Next: Convolution theorem, Previous: Fourier transform, Up: Frequency domain and Fourier operations [Contents][Index]

The Dirac \(\delta\) (delta) function (also known as an impulse) is the way that we convert a continuous function into a discrete one. It is defined to satisfy the following integral:

$$\int_{-\infty}^{\infty}\delta(l)dl=1$$

When integrated with another function, it gives that function’s value at \(l=0\):

$$\int_{-\infty}^{\infty}f(l)\delta(l)dt=f(0)$$

An impulse positioned at another point (say \(l_0\)) is written as \(\delta(l-l_0)\):

$$\int_{-\infty}^{\infty}f(l)\delta(l-l_0)dt=f(l_0)$$

The Dirac \(\delta\) function also operates similarly if we use summations instead of integrals. The Fourier transform of the delta function is:

$${\cal F}[\delta(l)]=\int_{-\infty}^{\infty}\delta(l)e^{-i{\omega}l}dl=e^{-i{\omega}0}=1$$

$${\cal F}[\delta(l-l_0)]=\int_{-\infty}^{\infty}\delta(l-l_0)e^{-i{\omega}l}dl=e^{-i{\omega}l_0}$$

From the definition of the Dirac \(\delta\) we can also define a Dirac comb (\({\rm III}_P\)) or an impulse train with infinite impulses separated by \(P\):

$${\rm III}_P(l)\equiv\displaystyle\sum_{k=-\infty}^{\infty}\delta(l-kP) $$

\(P\) is chosen to represent “pixel width” later in Sampling theorem. Therefore the Dirac comb is periodic with a period of \(P\). We have intentionally used a different name for the period of the Dirac comb compared to the input signal’s length of observation that we showed with \(L\) in Fourier series. This difference is highlighted here to avoid confusion later when these two periods are needed together in Discrete Fourier transform. The Fourier transform of the Dirac comb will be necessary in Sampling theorem, so let’s derive it. By its definition, it is periodic, with a period of \(P\), so the Fourier coefficients of its Fourier Series (Fourier series) can be calculated within one period:

$${\rm III}_P=\displaystyle\sum_{n=-\infty}^{\infty}c_ne^{i{2{\pi}n\over P}l}$$

We can now find the \(c_n\) from Fourier series:

$$c_n={1\over P}\int_{-P/2}^{P/2}\delta(l)e^{-i{2{\pi}n\over P}l} ={1\over P}\quad\quad \rightarrow \quad\quad {\rm III}_P={1\over P}\displaystyle\sum_{n=-\infty}^{\infty}e^{i{2{\pi}n\over P}l} $$

So we can write the Fourier transform of the Dirac comb as:

$${\cal F}[{\rm III}_P]=\int_{-\infty}^{\infty}{\rm III}_Pe^{-i{\omega}l}dl ={1\over P}\displaystyle\sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i(\omega-{2{\pi}n\over P})l}dl={1\over P}\displaystyle\sum_{n=-\infty}^{\infty}\delta\left(\omega-{2{\pi}n\over P}\right) $$

In the last step, we used the fact that the complex exponential is a periodic function, that \(n\) is an integer and that as we defined in Fourier transform, \(\omega{\equiv}m\omega_0\), where \(m\) was an integer. The integral will be zero for any \(\omega\) that is not equal to \(2{\pi}n/P\), a more complete explanation can be seen in Fourier series. Therefore, while in the spatial domain the impulses had spacing of \(P\) (meters for example), in the frequency space, the spacing between the different impulses are \(2\pi/P\) cycles per meters.

JavaScript license information

GNU Astronomy Utilities 0.7 manual, August 2018.