Fourier Transform#

The Fourier transform is an extension or generalization of the Fourier series. Instead of a finite period, we deal with an infinite period \(T\). Using this trick, we can represent aperiodic functions as well. Similar to the concept of the Fourier series we are looking in two directions:

  1. Fourier transform: going from the time domain \(y: \mathbb{R} \rightarrow \mathbb{C}\) to the frequency domain \(Y: \mathbb{R} \rightarrow \mathbb{C}\)

  2. Inverse Fourier transform: going from the frequency domain \(Y: \mathbb{R} \rightarrow \mathbb{C}\) to the time domain \(y: \mathbb{R} \rightarrow \mathbb{C}\).

Fourier series are limited to periodic functions. Musical signals are mostly periodic, but what if they are not? Well, the Fourier series is used to represent a periodic function by a discrete sum of complex exponentials, while the Fourier transform is then used to represent a general, non-periodic function by a continuous superposition or integral of complex exponentials.

The Fourier transform can be thought of as the Fourier series of a function with a period that approaches infinity. We switch from a discrete superposition to a continuous superposition. Of course, the Fourier transform can also deal with periodic functions. Therefore, one might say that it is more powerful.

Fourier Transform (TF)

The Fourier transform (FT) of an integrable function \(y : \mathbb{R} \rightarrow \mathbb{C}\) is defined by

(31)#\[Y(f) = \int\limits_{-\infty}^\infty y(t) e^{-i2\pi f t} dt, \quad \forall f \in \mathbb{R}.\]

The transform of function \(y(t)\) at frequency \(f\) is given by the complex number \(Y(f)\).

Let me give you an intuition of how this follows from the coefficients of the Fourier series. Consider an infinite period, i.e., \(T\) goes to infinity. Therefore, we get

\[\begin{equation*} \lim_{T \rightarrow \infty} c_n = \lim_{T \rightarrow \infty} \frac{1}{T} \int_T y(t) \cdot e^{-i2\pi n t / T}dt. \end{equation*}\]

Since \(T\) goes to infinity, the fundamental frequency \(f_0 = 1 / T\) goes to zero. Thus the discrete variables \(n \cdot f_0\) all become continuous since \(f_0 \rightarrow 0\). Remember that, for the exponential form, \(n\) goes from negative infinity to positive infinity. We can summarize:

\[\begin{equation*} \lim_{T \rightarrow \infty} \frac{n}{T} = \lim_{f_0 \rightarrow 0} n \cdot f_0 = f \end{equation*}\]

following

\[\begin{equation*} \lim_{T \rightarrow \infty} c_n = \left( \lim_{T \rightarrow \infty} \frac{1}{T}\right) \cdot \underbrace{\int_{-\infty}^\infty y(t) \cdot e^{-i2\pi f t}dt}_{Y(f)}. \end{equation*}\]

Consequently,

\[\begin{equation*} c_n = \frac{1}{T} Y\left( n \cdot f_0 \right) = \frac{1}{T} Y\left( n \cdot \frac{1}{T}\right) \end{equation*}\]

holds, since the value of each part of the integral \(Y(f)\) of length \(T\) is equal to \(T \cdot c_n\). Everything works out. Thus we can use the Fourier transform to compute the coefficients for the Fourier series.

Evaluating \(Y(f) \in \mathbb{C}\) for all values of \(f\) produces the frequency-domain function. Similar to \(Y(n)\) of the Fourier series, the complex number \(Y(f)\) (a phasor), conveys both apmplitude and phase of the frequency \(f\).

The effect of multiplying \(y(t)\) by \(e^{-i2\pi f t}\) is to subtract \(f\) from every frequency component of \(y(t)\). So the component that was at \(f\) ends up at zero herz. The integral produces its amplitude because all the other components are orthogonal and consequently integrate to zero over an infinite interval. In section Similarity of Periodic Functions, I tried to give an intuition for this phenomenon.

Inverse Fourier Transform (ITF)

Under suitable conditions \(y: \mathbb{R} \rightarrow \mathbb{C}\) can be represented as a recombination of complex exponentials of all possible frequencies, called inverse Fourier transform (IFT):

(32)#\[y(t) = \int\limits_{-\infty}^\infty Y(f) e^{i2\pi f t} df, \quad \forall t \in \mathbb{R}.\]

The inverse transform of function \(Y(f)\) at time \(t\) is given by the complex number \(y(t)\).

The pair \((y, Y)\) are called Fourier integral pair or Fourier transform pair and we define \(\mathcal{F}\) to be the Fourier transform operator with

\[\mathcal{F}\{y\} = Y, \quad \mathcal{F}^{-1}\{Y\} = y.\]

We are dealing with real-valued functions, but this is not a problem since \(\mathbb{R} \subset \mathbb{C}\). Real-valued functions are just a special case where \(y(t) = \overline{y(t)}\) which implies

\[Y(f) = Y(-f),\]

i.e., we do not have to remember or compute anything for negative frequencies.