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In a problem I'm working on, I have a signal that is sampled 256 times for 3.5 microseconds.

After performing a Discrete Fourier Transform on the data, what is the frequency step between the calculated values of Cn? The prompt states that this is also called the bandwidth of each point in the power spectrum.

How can I go about solving this problem?

Thanks

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Your specification of the particulars for your situation aren't clear to me, but I'll explain how you can arrive at the answer. The DFT is defined as:

$$ X[k] = \sum_{k=0}^{N-1}x[n] e^{-j2 \pi nk/N} $$

The (normalized) frequency associated with the $k$-th bin is $\frac{2\pi k}{N}$. Therefore, the spacing between the $k$-th and $(k+1)$-th bins is $\frac{2\pi}{N}$. Recall that the conversion between unnormalized frequencies (i.e. before sampling) and the normalized angular frequency $\omega$ used in the DTFT is:

$$ \omega = \frac{2\pi f}{f_s} $$

That is, a time-referenced frequency $f$ before sampling maps to $\omega$ in the normalized frequency domain, where $f_s$ is the sample rate of the signal (i.e. how many samples are taken of the signal per second). You can then map the bin spacing above into unnormalized frequency:

$$ \Delta f = \frac{2\pi}{N} \frac{f_s}{2\pi} = \frac{f_s}{N} $$

This simple relationship just says that the spacing between bins is equal to your signal's sample rate divided by the DFT length. It's unclear from your question whether the interval between each sample is 3.5 microseconds or whether your block of 256 samples covers an overall duration of 3.5 microseconds. Based on whichever is true, you can calculate the appropriate value.

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    $\begingroup$ If fs is just 256/(3.5 * 10^-6), then wouldn't the answer really just be 1/(3.5 * 10^-6), so just 1/t ? $\endgroup$ – Parthiv P Nov 11 '13 at 18:14
  • $\begingroup$ Yes, if that's the way that you specified the sample rate, that would be correct. Note that the relationship given above is applicable to the general case, where you only need to know the sample rate and DFT length. $\endgroup$ – Jason R Nov 12 '13 at 14:53

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