00:01
Hi, in this question we have an analog signal given as xat equals to sine of 480 pi t plus 3 sine of 720 pi t.
00:20
So this is sampled 600 times per second.
00:24
So in part a we have to determine the nyquist sampling rate for this signal and the folding frequency.
00:34
So here we have the value of fs given as 600 hertz.
00:43
So omega 1 is given as 480 pi therefore the value of frequency f1 can be calculated as omega 1 by 2 pi.
00:56
So it will be 480 pi divided by 2 pi which gives the value of frequency f1 as 240 hertz.
01:06
Now similarly we have the value of omega 2 as 720 pi which gives the of f2 as 720 pi divided by 2 pi which will be 360 hertz.
01:20
Now the value of fm is equals to maximum of f1 comma f2.
01:31
So here f1 is 240 f2 is 360.
01:35
Therefore the value of fm will be 360 hertz.
01:41
So now we can calculate the value of f minimum which will be twice of fm.
01:51
So we get this value as 720 hertz.
01:56
So the nyquist sampling rate is f minimum which is 720 hertz...