Frequency Offset Homework: 0.1s & 0.01s Time Vectors

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In summary, the given signal is s=cos(12*pi*t) with a time vector of 0 to 10 seconds. By changing the increments of the time vector from 0.1s to 0.01s, the plotted frequency appears to increase from 1Hz to 10Hz. This is because the smaller sampling increments allow for more points to be captured, making the signal appear more continuous and closer to its original function. This phenomenon can be further explained by the Nyquist theorem, which states that the sampling rate must be at least twice the highest frequency in order to accurately represent the original signal. Therefore, by sampling a 12Hz signal at a higher rate (0.01s), we are able to capture
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Homework Statement



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?


Homework Equations



f=1/T

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.
Thanks!
 
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  • #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:
  • #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.
 
  • #4
What course is this for?
Your explanation might be sufficient depending on the course, or might need a bit more :)
 
  • #5
Signals and systems. It seems like a sufficient answer to me but let me know if there's more to it!
 
  • #6
Btw I meant smaller sampling increments and higher sampling rate lol
 
  • #7
You should talk about the Nyquist theorm then and what happens when you sample a 12Hz signal at 10Hz vs 100Hz
 

FAQ: Frequency Offset Homework: 0.1s & 0.01s Time Vectors

What is frequency offset?

Frequency offset refers to the difference between the actual frequency of a signal and the desired frequency. It can be caused by factors such as interference, noise, or imperfections in electronic components.

How do you calculate frequency offset?

To calculate frequency offset, you need to measure the difference between the desired frequency and the actual frequency of a signal. This can be done by using a frequency counter or a spectrum analyzer. The result is typically expressed in Hertz (Hz) or parts per million (ppm).

What is the significance of a frequency offset of 0.1s and 0.01s?

A frequency offset of 0.1s and 0.01s indicates a small difference between the actual frequency and the desired frequency. This can result in a small amount of error in the signal, which may or may not be significant depending on the application.

How do you correct for frequency offset?

To correct for frequency offset, you can use a technique called frequency offset compensation. This involves adjusting the frequency of the signal to match the desired frequency using a phase-locked loop (PLL) or digital signal processing (DSP) algorithms.

Can frequency offset impact data transmission?

Yes, frequency offset can impact data transmission by causing errors or distortion in the signal. This is especially true in applications that require high precision and accuracy, such as wireless communication or satellite navigation systems. It is important to minimize frequency offset to ensure reliable data transmission.

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