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Homework Help: Frequency Offset

  1. Jul 28, 2014 #1
    1. The problem statement, all variables and given/known data

    The signal is s=cos(12*pi*t) and the time vector is 0 to 10s. One vector has increments of 0.1s and the other is 0.01s. What is the plotted frequency from these time scales? And why does it change by changing the increments?

    2. Relevant equations


    3. The attempt at a solution

    So we're supposed to find the frequencies from the plot of the graphs. For the 0.1s increments, the frequency seems to be 1Hz (T≈1 ∴ f=1/1). And for the 0.01s increments, the frequency seems to be 10Hz (T≈0.1 ∴ f=1/0.1). She wants us to explain this and I don't get it.

    Although, I think it has to do with frequency offset. Explain please?

    Hope you can help.
  2. jcsd
  3. Jul 28, 2014 #2
    Take a look at the nyquist theorem. That should point you in the right direction :)

    *edit* thinking about my statement it might not be immediately clear. By changing t from a continuous function to one that uses increments (of either 0.1s or 0.001s) you are essentially sampling the original function.
    Last edited: Jul 28, 2014
  4. Jul 28, 2014 #3
    Actually I think I've figured it out.
    In MATLAB, you are technically in discrete time since you are sampling times. Yes, the smaller the sampling rate, the more continuous it becomes, and this is exactly what's going on here.
    cos(12*pi*t) is such a compressed sinusoid that an increment of 0.1 will only get specific points that does not make it look as compressed as it actually is. But when you take increments of 0.01, it covers way more points, which allows it to look closer to its original function. This is why the frequency looks like it has increased when you give it increments of 0.01.
  5. Jul 28, 2014 #4
    What course is this for?
    Your explanation might be sufficient depending on the course, or might need a bit more :)
  6. Jul 28, 2014 #5
    Signals and systems. It seems like a sufficient answer to me but let me know if there's more to it!
  7. Jul 28, 2014 #6
    Btw I meant smaller sampling increments and higher sampling rate lol
  8. Jul 28, 2014 #7
    You should talk about the Nyquist theorm then and what happens when you sample a 12Hz signal at 10Hz vs 100Hz
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