00:01
So, here we are considering about an analog electrocardiogram signal which contains useful frequency up to 100 hz.
00:07
So, frequency is equal to 100 hz.
00:10
So, in the first part we have to find out what is the nyquist rate for this signal.
00:15
So, here we are given the value of f max that is equal to 100 hz.
00:20
So, the value of nyquist rate is equal to 2 f max that is equal to 200 samples per second.
00:32
This from here is equal to 200 samples per second.
00:36
Hence, the answer to the part a of the question.
00:38
Now, we are considering about the part b where f of s is given that is equal to 250 samples per second.
00:46
So, the value of f max from here is equal to f of s which is divided by 2 that is equal to 125 hz which is the maximum frequency that can be represented uniquely at the sampling rate.
01:00
So, this is the answer to the part b of the question.
01:02
Now, we are considering about the next question that is 1 .9 part a where we are having the value of x of a of t that is equal to sin of 480 pi t plus 3 sin of 720 pi t is the sample which is 600 times per second.
01:23
So, we have to determine the nyquist sample rate of q naught t.
01:28
So, we are considering about the first part where the value of x of a of t is equal to sin of 480 pi t plus 3 sin of 720 pi t.
01:41
So, this value from here is equal to sin of 2 pi which is multiplied by 240 that is further multiplied by t plus 3 sin of 2 pi which is multiplied by 360 that is further multiplied by t...