# Nyquist Sampling Rate Problem

I am really confused with the above problem. I doubt the solution.

According to me,

Sampling Frequency of x(t) = HCF(5,12.5) = HCF(5,25)/LCM(1,2) = 5/2 = 2.5Hz

Sampling Frequency of y(t) = 3x2.5 = 7.5Hz

Nyquist Sampling Rate = 2 * 7.5 = 15Hz

EDIT: I got confused because I was taught the following example in class and I tried to use the same here.

Q Find the fundamental time period for

x(t) = sin(22pit)+sin(7pit+30)

f = HCF(11,7/2) = HCF(11,7)/LCM(1,2) = 1/2

T = LCM(1/11,2/7) = LCM(1,2)/HCF(11,7) = 2

So, Fundamental Time Period = 2 using either of the above methods.

Is there a difference between sampling frequency and the fundamental frequency ?

• What's the bandwidth all the way from DC? – user253751 Feb 5 '18 at 3:24
• I don't get why they are calculating bandwidth. – Nikhil Kashyap Feb 5 '18 at 3:27
• What is "HCF" ? – Ale..chenski Feb 5 '18 at 4:46
• @AliChen Scratch what I just said. Maybe it's "Highest common factor" and LCM means "Lowest Common Multiple". – KingDuken Feb 5 '18 at 4:53
• @KingDuken, nothing still makes any sense. Highest FREQUENCY of x(t) signal is 12.5 Hz. So the Nyquist sampling must be 25 Hz. This is the first screw-up of OP. The rest follows. – Ale..chenski Feb 5 '18 at 5:05

The solution is correct, here's how I prefer to solve it.

Understand that if we can correctly identify frequency A, and A is greater than frequency B, then you can also correctly identify frequency B.

Here, $25\pi t$ is greater than $10\pi t$ so we can ignore the $10\pi t$ and only focus on the $25\pi t$.

Delays doesn't affect the frequency, so we can ignore the $+9$ in the $y(t)$ function.

At $t=1$, one second has passed and we can read the data straight out from the $e^{i25 \pi t}$, if we plug in the $3$ from $y(t)=x(3t)$ we get $e^{i75 \pi}$

One revolution is $2\pi$, this means we'll divide $75\pi$ by $2\pi$ to get $37.5 \text{ Hz}$. And then we multiply $37.5 \text{ Hz}$ by $2$ for the Nyquist frequency which is 75 Hz => 75 samples / sec.

• You surely mean 37.5 Hz, not 32.5. Right? :-) – Ale..chenski Feb 5 '18 at 6:01
• Hey mate, thanks for answering. I have edited the question, please check. – Nikhil Kashyap Feb 5 '18 at 6:19
• I feel so stupid now. I confused sampling frequency and periodic frequency. – Nikhil Kashyap Feb 5 '18 at 9:39