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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

Where I am doing the mistake? Please help me with this problem.

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 ?

  • \$\begingroup\$ What's the bandwidth all the way from DC? \$\endgroup\$ – user253751 Feb 5 '18 at 3:24
  • \$\begingroup\$ I don't get why they are calculating bandwidth. \$\endgroup\$ – Nikhil Kashyap Feb 5 '18 at 3:27
  • \$\begingroup\$ What is "HCF" ? \$\endgroup\$ – Ale..chenski Feb 5 '18 at 4:46
  • \$\begingroup\$ @AliChen Scratch what I just said. Maybe it's "Highest common factor" and LCM means "Lowest Common Multiple". \$\endgroup\$ – KingDuken Feb 5 '18 at 4:53
  • \$\begingroup\$ @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. \$\endgroup\$ – 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.

  • 1
    \$\begingroup\$ You surely mean 37.5 Hz, not 32.5. Right? :-) \$\endgroup\$ – Ale..chenski Feb 5 '18 at 6:01
  • \$\begingroup\$ Hey mate, thanks for answering. I have edited the question, please check. \$\endgroup\$ – Nikhil Kashyap Feb 5 '18 at 6:19
  • 1
    \$\begingroup\$ I feel so stupid now. I confused sampling frequency and periodic frequency. \$\endgroup\$ – Nikhil Kashyap Feb 5 '18 at 9:39

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