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

GATE-2011

Q.31 The frequency of a signal sampled at 200 samples per second appears as 75 Hz. If the signal was under-sampled, the frequency (in Hz) of the original signal would be

A) 100                       

B) 125                       

C) 150                         

D) 175 
 

Solution:

R (Sampling rate) = 200 samples/sec

f_s (Sampling frequency) = 75 Hz

f_a – aliased frequency 

f_a= |R*n – f_s|

  = | 200-75 |

  = 125 Hz