# What is the relation between analytical Fourier transform and DFT?

First of all let me state that I searched for this topic before asking. My question is as follows we have the Analytical Fourier Transform represented with an integral and Discrete Fourier Transform represented with a summation usually computed by a numerical software. My question is that although the relationship is exact at the sampled data points assuming they are equally spaced(say they are equally spaced in time) how can we say about the accuracy of the transform that is if we transform into frequency-domain both with DFT and AFT how will the coefficients for a particular frequency be related? Are they going to be the similar or the same under some conditions?

## 1 Answer

I assume you refer the continuous-time Fourier transform (CTFT)

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$

and to the discrete Fourier transform (DFT)

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-i2\pi nk/N}\tag{2}$$

where $x[n]$ and $X[k]$ are sequences, whereas $x(t)$ and $X(\omega)$ are functions. The DFT can be computed efficiently by the Fast Fourier Transform (FFT).

The DFT can be used to approximate the CTFT, but there are always two possible error sources, and you can never avoid both of them at the same time:

1. truncation error: since the DFT uses a sequence of finite length, any signal $x(t)$ which does not have finite length (or is too long for computer memory) must be truncated.

2. aliasing error: the DFT uses sample values $x[n]$ which can be equidistant samples of the function $x(t)$. Sampling can introduce aliasing if the sampling rate is smaller than twice the maximum frequency contained in $x(t)$.

You always have at least one of the two above errors because for avoiding the truncation error you need a finite length signal $x(t)$, which cannot be band-limited (i.e. you'll get an aliasing error). The aliasing error can be avoided if you have an ideally band-limited signal and if you choose the sampling frequency sufficiently high, but such a signal cannot have finite duration, so you get a truncation error.

Taking these two errors into account, you can approximate the CTFT by the DFT. Assuming equidistant frequency points with a spacing of $\Delta f$, and a time-domain sampling interval of $T$ you have

$$X(2\pi k\Delta f)\approx T\sum_{n=0}^{N-1}x(nT)e^{-i2\pi kn/N}$$

with $N=1/(\Delta f\cdot T)$, assuming that the relevant part of $x(t)$ has been shifted to the range $[0,NT]$.

• That was exactly what I was asking for in particular the relation ship between $X[k]$ and $X(w)$ with the assumption that n also represents time. So while sampling in time domain we got the exact representation(with infinitely precise instrument and a complete theory say) of the analytical time-domain function $x(t)$ sampled at particular times. I will study your answer in more detail. – Vesnog Jan 22 '15 at 18:00