diff --git a/_notes/fourier_series.md b/_notes/fourier_series.md index 9e6a35ecd9ebc681c37d566c1e49f6b9346c9186..8a628f909be7ecc0caa526d37052a2a46ca27ddc 100644 --- a/_notes/fourier_series.md +++ b/_notes/fourier_series.md @@ -70,9 +70,17 @@ $$f$$ has finite bandwidth then only a finite number of the samples will be nonz uncountable set of numbers is determined by a finite one. +As an aside, by taking $$x = 0$$ we can derive the +[Poisson summation formula](https://en.wikipedia.org/wiki/Poisson_summation_formula): + +$$ +\sum_{n \in \mathbb{Z}} f(nL) = \frac{1}{L} \sum_{n \in \mathbb{Z}} \hat{f}(n / L) +$$ + + ## Derivation of the Discrete Time Fourier Transform -We can apply the result above to $$\hat{f}$$ as well. Recall that the Fourier transform of +We can apply the above results to $$\hat{f}$$ as well. Recall that the Fourier transform of $$\hat{f}$$ is $$f(-x)$$. Thus the periodic summation of the Fourier transform of $$f$$ is $$ @@ -92,13 +100,65 @@ practice reasonably bandwidth limited signals can still be recovered quite well of samples. -## Poisson Resummation +## Interpretation of the Discrete Fourier Transform -By taking $$x = 0$$ we see that +### Forward DFT + +Suppose we take samples of a function $$f$$ at integer multiples of a time $$T$$ (or distance, +etc.). As we saw above, $$ -\sum_{n \in \mathbb{Z}} f(nL) = \frac{1}{L} \sum_{n \in \mathbb{Z}} \hat{f}(n / L) +\hat{f}_{1/T}(x) = T \sum_{n \in \mathbb{Z}} f(n T) e^{-2 \pi i n x T} +$$ + +So when $$f$$ is time limited such that only the $$N$$ samples $$f(0)$$, $$\ldots$$, +$$f((N - 1) T)$$ are nonzero, + +$$ +\hat{f}_{1/T}(x) = T \sum_{n = 0}^{N - 1} f(n T) e^{-2 \pi i n x T} +$$ + +In this case the discrete Fourier transform gives us $$N$$ evenly spaced samples from one period of +$$\hat{f}_{1/T}$$. Namely, for $$0 \leq k < N$$, + +$$ +\hat{f}_{1/T}(k / NT) = T \sum_{n = 0}^{N - 1} f(n T) e^{-2 \pi i n k / N} $$ -This is known as the -[Poisson summation formula](https://en.wikipedia.org/wiki/Poisson_summation_formula). +### Backward DFT + +Similarly, + +$$ +f_{NT}(x) = \frac{1}{NT} \sum_{n \in \mathbb{Z}} \hat{f}(n / NT) e^{2 \pi i n x / (NT)} +$$ + +So when $$f$$ is bandwidth limited such that only the $$N$$ samples $$\hat{f}(0)$$, $$\ldots$$, +$$\hat{f}((N - 1) / NT)$$ are nonzero, + +$$ +f_{NT}(x) = \frac{1}{NT} \sum_{n = 0}^{N - 1} \hat{f}(n / NT) e^{2 \pi i n x / (NT)} +$$ + +And in this case the inverse discrete Fourier transform gives us $$N$$ evenly spaced samples from +one period of $$f_{NT}$$. Namely, for $$0 \leq k < N$$, + +$$ +f_{NT}(kT) = \frac{1}{NT} \sum_{n = 0}^{N - 1} \hat{f}(n / NT) e^{2 \pi i n k / N} +$$ + +### Commentary + +If $$f$$ were both time and bandwidth limited as discussed above, then $$f_{NT} = f$$ and +$$\hat{f}_{1/T} = \hat{f}$$. So the samples of $$\hat{f}$$ we get from the forward DFT could be fed +into the backward DFT to exactly recover our original samples of $$f$$. Unfortunately no such +functions exist, but in practice it works pretty well for signals that strike a balance between time +and bandwidth limits. The penalty for the imprecision is some aliasing caused by overlapping tails +in the periodic summations. So make sure $$1/T$$ is large enough to avoid significant aliasing in +the frequency domain, and $$NT$$ is large enough to avoid significant aliasing in the time (or +space, etc. domain). + +We also see why the frequency spectra obtained from the DFT is periodic: we're not getting samples +of $$\hat{f}$$, but of its periodic summation. If $$\hat{f}$$ is localized near the origin, it +can be more instructive to view half the samples of $$\hat{f}_{1/T}$$ provided by the DFT as +representing positive frequencies, and the other half as negative frequencies.