The Fourier Transforms

  • Lokenath Debnath
  • Firdous A. Shah
Chapter
Part of the Compact Textbooks in Mathematics book series (CTM)

Abstract

Historically, Joseph Fourier (1770–1830) first introduced the remarkable idea of expansion of a function in terms of trigonometric series without giving any attention to rigorous mathematical analysis. The integral formulas for the coefficients of the Fourier expansion were already known to Leonhard Euler (1707–1783) and others. In fact, Fourier developed his new idea for finding the solution of heat (or Fourier) equation in terms of Fourier series so that the Fourier series can be used as a practical tool for determining the Fourier series solution of partial differential equations under prescribed boundary conditions. Thus, the Fourier series of a function f(t) defined on the interval (−L, L) is given by

Fourier’s theorem is not only one of the most beautiful results of modern analysis, but it may be said to furnish an indispensable instrument in the treatment of nearly every recondite question in modern physics.

 Lord Kelvin

Fourier was motivated by the study of heat diffusion, which is governed by a linear differential equation. However, the Fourier transform diagonalizes all linear time-invariant operators, which are building blocks of signal processing. It is therefore not only the starting point of our exploration but the basis of all further developments.

 Stéphane Mallat

1.1 Introduction

Historically, Joseph Fourier (1770–1830) first introduced the remarkable idea of expansion of a function in terms of trigonometric series without giving any attention to rigorous mathematical analysis (See Fourier 1822). The integral formulas for the coefficients of the Fourier expansion were already known to Leonhard Euler (1707–1783) and others. In fact, Fourier developed his new idea for finding the solution of heat (or Fourier) equation in terms of Fourier series so that the Fourier series can be used as a practical tool for determining the Fourier series solution of partial differential equations under prescribed boundary conditions. Thus, the Fourier series of a function f(t) defined on the interval (−L, L) is given by
$$\displaystyle{ f(t) =\sum _{n\in \mathbb{Z}}c_{n}\exp \left (\frac{in\pi t} {L} \right ), }$$
(1.1.1)
where the Fourier coefficients are
$$\displaystyle{ c_{n} = \frac{1} {2L}\int _{-L}^{L}f(t)\,\exp \left (-\frac{in\pi t} {L} \right )\,dt. }$$
(1.1.2)
In order to obtain a representation for a nonperiodic function defined for all real t, it seems desirable to take limit as L that leads to the formulation of the famous Fourier integral theorem:
$$\displaystyle{ f(t) = \frac{1} {2\pi }\int _{-\infty }^{\infty }e^{i\omega t}d\omega \int _{ -\infty }^{\infty }e^{-i\omega t}f(t)\,dt. }$$
(1.1.3)
Mathematically, this is a continuous version of the completeness property of Fourier series. Physically, this form (1.1.3) can be resolved into an infinite number of harmonic components with continuously varying frequency \(\left ( \dfrac{\omega } {2\pi }\right )\) and amplitude,
$$\displaystyle{ \frac{1} {2\pi }\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt, }$$
(1.1.4)
whereas the ordinary Fourier series represents a resolution of a given function into an infinite but discrete set of harmonic components. Thus, using the notation of an inner product, the Fourier transform of a continuous function f(t) can be expressed as
$$\displaystyle{ \hat{f}(\omega ) =\big\langle f,e^{i\omega t}\big\rangle =\int _{ -\infty }^{\infty }e^{-i\omega t}f(t)\,dt. }$$
(1.1.5)

This transform decomposes a signal into orthogonal trigonometric basis functions of different frequencies and phases, and it is often called the Fourier spectrum. It is generally believed that the theory of Fourier series and Fourier transforms is one of the most remarkable discoveries in mathematical sciences and has wide spread applications in mathematics, physics, and engineering.

This chapter deals with Fourier transforms in \(L^{1}(\mathbb{R})\) and \(L^{2}(\mathbb{R})\) and their basic properties. Several important results including the approximate identity theorem, general Parseval relation, and Plancherel theorem are proved. Discrete Fourier transform (DFT), fast Fourier transform (FFT) , and fractional Fourier transform (FrFT) are also discussed briefly for the purpose of comparing them with the continuous, discrete, and fractional wavelet transforms. Applications of the FrFT in solving generalized nonhomogeneous differential equations including the generalized wave and heat equations are also given. Special attention is also given to the Heisenberg uncertainty principle.

1.2 The Fourier Transform in \(L^{1}(\mathbb{R})\)

We begin by introducing some notation that will be used throughout this work. The set of natural numbers (positive integers) is denoted by \(\mathbb{N}\) and the set of integers by \(\mathbb{Z}\). The fields of real and complex numbers are denoted by \(\mathbb{R}\) and \(\mathbb{C}\), respectively. Elements of fields \(\mathbb{R}\) and \(\mathbb{C}\) are called scalars. We will deal with various spaces of functions defined on \(\mathbb{R}\). The simplest of these are the \(L^{p} = L^{p}(\mathbb{R})\) spaces, 1 ≤ p < , the vector space of all complex-valued Lebesgue integrable functions f defined on \(\mathbb{R}\) with a norm
$$\displaystyle{ \big\|f\big\|_{p} = \left [\int _{-\infty }^{\infty }\big\vert f(t)\big\vert ^{p}dt\right ]^{1/p} <\infty. }$$
(1.2.1)

The number \(\big\|f\big\|_{p}\) is called the L p -norm. These signal classes turn out to be Banach spaces, since Cauchy sequences of signals in L p converge to a limit signal also in L p . Since we do not require any knowledge of the Banach space for an understanding of wavelets in this introductory book, the reader needs to know some elementary properties of the L p -norms. The L p spaces for the cases p = 1, p = 2, 0 < p < 1, and 1 < p < are different in structure, importance, and technique, and these spaces play a very special role in many mathematical investigations. The case p = 2 is of special interest: L2 is a Hilbert space. That is, there is an inner product relation on square-integrable functions, which extends the idea of the vector space dot product to analog signals. When p = , the space \(L^{\infty }(\mathbb{R})\) is the collection of measurable functions which are bounded, after we neglect a set of measure zero. we neglect a set of measure zero (See Debnath and Bhatta, 2015; Debnath and Mikusinski, 1999).

In particular, \(L^{1}(\mathbb{R})\) is the space of all Lebesgue integrable functions defined on \(\mathbb{R}\) with the L1-norm given by
$$\displaystyle{ \big\|f\big\|_{1} =\int _{ -\infty }^{\infty }\big\vert f(t)\big\vert \,dt <\infty. }$$
Suppose f is a Lebesgue integrable function on \(\mathbb{R}\). Since eiωt is continuous and bounded, the product eiωt f(t) is locally integrable for any \(\omega \in \mathbb{R}\). Also, \(\big\vert e^{-i\omega t}\big\vert \leq 1\) for all ω and t on \(\mathbb{R}\). Consider the inner product
$$\displaystyle{ \big\langle f,e^{i\omega t}\big\rangle =\int _{ -\infty }^{\infty }f(t)\,e^{-i\omega t}dt,\quad \omega \in \mathbb{R}. }$$
(1.2.2)
Clearly,
$$\displaystyle{ \left \vert \int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt\right \vert \leq \int _{ -\infty }^{\infty }\big\vert f(t)\big\vert dt =\big\| f\big\|_{ 1} <\infty. }$$
(1.2.3)

This means that integral (1.2.2) exists for all \(\omega \in \mathbb{R}\). Thus, we give the following definition.

Definition 1.2.1 (The Fourier Transform in \(L^{1}(\mathbb{R})\)).

The Fourier transform of any function \(f \in L^{1}(\mathbb{R})\) is defined by
$$\displaystyle{ \hat{f}(\omega ) = \mathcal{F}\big\{f(t)\big\} =\int _{ -\infty }^{\infty }e^{-i\omega t}f(t)\,dt. }$$
(1.2.4)

Remarks.

 
  1. 1.

    Physically, the Fourier integral (1.2.4) measures oscillations of f at the frequency ω and \(\hat{f}(\omega )\) is called the frequency spectrum of a signal or waveform f(t).

     
  2. 2.

    The factor 1∕2π may be bundled with the Fourier transform or \(1/\sqrt{2\pi }\) can appear in front of the transform and the inversion formula to provide a symmetric appearance. All these approaches are found in the literature.

     
  3. 3.

    The Fourier transform is, in fact a continuous version of the Fourier series. A Fourier series decomposes a signal on [−π, π] into components that vibrate at integer frequencies. By contrast the Fourier transform decomposes a signal defined on an infinite time interval into a ω-frequency component, where ω can be any real number.

     
  4. 4.
    Another form for the Fourier transform of f used frequently in probability theory replaces the kernel exp(−iωt) by exp(iωt). In this case if f is the probability density function of the random variable x, then
    $$\displaystyle{g(x) =\int _{ -\infty }^{\infty }f(t)\,e^{ixt}\,dt,}$$

    is called the characteristic function of f.

     
  5. 5.
    In general, the Fourier transform \(\hat{f}(\omega )\) is a complex function of a real variable ω. From a physical point of view, the polar representation of the Fourier transform is often convenient. The Fourier transform \(\hat{f}(\omega )\) can be expressed in the polar form
    $$\displaystyle{ \hat{f}(\omega ) = R(\omega ) + iX(\omega ) = A(\omega )\exp \left \{i\theta (\omega )\right \}, }$$
    (1.2.5)

    where \(A(\omega ) = \left \vert \hat{f}(\omega )\right \vert\) is called the amplitude spectrum of the signal f(t), and \(\theta (\omega ) =\arg \left \{\hat{f}(\omega )\right \}\) is called the phase spectrum of f(t).

     

Example 1.2.1.

The Gaussian function is one of the most important functions in probability theory and analysis of random analysis. It plays a central role in Gabor transform. The Gaussian function with unit amplitude is given by
$$\displaystyle{ f(t) = e^{-a^{2}t^{2} },\quad a> 0. }$$
(1.2.6)
The Fourier transform of f is computed as
$$\displaystyle\begin{array}{rcl} \hat{f}(\omega )& =& \int _{-\infty }^{\infty }e^{-\big(i\omega t+a^{2}t^{2}\big) }dt =\int _{ -\infty }^{\infty }e^{-a^{2}\left (t+ \frac{i\omega } {2a^{2}} \right )^{2}- \frac{\omega ^{2}} {4a^{2}} }dt \\ & =& e^{-\omega ^{2}/4a^{2} }\int _{-\infty }^{\infty }e^{-a^{2}y^{2} }\,dy = \frac{\sqrt{\pi }} {a} e^{-\omega ^{2}/4a^{2} }, {}\end{array}$$
(1.2.7)
in which the change of variable \(y = \left (t + \dfrac{i\omega } {2a^{2}}\right )\) is used. Even though \(\left ( \dfrac{i\omega } {2a^{2}}\right )\) is a complex number, the above result is correct. The change of variable can be justified by the method of complex analysis. The graphs of f(t) and \(\hat{f}(\omega )\) are drawn in Figure 1.1.
Fig. 1.1

Graphs of \(f(t) = e^{-a^{2}t^{2} }\) and \(\hat{f}(\omega )\) with a = 1

It is interesting to note that the Fourier transform of a Gaussian function is also a Gaussian function (see Figure 1.1). The parameter a can be used to control the width of the Gaussian pulse. It is evident from relations (1.2.6) and (1.2.7) that large values of a produce a narrow pulse but its spectrum spreads wider on ω-axis.

In particular, when \(a^{2} = \dfrac{1} {2}\) and a = 1, we obtain the following results
$$\displaystyle{ (a)\qquad \qquad \qquad \mathcal{F}\left \{e^{-t^{2}/2 }\right \} = \sqrt{2\pi }\,e^{-\omega ^{2}/2 },\qquad \qquad }$$
(1.2.8)
$$\displaystyle{ (b)\qquad \qquad \quad \mathcal{F}\left \{e^{-t^{2} }\right \} = \sqrt{\pi }\,e^{-\omega ^{2}/4 }.\qquad \qquad \qquad }$$
(1.2.9)

Example 1.2.2 (Characteristic Function).

This function is defined by
$$\displaystyle{ f(t) = \left \{\begin{array}{ll} 1,& - a <t <a\\ 0, &\text{otherwise.}\end{array} \right. }$$
(1.2.10)
In science and engineering, this function is often called a rectangular pulse or gate function. Its Fourier transform is
$$\displaystyle{ \hat{f}(\omega ) = \mathcal{F}\big\{f(t)\big\} = \left (\frac{2} {\omega } \right )\,\sin (a\omega ). }$$
(1.2.11)
We have
$$\displaystyle{\hat{f}(\omega ) =\int _{ -\infty }^{\infty }f(t)\,e^{-i\omega t}\,dt =\int _{ -a}^{a}e^{-i\omega t}\,dt = \left (\frac{2} {\omega } \right )\,\sin (a\omega ).}$$
Note that the Fourier transform \(\hat{f}(\omega )\) vibrates with a zero frequency and hence we should expect that the larger values \(\hat{f}(\omega )\) occur when ω is near zero (see Figure 1.2). Also, it should be noted that \(f(t) \in L^{1}(\mathbb{R})\), but its Fourier transform \(\hat{f}(\omega )\not\in L^{1}(\mathbb{R})\).
Fig. 1.2

Graphs of f(t) and \(\hat{f}(\omega )\) with a = 1

Example 1.2.3.

Find the Fourier transform of
$$\displaystyle{ f(t) = \left (1 -\dfrac{\vert t\vert } {a} \right )H\left (1 -\dfrac{\vert t\vert } {a} \right ) }$$
where H(t) is the Heaviside unit step function defined by
$$\displaystyle{ H(t) = \left \{\begin{array}{ll} 1,&t> 0,\\ 0, &t <0.\end{array} \right. }$$
(1.2.12)
Or, more generally,
$$\displaystyle{ H(t-a) = \left \{\begin{array}{ll} 1,&t> a,\\ 0, &t <a,\end{array} \right. }$$

where a is a fixed real number. So the Heaviside function H(ta) has a finite discontinuity at t = a. Then, it can easily be verified that

$$\displaystyle{ \hat{f}(\omega ) = a\, \cdot \frac{\sin ^{2}\left (\dfrac{a\omega } {2}\right )} {\left (\dfrac{a\omega } {2}\right )^{2}}. }$$
(1.2.13)

Example 1.2.4.

Find the Fourier transform of f(t) = ea | t |, a > 0.

We have
$$\displaystyle\begin{array}{rcl} \mathcal{F}\big\{e^{-a\vert t\vert }\big\}& =& \int _{ -\infty }^{\infty }e^{-a\vert t\vert -i\omega t}dt {}\\ & =& \int _{-\infty }^{0}e^{(a-i\omega )t}dt +\int _{ 0}^{\infty }e^{-(a-i\omega )t}dt {}\\ & =& \dfrac{1} {a - i\omega } + \dfrac{1} {a + i\omega } = \dfrac{2a} {a^{2} +\omega ^{2}}. {}\\ \end{array}$$
We note that f(t) = ea | t | decreases rapidly at infinity and it is not differentiable at t = 0 (Figure 1.3).
Fig. 1.3

Graphs of f(t) = ea | t | and \(\hat{f}(\omega )\) with a = 1

Example 1.2.5.

\(\mathcal{F}\big\{f(t)\big\} = \mathcal{F}\left \{\big(a^{2} + t^{2}\big)^{-1}\right \} = \frac{\pi } {a}\,e^{-a\vert \omega \vert },\ \ a> 0.\)

Example 1.2.5 can easily be verified and hence left to the reader.

Before we discuss the basic properties of Fourier transforms, we define the translation, modulation, and dilation operators respectively, by
$$\displaystyle\begin{array}{rcl} T_{a}f(t)& =& f(t - a)\qquad \qquad \text{(Translation),} {}\\ M_{b}f(t)& =& e^{ibt}\,f(t)\qquad \qquad \text{(Modulation),} {}\\ D_{c}f(t)& =& \frac{1} {\sqrt{\vert c\vert }}\,f\left (\frac{t} {c}\right )\qquad \text{(Dilation),} {}\\ \end{array}$$
where \(a,b,c \in \mathbb{R}\) and c ≠ 0. Each of these operators is a unitary operator from \(L^{2}(\mathbb{R})\) onto itself. The following results can easily be verified:
$$\displaystyle\begin{array}{rcl} T_{a}\,M_{b}\,f(t)& =& e^{ib(t-a)}f(t - a), {}\\ M_{b}\,T_{a}\,f(t)& =& e^{ibt}\,f(t - a), {}\\ D_{c}\,T_{a}\,f(t)& =& \dfrac{1} {\sqrt{\vert c\vert }}\,f\left (\dfrac{t - a} {c} \right ), {}\\ T_{a}\,D_{c}\,f(t)& =& \dfrac{1} {\sqrt{\vert c\vert }}\,f\left (\dfrac{t - a} {c} \right ), {}\\ M_{b}\,D_{c}\,f(t)& =& \dfrac{1} {\sqrt{\vert c\vert }}\,\exp \left (i\,\dfrac{b} {c}\,t\right )f\left (\dfrac{t} {c}\right ), {}\\ D_{c}\,M_{b}\,f(t)& =& \dfrac{1} {\sqrt{\vert c\vert }}\,\exp \left (i\,\dfrac{b} {c}\,t\right )f\left (\dfrac{t} {c}\right ). {}\\ \end{array}$$

Theorem 1.2.1.

