The Fourier Transform

We define the forward Fourier transform $\tilde{h}(f)$ of a time domain quantity $h(t)$ to be

$$\tilde{h}(f)=\int_{-\infty}^\infty dt h(t) e^{- 2 \pi i f t}$$ (16.1)

and the inverse Fourier transform to be

$$h(t)=\int_{-\infty}^\infty df \tilde{h}(f) e^{2 \pi i f t}.$$ (16.2)

If the function $h(t)$ is sampled at $N$ consecitive points with sampling interval $\Delta t$, that is

$$h_j \equiv h(t_j) ,\quad t_j = j\Delta t,$$ (16.3)

we only have $N$ input values, so we can only produce $N$ independent values for the Fourier transform. Further, we can only produce values in the interval $(-f_c,f_c)$ where $f_c$ is the Nyquist critical frequency

$$f_c = \frac{1}{2\Delta t}.$$ (16.4)

We compute estimates of the Fourier transform at the $N + 1$ discrete values

$$f_k \equiv \frac{k}{N\Delta t}\quad k = -\frac{N}{2},\ldots,\frac{N}{2}.$$ (16.5)

There are only $N$ independent values here as the extremevales of $k$ correspond to the upper and lower limites of the Nyquist critical frequency range and are equal. We now proceed to define the discrete Fourier transform. Consider

$$\tilde{h}(f_k) = \int_{-\infty}^\infty dt h(t) e^{-2 \pi i f t}$$ (16.6)

$$\approx \sum_{j=0}^{N-1} \Delta t h(t_j) e^{-2 \pi i f_k t_j}$$ (16.7)

$$\approx \Delta t \sum_{j=0}^{N-1} h_j e^{-2 \pi i j k / N} .$$ (16.8)

According to T010095, we define the discrete Fourier transform (DFT) to be

$$\tilde{h}_k = \sum_{j=0}^{N-1} h_j e^{-i 2 \pi j k / N}$$ (16.9)

and then we can recover $\tilde{h}(f_k)$ by

$$\tilde{h}(f_k) = \Delta t \tilde{h}_k.$$ (16.10)

The inverse Fourier transform is

$$h(t)=\int_{-\infty}^\infty dt \tilde{h}(f) e^{2 \pi i f t} .$$ (16.11)

Using

$$\Delta f = f_{k+1} - f_k = \frac{k+1}{N\Delta t} - \frac{k}{N\Delta t} = \frac{1}{N\Delta t}$$ (16.12)

we may write

$$h(t_j) \approx \sum_{k=0}^{N-1} \tilde{h}(f_k) e^{2 \pi i f_k t_j / N} \Delta f$$ (16.13)

$$= \sum_{k=0}^{N-1} \Delta \tilde{h}_k e^{2 \pi i j k / N}\frac{1}{N\Delta t}$$ (16.14)

$$= \frac{1}{N} \sum_{k=0}^{N-1} \tilde{h}_k e^{2 \pi i j k / N}$$ (16.15)

which is the inverse DFT according to T010095. Note that the LAL ``reverse'' DFT functions do not include the factor $1/N$ in their output.