6. Consider the following sound wave, where t is in milliseconds: xlt) sin(10xt)
ID: 2291219 • Letter: 6
Question
6. Consider the following sound wave, where t is in milliseconds: xlt) sin(10xt) sin(20t) +sin(60nt)+ sin(90tt) This signal is prefiltered by an analog antialiasing prefilter H(0 and then sampled at an audio rate of 40 kHz. The resulting samples are immediately reconstructed using an ideal reconstructor. Determine the output yalt) of the reconstructor in the following cases and compare it with the audible part of x(t): a. When there is no prefilter, that is, H()1 b. When H(0 is an ideal prefilter with cut off of 20 kHz. c. When H() is a practical prefilter that has a flat passband up to 20 kHz and attenuates at a rate of 48 dB/octave beyond 20 kHz. (You may ignore the effects of t response of the filter.)Explanation / Answer
The input signal has 4 frequency components at 5KHz, 10KHz, 30KHz and 45KHz
a)
fs=40KHz
When there is no prefilter, we will get sampled signal components at
f_sampled=5KHz, 40KHz+5KHz=45KHz, 40KHz-5KHz=35KHz, 10KHz, 40KHz+10KHz=55KHz,40KHz-10KHz=35KHz
30KHz, 40KHz+30KHz=70KHz, 40KHz-30KHz=10KHz, 45KHz, 40KHz+45KHz=95KHz, 40KHz-45KHz=-5KHz and so on
but the ideal reconstructor removes all the component above 20KHz
therefore at the output we will have 5KHz, 10KHz. Since we didnt use prefilter the component of 30KHz and 45KHz aliases back into frequency at 10KHz and 5KHz respectively.
b) when we use 20KHz prefilter, it will remove 30KHz and 45KHz from the input
therefore sampled signal will have frequency component at
f_sampled=5KHz, 40KHz+5KHz=45KHz, 40KHz-5KHz=35KHz, 10KHz, 40KHz+10KHz=55KHz,40KHz-10KHz=35KHz, and so on
but the ideal reconstructor removes all the component above 20KHz therefore at the output we will have 5KHz, 10KHz.
C) in this case the a very small amount of 30KHz and 45KHz aliases back to output since filter is not a brickwall filter.
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