If \(f(t),g(t) \in L^{1}(\mathbb{R})\) and α, β are any two complex constants, then
  1. (a)

    Linearity: \(\mathcal{F}\big\{\alpha f(t) +\beta g(t)\big\} =\alpha \, \mathcal{F}\big\{f(t)\big\} +\beta \, \mathcal{F}\big\{g(t)\big\}.\)

     
  2. (b)

    Shifting: \(\mathcal{F}\big\{T_{a}f(t)\big\} = M_{-a}\hat{f}(\omega ),\)

     
  3. (c)

    Scaling: \(\mathcal{F}\left \{D_{\frac{1} {a} }f(t)\right \} = D_{a}\hat{f}(\omega ),\)

     
  4. (d)

    Conjugation: \(\mathcal{F}\left \{\overline{D_{-1}f(t)}\right \} = \overline{\hat{f}(\omega )},\)

     
  5. (e)

    Modulation: \(\mathcal{F}\big\{M_{a}f(t)\big\} = T_{a}\hat{f}(\omega ).\)

     

The proof follows readily from Definition 1.2.1 and is left as an exercise.

Theorem 1.2.2 (Continuity).

If \(f(t) \in L^{1}(\mathbb{R})\), then \(\hat{f}(\omega )\) is continuous on \(\mathbb{R}\).

We shall discuss the derivative of the Fourier transform. As we know that smoother the function f, the more rapidly \(\hat{f}\) will decay at infinity and conversely. Therefore, more rapidly f decays at infinity, the smoother \(\hat{f}\) will be. There are various ways to measure the smoothness of a given function f but here we will measure the smoothness of f by counting the number of derivatives it has.

Theorem 1.2.3 (Differentiation Theorem).

If both f(t) and tf(t) belong to \(L^{1}(\mathbb{R})\), then \(\dfrac{d} {d\omega }\,\hat{f}(\omega )\) exists and is given by
$$\displaystyle{ \frac{d} {d\omega }\,\hat{f}(\omega ) = (-i)\,\mathcal{F}\big\{tf(t)\big\}. }$$
(1.2.14)

Proof.

We have
$$\displaystyle{ \frac{d\hat{f}} {d\omega } =\lim _{h\rightarrow 0} \frac{1} {h}\left [\hat{f}(\omega +h) -\hat{ f}(\omega )\right ] =\lim _{h\rightarrow 0}\left [\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\left (\frac{e^{-iht} - 1} {h} \right )dt\right ]. }$$
(1.2.15)
Note that
$$\displaystyle{\left \vert \frac{1} {h}(e^{-iht} - 1)\right \vert = \frac{1} {\vert h\vert }\left \vert e^{-\frac{iht} {2} }\left (e^{\frac{iht} {2} } - e^{-\frac{iht} {2} }\right )\right \vert = 2\left \vert \frac{\sin \left (\dfrac{ht} {2} \right )} {h} \right \vert \leq \vert t\vert.}$$
Also,
$$\displaystyle{\lim _{h\rightarrow 0}\left (\dfrac{e^{-iht} - 1} {h} \right ) = -it.}$$
Thus, result (1.2.15) becomes
$$\displaystyle{\begin{array}{rcl} \frac{d\hat{f}} {d\omega } & =&\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\lim _{ h\rightarrow 0}\left (\frac{e^{-iht} - 1} {h} \right )dt\\ \\ & =&(-i)\int _{-\infty }^{\infty }tf(t)\,e^{-i\omega t}\,dt = (-i)\,\mathcal{F}\big\{tf(t)\big\}.\end{array} }$$

This proves the theorem.

Corollary 1.2.1.

If \(f \in L^{1}(\mathbb{R})\) such that t n f(t) is integrable for finite \(n \in \mathbb{N}\), then the nth derivative of \(\hat{f}(\omega )\) exists and is given by
$$\displaystyle{ \frac{d^{n}\hat{f}} {d\omega ^{n}} = (-i)^{n}\mathcal{F}\big\{t^{n}f(t)\big\}. }$$
(1.2.16)

Proof.

This corollary follows from Theorem 1.2.3 combined with the mathematical induction principle. In particular, putting ω = 0 in (1.2.16) gives
$$\displaystyle{ \left [\frac{d^{n}\hat{f}(\omega )} {d\omega ^{n}} \right ]_{\omega =0} = (-i)^{n}\int _{ -\infty }^{\infty }t^{n}f(t)\,dt = (-i)^{n}m_{ n}, }$$
(1.2.17)

where m n represents the nth moment of f(t). Thus, the moments m1, m2, , m n can be calculated from (1.2.17)

Theorem 1.2.4 (The Riemann-Lebesgue Lemma).

If \(f \in L^{1}(\mathbb{R})\) , then
$$\displaystyle{ \lim _{\vert \omega \vert \rightarrow \infty }\left \vert \hat{f}(\omega )\right \vert = 0. }$$
(1.2.18)

Proof.

Since \(e^{-i\omega t} = -e^{-i\omega \left (t+\frac{\pi }{\omega }\right )}\), we have
$$\displaystyle{\hat{f}(\omega ) = -\int _{-\infty }^{\infty }e^{-i\omega \left (t+\frac{\pi }{\omega }\right )}f(t)\,dt = -\int _{ -\infty }^{\infty }e^{-i\omega x}f\left (x -\frac{\pi } {\omega }\right )dx.}$$
Thus,
$$\displaystyle\begin{array}{rcl} \hat{f}(\omega )& =& \dfrac{1} {2}\left [\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt -\int _{ -\infty }^{\infty }e^{-i\omega t}f\left (t -\frac{\pi } {\omega }\right )\,dt\right ] {}\\ & & {}\\ & =& \dfrac{1} {2}\int _{-\infty }^{\infty }e^{-i\omega t}\left [f(t) - f\left (t -\frac{\pi } {\omega }\right )\right ]dt. {}\\ \end{array}$$
Clearly,
$$\displaystyle{\lim _{\vert \omega \vert \rightarrow \infty }\left \vert \hat{f}(\omega )\right \vert \leq \frac{1} {2}\lim _{\vert \omega \vert \rightarrow \infty }\int _{-\infty }^{\infty }\left \vert f(t) - f\left (t -\frac{\pi } {\omega }\right )\right \vert dt = 0.}$$

This completes the proof.

Observe that the space \(C_{0}(\mathbb{R})\) of all continuous functions on \(\mathbb{R}\) which decay at infinity, that is, f(t) → 0 as | t | → , is a normed space with respect to the norm defined by
$$\displaystyle{ \big\|f\big\| =\sup _{t\in \mathbb{R}}\big\vert f(t)\big\vert. }$$
(1.2.19)

It follows from above theorems that the Fourier transform is a continuous linear operator from \(L^{1}(\mathbb{R})\) into \(C_{0}(\mathbb{R})\). Theorem 1.2.4 gives a necessary condition for a function f to have a Fourier transform. However, that belonging to \(C_{0}(\mathbb{R})\) is not a sufficient condition for being the Fourier transform of an integrable function.

Theorem 1.2.5.

 
  1. (a)
    If f(t) is a continuously differentiable function, \(\lim _{\vert t\vert \rightarrow \infty }f(t) = 0\) and both \(f,f^{{\prime}}\in L^{1}(\mathbb{R})\), then
    $$\displaystyle{ \mathcal{F}\big\{f^{{\prime}}(t)\big\} = i\omega \,\mathcal{F}\big\{f(t)\big\} = (i\omega )\hat{f}(\omega ). }$$
    (1.2.20)
     
  2. (b)
    If f(t) is continuously n-times differentiable, \(f,f^{{\prime}},\ldots,f^{(n)} \in L^{1}(\mathbb{R})\) and  
    $$\displaystyle{\lim _{\vert t\vert \rightarrow \infty }f^{(r)}(t) = 0\qquad \mathit{\text{for}}\ r = 1,2,\ldots,n - 1,}$$
    then
    $$\displaystyle{ \mathcal{F}\left \{f^{(n)}(t)\right \} = (i\omega )^{n}\,\mathcal{F}\big\{f(t)\big\} = (i\omega )^{n}\hat{f}(\omega ). }$$
    (1.2.21)
     

Proof.

We have, by definition,
$$\displaystyle{\mathcal{F}\big\{f^{{\prime}}(t)\big\} =\int _{ -\infty }^{\infty }e^{-i\omega t}f^{{\prime}}(t)\,dt,}$$
which is, integrating by parts,
$$\displaystyle\begin{array}{rcl} & =& \left [e^{-i\omega t}f(t)\right ]_{ -\infty }^{\infty } + (i\omega )\int _{ -\infty }^{\infty }e^{-i\omega t}f(t)\,dt {}\\ & =& (i\omega )\hat{f}(\omega ). {}\\ \end{array}$$

This proves part (a) of the theorem.

A repeated application of (1.2.16) to higher-order derivatives gives result (1.2.21).

We next calculate the Fourier transform of partial derivatives. If u(x, t) is continuously n times differentiable and \(\dfrac{\partial ^{r}u} {\partial x^{r}} \rightarrow 0\) as | x | → for r = 1, 2, 3, , (n − 1), then, the Fourier transform of \(\dfrac{\partial ^{n}u} {\partial x^{n}}\) with respect to x is

$$\displaystyle{ \mathcal{F}\left \{\dfrac{\partial ^{n}u} {\partial x^{n}}\right \} = (ik)^{n}\mathcal{F}\left \{u(x,t)\right \} = (ik)^{n}\,\hat{u}(k,t). }$$
(1.2.22)
It also follows from the Definition 1.2.1 that
$$\displaystyle{ \mathcal{F}\left \{\frac{\partial u} {\partial t} \right \} = \frac{d\hat{u}} {dt},\ \mathcal{F}\left \{\frac{\partial ^{2}u} {\partial t^{2}} \right \} = \frac{d^{2}\hat{u}} {dt^{2}},\ldots,\mathcal{F}\left \{\frac{\partial ^{n}u} {\partial t^{n}} \right \} = \frac{d^{n}\hat{u}} {dt^{n}}. }$$
(1.2.23)

Remark.

This result has an important consequence: The smoothness of f is manifested in the rate of decay of its Fourier transform \(\hat{f}\). We have already noted that the Fourier transform of a \(L^{1}(\mathbb{R})\) function must decay to zero at large frequencies \(\hat{f}(\omega ) \rightarrow 0\) as ω. If the nth derivative f n is also in \(L^{1}(\mathbb{R})\) and the derivatives vanish at infinity, then its Fourier transform \(\mathcal{F}\big[f^{n}(t)\big] = (i\omega )^{n}\hat{f}(\omega )\) must go to zero as ω. This requires that \(\hat{f}(\omega )\) go to zero more rapidly than | ω |n . Thus, the smoother f, the more rapid the decay of its Fourier transform. As a general rule of thumb, local features of f such as smoothness are manifested by global features of \(\hat{f}(\omega )\), such as decay for large | ω |. The symmetry principle implies that reverse is also true: global features of f correspond to local features of \(\hat{f}(\omega )\). This local-global duality is one of the major themes of Fourier theory.

We now introduce the concept of convolution fg of two functions \(f,g \in L^{1}(\mathbb{R})\). Recall that if f and g are integrable functions on \(\mathbb{R}\), then the convolution is defined by
$$\displaystyle{ \big(f {\ast} g\big)(t) =\int _{ -\infty }^{\infty }f(t-\tau )\,g(\tau )\,d\tau. }$$
(1.2.24)
The existence of the integral is justified by the following argument:
$$\displaystyle{\int _{-\infty }^{\infty }\int _{ -\infty }^{\infty }\big\vert f(t-\tau )\,g(\tau )\big\vert dt\,d\tau =\int _{ -\infty }^{\infty }\big\vert g(\tau )\big\vert d\tau \int _{ -\infty }^{\infty }\big\vert f(t)\big\vert dt =\big\| g\big\|_{ 1}\big\|f\big\|_{1}.}$$
It is clear that \((f {\ast} g)(t) \in L^{1}(\mathbb{R})\) and in fact, we have
$$\displaystyle{\big\|f {\ast} g\big\|_{1} \leq \big\| g\big\|_{1}\big\|f\big\|_{1}.}$$
It is easy to verify that convolution is commutative, associative and distributive. That is;
$$\displaystyle{\begin{array}{lcr} f(t) {\ast} g(t) = g(t) {\ast} f(t), \\ f(t) {\ast}\big (g(t) {\ast} h(t)\big) =\big (f(t) {\ast} g(t)\big) {\ast} h(t), \\ f(t) {\ast}\big (g(t) + h(t)\big) = f(t) {\ast} g(t) + f(t) {\ast} h(t).\end{array} }$$

Proposition 1.2.1.

If \(f \in L^{1}(\mathbb{R})\) and \(g \in L^{\infty }(\mathbb{R})\), then the convolution fg is continuous on \(\mathbb{R}\).

Proposition 1.2.2 (Young’s Inequality).

If the exponents p, q and s satisfy 1∕s = 1∕p + 1∕q − 1, then
$$\displaystyle{ \big\|f {\ast} g\big\|_{s} \leq \big\| f\big\|_{p}\big\|g\big\|_{q}. }$$
(1.2.25)

Theorem 1.2.6 (Convolution Theorem).

If \(f,g \in L^{1}(\mathbb{R})\) , then
$$\displaystyle{ \mathcal{F}\left \{\big(f {\ast} g\big)(t)\right \} = \mathcal{F}\big\{f(t)\big\}\mathcal{F}\big\{g(t)\big\} =\hat{ f}(\omega )\,\hat{g}(\omega ). }$$
(1.2.26)

Proof.

Since \(f {\ast} g \in L^{1}(\mathbb{R})\), we apply the definition of the Fourier transform to obtain
$$\displaystyle{\begin{array}{rcl} \mathcal{F}\left \{\big(f {\ast} g\big)(t)\right \}& =&\int _{-\infty }^{\infty }e^{-i\omega t}dt\int _{ -\infty }^{\infty }f(t-\tau )\,g(\tau )\,d\tau \\ \\ & =&\int _{-\infty }^{\infty }g(\tau )\int _{ -\infty }^{\infty }e^{-i\omega t}f(t-\tau )\,dt\,d\tau \\ \\ & =&\int _{-\infty }^{\infty }e^{-i\omega \tau }\,g(\tau )d\tau \int _{ -\infty }^{\infty }e^{-i\omega u}f(u)\,du,\quad (t-\tau = u)\\ \\ & =&\hat{f}(\omega )\,\hat{g}(\omega ),\end{array} }$$

in which Fubini’s theorem was utilized.

Corollary 1.2.2.

If \(f,g,h \in L^{1}(\mathbb{R})\) such that
$$\displaystyle{ h(x) =\int _{ -\infty }^{\infty }g(\omega )\,e^{i\omega x}d\omega,then }$$
(1.2.27)
$$\displaystyle{\big(f {\ast} h\big)(x) =\int _{ -\infty }^{\infty }g(\omega )\hat{f}(\omega )\,e^{i\omega x}d\omega.}$$
We now turn to the problem of inverting the Fourier transform. That is, we shall consider the question: Given the Fourier transform of an integrable function f(t), how do we obtain f(t) back again from \(\hat{f}(\omega )\)? The reader, who is familiar with the elementary theory of Fourier series and integrals, would expect
$$\displaystyle{ f(t) = \frac{1} {2\pi }\int _{-\infty }^{\infty }e^{i\omega t}\hat{f}(\omega )\,d\omega. }$$
(1.2.28)
Unfortunately, it is not necessary that if \(f \in L^{1}(\mathbb{R})\), then its Fourier transform \(\hat{f}\) also belongs to \(L^{1}(\mathbb{R})\) (see Example 1.2.2), so that the Fourier integral \(\int _{-\infty }^{\infty }\hat{f}(\omega )\,e^{i\omega t}d\omega\) may not exist as a Lebesgue integral. However, we can introduce a function K λ (ω) in the integrand and formulate general conditions on K λ (ω) and its Fourier transform so that the following result holds:
$$\displaystyle{ \lim _{\lambda \rightarrow \infty }\int _{-\lambda }^{\lambda }\hat{f}(\omega )K_{\lambda }(\omega )\,e^{i\omega t}d\omega = f(t) }$$
(1.2.29)

for almost every t. This kernel K λ (ω) is called a convergent factor or a summability kernel on \(\mathbb{R}\) which can formally be defined as follows.

Definition 1.2.2 (Summability Kernel).

A summability kernel on \(\mathbb{R}\) is a family \(\big\{K_{\lambda },\lambda> 0\big\}\) of continuous functions with the following properties:
  1. (i)

    \(\int _{\mathbb{R}}K_{\lambda }(x)\,dx = 1,\quad \text{ for all}\ \lambda> 0,\)

     
  2. (ii)

    \(\int _{\mathbb{R}}\big\vert K_{\lambda }(x)\,dx\big\vert \leq M,\quad \text{for all}\ \lambda> 0\ \text{and for a constant}\ M,\)

     
  3. (iii)

    \(\lim _{\lambda \rightarrow \infty }\int _{\vert x\vert>\delta }\big\vert K_{\lambda }(x)\big\vert dx = 0,\quad \text{ for all}\ \delta> 0.\)

     

The idea of a summability kernel helps to establish the so-called approximate identity theorem.

Theorem 1.2.7 (Approximate Identity Theorem).

If \(f \in L^{1}(\mathbb{R})\) and \(\left \{K_{\lambda }\right.\), \(\left.\lambda> 0\right \} \in L^{1}(\mathbb{R})\) is a summability kernel, then
$$\displaystyle{ \lim _{\lambda \rightarrow \infty }\big\|\big(f {\ast} K_{\lambda }\big) - f\big\| = 0. }$$
(1.2.30)

Proof.

