## The Discrete Fourier Transform

## The Discrete-Time Fourier Transform

Let's "get rid of" the infinities by just taking limits

$$ \hat{f}(\omega) = \lim_{W \rightarrow \infty} \int_{-W}^{W} f(t) e^{-i \omega t} dt $$

A very long signal relative to its period should be well-approximated

Now a fiddly argument ensues. If the signal is periodic, we should be able to replace the limits by the period: after all, it doesn't change anything

$$ \hat{f}(\omega) = \int_{0}^{P} f(t) e^{-i \omega t} dt $$

Let's turn the integral into a sum by treating \(f(t)\) as a series of

*impulses*: nonzero only at discrete timesteps$$ X[k] = \sum_{n=0}^{N-1} x[n] e^{-i k n / N} $$

The frequency-domain value at

*unit frequency k*, namely \(X[k]\) has been written as a weighted sum of the discrete time-domain values \(x[i]\) over the period \(N\)The inverse transformation works similarly

$$ x[n] = \frac{1}{2 \pi N} \sum_{k=0}^{N-1} X[k] e^{i k n / N} $$

## Notes On The DFT

The \(X\)s are still complex (!)

The inverse is computed only as the sum of discrete frequencies

Be careful about the amplitude. The bin amplitudes grow like \(N\): there's different conventions about where to put the \(1/N\) to scale them back. I am following Wikipedia here, and putting the scaling in the inverse transform.

This math is so full of fiddly that it is really easy to get it screwed up

Frequency resolution is a function of sampled period

The sampling rate is handled here by unit frequency. Frequency needs to be scaled by the sample rate to be meaningful: if you do a 1024-point DFT of a signal sampled at 1024 × 48 samples per second, each bin of the DFT result represents about 24Hz of signal

Windowing is a thing: next lecture. In brief, because the signal is treated as cyclic, truncating the signal so that the left and right ends don't match up (at all frequencies!) is… bad