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by Pravegaa Education

Chapter 8: Fourier Transformation

Mathematical Physics, Part 2. Theory, numbered worked examples and 5 previous years’ questions (CSIR NET-JRF, GATE, IIT-JAM and JEST, 2020–2026) with full solutions. Click Show solution under any question.

Theory

Introduction

  1. The Fourier transform is a principal analysis tool in modern science.

  2. A Fourier series is a discrete sum; the Fourier transform is continuous.

  3. Its best-known application is the analysis of linear time-invariant systems.

  4. It is a universal problem-solving technique.

  5. Viewing a function and its transform side by side is the key to the analysis.

The Fourier transform has several definitions, chosen by convenience. For a function \(h(t)\),

\(h(t)\) and \(H(f)\) are known as a Fourier transform pair.

Uses

  1. Image processing.

  2. Solving partial differential equations in electromagnetic theory.

  3. Quantum mechanics.

Throughout this chapter \(\jj=\sqrt{-1}\) (engineering notation), the same as \(\ii\).

Basic Definitions of the Fourier Transform

\begin{align*} H(f)&=\int_{-\infty}^{\infty}h(t)\,\ee^{-\jj2\pi ft}\,\dd t &&\text{(Fourier transform)},\\ h(t)&=\int_{-\infty}^{\infty}H(f)\,\ee^{\jj2\pi ft}\,\dd f &&\text{(inverse Fourier transform)}. \end{align*}

In general \(H(f)=R(f)+\jj I(f)\), where

  • \(H(f)\) is a complex function;

  • \(R(f)\) is the real part of \(H(f)\);

  • \(I(f)\) is the imaginary part of \(H(f)\).

Then \[ \abs{H(f)}=\sqrt{R^2+I^2},\qquad \theta(f)=\tan^{-1}\frac IR. \]

Some other useful definitions In terms of \(\omega=2\pi f\), \[ H(\omega)=a_1\int_{-\infty}^{\infty}h(t)\,\ee^{-\jj\omega t}\,\dd t\qquad\&\qquad h(t)=a_2\int_{-\infty}^{\infty}H(\omega)\,\ee^{\jj\omega t}\,\dd\omega, \] with \(a_1a_2=\dfrac1{2\pi}\). Common choices are \[ a_1=\frac1{2\pi},\ a_2=1;\qquad a_1=1,\ a_2=\frac1{2\pi};\qquad a_1=a_2=\frac1{\sqrt{2\pi}}. \] (The choice \(a_1=a_2=1\) is possible only with \(f\) as the variable, as above.)

  • Time and frequency are reciprocal variables.

  • So are position and wave vector.

Fourier transformation in position (\(x\)) and wave-vector (\(k\)) space \[ F(k)=\int_{-\infty}^{\infty}f(x)\,\ee^{\ii kx}\,\dd x,\qquad f(x)=\frac1{2\pi}\int_{-\infty}^{\infty}F(k)\,\ee^{-\ii kx}\,\dd k. \]

Dirac Delta Function

Consider a rectangular pulse of width \(2k\) and height \(1/2k\) centred at \(x_0\). Its area is \[ A=\frac1{2k}\times2k=1. \] Let \(2k\to0\): the height \(\dfrac1{2k}\to\infty\) while the area stays \(1\). Then \[ f(x)=\delta(x-x_0). \]

Properties

  • The area under the curve is unity: \(\displaystyle\int_{-\infty}^{\infty}f(x)\,\dd x=\int_{-\infty}^{\infty}\delta(x-x_0)\,\dd x=1\).

  • \(\displaystyle\int_{-\infty}^{\infty}f(x)\,\delta(x-x_0)\,\dd x=f(x_0)\int_{-\infty}^{\infty}\delta(x-x_0)\,\dd x=f(x_0)\),   \(x_0\in\,]-\infty,\infty[\).

  • \(\displaystyle\int_{-\infty}^{\infty}\ee^{\ii x}\,\delta(x-x_0)\,\dd x=\ee^{\ii x_0}\).

In suitable limits many functions behave like the delta function:

  • \(\displaystyle\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f-f_0)t}\,\dd t=\delta(f-f_0)\),

  • \(\displaystyle\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f+f_0)t}\,\dd t=\delta(f+f_0)\).

The oscillating sine and cosine parts integrate to zero except at \(f=f_0\) (respectively \(f=-f_0\)).