We have, by definition of the convolution (1.2.24),
$$\displaystyle{\big(f {\ast} K_{\lambda }\big)(t) =\int _{ -\infty }^{\infty }f(t - u)K_{\lambda }(u)\,du,}$$
so that
$$\displaystyle{\begin{array}{rcl} \big\vert \big(f {\ast} K_{\lambda }\big)(t) - f(t)\big\vert & =&\left \vert \int _{-\infty }^{\infty }\big\{f(t - u)K_{\lambda }(u)\,du - f(t)\big\}\right \vert \\ \\ & =&\left \vert \int _{-\infty }^{\infty }\big\{f(t - u) - f(t)\big\}K_{\lambda }(u)\,du\right \vert,\quad \text{by Definition1.2.2(i)},\\ \\ & \leq &\int _{-\infty }^{\infty }\big\vert K_{\lambda }(u)\big\vert \big\vert f(t - u) - f(t)\big\vert \,du.\end{array} }$$
Given ɛ > 0, we can choose δ > 0 such that if 0 ≤ | u | < δ, then \(\big\vert f(t - u)\) \(-f(t)\big\vert <\dfrac{\varepsilon } {M}\), where \(\big\|K_{\lambda }\big\|_{1} \leq M\). Consequently,
$$\displaystyle{\begin{array}{rcl} \big\|\big(f {\ast} K_{\lambda }\big)(t) - f(t)\big\|& =&\int _{\mathbb{R}}\big\vert f {\ast} K_{\lambda }(t) - f(t)\big\vert \,dt\\ \\ & \leq &\int _{\mathbb{R}}dt\int _{-\infty }^{\infty }\big\vert K_{\lambda }(u)\big\vert \big\vert f(t - u) - f(t)\big\vert \,du\\ \\ & =&\int _{-\infty }^{\infty }\big\vert K_{\lambda }(u)\big\vert \,\sigma _{ f}(u)\,du,\end{array} }$$
where
$$\displaystyle{\sigma _{f}(u) =\int _{\mathbb{R}}\big\vert f(t - u) - f(t)\big\vert \,dt \leq C.}$$
Thus,
$$\displaystyle\begin{array}{rcl} \left \|\big(f {\ast} K_{\lambda }\big)(t) - f(t)\right \|& \leq & \int _{\vert u\vert <\delta }\big\vert K_{\lambda }(u)\big\vert \,\sigma _{f}(u)\,du +\int _{\vert u\vert>\delta }\big\vert K_{\lambda }(u)\big\vert \,\sigma _{f}(u)\,du {}\\ \ & \leq & \varepsilon +C\int _{\vert u\vert>\delta }\big\vert K_{\lambda }(u)\big\vert \,du =\varepsilon, {}\\ \end{array}$$

since the integral on the right-hand side tends to zero for δ > 0 by Definition 1.2.2 (iii). This completes the proof.

We have the following convolution theorem with respect to the frequency variable.

Theorem 1.2.8 (General Modulation).

If \(\mathcal{F}\left \{f(t)\right \} =\hat{ f}(\omega )\) and \(\mathcal{F}\left \{g(t)\right \} =\hat{ g}(\omega )\) , where \(\hat{f}\) and \(\hat{g}\) belong to \(L^{1}(\mathbb{R})\) , then
$$\displaystyle{ \mathcal{F}\big\{f(t)\,g(t)\big\} = \frac{1} {2\pi }\left (\hat{f} {\ast}\hat{ g}\right )(\omega ). }$$
(1.2.31)

Proof.

Using the inverse Fourier transform, we can rewrite the left-hand side of (1.2.31) as
$$\displaystyle{\begin{array}{rcl} \mathcal{F}\big\{f(t)\,g(t)\big\}& =&\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,g(t)\,dt\\ \\ & =&\frac{1} {2\pi }\int _{-\infty }^{\infty }e^{-i\omega t}g(t)\,dt\int _{ -\infty }^{\infty }e^{ixt}\hat{f}(x)\,dx \\ \\ & =&\frac{1} {2\pi }\int _{-\infty }^{\infty }\hat{f}(x)\,dx\int _{ -\infty }^{\infty }e^{-it(\omega -x)}g(t)\,dt \\ \\ & =&\frac{1} {2\pi }\int _{-\infty }^{\infty }\hat{f}(x)\,\hat{g}(\omega -x)\,dx \\ \\ & =&\frac{1} {2\pi }\left (\hat{f} {\ast}\hat{ g}\right )(\omega ).\end{array} }$$

This completes the proof.

1.3 The Fourier Transform in \(L^{2}(\mathbb{R})\)

In this section, we discuss the extension of the Fourier transform onto \(L^{2}(\mathbb{R})\). It turns out that if \(f \in L^{2}(\mathbb{R})\), then the Fourier transform \(\hat{f}\) of f is also in \(L^{2}(\mathbb{R})\) and \(\left \|\hat{f}\right \|_{2} = \sqrt{2\pi }\big\|f\big\|_{2},\) where
$$\displaystyle{ \big\|f\big\|_{2} = \left \{\int _{-\infty }^{\infty }\big\vert f(t)\big\vert ^{2}dt\right \}^{\frac{1} {2} }. }$$
(1.3.1)
The factor \(\sqrt{2\pi }\) involved in the above result can be avoided by defining the Fourier transform as
$$\displaystyle{ \hat{f}(\omega ) = \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt. }$$
(1.3.2)

Theorem 1.3.1.

Suppose f is a continuous function on \(\mathbb{R}\) vanishing outside a bounded interval. Then, \(\hat{f} \in L^{2}(\mathbb{R})\) and
$$\displaystyle{ \big\|f\big\|_{2} = \left \|\hat{f}\right \|_{2}. }$$
(1.3.3)

Proof.

We assume that f vanishes outside the interval [−π, π]. We use the Parseval formula for the orthonormal sequence of functions on [−π, π],  
$$\displaystyle{\phi _{n}(t) = \frac{1} {\sqrt{2\pi }}\,e^{int},\ \ \ n = 0,\pm 1,\pm 2,\ldots,}$$
to obtain
$$\displaystyle{\big\|f\big\|_{2}^{2} =\sum _{ n\in \mathbb{Z}}\left \vert \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-int}f(t)\,dt\right \vert ^{2} =\sum _{ n\in \mathbb{Z}}\left \vert \hat{f}(n)\right \vert ^{2}.}$$
Since this result also holds for g(t) = eixt f(t) instead of f(t), and \(\big\|f\big\|_{2}^{2} =\big\| g\big\|_{2}^{2}\), then
$$\displaystyle{\big\|f\big\|_{2}^{2} =\sum _{ n\in \mathbb{Z}}\left \vert \hat{f}(n + x)\right \vert ^{2}.}$$
Integrating this result with respect to x from 0 to 1 gives
$$\displaystyle{\begin{array}{rcl} \big\|f\big\|_{2}^{2} & =&\sum _{n\in \mathbb{Z}}\int _{0}^{1}\left \vert \hat{f}(n + x)\right \vert ^{2}dx =\sum _{ n\in \mathbb{Z}}\int _{n}^{n+1}\left \vert \hat{f}(y)\right \vert ^{2}dy,\quad (y = n + x)\\ \\ & =&\int _{-\infty }^{\infty }\left \vert \hat{f}(y)\right \vert ^{2}dy = \left \|\hat{f}\right \|_{ 2}^{2}.\end{array} }$$
If f does not vanish outside [−π, π], then we take a positive number a for which the function g(t) = f(at) vanishes outside [−π, π]. Then,
$$\displaystyle{\hat{g}(\omega ) = \frac{1} {a}\,\hat{f}\left ( \frac{\omega } {a}\right ).}$$
Thus, it turns out that
$$\displaystyle{\big\|f\big\|_{2}^{2} = a\big\|g\big\|_{ 2}^{2} = a\int _{ -\infty }^{\infty }\left \vert \frac{1} {a}\,\hat{f}\left ( \frac{\omega } {a}\right )\right \vert ^{2}d\omega =\int _{ -\infty }^{\infty }\left \vert \hat{f}(\omega )\right \vert ^{2}d\omega = \left \|\hat{f}\right \|_{ 2}^{2}.}$$

This completes the proof.

The space of all continuous functions on \(\mathbb{R}\) with compact support is dense in \(L^{2}(\mathbb{R})\). Theorem 1.3.1 shows that the Fourier transform is a continuous mapping from that space into \(L^{2}(\mathbb{R})\). Since the mapping is linear, it has a unique extension to a linear mapping from \(L^{2}(\mathbb{R})\) into itself. This extension will be called the Fourier transform on \(L^{2}(\mathbb{R})\).

Definition 1.3.1 (Fourier Transform in \(L^{2}(\mathbb{R})\)).

If \(f \in L^{2}(\mathbb{R})\) and \(\left \{f_{n}\right \}\) is a sequence of continuous functions with compact support convergent to f in \(L^{2}(\mathbb{R})\), that is, \(\big\|f - f_{n}\big\| \rightarrow 0\) as n, then the Fourier transform of f is defined by
$$\displaystyle{ \hat{f} =\lim _{n\rightarrow \infty }f_{n}, }$$
(1.3.4)
where the limit is taken with respect to the norm in \(L^{2}(\mathbb{R})\).

Theorem 1.3.1 ensures that the limit exists and is independent of a particular sequence approximating f. It is important to note that the convergence in \(L^{2}(\mathbb{R})\) does not imply pointwise convergence, and therefore the Fourier transform of a square-integrable function is not defined at a point, unlike the Fourier transform of an integrable function. We can assert that the Fourier transform \(\hat{f}\) of \(f \in L^{2}(\mathbb{R})\) is defined almost everywhere on \(\mathbb{R}\) and \(\hat{f} \in L^{2}(\mathbb{R})\). For this reason, we cannot say that if \(f \in L^{1}(\mathbb{R}) \cap L^{2}(\mathbb{R})\), the Fourier transform defined by (1.2.4) and the one defined by (1.3.4) are equal.

An immediate consequence of Definition 1.2.2 and Theorem 1.3.1 leads to the following theorem.

Theorem 1.3.2 (Parseval’s Identity).

If \(f \in L^{2}(\mathbb{R})\) , then
$$\displaystyle{ \big\|f\big\|_{2} = \left \|\hat{f}\right \|_{2}. }$$
(1.3.5)

Theorem 1.3.3.

If \(f \in L^{2}(\mathbb{R})\) , then
$$\displaystyle{ \hat{f}(\omega ) =\lim _{n\rightarrow \infty } \frac{1} {\sqrt{2\pi }}\int _{-n}^{n}e^{-i\omega t}f(t)\,dt, }$$
(1.3.6)
where the convergence is with respect to the L 2 -norm.

Proof.

For n = 1, 2, 3, , we define
$$\displaystyle{ f_{n}(t) = \left \{\begin{array}{ll} f(t),&\text{for}\ \vert t\vert <n\\ \quad 0, &\text{for} \ \vert t\vert \geq n. \\ \end{array} \right. }$$
(1.3.7)

Clearly, \(\big\|f - f_{n}\big\|_{2} \rightarrow 0\) and, hence, \(\left \|\hat{f} -\hat{ f}_{n}\right \|_{2} \rightarrow 0\) as n.

Theorem 1.3.4 (Change of Roof).

If \(f,g \in L^{2}(\mathbb{R})\) , then
$$\displaystyle{ \left \langle f,\bar{\hat{g}}\right \rangle =\int _{ -\infty }^{\infty }f(t)\,\hat{g}(t)\,dt =\int _{ -\infty }^{\infty }\hat{f}(t)\,g(t)\,dt = \left \langle \hat{f},\bar{g}\right \rangle. }$$
(1.3.8)

Proof.

We define both f n (t) and g n (t) by (1.3.7) for n = 1, 2, 3, . Since
$$\displaystyle{\hat{f}_{m}(t) = \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-ixt}f_{ m}(x)\,dx,}$$
we obtain
$$\displaystyle{\int _{-\infty }^{\infty }\hat{f}_{ m}(t)\,g_{n}(t)\,dt = \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }g_{ n}(t)\int _{-\infty }^{\infty }e^{-ixt}f_{ m}(x)\,dx\,dt.}$$
The function exp(−ixt) g n (t) f m (x) is integrable over \(\mathbb{R}^{2}\), and hence, the Fubini theorem can be used to rewrite the above integral in the form
$$\displaystyle{\begin{array}{rcl} \int _{-\infty }^{\infty }\hat{f}_{ m}(t)g_{n}(t)\,dt& =& \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }f_{ m}(x)\int _{-\infty }^{\infty }e^{-ixt}g_{ n}(t)\,dt\,dx \\ \\ & =&\int _{-\infty }^{\infty }f_{ m}(x)\,\hat{g}_{n}(x)\,dx.\end{array} }$$
Since \(\big\|g - g_{n}\big\|_{2} \rightarrow 0\) and \(\big\|\hat{g} -\hat{ g}_{n}\big\|_{2} \rightarrow 0\), letting n with the continuity of the inner product yields
$$\displaystyle{\int _{-\infty }^{\infty }\hat{f}_{ m}(t)\,g_{n}(t)\,dt =\int _{ -\infty }^{\infty }f_{ m}(t)\,\hat{g}_{n}(t)\,dt.}$$

Similarly, letting m gives the desired result (1.3.8).

Lemma 1.3.1.

If \(f \in L^{2}(\mathbb{R})\) and \(g =\bar{\hat{ f}}\), then \(f =\bar{\hat{ g}}\).

Proof.

In view of Theorems 1.3.2 and 1.3.3 and the assumption \(g =\bar{\hat{ f}}\), we find
$$\displaystyle{ \left \langle f,\bar{\hat{g}}\right \rangle = \left \langle \hat{f},\bar{g}\right \rangle = \left \langle \hat{f},\hat{f}\right \rangle = \left \|\hat{f}\right \|_{2}^{2} =\big\| f\big\|_{ 2}^{2}. }$$
(1.3.9)
Also, we have
$$\displaystyle{ \overline{\big\langle f,\bar{\hat{g}}\big\rangle } = \overline{\left \langle \hat{f},\hat{f}\right \rangle } =\big\| f\big\|_{2}^{2}. }$$
(1.3.10)
Finally, by Parseval’s relation,
$$\displaystyle{ \big\|\hat{g}\big\|_{2}^{2} =\big\| g\big\|_{ 2}^{2} = \left \|\hat{f}\right \|_{ 2}^{2} =\big\| f\big\|_{ 2}^{2}. }$$
(1.3.11)
Using (1.3.9)– (1.3.11) gives the following:
$$\displaystyle{\Big\|f -\bar{\hat{ g}}\Big\|_{2}^{2} =\big\langle f -\bar{\hat{ g}},f -\bar{\hat{ g}}\big\rangle =\big\| f\big\|_{ 2}^{2} -\big\langle f,\bar{\hat{g}}\big\rangle -\overline{\big\langle f,\bar{\hat{g}}\big\rangle } +\big\|\hat{ g}\big\|_{ 2}^{2} = 0.}$$

This proves the result \(f =\bar{\hat{ g}}\).

Example 1.3.1 (The Haar Function).

The Haar function is defined by
$$\displaystyle{ f(t) = \left \{\begin{array}{ll} 1, &\text{for}\ \ 0 \leq t <\frac{1} {2}, \\ - 1,&\text{for}\ \ \frac{1} {2} \leq t <1, \\ 0, &\text{otherwise}.\end{array} \right. }$$
(1.3.12)
The Fourier transform of f(t) is given as
$$\displaystyle\begin{array}{rcl} \hat{f}(\omega )& =& \int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt = \left [\int _{ 0}^{\frac{1} {2} }e^{-i\omega t}\,dt -\int _{\frac{1} {2} }^{1}e^{-i\omega t}\,dt\right ] \\ & =& \frac{1} {i\omega } \left (1 - 2e^{-\frac{i\omega } {2} } + e^{-i\omega }\right ) \\ & =& \frac{e^{-\frac{i\omega } {2} }} {(i\omega )} \left (e^{\frac{i\omega } {2} } - 2 + e^{-\frac{i\omega } {2} }\right ) \\ & =& e^{-\frac{i\omega } {2} } \frac{\sin ^{2}\left ( \frac{\omega } {4}\right )} { \frac{\omega } {4}}. {}\end{array}$$
(1.3.13)
The graphs of f(t) and \(\hat{f}(\omega )\) are shown in Figure 1.4.
Fig. 1.4

The graphs of f(t) and \(\hat{f}(\omega )\)

Example 1.3.2 (The Second Derivative of the Gaussian Function).

If
$$\displaystyle{ f(t) =\big (1 - t^{2}\big)e^{-t^{2}/2 }, }$$
(1.3.14)
then the Fourier transform of f(t) can be computed as
$$\displaystyle{\begin{array}{rcl} \hat{f}(\omega )& =&\mathcal{F}\left \{\big(1 - t^{2}\big)e^{-t^{2}/2 }\right \} \\ & =& -\mathcal{F}\left \{ \frac{d^{2}} {dt^{2}}e^{-t^{2}/2 }\right \} \\ & =& - (i\omega )^{2}\mathcal{F}\left \{e^{-t^{2}/2 }\right \} \\ & =&\omega ^{2}\,e^{-t^{2}/2 }.\end{array} }$$
Both f(t) and \(\hat{f}(\omega )\) are plotted in Figure 1.5.
Fig. 1.5

Graphs of f(t) and \(\hat{f}(\omega )\)

Theorem 1.3.5 (Inversion Formula).

