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.