Why DFT is used for approximating CTFT when you can approximate CTFT-integral itself?

Question

I was using MATLAB for approximating FTs. Why DFT is used if we can approximate the transform-integration using summation.

Answer 1

Assuming that future vistors won't take the time to read all the comments, I'd like to give a very simple and straightforward interpretation of the discrete Fourier transform (DFT) as an approximation of the continuous-time Fourier transform (CTFT).

First, we need to assume that the signal to be transformed decays sufficiently for large value of $|t|$. (If it doesn't, we can still use the DFT to approximate the CTFT, but the truncation error will be large.) Furthermore, we choose the origin $t=0$ such that the relevant part of the signal appears for $t\ge 0$. Then we can approximate the Fourier integral in the following way:

$$X(j\Omega)=\int_{-\infty}^{\infty}x(t)e^{-j\Omega t}dt\approx \int_{0}^{T}x(t)e^{-j\Omega t}dt\tag{1}$$

with some sufficiently large $T$. The approximation $(1)$ generally results in a truncation error, which can be made small by an appropriate choice of $T$.

Now we can approximate the right-hand side of $(1)$ by a Riemann sum:

$$\int_{0}^{T}x(t)e^{-j\Omega t}dt\approx\sum_{n=0}^{N-1}x(n\Delta t)e^{-j\Omega n\Delta t}\Delta t\tag{2}$$

where $\Delta t$ is the sampling interval and $T=N\Delta t$. The approximation $(2)$ introduces an aliasing error, which can be made small by choosing $\Delta t$ sufficiently small.

Computing the right-hand side of $(2)$ at discrete frequencies $\Omega_k=2\pi k f_s/N$, $k=0,1,\ldots,N-1$, where $f_s=1/\Delta t$ is the sampling frequency, finally results in

$$\begin{align}X(j\Omega_k)&\approx\sum_{n=0}^{N-1}x(n\Delta t)e^{-j\Omega_k n\Delta t}\Delta t\\&=\frac{1}{f_s}\sum_{n=0}^{N-1}x(n\Delta t)e^{-j\frac{2\pi}{N}nk}\\&=\frac{1}{f_s}\textrm{DFT}\big\{x_d[n]\big\},\qquad x_d[n]=x(n\Delta t)\tag{3}\end{align}$$

Eq. $(3)$ says that the CTFT of $x(t)$ can be approximated on a set of discrete frequencies by the DFT of a sampled version of $x(t)$. Two types of error have to be taken into account: 1. the truncation error, from truncating the signal, and 2. the aliasing error, from sampling the signal.

Note that we can never avoid both errors, because for finite length signals (with zero truncation error) we always get an aliasing error due to the infinite bandwidth of finite length signals, and for perfectly band-limited signals (with zero aliasing error), we always have a truncation error because they can't have finite length. In practice we always have to deal with both types of error.

In sum, answering the question in your question title: under the aforementioned conditions, the DFT is an approximation of the CTFT integral, and it has the advantage that there exist very efficient algorithms (FFT) for its computation.