If \(f \in L^{2}(\mathbb{R})\) , then
$$\displaystyle{ f(t) =\lim _{n\rightarrow \infty }\frac{1} {2\pi }\int _{-n}^{n}e^{i\omega t}\hat{f}(\omega )\,d\omega, }$$
(1.3.15)
where the convergence is with respect to the \(L^{2}(\mathbb{R})\) -norm.

Proof.

If \(f \in L^{2}(\mathbb{R})\) and \(g =\bar{\hat{ f}}\), then, by Lemma 1.3.1,
$$\displaystyle{\begin{array}{rcl} f(t) = \overline{\hat{g}(t)}& =&\lim _{n\rightarrow \infty }\frac{1} {2\pi }\overline{\int _{-n}^{n}e^{-i\omega t}\,g(\omega )\,d\omega }\\ \\ & =&\lim _{n\rightarrow \infty }\frac{1} {2\pi }\int _{-n}^{n}\overline{e^{-i\omega t}\,g(\omega )}\,d\omega \\ \\ & =&\lim _{n\rightarrow \infty }\frac{1} {2\pi }\int _{-n}^{n}e^{i\omega t}\,\overline{g(\omega )}\,d\omega \\ \\ & =&\lim _{n\rightarrow \infty }\frac{1} {2\pi }\int _{-n}^{n}e^{i\omega t}\,\hat{f}(\omega )\,d\omega. \end{array} }$$
The formula (1.3.15) is called the inverse Fourier transform. If we use the factor \(\left (1/\sqrt{2\pi }\right )\) in the definition of the Fourier transform, then the Fourier transform and its inverse are symmetrical in form, that is,
$$\displaystyle{ \hat{f}(\omega ) = \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-i\omega t}\,f(t)\,dt,\quad f(t) = \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{i\omega t}\,\hat{f}(\omega )\,d\omega. }$$
(1.3.16)

Theorem 1.3.6 (General Parseval’s Relation).

If \(f,g \in L^{2}(\mathbb{R})\) , then
$$\displaystyle{ \big\langle f,g\big\rangle =\int _{ -\infty }^{\infty }f(t)\,\overline{g(t)}\,dt =\int _{ -\infty }^{\infty }\hat{f}(\omega )\,\overline{\hat{g}(\omega )}\,d\omega =\big\langle \hat{ f},\hat{g}\big\rangle, }$$
(1.3.17)

where the symmetrical form (1.3.16) of the Fourier transform and its inverse is used.

Proof.

It follows from (1.3.3) that
$$\displaystyle{\big\|f + g\big\|_{2}^{2} = \left \|\hat{f} +\hat{ g}\right \|_{ 2}^{2}.}$$
Or, equivalently,
$$\displaystyle{\int _{-\infty }^{\infty }\big\vert f + g\big\vert ^{2}dt =\int _{ -\infty }^{\infty }\left \vert \hat{f} +\hat{ g}\right \vert ^{2}d\omega,\quad \qquad \qquad }$$
$$\displaystyle{\qquad \int _{-\infty }^{\infty }\big(f + g\big)\big(\bar{f} +\bar{ g}\big)\,dt =\int _{ -\infty }^{\infty }\left (\hat{f} +\hat{ g}\right )\left (\bar{\hat{f}} +\bar{\hat{ g}}\right )d\omega.}$$
Simplifying both sides gives
$$\displaystyle{\int _{-\infty }^{\infty }\big\vert f\big\vert ^{2}dt +\int _{ -\infty }^{\infty }\big(f\bar{g} + g\bar{f}\,\big)dt +\int _{ -\infty }^{\infty }\big\vert g\big\vert ^{2}dt\qquad \qquad \qquad \qquad }$$
$$\displaystyle{\qquad \qquad =\int _{ -\infty }^{\infty }\left \vert \hat{f}\right \vert ^{2}d\omega +\int _{ -\infty }^{\infty }\left (\hat{f}\,\bar{\hat{g}} +\hat{ g}\,\bar{\hat{f}}\,\right )d\omega +\int _{ -\infty }^{\infty }\left \vert \hat{g}\right \vert ^{2}d\omega.}$$
Applying (1.3.3) to the above identity leads to
$$\displaystyle{ \int _{-\infty }^{\infty }\big(f\bar{g} + g\bar{f}\,\big)dt =\int _{ -\infty }^{\infty }\left (\hat{f}\,\bar{\hat{g}} +\hat{ g}\,\bar{\hat{f}}\,\right )d\omega. }$$
(1.3.18)
Since g is an arbitrary element of \(L^{2}(\mathbb{R})\), we can replace \(g,\hat{g}\) by \(ig,i\hat{g}\) respectively, in (1.3.18) to obtain
$$\displaystyle{\int _{-\infty }^{\infty }\big[f(\overline{ig}) + (ig)\bar{f}\,\big]dt =\int _{ -\infty }^{\infty }\left [\hat{f}\,\left (\overline{i\hat{g}}\right ) + (i\hat{g})\,\bar{\hat{f}}\,\right ]d\omega.}$$
Or
$$\displaystyle{-i\int _{-\infty }^{\infty }f\,\bar{g}\,dt + i\int _{ -\infty }^{\infty }g\,\bar{f}\,dt = -i\int _{ -\infty }^{\infty }\hat{f}\,\bar{\hat{g}}\,d\omega + i\int _{ -\infty }^{\infty }\hat{g}\,\bar{\hat{f}}\,d\omega,\ \ }$$
which is, multiplying by i,
$$\displaystyle{ \int _{-\infty }^{\infty }f\,\bar{g}\,dt -\int _{ -\infty }^{\infty }g\,\bar{f}\,dt =\int _{ -\infty }^{\infty }\hat{f}\,\bar{\hat{g}}\,d\omega -\int _{ -\infty }^{\infty }\hat{g}\,\bar{\hat{f}}\,d\omega. }$$
(1.3.19)
Adding (1.3.18) and (1.3.19) gives
$$\displaystyle{\int _{-\infty }^{\infty }f(t)\,\overline{g(t)}\,dt =\int _{ -\infty }^{\infty }\hat{f}(\omega )\,\overline{\hat{g}(\omega )}\,d\omega.}$$

This completes the proof.

The following theorem summaries the major results of this section and is usually called Plancherel theorem.

Theorem 1.3.7 (Plancherel’s Theorem).

For every \(f \in L^{2}(\mathbb{R})\) , there exists \(\hat{f} \in L^{2}(\mathbb{R})\) such that
  1. (i)

    If \(f \in L^{1}(\mathbb{R}) \cap L^{2}(\mathbb{R})\) , then \(\hat{f}(\omega ) = \frac{1} {\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt,\)

     
  2. (ii)

    \(\hat{f}(\omega ) =\lim _{n\rightarrow \infty } \dfrac{1} {\sqrt{2\pi }}\int _{-n}^{n}e^{-i\omega t}f(t)\,dt,\)

     
  3. (iii)

    \(f(t) =\lim _{n\rightarrow \infty } \frac{1} {\sqrt{2\pi }}\int _{-n}^{n}e^{i\omega t}\hat{f}(\omega )\,d\omega\),

     
  4. (iv)

    \(\big\langle f,g\big\rangle =\big\langle \hat{ f},\hat{g}\big\rangle,\)

     
  5. (v)

    \(\big\|f\big\|_{2} = \left \|\hat{f}\right \|_{2},\)

     
  6. (vi)

    The mapping \(f \rightarrow \hat{ f}\) is a Hilbert space isomorphism of \(L^{2}(\mathbb{R})\) onto \(L^{2}(\mathbb{R})\).

     

1.4 The Discrete Fourier Transform

The Fourier transform deals generally with continuous functions, i.e., functions which are defined at all values of the time t. However, for many applications, we require functions which are discrete in nature rather than continuous. In modern digital media audio, still images or video-continuous signals are sampled at discrete time intervals before being processed. Fourier analysis decomposes the sampled signal into its fundamental periodic constituents sines and cosines or, more conveniently, complex exponentials. The crucial fact, upon which all modern signal processing is based, is that the sampled complex exponentials form an orthogonal basis. To meet the needs of both the automated and experimental computations, the discrete Fourier transform (DFT) has been introduced.

To motivate the idea behind the discrete Fourier transform, we take two approaches, one from the approximation point of view and other one from discrete point of view.

Consider the function f with Fourier transform
$$\displaystyle{ \hat{f}(\omega ) =\int _{ -\infty }^{\infty }f(t)\,e^{-i\omega t}\,dt. }$$
(1.4.1)
For some functions f, it is not always possible to evaluate the Fourier transform (1.4.1), and for such functions, one needs to truncate the range of integration to an interval [a, b] and approximate the integral for \(\hat{f}\) by a finite sum as
$$\displaystyle{\hat{f}(\omega ) =\sum _{ k=0}^{N-1}f\big(t_{ k}\big)\,e^{-i\omega t_{k} }h.}$$
Now, for sufficiently large a < 0 and b > 0
$$\displaystyle{\int _{a}^{b}f(t)\,e^{-i\omega t}\,dt}$$
is a good approximation to \(\hat{f}\). Therefore, in order to approximate this integral, we sample the signal f at a finite number of sample points, say
$$\displaystyle{t_{0} = a <t_{1} <t_{2} <\ldots <t_{N-1} = b,\quad a <0,\,b> 0.}$$
For simplicity the sample points are equally spaced and so
$$\displaystyle{t_{k} = a + kh,\qquad k = 0,1,2,\ldots,N}$$

where \(h = \dfrac{b - a} {N}\) indicates the sample rate. In signal processing applications, t represents time instead of space and t k are the times at which we sample the signal f. This sample rate can be very high, e.g., every 10 −−20 milliseconds in current speech recognition systems.

Thus, the approximation g of \(\hat{f}\) is given by
$$\displaystyle{\begin{array}{rcl} g(\omega )& =&\sum _{k=0}^{N-1}f\big(t_{ k}\big)\,e^{-i\omega t_{k} }h \\ & =&e^{-i\omega a}\sum _{ k=0}^{N-1}f\big(t_{ k}\big)\,e^{-i\omega (b-a)k/N}h. \end{array} }$$
We now take the time duration [a, b] into account by focusing attention on the points (frequencies), \(\omega _{n} = \dfrac{2\pi n} {b - a},\) where n is an integer. Then, the approximation g of \(\hat{f}\) at these points becomes
$$\displaystyle{g(\omega _{n}) = e^{-ia\omega _{n} }\sum _{k=0}^{N-1}f\big(t_{ k}\big)\,e^{-i2nk\pi /N}h.}$$
By neglecting the term \(e^{-ia\omega _{n}}h\) in the R.H.S. of the above expression and focusing attention on the N-periodic function \(\hat{f}: \mathbb{Z} \rightarrow \mathbb{C}\), we obtain
$$\displaystyle\begin{array}{rcl} \hat{f}(n)& =& \sum _{k=0}^{N-1}f\big(t_{ k}\big)\,e^{-i2nk\pi /N}h,\quad n \in \mathbb{Z} \\ & =& \sum _{k=0}^{N-1}f\big(t_{ k}\big)\,w^{-nk}, {}\end{array}$$
(1.4.2)

where w = e2πiN . Equation (1.4.2) is known as discrete Fourier transform (DFT).

From a discrete perspective, one is dealing with the values of f at only a finite number of points \(\big\{0,1,2,\ldots,N - 1\big\}\). Consider f as defined on the cyclic group of integers modulo the positive integer N:
$$\displaystyle{\mathbb{Z}_{N} = \dfrac{\mathbb{Z}} {\big(N\big)}\,\big(\mathbb{Z}\,\text{modulo}\,N\big)}$$
where \(\big(N\big) =\big\{ kN: k \in \mathbb{Z}\big\}\) and \(f: \mathbb{Z}_{N} \rightarrow \mathbb{C}\). This function f can be viewed as the N-periodic function defined on \(\mathbb{Z}\) by taking
$$\displaystyle{ f\big(k + nN\big) = f(k),\quad \forall \;n \in \mathbb{Z},\;k = 0,1,\ldots,N - 1. }$$
But \(\mathbb{Z}_{N}\) is finite. Therefore, any function defined on it is integrable. Thus, \(L^{1}\big(\mathbb{Z}_{N}\big) = L^{2}\big(\mathbb{Z}_{N}\big) = \mathbb{C}_{N}\) is the collection of all functions \(f: \mathbb{Z}_{N} \rightarrow \mathbb{C}\). One gets the discrete Fourier transform for \(f: \mathbb{Z}_{N} \rightarrow \mathbb{C}\) as
$$\displaystyle{ \hat{f}(n) =\sum _{ k=0}^{N-1}f\big(k\big)\,e^{-i2nk\pi /N},\quad n \in \mathbb{Z}_{ N}. }$$

This formula for the discrete Fourier transform is analogous to the formula for the nth Fourier coefficient with the sum over k taking the place of the integral over t.

In matrix notation, the above discussion can be summarized by the following.

Here we replace \(\mathbb{Z}_{N}\) by the cyclic group of Nth root of unity. Therefore, f and its discrete Fourier transform \(\hat{f}\) can be viewed as vectors.
$$\displaystyle{f = \left (\begin{array}{*{10}c} f(0)\\ \vdots\\ f(N - 1) \end{array} \right )\quad \text{and}\quad \hat{f} = \left (\begin{array}{*{10}c} \hat{f}(0)\\ \vdots\\ \hat{f}(N - 1) \end{array} \right )}$$
with
$$\displaystyle{W_{N} = \left (\begin{array}{*{10}c} 1& 1 & 1 &\cdots & 1 \\ 1& e^{-1\cdot 2\pi i/N} & e^{-2\cdot 2\pi i/N} &\cdots & e^{-(N-1)\cdot 2\pi i/N} \\ 1& e^{-2\cdot 2\pi i/N} & e^{-2\cdot 2\cdot 2\pi i/N} &\cdots &e^{-2(N-1)\cdot 2\pi i/N}\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 1&e^{-(N-1)\cdot 2\pi i/N}&e^{-2(N-1)\cdot 2\pi i/N}&\cdots &e^{-(N-1)^{2}\cdot 2\pi i/N }\\ \end{array} \right ).}$$

Then, clearly \(\hat{f} = W_{N}f\). Here, W N is also called the Nth-order DFT matrix.

Example 1.4.1.

Let \(f: \mathbb{Z}_{4} \rightarrow \mathbb{C}\) be defined by
$$\displaystyle{f(n) = 1,\quad \text{for all}\ \ n = 0,1,2,3.}$$
Then,
$$\displaystyle{\hat{f}(n) = W_{4}f = \left (\begin{array}{*{10}c} 1& 1 & 1 & & 1\\ 1 & -i &-1 & & i \\ 1&-1& 1 & &-1 \\ 1& i &-1& & -i\\ \end{array} \right )\left (\begin{array}{*{10}c} 1\\ 1 \\ 1\\ 1\end{array} \right ) = \left (\begin{array}{*{10}c} 4\\ 0 \\ 0\\ 0\end{array} \right ).}$$

Here some properties are analogous to the corresponding properties for the Fourier transform given in Theorem 1.2.1.

Theorem 1.4.1.

The following properties hold for the discrete Fourier transform:
  1. (a)

    Time shift: \(\hat{f}(n - j) =\hat{ f}(n)\,e^{-2\pi ijn/N},\)

     
  2. (b)

    Frequency shift: \(\Big(f(n)\,e^{2\pi ijn/N}\Big)^{\wedge } =\hat{ f}(n - j),\)

     
  3. (c)

    Modulation: \(\left (f(n)\cos \dfrac{2\pi nj} {N} \right )^{\wedge } = \dfrac{1} {2}\,\left [\hat{f}(n - j) +\hat{ f}(n + j)\right ].\)

     

Proof.

Let g(n) = f(nj). Then, by using the fact that f is N-periodic, we obtain
$$\displaystyle{\begin{array}{rcl} \hat{g}(n)& =&\sum _{k=0}^{N-1}g(k)\,e^{-2n\pi ik/N} \\ & =&\sum _{k=0}^{N-1}f(k - j)\,e^{-2n\pi ik/N} \\ & =&\sum _{m=-j}^{N-1-j}f(m)\,e^{-2n\pi i(m+j)/N} \\ & =&\sum _{m=-j}^{-1}f(m)\,e^{-2n\pi i(m+j)/N} +\sum _{ m=0}^{N-1-j}f(m)\,e^{-2n\pi i(m+j)/N} \\ & =&\sum _{m=N-j}^{N-1}f(m)\,e^{-2n\pi i(m+j)/N} +\sum _{ m=0}^{N-1-j}f(m)\,e^{-2n\pi i(m+j)/N} \\ & =&\left (\sum _{m=0}^{N-1}f(m)\,e^{-2n\pi im/N}\right )e^{-2n\pi ij/N} \\ \\ & =&\hat{f}(n)\,e^{-2n\pi ij/N}.\end{array} }$$
This proves part (a).
  1. (b)
    Let g(n) = f(n) e−2nπijN . Then
    $$\displaystyle{\begin{array}{rcl} \hat{g}(n)& =&\sum _{k=0}^{N-1}g(k)\,e^{-2n\pi ik/N} \\ & =&\sum _{k=0}^{N-1}f(k)\,e^{2\pi ijk/N} \cdot e^{-2n\pi ik/N} \\ & =&\sum _{k=0}^{N-1}f(k)\,e^{-2\pi ik(n-j)/N} =\hat{ f}(n - j).\end{array} }$$
    Hence
    $$\displaystyle{\Big(f(n)\,e^{2\pi ijn/N}\Big)^{\wedge } =\hat{ f}(n - j).}$$
     
  2. (c)
    Since
    $$\displaystyle{f(n)\cos \dfrac{2\pi in} {N} = \dfrac{1} {2}\,\Big[f(n)\,e^{2\pi ijn/N} + f(n)\,e^{-2\pi ijn/N}\Big]}$$

    the desired result is obtained by using part (b).