  • \(\delta(x+a)+\delta(x-a)\) for \(a>0\):

  • \(-\delta(x+a)\) and \(-\delta(x-a)\) for \(a>0\):

  • \(\displaystyle\int_{-\infty}^{\infty}\delta(ax)\,\dd x\): put \(ax=y\), \(a\,\dd x=\dd y\) (\(a>0\)). Then \[ \int_{-\infty}^{\infty}\delta(ax)\,\dd x=\frac1a\int_{-\infty}^{\infty}\delta(y)\,\dd y=\frac1a. \]

The delta function has infinite height but finite (unit) area.

Example 8.1

If \(h(t)=k\) (a constant), find its Fourier transform.

Solution.

\begin{align*} H(f)&=\int_{-\infty}^{\infty}h(t)\,\ee^{-\jj2\pi ft}\,\dd t=k\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f-0)t}\,\dd t =k\,\delta(f-0)=k\,\delta(f). \end{align*}

Example 8.2

If \(h(t)=A\sin2\pi f_0t=\dfrac A{2\jj}\bigl[\ee^{\jj2\pi f_0t}-\ee^{-\jj2\pi f_0t}\bigr]\), where \(f_0\) is the frequency of the signal and \(f\) is the frequency variable of the Fourier transform, find its Fourier transform.

Solution.

\begin{align*} H(f)&=\int_{-\infty}^{\infty}\frac A{2\jj}\bigl[\ee^{\jj2\pi f_0t}-\ee^{-\jj2\pi f_0t}\bigr]\ee^{-\jj2\pi ft}\,\dd t\\ &=\frac A{2\jj}\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f-f_0)t}\,\dd t-\frac A{2\jj}\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f+f_0)t}\,\dd t,\\ H(f)&=\frac A{2\jj}\bigl[\delta(f-f_0)-\delta(f+f_0)\bigr]=-\jj\frac A2\bigl[\delta(f-f_0)-\delta(f+f_0)\bigr]\\ &=\jj\frac A2\bigl[\delta(f+f_0)-\delta(f-f_0)\bigr]. \end{align*} Both \(h(t)\) and the imaginary part of \(H(f)\) (plotted as \(\jj H(f)\) below) are odd:

Symmetry of the Fourier Transform

  1. If a function is odd, its Fourier transform is imaginary.

  2. If a function is even, its Fourier transform is real.

  3. If a function is mixed (neither even nor odd), its Fourier transform is complex.

For the sine function above, the imaginary part of \(H(f)\) is two opposite impulses at \(\mp f_0\):

To resolve the peak at \(f_0\), the sampling frequency must exceed \(2f_0\) (Nyquist criterion).

Analogy between Fourier series and Fourier transformation \[ H(f)=\int_{-\infty}^{\infty}h(t)\,\ee^{-\jj2\pi ft}\,\dd t. \] If \(h(t)\) is even, then, since \(\ee^{-\jj2\pi ft}=\cos2\pi ft-\jj\sin2\pi ft\), \[ H(f)=\int_{-\infty}^{\infty}\Bigl\{\underbrace{h(t)\cos2\pi ft}_{\text{even integrand}\,\neq0} -\underbrace{\jj\sin2\pi ft\times h(t)}_{\text{odd integrand}\,=0}\Bigr\}\dd t, \qquad H(f)=\text{real}. \] Similarly, if \(h(t)\) is odd, then \(H(f)\) is imaginary.

Example 8.3

If \(h(t)=A\cos2\pi f_0t\), where \(f_0\) is the frequency of the signal and \(f\) is the frequency variable of the Fourier transform, find its Fourier transform.

Solution.

\begin{align*} H(f)&=\int_{-\infty}^{\infty}\frac A2\bigl[\ee^{\jj2\pi f_0t}+\ee^{-\jj2\pi f_0t}\bigr]\ee^{-\jj2\pi ft}\,\dd t =\frac A2\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f-f_0)t}\,\dd t+\frac A2\int_{-\infty}^{\infty}\ee^{-\jj2\pi(f+f_0)t}\,\dd t,\\ H(f)&=\frac A2\bigl[\delta(f-f_0)+\delta(f+f_0)\bigr]. \end{align*}

Figure 8.1 Time domain (left) and frequency domain (right) for \(h(t)=A\cos2\pi f_0t\): both are symmetric about the \(y\)-axis (even); each impulse has area \(A/2\).

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