = 30 Hz
-  A continuous-time LTI system with system function H(ω) has the following pole-zero plot. For this system, which of the alternatives is TRUE? 
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                        View Hint View Answer Discuss in Forum  
 The transfer function can be written asH(s) = (s - z1)(s - z1*)(s - z2)(s - z2*) (s - P1)(s - P1*)(s - P2)(s - P2*) |H(jω)| = √ω² + |z1|²√ω² + |z1|²√ω² + |z2|²√ω² + |z2|² √s² + |p1|²√ω² + |p1|²√ω² + |p2|²√ω² + |p2|² 
 from figure |z1| = |p2|
 |z2| = |p1|
 ⇒ (H(jω)) = K [constant]Correct Option: D 
 The transfer function can be written asH(s) = (s - z1)(s - z1*)(s - z2)(s - z2*) (s - P1)(s - P1*)(s - P2)(s - P2*) |H(jω)| = √ω² + |z1|²√ω² + |z1|²√ω² + |z2|²√ω² + |z2|² √s² + |p1|²√ω² + |p1|²√ω² + |p2|²√ω² + |p2|² 
 from figure |z1| = |p2|
 |z2| = |p1|
 ⇒ (H(jω)) = K [constant]
-  A signal is represented byx(t) =  1 |t| < 1 0 |t| > 1 
 The Fourier transform of the convolved single y(t) = x(2t)*x(t/2) is
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                        View Hint View Answer Discuss in Forum  Correct Option: A 
-  A function ƒ(t) is shown in the figure. 
 The Fourier transform F(ω) of ƒ(t) is
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                        View Hint View Answer Discuss in Forum Since ƒ(t) is odd and real 
 ƒ(t) = – ƒ(– t)
 ∮ F(ω) is imaginary and odd [symmetry property of fourier Transform]Correct Option: CSince ƒ(t) is odd and real 
 ƒ(t) = – ƒ(– t)
 ∮ F(ω) is imaginary and odd [symmetry property of fourier Transform]
-  A 10 kHz even-symmetric square wave is passed through a bandpass filter with centre frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is
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                        View Hint View Answer Discuss in Forum NA Correct Option: CNA 
 
	