    Our next task is to compute the inverse of the discrete Fourier transform. We have already computed the inverse Fourier transform, since Theorem 1.3.5 tells how to recover the function f from its Fourier transform. The inverse discrete Fourier transform is analogous and it allows us to recover the original discrete signal f from its discrete Fourier transform \(\hat{f}\).

     

Theorem 1.4.2 (Inversion Formula).

Let \(f: \mathbb{Z}_{N} \rightarrow \mathbb{C}\) be such that
$$\displaystyle{\hat{f}(n) =\sum _{ k=0}^{N-1}f(k)\,e^{-2\pi ikn/N} =\sum _{ k=0}^{N-1}f(k)\,w^{-kn}}$$
where w = e2πiN . Then
$$\displaystyle{ f(n) = \dfrac{1} {N}\sum _{k=0}^{N-1}\hat{f}(k)\,e^{2\pi ikn/N} = \dfrac{1} {N}\sum _{k=0}^{N-1}\hat{f}(k)\,w^{kn}. }$$
(1.4.3)

Proof.

In order to establish the result (1.4.3), we must show that
$$\displaystyle{ \sum _{k=0}^{N-1}w^{(n-j)k} = \left \{\begin{array}{ll} 1&\text{if}\;n = j \\ 0&\text{if}\;n\neq j\end{array} \right. }$$
(1.4.4)

Since w = e2πiN , w kn ⋅ wkj = w k(nj). Therefore, in order to sum this expression over k = 0, 1, 2, , N − 1, we use the following elementary observation that

$$\displaystyle{1+x+x^{2}+\ldots +x^{N-1} = \left \{\begin{array}{ll} \ \ N &\text{if}\;x = 1 \\ \dfrac{1 - x^{N}} {1 - x} &\text{if}\;x\neq 1 \end{array} \right.}$$
Set x = w nj and note that x N = 1 because w N = 1. Also, we note that w nj ≠ 1 unless n = j for 0 ≤ n, j ≤ 1. Thus
$$\displaystyle{\sum _{k=0}^{N-1}w^{(n-j)k} = \left \{\begin{array}{ll} 1&\text{if}\;n = j \\ 0&\text{if}\;n\neq j\end{array} \right.}$$
Hence
$$\displaystyle{\begin{array}{rcl} \dfrac{1} {N}\sum _{k=0}^{N-1}\hat{f}(k)\,e^{2\pi ikn/N}& =& \dfrac{1} {N}\sum _{k=0}^{N-1}\hat{f}(k)\,w^{kn} \\ \\ & =& \dfrac{1} {N}\sum _{k=0}^{N-1}\left (\sum _{ j=0}^{N-1}f(j)\,w^{-kj}\right )w^{kn} \\ \\ & =& \dfrac{1} {N}\sum _{j=0}^{N-1}f(j)\left (\sum _{ k=0}^{N-1}w^{(n-j)k}\right ). \end{array} }$$
Therefore, by using (1.4.4), we get
$$\displaystyle{ \dfrac{1} {N}\sum _{k=0}^{N-1}\hat{f}(k)\,e^{2\pi ikn/N} = \dfrac{1} {N}\,f(n) \cdot N = f(n).}$$

Example 1.4.2.

Consider the discrete sinusoid f(n) = cos(ωn). 

Note that f(n) is periodic if and only if ω is a rational multiple of 2π: ω = 2πpq, for some \(p,q \in \mathbb{Z}\). If p = 1 and q = N, then ω = 2πN, and f(n) is periodic on [0, N − 1] with fundamental period N. Therefore, the functions of the form cos(2πknN) are also periodic on [0, N − 1], but since \(\cos (2\pi kn/N) =\cos \big (2\pi (N - k)n/N\big)\), they are different only for k = 1, 2, , ⌊N∕2⌋. Similar results hold for g(n) = sin(ωn), except that \(\sin (2\pi kn/N) = -\sin \big(2\pi (N - k)n/N\big)\).

Note that for k = 0, 1, , ⌊N∕2⌋, we have
$$\displaystyle\begin{array}{rcl} \cos \left (\dfrac{2\pi nk} {N} \right )& = \dfrac{1} {2}\exp \left (\dfrac{2\pi ink} {N} \right ) + \dfrac{1} {2}\exp \left (\dfrac{2\pi in(N - k)} {N} \right ),&{}\end{array}$$
(1.4.5)
$$\displaystyle\begin{array}{rcl} \sin \left (\dfrac{2\pi nk} {N} \right )& = \dfrac{1} {2i}\exp \left (\dfrac{2\pi ink} {N} \right ) - \dfrac{1} {2i}\exp \left (\dfrac{2\pi in(N - k)} {N} \right ).&{}\end{array}$$
(1.4.6)
Equations (1.4.5) and (1.4.6) thereby imply that
$$\displaystyle\begin{array}{rcl} \hat{f}(k)& =& \hat{f}(N - k) = N/2\ \ \text{with}\ \ \hat{f}(m) = 0,\ \ \text{ for}\ \ 0 \leq m \leq N - 1,m\neq k;\ \ \text{and} {}\\ \hat{g}(k)& =& -\hat{g}(N - k) = (-iN)/2\ \ \text{with}\ \ \hat{g}(m) = 0,\ \ \text{ for}\ \ 0 \leq m \leq N - 1,m\neq k. {}\\ \end{array}$$

The factor of N in the expressions for \(\hat{f}(k)\) and \(\hat{g}(k)\) ensures that the inverse DFT relation (1.4.3) holds for f(n) and g(n), respectively.

For more about DFT, the reader is referred to Sundararajan (2001), Butz (2006), and Wong (2010).

1.5 The Fast Fourier Transform

Although the ability of the discrete Fourier transform to provide information about the frequency components of a signal is extremely valuable, the huge computational effort involved meant that until the 1960’s, it was rarely used in practical applications. Two important advances have changed the situation completely. The first was the development of the digital computer with its ability to perform numerical calculations rapidly and accurately. The second was the discovery by Cooley and Tukey of a numerical algorithm which allows the DFT to be evaluated with a significant reduction in the amount of calculations required. This algorithm, called the fast Fourier transform (FFT), allows the DFT of a sampled signal to be obtained rapidly and efficiently.

Actually the idea of this algorithm goes back to Carl Friedrich Gauss (1777–1855) in 1805, but this early work was forgotten because it lacked the tool to make it practical: the digital computer. Cooley and Tukey are honored because they discovered the FFT at the right time, the beginning of the computer revolution. The publication of the FFT algorithm by Cooley and Tukey in 1965 was the turning point in digital signal processing and in certain areas of numerical analysis. Nowadays, the FFT is used in many areas, from the identification of characteristic mechanical vibration frequencies to image enhancement. Standard routines are available to perform the FFT by computer in programming languages such as Pascal, Fortran, and C, and many spreadsheet and other software packages for the analysis of numerical data allow the FFT of a set of data values to be determined readily. For more about FFTs and their applications, we refer to the monographs (Brigham, 1998; Bracewell, 2000; Butz, 2006; Duhamel and Vetterli, 1990; Rao et al., 2010).

The DFT of an N times sampled signal requires a total of N2 multiplications and N(N − 1) additions. The FFT algorithm for N = 2 k reduces the N2 multiplications to something proportional to Nlog2N. If the calculations are handmade, then N is necessarily small and this is not that significant, but in case N is large, the number of operations is drastically reduced. For example, a 3-minute song may contain N = (44, 000) × 180 = 7, 920, 000 samples. Thus, the DFT would take 63, 000, 000, 000, 000 multiplications. The FFT on the other hand would only take approximately 90, 751, 593 multiplications. This result is a significant difference in computing time.

How do we go about achieving this reduction in computing time? Where do we start? The idea is to look at even entries and odd separately, and we can then piece them back together. This, seemingly simple, idea will allow certain multiplications that are normally repeated to only be done once.

Let \(N \in \mathbb{N}\) with N even and w N = e−2πiN . If \(N = 2M,\,M \in \mathbb{N}\), then
$$\displaystyle{w_{N}^{2} = e^{-2\pi i2/N} = e^{-2\pi i/(N/2)} = e^{-2\pi i/M} = w_{ M}.}$$
Suppose \(f \in \ell^{2}\big(\mathbb{Z}_{N}\big) = \mathbb{C}_{N},\) define \(a,b \in \ell^{2}\big(\mathbb{Z}_{M}\big) = \mathbb{C}_{M} = \mathbb{C}_{N/2}\) by
$$\displaystyle\begin{array}{rcl} a(k)& =& f(2k)\ \ \quad \quad \text{for}\ k = 0,1,2,\ldots,M - 1, {}\\ b(k)& =& f(2k + 1)\quad \text{for}\ k = 0,1,2,\ldots,M - 1. {}\\ \end{array}$$
Here a is the vector of the even entries of f and b is the vector of the odd entries. Thus
$$\displaystyle{f(k) =\big (a(0),b(0),a(1),b(1),\ldots,a(M - 1),b(M - 1)\big).}$$
Then, breaking the definition of the DFT of f into even and its odd parts, we obtain
$$\displaystyle{\begin{array}{rcl} \hat{f}(n)& =&\sum _{k=0}^{N-1}f(k)\,e^{-2\pi ink/N} \\ & =&\sum _{k=0}^{N-1}f(k)\,w_{ N}^{nk} \\ & =&\sum _{k=0}^{M-1}f(2k)\,w_{ N}^{2kn} +\sum _{ k=0}^{M-1}f(2k + 1)\,w_{ N}^{(2k+1)n} \\ & =&\sum _{k=0}^{M-1}a(k)\,\big(w_{ N}^{2}\big)^{nk} + w_{ N}^{n}\sum _{ k=0}^{M-1}b(k)\,\big(w_{ N}^{2}\big)^{nk} \\ & =&\sum _{k=0}^{M-1}a(k)\,w_{ M}^{nk} + w_{ N}^{n}\sum _{ k=0}^{M-1}b(k)\,w_{ M}^{nk} \\ & =&\hat{a}(n) + w_{N}^{n}\,\hat{b}(n)\end{array} }$$
where \(\hat{a}(n)\) and \(\hat{a}(n)\) are the M-points, DFTs of a and b, respectively. But
$$\displaystyle{w_{N}^{-M} =\big (e^{-2\pi i/N}\big)^{-M} = e^{2\pi iM/N} = e^{\pi i} = -1.}$$
Therefore, if 0 ≤ nM − 1, then
$$\displaystyle{ \hat{f}(n) =\hat{ a}(n) + w_{N}^{n}\,\hat{b}(n). }$$
(1.5.1)
Moreover, if MnN − 1, then
$$\displaystyle\begin{array}{rcl} \hat{f}(n)& =& \hat{a}(n - M) + w_{N}^{(n-M)}\,\hat{b}(n - M) \\ & =& \hat{a}(n) + w_{N}^{-M} \cdot w_{ N}^{n}\,\hat{b}(n) \\ & =& \hat{a}(n) - w_{N}^{n}\,\hat{b}(n). {}\end{array}$$
(1.5.2)
Let us now look at the number of multiplications that it takes to compute \(\hat{f}(n)\) in this way when \(f \in \mathbb{C}_{N} =\ell ^{2}\big(\mathbb{Z}_{N}\big)\) and N = 2M. Computing \(\hat{a}\) and \(\hat{b}\) each takes \(M^{2} = \dfrac{N^{2}} {4}\) multiplications. In addition, for each entry of \(\hat{b}\), we need \(\hat{b}(n)\,w_{N}^{n}\). This gives an additional M = N∕2 multiplication. Thus, if m(N) denotes the number of multiplications required to compute an N-point DFT using the FFT algorithm, then our total is
$$\displaystyle{ m(N) = 2M^{2} + M = 2m\big(N/2\big) +\big (N/2\big). }$$
(1.5.3)

Here we have essentially cut the half time. This is good but not great; half of the ten thousand years in a previous example is still a long time. The order of magnitude really has not changed. How can we get a more significant change? The big improvement is going to come when \(N = 2^{n},\,n \in \mathbb{N}\). Now, with this assumption N = 2 n , the M defined above will also be even. Thus, we can use the same method recursively, when computing \(\hat{a}\) and \(\hat{b}\). Therefore, we have the following theorem in this regard.

Theorem 1.5.1.

If N = 2 n for some \(n \in \mathbb{N}\), then
$$\displaystyle{ m(N) =\big (N/2\big)\log _{2}N. }$$
(1.5.4)

Proof.

We prove the theorem by using the method of induction on n. Let n = 2. Then, m(22) = m(4) = 4 and hence the result holds for n = 2. Assume that the result holds for n = k − 1. Then
$$\displaystyle\begin{array}{rcl} m\big(2^{k}\big)& =& 2 \cdot m\big(2^{k-1}\big) + 2^{k-1} {}\\ & =& 2\big(2^{k-2}(k - 1)\big) + 2^{k-1} {}\\ & =& 2^{k-1}\big((k - 1) + 1\big) {}\\ & =& 2^{k-1}k = \left (\dfrac{2^{k}} {2} \right )\big(\log _{2}2^{k}\big). {}\\ \end{array}$$

This proves that the result holds for n = k also. Thus, by induction the result must hold for any \(n \in \mathbb{N}\).

1.6 The Fractional Fourier Transform

Undoubtedly, one of the most recognized tools in signal and image processing is the Fourier transform which generally converts a signal from time versus amplitude to frequency versus amplitude. The classical Fourier transform can be visualized as a change in representation of the signal corresponding to a counterclockwise rotation of the axis by an angle \(\dfrac{\pi } {2}\). Two successive rotations of the signal through \(\dfrac{\pi } {2}\) will result in an inversion of the time axis. In spite of some remarkable success, Fourier transform seems to be inadequate for studying nonstationary signals for at least two reasons. First, the Fourier transform of a signal does not contain any local information in the sense that it does not reflect the change of wave number with space or of frequency with time. Second, the Fourier transform method enables us to investigate problems either in time (space) domain or the frequency (wave number) domain, but not simultaneously in both domains. These are probably the major weaknesses of the Fourier transform analysis. To overcome these problems, Victor Namias (1980) proposed the fractional Fourier transform (FrFT) as a generalization of the conventional Fourier transform to solve certain problems arising in quantum mechanics from classical quadratic Hamiltonians. It is also called rotational Fourier transform or angular Fourier transform since it depends on a parameter α which is interpreted as a rotation by an angle α in the time-frequency plane. Like the ordinary Fourier transform that corresponds to a rotation in the time-frequency plane over an angle \(\alpha = 1 \times \dfrac{\pi } {2}\), the FrFT corresponds to a rotation over an arbitrary angle \(\alpha = a \times \dfrac{\pi } {2}\) with \(a \in \mathbb{R}\).

The fractional Fourier transforms are relatively recent developments that have fascinated the scientific, engineering, and mathematics community with their versatile applicability. The application areas for FrFT have been growing for two decades at a very rapid rate. They have been applied in a number of fields including signal processing, image processing, quantum mechanics, neural networks, differential equations, time-frequency distributions, optical systems, statistical optics, signal detectors and pattern recognition, radar, sonar, and communications. A comprehensive overview of FrFTs and their applications can also be found in Mendlovic and Ozaktas (1993a,b), Almeida (1994), Zayed (1998), Atakishiyev et al. (1999), Candan et al. (2000), Ozaktas et al. (20002010), Ran et al. (2006), Tao et al. (2009), and Sejdić et al. (2011).

Definition 1.6.1 (Fractional Fourier Transform).

The fractional Fourier transform with parameter α of signal f(t) is defined by
$$\displaystyle{ \mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega ) =\hat{ f}^{\alpha }(\omega ) =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )f(t)\,dt, }$$
(1.6.1)
where \(\mathcal{K}_{\alpha }(t,\omega )\) is the so-called kernel of the FrFT given by
$$\displaystyle{ \mathcal{K}_{\alpha }(t,\omega ) = \left \{\begin{array}{ll} C_{\alpha }\exp \Big\{i(t^{2} +\omega ^{2}) \dfrac{\cot \alpha } {2} - it\omega \ \mbox{csc} \,\alpha \Big\},&\alpha \neq n\pi, \\ \delta (t-\omega ), &\alpha = 2n\pi, \\ \delta (t+\omega ), &\alpha = (2n \pm 1)\pi,\\ \end{array} \right. }$$
(1.6.2)
\(\alpha = \dfrac{a\pi } {2}\) denotes the rotation angle of the transformed signal for FrFT, the FrFT operator is designated by \(\mathcal{F}^{\alpha }\), and
$$\displaystyle{ C_{\alpha } = \left (2\pi i\sin \alpha \right )^{-1/2}e^{i\alpha /2} = \sqrt{\dfrac{1 - i\cot \alpha } {2\pi }}.\qquad \qquad }$$
(1.6.3)
The corresponding inversion formula is given by
$$\displaystyle{ f(t) = \frac{1} {2\pi }\int _{-\infty }^{\infty }\overline{\mathcal{K}_{\alpha }(t,\omega )}\,\hat{f}^{\alpha }(\omega )\,d\omega, }$$
(1.6.4)
where
$$\displaystyle\begin{array}{rcl} \mathcal{K}_{\alpha }(t,\omega )& =& \frac{(2\pi i\sin \alpha )^{1/2}\,e^{-i\alpha /2}} {\sin \alpha } \cdot \exp \left \{\frac{-i(t^{2} +\omega ^{2})\cot \alpha } {2} + it\omega \ \mbox{csc} \,\alpha \right \} \\ & =& \overline{C_{\alpha }}\exp \left \{\frac{-i(t^{2} +\omega ^{2})\cot \alpha } {2} + it\omega \ \mbox{csc} \,\alpha \right \} \\ & =& \mathcal{K}_{-\alpha }(t,\omega ) {}\end{array}$$
(1.6.5)
and
$$\displaystyle{ C_{\alpha } = \frac{(2\pi i\sin \alpha )^{1/2}e^{-i\alpha /2}} {2\pi \sin \alpha } = \sqrt{\dfrac{1 + i\cot \alpha } {2\pi }} = C_{-\alpha }. }$$
(1.6.6)

Remarks:

 
  1. 1.

