GATE-2018 (24) :Wyllie – Time average equation

GATE-2012

A signal having duration of 10 seconds is sampled at a rate of 1000samples per second. The maximum frequency of the sampled signal is 475 Hz .

48. If the signal has been under-sampled, the maximum frequency (in Hz) of the original signal would have been 

A) 475                          

B) 500                    

C) 525                       

D) 550

Solution:

R (Sampling rate)= 1000 samples/second

f_s (Sampling frequency) = 475 Hz

f_a =aliased frequency

In under-sampling sample rate below its Nyquist 

rate, in under-sampling frequency decreasing 

(aliasing) 

f_a=| R*n – f_s|

  = |1000 – 475|

  = 525 Hz

49. What is the frequency interval (in Hz) at which the spectrum of the above signal is evaluated?

A) 0.08                    

B) 0.10                  

C) 0.12                              

D)0.14

 

Solution:

Duration of the signal = 10 seconds

Frequency interval = 1/ Time (Duration)

                               = 1/ 10

                              =0.1 Hz