    It is important to point out that when a = 1, the kernel \(\mathcal{K}_{\alpha }(t,\omega )\) given by (1.6.2) reduces to \(e^{-it\omega }/\sqrt{2\pi }\), corresponding to the ordinary Fourier transform, and that when a = −1, the kernel \(\mathcal{K}_{\alpha }(t,\omega )\) reduces to \(e^{it\omega }/\sqrt{2\pi }\), corresponding to the ordinary inverse Fourier transform.

     
  2. 2.

    The zeroth-order FrFT operator \(\mathcal{F}^{0}\) is equal to the identity operator \(\mathcal{I}\), whereas the integer values of α correspond to repeated application of the Fourier transform; for instance, \(\mathcal{F}^{2}\) corresponds to the Fourier transform of the Fourier transform. The order α may assume any real value; however the operator \(\mathcal{F}^{\alpha }\) is periodic in α with period 4, that \(\mathcal{F}^{\alpha +4n} = \mathcal{F}^{\alpha },\) where n is any integer. This is because \(\mathcal{F}^{2}\) equals the parity operator \(\mathcal{P}\) which maps f(t) to f(−t) and \(\mathcal{F}^{4}\) equals the identity operator. Therefore, the range of a is usually restricted to [0, 4). These facts can be restated in operator notation:

    $$\displaystyle{ \mathcal{F}^{0} = \mathcal{I},\mathcal{F}^{1} = \mathcal{F},\mathcal{F}^{2} = \mathcal{P},\mathcal{F}^{3} = \mathcal{F}\mathcal{P},\mathcal{F}^{4} = \mathcal{F}^{0} = \mathcal{I},\mathcal{F}^{\alpha +4n} = \mathcal{F}^{\alpha +4m},m,n \in \mathbb{Z}. }$$
     
  3. 3.
    Let \(\mathcal{F}^{\alpha }(\mathbb{R})\) denote the class of all FrFT on \(\mathbb{R}\) parameterized by the parameter α; then it has a group structure called the elliptic group. If \(\mathcal{R}^{\alpha }\) denotes a rotation operator whose action is governed by
    $$\displaystyle{ \mathcal{R}^{\alpha }f = \mathcal{F}^{\alpha }, }$$
    (1.6.7)
    where
    $$\displaystyle{ \mathcal{R}^{\alpha } = \left [\begin{array}{*{10}c} \cos \alpha &\sin \alpha \\ -\sin \alpha &\cos \alpha \end{array} \right ],\quad \alpha = \dfrac{a\pi } {2}. }$$
    Then, \(\mathcal{R}^{\alpha }\) is required to satisfy the following properties:
    $$\displaystyle{ \mathcal{R}^{0} = \mathcal{I},\ \mathcal{R}^{2\pi } = \mathcal{I},\ \mathcal{R}^{\pi /2} = \mathcal{F},\quad \text{and}\quad \mathcal{R}^{\alpha +\beta } =\mathcal{ R}^{\alpha } \cdot \mathcal{ R}^{\beta }. }$$
    The first two of these requirements involve the identity operator and they follow directly from the definition of \(\mathcal{K}_{\alpha }(t,\omega )\), whereas the third one is obvious. The fourth one is the index additive property which can be obtained as follows:
    $$\displaystyle\begin{array}{rcl} \mathcal{R}^{\alpha }\,\mathcal{R}^{\beta }f(\omega )& =& \int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,dt\int _{ -\infty }^{\infty }f(\xi )\,\mathcal{K}_{\beta }(\xi,t)\,d\xi {}\\ & =& \int _{-\infty }^{\infty }f(\xi )\,d\xi \int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,\mathcal{K}_{\beta }(\xi,t)\,dt {}\\ & =& \int _{-\infty }^{\infty }f(\xi )\,\mathcal{K}_{\alpha +\beta }(\xi,\omega )\,d\xi {}\\ & =& \mathcal{R}^{\alpha +\beta }f(\omega ). {}\\ \end{array}$$
    Thus, we have
    $$\displaystyle{ \mathcal{F}^{\alpha } \cdot \mathcal{ F}^{\beta } =\mathcal{ F}^{\alpha +\beta }. }$$
    From the index additive property, we can deduce the inverse of the αth-order fractional Fourier operator \(\mathcal{F}^{-\alpha }\) as
    $$\displaystyle\begin{array}{rcl} \int _{-\infty }^{\infty }\mathcal{F}^{\alpha }(\omega )\,\mathcal{K}_{ -\alpha }(t,\omega )\,d\omega & =& \int _{-\infty }^{\infty }\mathcal{K}_{ -\alpha }(t,\omega )\,d\omega \int _{-\infty }^{\infty }f(t^{{\prime}})\,\mathcal{K}_{\alpha }(t^{{\prime}},\omega )\,dt^{{\prime}} {}\\ & =& \int _{-\infty }^{\infty }f(t^{{\prime}})\,dt^{{\prime}}\int _{ -\infty }^{\infty }\mathcal{K}_{ -\alpha }(t,\omega )\,\mathcal{K}_{\alpha }(\omega,t^{{\prime}})\,d\omega {}\\ & =& \int _{-\infty }^{\infty }f(t^{{\prime}})\,\mathcal{K}_{ 0}(t,t^{{\prime}})\,dt^{{\prime}} {}\\ & =& \int _{-\infty }^{\infty }f(t^{{\prime}})\,\delta (t - t^{{\prime}})\,dt^{{\prime}} {}\\ & =& f(t), {}\\ \end{array}$$

    where it has been assumed that the integration order can be inverted.

     
  4. 4.
    For any nonnegative integer r and \(f \in L^{1}(\mathbb{R})\), we have
    $$\displaystyle{ \mathcal{F}^{\alpha }\left \{ \dfrac{d^{r}} {dt^{r}}f(t)\right \}(\omega ) = \left \{i\omega \sin \alpha +\cos \alpha \dfrac{d} {d\omega }\right \}^{r}\mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega ). }$$
    (1.6.8)
     

Example 1.6.1.

The FrFT of the Gaussian function \(f(t) = e^{-bt^{2} },\ b> 0\) is
$$\displaystyle{\mathcal{F}^{\alpha }\left \{e^{-bt^{2} }\right \}(\omega ) = \sqrt{ \dfrac{1 - i\cot \alpha } {2b - i\cot \alpha }}\exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} - \dfrac{(\omega \ \mbox{csc} \,\alpha )^{2}} {2(2b - i\cot \alpha )}\right \}.}$$

Example 1.6.2 (The Second Derivative of the Gaussian Function).

This function is defined by
$$\displaystyle{ f(t) = (1 - t^{2})\,e^{-t^{2}/2 } = -\dfrac{d^{2}} {dt^{2}}\,e^{-t^{2}/2 }. }$$
(1.6.9)
Applying the definition of FrFT (1.6.1) and using (1.6.8) with r = 2, we obtain
$$\displaystyle\begin{array}{rcl} & & \mathcal{F}^{\alpha }\Big\{f(t)\Big\}(\omega ) {}\\ & & \qquad = -\mathcal{F}^{\alpha }\left \{ \dfrac{d^{2}} {dt^{2}}\,e^{-t^{2}/2 }\right \}(\omega ) {}\\ & & \qquad = -\left \{i\omega \sin \alpha +\cos \alpha \dfrac{d} {d\omega }\right \}^{2}\,\mathcal{F}^{\alpha }\left \{e^{-t^{2}/2 }\right \}(\omega ) {}\\ & & \qquad =\exp \left \{ \dfrac{i} {2}\omega ^{2}\cot \alpha - \dfrac{\omega ^{2} \ \mbox{csc} ^{2}\alpha } {2(1 - i\cot \alpha )}\right \}\left \{i\sin \alpha + i\cot \alpha \cos \alpha - \dfrac{\cos \alpha \ \mbox{csc} ^{2}\alpha } {(1 - i\cot \alpha )}\right \} {}\\ & & \quad \qquad \times \left \{-i\omega ^{2}\sin \alpha -\cos \alpha \left [1 +\omega ^{2}\left (i\cot \alpha - \dfrac{csc^{2}\alpha } {(1 - i\cot \alpha )}\right )\right ]\right \} {}\\ & & \qquad =\exp \left \{ \dfrac{i} {2}\,\omega ^{2}\cot \alpha - \dfrac{\omega ^{2} \ \mbox{csc} ^{2}\alpha } {2(1 +\cot ^{2}\alpha )}\,(1 + i\cot \alpha )\right \} {}\\ & & \quad \qquad \times \left \{i\sin \alpha + i\cot \alpha \cos \alpha -\dfrac{\cos \alpha \ \mbox{csc} ^{2}\alpha (1 + i\cot \alpha )} {(1 +\cot ^{2}\alpha )} \right \} {}\\ & & \quad \qquad \times \left \{-i\omega ^{2}\sin \alpha -\cos \alpha \left [1 +\omega ^{2}\left (i\cot \alpha -\dfrac{ \ \mbox{csc} ^{2}\alpha (1 + i\cot \alpha )} {(1 +\cot ^{2}\alpha )} \right )\right ]\right \} {}\\ & & \qquad = e^{-\omega ^{2}/2 }\Big\{\omega ^{2}\sin ^{2}\alpha +\cos ^{2}\alpha (1 -\omega ^{2})\Big\} + ie^{-\omega ^{2}/2 }\Big\{\sin \alpha \cos \alpha (2\,\omega ^{2} - 1)\Big\}. {}\\ \end{array}$$
Both f(t) and \(\hat{f}^{\alpha }(\omega )\) are plotted in Figure 1.6.
Fig. 1.6

Graphs of f(t) and \(\hat{f}^{\alpha }(\omega )\) with α = π∕6, π∕4 and π∕2

The FrFT is a generalization of the Fourier transform, so most of the properties of the Fourier transforms have their corresponding generalization versions of the FrFT. However, in order to discuss some properties of the FrFT, we shall first recall the definition of the Schwartz space \(\mathcal{S}(\mathbb{R})\).

Definition 1.6.2.

An infinitely differential complex-valued function f is member of \(\mathcal{S}(\mathbb{R})\) iff for every choice of β and γ of nonnegative integers, it satisfies
$$\displaystyle{ \Gamma _{\beta,\gamma }(f) =\sup _{t\in \mathbb{R}}\Big\vert t^{\beta }\mathfrak{D}^{\gamma }f(t)\Big\vert <\infty. }$$
(1.6.10)

Definition 1.6.3.

The space \(\mathcal{S}_{\alpha }(\mathbb{R})\) consists of all infinitely differentiable functions f(t) that vanish at infinity and satisfying
$$\displaystyle{ \Omega _{\beta }^{\gamma }(f) =\sup _{t\in \mathbb{R}}\Big\vert t^{\beta }\Delta _{t}^{\gamma }f(t)\Big\vert <\infty,\quad \beta,\gamma \in \mathbb{N}_{ 0} }$$
(1.6.11)

where

$$\displaystyle{ \Delta _{t}^{\gamma } = \left ( \dfrac{d} {dt} - it\cot \alpha \right )^{\gamma }. }$$
(1.6.12)

Proposition 1.6.1.

Let \(\mathcal{K}_{\alpha }(t,\omega )\) be the kernel of FrFT (1.6.1) and \(\Delta _{t}^{r}\) be as in (1.6.12); then
$$\displaystyle{ \Delta _{t}^{r}\,\Big\{\mathcal{K}_{\alpha }(t,\omega )\Big\} =\big (-i\omega \ \mbox{csc} \,\alpha \big)^{r}\,\mathcal{K}_{\alpha }(t,\omega ),\quad r \in \mathbb{N}_{ 0}. }$$
(1.6.13)

Proof.

Differentiating the kernel \(\mathcal{K}_{\alpha }(t,\omega )\) with respect to t, we obtain
$$\displaystyle\begin{array}{rcl} \dfrac{d} {dt}\,\mathcal{K}_{\alpha }(t,\omega )& =& C_{\alpha } \dfrac{d} {dt}\left [\exp \left \{-\dfrac{i(t^{2} +\omega ^{2})\cot \alpha } {2} - it\omega \ \mbox{csc} \,\alpha \right \}\right ] {}\\ & =& \mathcal{K}_{\alpha }\left (t,\omega \right )\,i\left (t\cos \alpha -\omega \ \mbox{csc} \,\alpha \right ) {}\\ \end{array}$$
so that
$$\displaystyle{\left ( \dfrac{d} {dt} - it\cot \alpha \right )\mathcal{K}_{\alpha }(t,\omega ) =\big (-i\omega \ \mbox{csc} \,\alpha \big)\mathcal{K}_{\alpha }(t,\omega ).}$$
Continuing this process r-times, we obtain
$$\displaystyle{\left ( \dfrac{d} {dt} - it\cot \alpha \right )^{r}\,\mathcal{K}_{\alpha }(t,\omega ) =\big (-i\omega \ \mbox{csc} \,\alpha \big)^{r}\,\mathcal{K}_{\alpha }(t,\omega ).}$$

This completes the proof.

Proposition 1.6.2.

If \(f \in S(\mathbb{R}) \subset L^{1}(\mathbb{R}),\) then
$$\displaystyle{ \int _{-\infty }^{\infty }\Delta _{ t}^{r}\mathcal{K}_{\alpha }(t,\omega )f(t)dt =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,\big(\Delta _{ t}^{{\prime}}\big)^{r}\,f(t)\,dt, }$$
(1.6.14)

where \(\Delta _{t}^{{\prime}} = -\left ( \dfrac{d} {dt} + it\cot \alpha \right ).\)

Proof.

We shall first prove the result for r = 1, that is,
$$\displaystyle{\int _{-\infty }^{\infty }\Delta _{ t}\,\mathcal{K}_{\alpha }(t,\omega )f(t)\,dt =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,(\Delta _{ t}^{{\prime}})f(t)\,dt.}$$
Integrating by parts, we have
$$\displaystyle{\int _{-\infty }^{\infty }\left ( \dfrac{d} {dt} - it\cot \alpha \right )\,\mathcal{K}_{\alpha }(t,\omega )\,dt =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\left ( \dfrac{d} {dt} + it\cot \alpha \right )f(t)\,dt.}$$
Thus, we have
$$\displaystyle{\int _{-\infty }^{\infty }\Delta _{ t}\,\mathcal{K}_{\alpha }(t,\omega )f(t)\,dt =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,(\Delta _{ t}^{{\prime}})f(t)\,dt.}$$
In general, we can obtain
$$\displaystyle{\int _{-\infty }^{\infty }\Delta _{ t}^{r}\mathcal{K}_{\alpha }(t,\omega )f(t)\,dt =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )(\Delta _{ t}^{{\prime}})^{r}\,f(t)\,dt.}$$

Proposition 1.6.3.

For all \(f \in S(\mathbb{R}) \subset L^{1}(\mathbb{R})\) and \(r \in \mathbb{N}_{0}\) , we have
  1. (a)

    \(\Big\{\mathcal{F}^{\alpha }(\Delta _{t}^{{\prime}})^{r}\,f(t)\Big\} =\big (-i\omega \ \mbox{csc} \,\alpha \big)^{r}\,\mathcal{F}^{\alpha }\big\{f(t)\big\}\)

     
  2. (b)

    \(\Big\{\Delta _{\omega }^{r}\mathcal{F}^{\alpha }f(t)\Big\} =\mathcal{ F}^{\alpha }\Big\{(-it \ \mbox{csc} \,\alpha )^{r}f(t)\Big\}.\)

     

Proof.

 
  1. (a)
    By invoking Propositions 1.6.1 and 1.6.2, we obtain
    $$\displaystyle\begin{array}{rcl} \Big\{\mathcal{F}^{\alpha }(\Delta _{t}^{{\prime}})^{r}\,f(t)\Big\}(\omega )& =& \int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\left \{(\Delta _{ t}^{{\prime}})^{r}f(t)\right \}dt {}\\ & =& \int _{-\infty }^{\infty }\Delta _{ t}^{r}\,\mathcal{K}_{\alpha }(t,\omega )\,f(t)\,dt {}\\ & =& (-i\omega \ \mbox{csc} \,\alpha )^{r}\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,f(t)\,dt {}\\ & =& \big(-i\omega \ \mbox{csc} \,\alpha \big)^{r}\,\mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega ). {}\\ \end{array}$$
     
  2. (b)
    Since \(f \in S(\mathbb{R})\) and the integral defining the FrFT is uniformly convergent for \(\omega \in \mathbb{R}\), we can differentiate within the integral sign, and using (1.6.13), we obtain
    $$\displaystyle\begin{array}{rcl} \Big\{\Delta _{\omega }^{r}\mathcal{F}^{\alpha }f(t)\Big\}& =& \int _{ -\infty }^{\infty }\Delta _{\omega }^{r}\,\mathcal{K}_{\alpha }(t,\omega )f(t)\,dt {}\\ & =& \int _{-\infty }^{\infty }(-itcsc\,\alpha )^{r}\,\mathcal{K}_{\alpha }(t,\omega )f(t)\,dt {}\\ & =& \int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\Big\{(-itcsc\,\alpha )^{r}f(t)\Big\}\,dt {}\\ & =& \mathcal{F}^{\alpha }\Big\{(-it \ \mbox{csc} \,\alpha )^{r}f(t)\Big\}. {}\\ \end{array}$$
     

Proposition 1.6.4.

The mapping \(\mathcal{F}^{\alpha }: \mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}_{\alpha }(\mathbb{R})\) is linear and continuous.

Proof.

The proof is left to the reader as an exercise.

Proposition 1.6.5.

If \(f \in L^{1}(\mathbb{R})\) , then \(\hat{f}^{\alpha }\) satisfies the following properties:
  1. (a)

    \(\hat{f}^{\alpha }(\omega ) \in L^{\infty }(\mathbb{R})\),

     
  2. (b)

    \(\hat{f}^{\alpha }(\omega )\) continuous on \(\mathbb{R}\),

     
  3. (c)

    \(\hat{f}^{\alpha }(\omega ) \rightarrow 0\) as ω → ±. 

     

Proof.

Part (a) follows directly by using the definition of FrFT and the fact that
$$\displaystyle{\left \|\hat{f}^{\alpha }\right \|_{\infty }\leq \vert C_{\alpha }\vert \,\big\|f\big\|_{1}.}$$
(b) For any h > 0, we have
$$\displaystyle\begin{array}{rcl} & & \sup _{\omega \in \mathbb{R}}\Big\vert \hat{f}^{\alpha }(\omega +h) -\hat{ f}^{\alpha }(\omega )\Big\vert {}\\ & &\quad \qquad =\sup _{\omega \in \mathbb{R}}\Bigg\vert \int _{-\infty }^{\infty }C_{\alpha }\left [\exp \Big\{i(t^{2} + (\omega +h)^{2}) \dfrac{\cot \alpha } {2} - it(\omega +h) \ \mbox{csc} \,\alpha \Big\}\right. {}\\ & & \qquad \qquad \times \left.-\exp \Big\{- i(t^{2} +\omega ^{2}) \dfrac{\cot \alpha } {2} - it\omega \ \mbox{csc} \,\alpha \Big\}\right ]f(t)\,dt\Bigg\vert {}\\ & &\quad \qquad \leq \sup _{\omega \in \mathbb{R}}\int _{-\infty }^{\infty }\left \vert C_{\alpha }\right \vert \left \vert \exp \left \{ih\left (\frac{h\cot \alpha } {2} +\omega \cot \alpha -t \ \mbox{csc} \,\alpha \right )\right \}\right. - 1\bigg\vert \big\vert f(t)\big\vert \,dt. {}\\ \end{array}$$

Since

$$\displaystyle{\left \vert \exp \left \{ih\left (\frac{h\cot \alpha } {2} +\omega \cot \alpha -t \ \mbox{csc} \,\alpha \right )\right \} - 1\right \vert \rightarrow 0,\quad \text{as}\ h \rightarrow 0.}$$

Therefore, it follows that the R.H.S of above inequality tends to zero as h → 0, that is,

$$\displaystyle{\lim _{h\rightarrow 0}\sup _{\omega \in \mathbb{R}}\Big\vert \hat{f}^{\alpha }(\omega +h) -\hat{ f}^{\alpha }(\omega )\Big\vert = 0.}$$

This shows that \(\hat{f}^{\alpha }(\omega )\) is continuous on \(\mathbb{R}\).

(c) By taking r = 1 in Proposition 1.6.3(a), we have

$$\displaystyle{\Big\{\mathcal{F}^{\alpha }(\Delta _{t}^{{\prime}})\,f(t)\Big\} =\big (-i\omega \ \mbox{csc} \,\alpha \big)\,\mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega ) =\big (-i\omega \ \mbox{csc} \,\alpha \big)\hat{f}^{\alpha }(\omega )}$$
so that
$$\displaystyle{ \Big\vert \hat{f}^{\alpha }(\omega )\Big\vert = \dfrac{1} {\vert \omega \ \mbox{csc} \,\theta \vert }\,\Big\vert \Big\{\hat{\mathcal{F}^{\alpha }}(\Delta _{t}^{{\prime}})f(t)\Big\}\Big\vert \rightarrow 0\quad \text{as}\quad \,\omega \rightarrow \pm \infty. }$$

Remark.

As we have seen in Proposition 1.6.5(c) that \(\hat{f}^{\alpha }(\omega ) \rightarrow 0\) as ω → ± for every \(f \in L^{1}(\mathbb{R})\), it does not necessarily mean that \(\hat{f}^{\alpha }(\omega ) \in L^{1}(\mathbb{R})\). For example, consider the Heaviside unit step function given by
$$\displaystyle{ h(t) = \left \{\begin{array}{ll} 1,&t \geq 0,\\ 0, &t <0.\end{array} \right. }$$

Then, it is easy to verify that the product \(f^{\alpha }(t) =\exp \Big\{ -\dfrac{it^{2}\cot \alpha } {2} - t\Big\}h(t) \in L^{1}(\mathbb{R})\),but its FrFT is not in \(L^{1}(\mathbb{R})\).

Before we discuss basic properties of FrFT, we define the fractional translation, modulation, and dilation operators, respectively, by
$$\displaystyle\begin{array}{rcl} T_{\omega }^{\alpha }f(t)& =& f\left (t+\omega \right )\exp \{it\omega \cot \alpha \},\quad \qquad \qquad \text{(Translation),} {}\\ M_{\omega }^{\alpha }f(t)& =& \exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} + it\omega \ \mbox{csc} \,\alpha \right \}f(t),\quad \text{(Modulation),} {}\\ D_{a}f(t)& =& f(at),\qquad \qquad \qquad \qquad \qquad \qquad \quad \text{(Dilation),} {}\\ \end{array}$$
where \(t,\omega,a \in \mathbb{R}\) and a ≠ 0. The following results can easily be verified:
$$\displaystyle\begin{array}{rcl} \mathcal{F}^{\alpha }\big\{T_{\omega }^{\alpha }f(t)\big\}(\omega )& =& M_{-\omega }^{-\alpha }\mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega ) {}\\ \mathcal{F}^{\alpha }\big\{M_{\omega }^{\alpha }f(t)\big\}(\omega )& =& T_{-\omega }^{-\alpha }\mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega ) {}\\ \mathcal{F}^{\alpha }\big\{D_{-1}f(t)\big\}(\omega )& =& \mathcal{F}^{\alpha }\big\{f(t)\big\}(-\omega ) {}\\ \int _{-\infty }^{\infty }\hat{f}^{\alpha }(t)\,g(t)\,dt& =& \int _{ -\infty }^{\infty }f(\omega )\,\hat{g}^{\alpha }(\omega )\,d\omega. {}\\ \end{array}$$

Theorem 1.6.1.

The fractional Fourier transform \(\mathcal{F}^{\alpha }\) is a continuous linear operator from \(\mathcal{S}(\mathbb{R})\) onto itself.

Proof.

For any \(f(t) \in \mathcal{S}(\mathbb{R}) \subset L^{1}(\mathbb{R})\), we have
$$\displaystyle\begin{array}{rcl} \mathcal{F}^{\alpha }\big\{f(t)\big\}(\omega )& =& \hat{f}^{\alpha }(\omega ) \\ & =& \sqrt{2\pi }\,C_{\alpha }\exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} \right \}\mathcal{F}\left [\exp \left \{\dfrac{it^{2}\cot \alpha } {2} \right \}f(t)\right ](\omega \ \mbox{csc} \,\alpha ) \\ & =& C_{\alpha }\exp \Big\{\dfrac{i\omega ^{2}\cot \alpha } {2} \Big\}\,F^{\alpha }(\omega ) {}\end{array}$$
(1.6.15)
where
$$\displaystyle{F^{\alpha }(\omega ) = \sqrt{2\pi }\,\mathcal{F}\left [\exp \left \{\dfrac{it^{2}\cot \alpha } {2} \right \}f(t)\right ](\omega \ \mbox{csc} \,\alpha ) \in \mathcal{S}(\mathbb{R}).}$$
We have
$$\displaystyle\begin{array}{rcl} \mathfrak{D}_{\omega }^{\beta }\hat{f}^{\alpha }(\omega )& =& C_{\alpha }\,\mathfrak{D}_{\omega }^{\beta }\left [\exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} \right \}\hat{f}^{\alpha }(\omega )\right ] {}\\ & =& C_{\alpha }\sum _{\beta ^{{\prime}}=0}^{\beta }\binom{\beta }{\beta ^{{\prime}}}\,\mathfrak{D}_{\omega }^{\beta ^{{\prime}} }\left [\exp \Big\{\dfrac{i\omega ^{2}\cot \alpha } {2} \Big\}\right ]\mathfrak{D}_{\omega }^{\beta -\beta ^{{\prime}} }F^{\alpha }(\omega ) {}\\ & =& C_{\alpha }\sum _{\beta ^{{\prime}}=0}^{\beta }\binom{\beta }{\beta ^{{\prime}}}\exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} \right \}P_{\beta ^{{\prime}}}\Big(\omega, \dfrac{i\cot \alpha } {2}\Big)\mathfrak{D}_{\omega }^{\beta -\beta ^{{\prime}} }F^{\alpha }(\omega ) {}\\ \end{array}$$
where \(P_{\beta ^{'}}\Big(\omega, \dfrac{i\cot \alpha } {2}\Big)\) is a polynomial. Thus
$$\displaystyle{\mathfrak{D}_{\omega }^{\beta }\hat{f}^{\alpha }(\omega ) = C_{\alpha }\sum _{\beta ^{{\prime}}=0}^{\beta }\binom{\beta }{\beta ^{{\prime}}}\exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} \right \}\sum _{\beta ^{''}=0}^{\beta ^{'}}a_{\beta ^{''}}(\cot \alpha )\,\omega ^{\beta ^{''}}\,\mathfrak{D}_{\omega }^{\beta -\beta ^{'}}F^{\alpha }(\omega ).}$$

Therefore, we have

$$\displaystyle\begin{array}{rcl} \left \vert \omega ^{t}\,\mathfrak{D}_{\omega }^{\beta }\hat{f}^{\alpha }(\omega )\right \vert & =& \left \vert C_{\alpha }\sum _{\beta ^{{\prime} }=0}^{\beta }\binom{\beta }{\beta ^{{\prime}}}\sum _{ \beta ^{''}=0}^{\beta ^{{\prime}} }\exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} \right \}\,a_{\beta ^{''}}(\cot \alpha )\,\omega ^{\beta ^{''}+t }\,\mathfrak{D}_{\omega }^{\beta -\beta ^{{\prime}} }F^{\alpha }(\omega )\right \vert {}\\ &\leq & \big\vert C_{\alpha }\big\vert \sum _{\beta ^{{\prime}}=0}^{\beta }\binom{\beta }{\beta ^{{\prime}}}\sum _{ \beta ^{''}=0}^{\beta ^{{\prime}} }\Big\vert a_{\beta ^{''}}(\cot \alpha )\Big\vert \,\Big\vert \omega ^{\beta ^{''}+t }\,\mathfrak{D}_{\omega }^{\beta -\beta ^{{\prime}} }F^{\alpha }(\omega )\Big\vert. {}\\ \end{array}$$

Taking supremum over ω on both sides of above inequality and using the fact that \(F^{\alpha }(\omega ) \in S(\mathbb{R})\), we obtain

$$\displaystyle{ \sup _{\omega \in \mathbb{R}}\left \vert \omega ^{t}\,\mathfrak{D}_{\omega }^{\beta }\hat{f}^{\alpha }(\omega )\right \vert \leq \big\vert C_{\alpha }\big\vert \sum _{\beta ^{{\prime} }=0}^{\beta }\binom{\beta }{\beta ^{{\prime}}}\sum _{ \beta ^{''}=0}^{\beta ^{{\prime}} }\Big\vert a_{\beta ^{''}}(\cot \alpha )\Big\vert \,\sup _{\omega \in \mathbb{R}}\Big\vert \omega ^{\beta ^{''}+t }\,\mathfrak{D}_{\omega }^{\beta -\beta ^{{\prime}} }F^{\alpha }(\omega )\Big\vert <\infty. }$$
(1.6.16)
Hence, \(\hat{f^{\alpha }}(\omega ) \in \mathcal{S}(\mathbb{R}).\) In view of (1.6.1) and (1.6.4), we see that for all \(f \in \mathcal{S}(\mathbb{R})\), we have
$$\displaystyle{ f(t) =\mathcal{ F}^{-\alpha }\Big[\mathcal{F}^{\alpha }f(t)\Big] =\mathcal{ F}^{\alpha }\Big[\mathcal{F}^{-\alpha }f(\omega )\Big]. }$$
(1.6.17)

Therefore, it follows that \(\mathcal{F}^{\alpha }\) is a one-one map from \(\mathcal{S}(\mathbb{R})\) onto itself. To show that \(\mathcal{F}^{\alpha }\) is continuous on \(\mathbb{R}\), assume that there exist a null sequence \(\big\{f_{n}\big\}_{n\in \mathbb{N}}\) in \(\mathcal{S}(\mathbb{R})\); then from (1.6.16), we infer that \(\mathcal{F}^{\alpha }\left \{f_{n}(t)\right \} \rightarrow 0\) in \(\mathcal{S}(\mathbb{R})\); and hence, the continuity of FrFT follows.

Theorem 1.6.2 (Parseval’s Identity for FrFT).

If \(f,g \in L^{1}(\mathbb{R})\) , then
$$\displaystyle{ \big\langle f,g\big\rangle =\int _{ -\infty }^{\infty }\hat{f}^{\alpha }(\omega )\,\overline{\hat{g}^{\alpha }(\omega )}\,d\omega = \left \langle \hat{f}^{\alpha },\hat{g}^{\alpha }\right \rangle. }$$
(1.6.18)

Proof.

The proof of the theorem follows immediately from the Parseval formula of the Fourier transforms in \(L^{1}(\mathbb{R})\) and equation (1.6.15).

The rest of this section is devoted to find out the solution of some well-known differential equations by using the fractional Fourier transform method.

We consider the nth-order linear nonhomogeneous ordinary differential equations with constant coefficients:
$$\displaystyle{ Ly(t) = f(t) }$$
(1.6.19)
where L is the nth-order differential operator given by
$$\displaystyle{ L = a_{n}(\Delta _{t}^{{\prime}})^{n} + a_{ n-1}(\Delta _{t}^{{\prime}})^{n-1} +\ldots +a_{ 1}(\Delta _{t}^{{\prime}}) + a_{ 0} }$$
(1.6.20)

where a n , a n−1, , a1, a0 are constants, \(\Delta _{t}^{{\prime}}\) is the same as defined in Proposition 1.6.2, and f(t) is a given function.

Application of the fractional Fourier transform to both sides of (1.6.19) gives

$$\displaystyle{\int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,Ly(t)\,dt =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(t,\omega )\,f(t)\,dt,}$$
that is,
$$\displaystyle{\Big[a_{0}(-i\omega \ \mbox{csc} \,\alpha )^{n} + a_{ n-1}(-i\omega \ \mbox{csc} \,\alpha )^{n-1} +\ldots +a_{ 1}(-i\omega \ \mbox{csc} \,\alpha ) + a_{0}\Big]\hat{y}^{\alpha }(\omega ) =\hat{ f^{\alpha }}(\omega ).}$$
Or, equivalently,
$$\displaystyle{P(-i\omega \ \mbox{csc} \,\alpha )\,\hat{y}^{\alpha }(\omega ) =\hat{ f^{\alpha }}(\omega )}$$
where \(P(z) =\sum _{ m=0}^{n}\,a_{ m}z^{m}\). Therefore, it follows that
$$\displaystyle{ \hat{y}^{\alpha }(\omega ) = \dfrac{\hat{f^{\alpha }}(\omega )} {P(-i\omega \ \mbox{csc} \,\alpha )}. }$$
(1.6.21)

Applying the inverse FrFT to (1.6.21) gives the formal solution

$$\displaystyle{ y(t) =\mathcal{ F}^{-\alpha }\left [ \dfrac{\hat{f^{\alpha }}(\omega )} {P(-i\omega \ \mbox{csc} \,\alpha )}\right ]. }$$
(1.6.22)

Example 1.6.3.

Consider the generalized wave equation
$$\displaystyle{ \dfrac{\partial ^{2}y(x,t)} {\partial t^{2}} = k^{2}(\Delta _{ x}^{{\prime}})^{2}y(x,t),\quad -\infty <x <\infty,\,t> 0 }$$
(1.6.23)
with the initial data \(y(x,0) = f(x),\ \dfrac{\partial y(x,0)} {\partial t} = g(x)\), where k is a constant and \(\Delta _{x}^{{\prime}}\) is the same as in Proposition 1.6.2.
Application of the fractional Fourier transform to (1.6.23) gives
$$\displaystyle{\int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(x,\omega )\dfrac{\partial ^{2}y(x,t)} {\partial t^{2}} \,dx = k^{2}\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(x,\omega )(\Delta _{ x}^{{\prime}})^{2}y(x,t)\,dx}$$
so that
$$\displaystyle{\dfrac{\partial ^{2}\hat{y}^{\alpha }(\omega,t)} {\partial t^{2}} = k^{2}\int _{ -\infty }^{\infty }(\Delta _{ x})^{2}\,\mathcal{K}_{\alpha }(x,\omega )\,y(x,t)\,dx = -k^{2}\omega ^{2} \ \mbox{csc} ^{2}\alpha \,\hat{y}^{\alpha }(\omega,t).}$$
Therefore, it follows that
$$\displaystyle{ \hat{y}^{\alpha }(\omega,t) =\big\{\mathcal{ F}^{\alpha }f\big\}(\omega )\cos \big[(c\omega \ \mbox{csc} \,\alpha )t\big] + \dfrac{\big\{\mathcal{F}^{\alpha }g\big\}(\omega )} {c\omega \ \mbox{csc} \,\alpha } \sin \big[(c\omega \ \mbox{csc} \,\alpha )t\big]. }$$
(1.6.24)
This can readily be inverted by the inverse FrFT (1.6.4) to obtain
$$\displaystyle\begin{array}{rcl} y(x,t)& =& \mathcal{F}^{-\alpha }\Big[\big\{\mathcal{F}^{\alpha }f\big\}(\omega )\cos \big[(c\omega \ \mbox{csc} \,\alpha )t\big]\Big](x) \\ & & \quad +\mathcal{ F}^{-\alpha }\left [\dfrac{\big\{\mathcal{F}^{\alpha }g\big\}(\omega )} {c\omega \ \mbox{csc} \,\alpha } \sin \big[(c\omega \ \mbox{csc} \,\alpha )t\big]\right ](x) \\ & =& R(x,t) + S(x,t){}\end{array}$$
(1.6.25)
where
$$\displaystyle{ R(x,t) =\mathcal{ F}^{-\alpha }\Big[\big\{\mathcal{F}^{\alpha }f\big\}(\omega )\cos \big[(c\omega \ \mbox{csc} \,\alpha )t\big]\Big](x)\qquad \qquad }$$
(1.6.26)
and
$$\displaystyle{ S(x,t) =\mathcal{ F}^{-\alpha }\left [\dfrac{\big\{\mathcal{F}^{\alpha }g\big\}(\omega )} {c\omega \ \mbox{csc} \,\alpha } \sin \big[(c\omega \ \mbox{csc} \,\alpha )t\big]\right ](x).\qquad \ \ \ }$$
(1.6.27)
We now estimate R(x, t) as
$$\displaystyle\begin{array}{rcl} R(x,t)& =& \overline{C}_{\alpha }\int _{-\infty }^{\infty }\exp \left \{-i(x^{2} +\omega ^{2})\, \frac{\cot \alpha } {2} + ix\omega \ \mbox{csc} \,\alpha \right \} \\ & & \qquad \big\{\mathcal{F}^{\alpha }f\big\}(\omega )\cos \big\{(c\omega \ \mbox{csc} \,\alpha )t\big\}d\omega \\ & =& \frac{1} {2\pi \sin \alpha }\int _{-\infty }^{\infty }\exp \left \{-i(x^{2} +\omega ^{2}) \frac{\cot \alpha } {2} + ix\omega \ \mbox{csc} \omega \right \} \\ & & \quad \times \left [\int _{-\infty }^{\infty }\exp \left \{i(z^{2} +\omega ^{2}) \frac{\cot \alpha } {2} - iz\omega \ \mbox{csc} \,\alpha \right \}f(z)\,dz\right ]\cos \big\{(c\omega \ \mbox{csc} \,\alpha )t\big\}d\omega.{}\end{array}$$
(1.6.28)

Setting \(H_{\alpha }(z) =\exp \left \{\frac{iz^{2}\cot \alpha } {2} \right \}f(z)\) and ωcsc α = γ, equation (1.6.28) becomes

$$\displaystyle\begin{array}{rcl} R(x,t)& =& \frac{1} {2\pi \sin \alpha }\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\sin \alpha \left [\int _{-\infty }^{\infty }e^{ix\gamma }\cos (c\gamma t)\,d\gamma \int _{ -\infty }^{\infty }e^{-iz\gamma }H_{\alpha }(z)\,dz\right ] \\ & =& \frac{1} {2\sqrt{2\pi }}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\left [\int _{-\infty }^{\infty }\left [e^{i(x+ct)\gamma } + e^{i(x-ct)\gamma }\right ]\mathcal{F}\big\{H_{\alpha }(z)\big\}(\gamma )\,d\gamma \right ] \\ & =& \frac{1} {2}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\Big[H_{\alpha }(x + ct) + H_{\alpha }(x - ct)\Big] \\ & =& \frac{1} {2}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\left [e^{i(x+ct)^{2} \frac{\cot \alpha } {2} }f(x + ct) + e^{i(x-ct)^{2} \frac{\cot \alpha } {2} }f(x - ct)\right ].{}\end{array}$$
(1.6.29)
Similarly, assume that \(G_{\alpha }(z) =\exp \left \{\frac{iz^{2}\cot \alpha } {2} \right \}g(z)\), and again setting ωcsc α = γ, we obtain
$$\displaystyle{ S(x,t) = \frac{1} {\sqrt{2\pi }}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\dfrac{e^{ix\gamma }\sin (c\gamma t)} {c\gamma } \,\mathcal{F}\big\{G_{\alpha }(z)\big\}(\gamma )\,d\gamma. }$$
(1.6.30)
Differentiating equation (1.6.30) with respect to t, we obtain the same result as that of R(x, t), and then by integrating, we have
$$\displaystyle{ S(x,t) = \frac{1} {2c}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{x-ct}^{x+ct}\exp \left \{\dfrac{i\omega ^{2}\cos \alpha } {2} \right \}g(\omega )\,d\omega. }$$
(1.6.31)
After substituting equations (1.6.29) and (1.6.31) in (1.6.25), the solution of the given problem (1.6.23) can be obtained in the form
$$\displaystyle\begin{array}{rcl} y(x,t)& =& \frac{1} {2}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\Bigg[e^{i(x+ct)^{2} \frac{\cot \alpha } {2} }f(x + ct) + e^{i(x-ct)^{2} \frac{\cot \alpha } {2} }f(x - ct) + {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\frac{1} {c}\int _{x-ct}^{x+ct}\exp \left \{\frac{i\omega ^{2}\cot \alpha } {2} \right \}g(\omega )d\omega \Bigg]. {}\\ \end{array}$$

Example 1.6.4.

Consider the generalized heat equation
$$\displaystyle{ \frac{\partial y(x,t)} {\partial t} =\big (\Delta _{x}^{{\prime}}\big)^{2}y(x,t),\quad -\infty <x <\infty,\,t> 0 }$$
(1.6.32)
where \(\Delta _{x}^{{\prime}}\) is the same as given in Proposition 1.6.2 and y(x, 0) = f(x).
Application of the fractional Fourier transform to (1.6.32) gives
$$\displaystyle{\int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(x,\omega )\frac{\partial y(x,t)} {\partial t} \,dx =\int _{ -\infty }^{\infty }\mathcal{K}_{\alpha }(x,\omega )(\Delta _{ x}^{{\prime}})^{2}\,y(x,t)\,dx}$$
so that
$$\displaystyle{\frac{\partial \hat{y}^{\alpha }(\omega,t)} {\partial t} =\int _{ -\infty }^{\infty }(\Delta _{ x})^{2}\,\mathcal{K}_{\alpha }(x,\omega )\,y(x,t)\,dx = -\omega ^{2} \ \mbox{csc} ^{2}\alpha \,\hat{y}^{\alpha }(\omega,t).}$$
Therefore, it follows that
$$\displaystyle{ \hat{y}^{\alpha }(\omega,t) = C(\omega )\exp \big\{ - (\omega ^{2}csc^{2}\alpha )t\big\} }$$
(1.6.33)
which gives \(\hat{y}^{\alpha }(\omega,0) = C(\omega )\). Since
$$\displaystyle\begin{array}{rcl} \hat{y}^{\alpha }(\omega,0)& =& \int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(x,\omega )\,y(x,0)\,dx {}\\ & =& \int _{-\infty }^{\infty }\mathcal{K}_{\alpha }(x,\omega )f(x)\,dx {}\\ & =& \mathcal{F}^{\alpha }\big\{f(x)\big\}(\omega ), {}\\ \end{array}$$
hence,
$$\displaystyle{ C(\omega ) =\mathcal{ F}^{\alpha }\big\{f(x)\big\}(\omega ). }$$
(1.6.34)
Thus, equation (1.6.33) becomes
$$\displaystyle{ \hat{y}^{\alpha }(\omega,t) = \left \{\mathcal{F}^{\alpha }f\right \}(\omega )\exp \left \{-(\omega ^{2} \ \mbox{csc} ^{2}\alpha )t\right \}. }$$
(1.6.35)

Applying the inverse FrFT on both sides of (1.6.35), we obtain

$$\displaystyle\begin{array}{rcl} & & y(x,t) {}\\ & & =\mathcal{ F}^{-\alpha }\Big\{(\mathcal{F}^{\alpha }f)(\omega )\exp \left \{-(\omega ^{2} \ \mbox{csc} ^{2}\alpha )t\right \}\Big\}(x) {}\\ & & = \frac{E_{\alpha }} {2\pi } \int _{-\infty }^{\infty }\exp \left \{-\frac{i(x^{2} +\omega ^{2})\cot \alpha } {2} + ix\omega \ \mbox{csc} \,\alpha \right \}\!(\mathcal{F}^{\alpha }f)(\omega )\exp \big\{-(\omega ^{2} \ \mbox{csc} ^{2}\alpha )t\big\}d\omega {}\\ & & = \frac{E_{\alpha }} {2\pi } \int _{-\infty }^{\infty }\exp \left \{-\frac{i(x^{2} +\omega ^{2})\cot \alpha } {2} + ix\omega \ \mbox{csc} \,\alpha \right \}\, {}\\ & & \quad \times \left [C_{\alpha }\int _{-\infty }^{\infty }\exp \left \{\frac{i(z^{2} +\omega ^{2})\cot \alpha } {2} - iz\omega \ \mbox{csc} \,\alpha \right \}\,f(z)dz\right ]\exp \left \{-(\omega ^{2} \ \mbox{csc} ^{2}\alpha )t\right \}d\omega {}\\ & & = \dfrac{1} {2\pi \sin \alpha }\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\exp \big\{ix\omega \ \mbox{csc} \,\alpha \big\}\exp \big\{ - (\omega ^{2} \ \mbox{csc} ^{2}\alpha )t\big\}d\omega \, {}\\ & & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \times \int _{-\infty }^{\infty }\exp \big\{- iz\omega \ \mbox{csc} \,\alpha \big\}\,\exp \left \{\frac{iz^{2}\cot \alpha } {2} \right \}f(z)dz. {}\\ \end{array}$$

Let us assume that \(H_{\alpha }(z) =\exp \left \{\dfrac{iz^{2}\cot \alpha } {2} \right \}f(z)\); then

$$\displaystyle\begin{array}{rcl} & & y(x,t) {}\\ & & \quad = \dfrac{1} {2\pi \sin \alpha }\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }e^{ix\omega \ \mbox{csc} \,\alpha }e^{-(\omega ^{2} \ \mbox{csc} ^{2}\alpha )t }d\omega \int _{-\infty }^{\infty }e^{-iz\omega \ \mbox{csc} \,\alpha }H_{\alpha }(z)dz. {}\\ \end{array}$$

Setting ωcsc α = γ, the above relation becomes

$$\displaystyle\begin{array}{rcl} y(x,t)& =& \frac{1} {2\pi }\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\exp \left \{ix\gamma \right \}\exp \left \{-\gamma ^{2}t\right \}d\gamma \int _{ -\infty }^{\infty }\exp \{- i\gamma z\}H_{\alpha }(z)\,dz {}\\ & =& \frac{1} {2\sqrt{\pi t}}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\exp \left \{ix\gamma \right \}\mathcal{F}\Big\{e^{-x^{2} /4t}\Big\}(\gamma )\mathcal{F}\Big\{H_{\alpha }(z)\Big\}(\gamma )\,d\gamma {}\\ & =& \frac{1} {2\sqrt{\pi t} \cdot \sqrt{2\pi }}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\exp \left \{ix\gamma \right \}\mathcal{F}\Big[e^{-x^{2} /4t} {\ast} H_{\alpha }(x)\Big](\gamma )d\gamma {}\\ & =& \frac{1} {2\sqrt{\pi t}}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\exp \left \{\dfrac{-(x-\omega )^{2}} {4t} \right \}H_{\alpha }(\omega )\,d\omega. {}\\ \end{array}$$
Therefore, the solution of the generalized heat equation (1.6.32) is
$$\displaystyle\begin{array}{rcl} y(x,t)& = \frac{1} {2\sqrt{\pi t}}\exp \left \{\frac{-ix^{2}\cot \alpha } {2} \right \}\int _{-\infty }^{\infty }\exp \left \{\frac{-(x-\omega )^{2}} {4t} \right \} \cdot \exp \left \{\dfrac{i\omega ^{2}\cot \alpha } {2} \right \}f(\omega )\,d\omega.& {}\\ \end{array}$$

1.7 The Uncertainty Principle

Heisenberg first formulated the uncertainty principle between the position and momentum in quantum mechanics. This principle has an important interpretation as an uncertainty of both the position and momentum of a particle described by a wave function \(\psi \in L^{2}(\mathbb{R})\). In other words, it is impossible to determine the position and momentum of a particle exactly and simultaneously (See Heisenberg, 1948a,b).

In signal processing, time and frequency concentrations of energy of a signal f are also governed by the Heisenberg uncertainty principle. The average or expectation values of time t and frequency ω are, respectively, defined by
$$\displaystyle{ t^{{\ast}} = \frac{1} {\big\|f\big\|_{2}^{2}}\int _{-\infty }^{\infty }t\big\vert f(t)\big\vert ^{2}dt,\quad \omega ^{{\ast}} = \frac{1} {\left \|\hat{f}\right \|_{2}^{2}}\int _{-\infty }^{\infty }\omega \left \vert \hat{f}(\omega )\right \vert ^{2}d\omega, }$$
(1.7.1)

where the energy of a signal f is well localized in time, and its Fourier transform \(\hat{f}\) has an energy concentrated in a small frequency domain.

The variances around these average values are given respectively by

$$\displaystyle{ \sigma _{t}^{2} = \frac{1} {\big\|f\big\|_{2}^{2}}\int _{-\infty }^{\infty }\big(t - t^{{\ast}}\big)^{2}\big\vert f(t)\big\vert ^{2}dt,\quad \sigma _{\omega }^{2} = \frac{1} {2\pi \left \|\hat{f}\right \|_{2}^{2}}\int _{-\infty }^{\infty }\big(\omega -\omega ^{{\ast}}\big)^{2}\left \vert \hat{f}(\omega )\right \vert ^{2}d\omega. }$$
(1.7.2)

Theorem 1.7.1 (Heisenberg’s Inequality).

If f(t), tf(t), and \(\omega \hat{f}(\omega )\) belong to \(L^{2}(\mathbb{R})\) and \(\sqrt{ t}\,\vert f(t)\vert \rightarrow 0\) as | t | → ∞, then
$$\displaystyle{ \sigma _{t}^{2}\,\sigma _{ \omega }^{2} \geq \frac{1} {4}, }$$
(1.7.3)

where σ t is defined as a measure of time duration of a signal f and σ ω is a measure of frequency dispersion (or bandwidth) of its Fourier transform \(\hat{f}\).

Equality in (1.7.3) holds only if f(t) is a Gaussian signal given by \(f(t) = C\,e^{-bt^{2} },b> 0\).

Proof.

If the average time and frequency localization of a signal f are 〈t〉 and 〈ω〉, then the average time and frequency location of \(e^{-i\omega ^{{\ast}}t }f\big(t + t^{{\ast}}\big)\) is zero. Hence, it is sufficient to prove the theorem around the zero mean values, that is, 〈t〉 = 〈ω〉 = 0. 

Since \(\big\|f\big\|_{2} = \left \|\hat{f}\right \|_{2}\), we have
$$\displaystyle{\big\|f\big\|_{2}^{4}\,\sigma _{ t}^{2}\,\sigma _{ \omega }^{2} = \frac{1} {2\pi }\int _{-\infty }^{\infty }\big\vert tf(t)\big\vert ^{2}dt\int _{ -\infty }^{\infty }\left \vert \omega \hat{f}(\omega )\right \vert ^{2}d\omega.}$$

Using \(i\omega \hat{f}(\omega ) = \mathcal{F}\big\{f^{{\prime}}(t)\big\}\) and Parseval’s formula

$$\displaystyle{\big\|f^{{\prime}}(t)\big\|_{ 2} = \frac{1} {2\pi }\left \|i\omega \hat{f}(\omega )\right \|_{2},}$$
we obtain
$$\displaystyle{\begin{array}{rcl} \big\|f\big\|_{2}^{4}\,\sigma _{t}^{2}\,\sigma _{\omega }^{2} & =&\int _{-\infty }^{\infty }\big\vert tf(t)\big\vert ^{2}dt\int _{ -\infty }^{\infty }\left \vert f^{{\prime}}(t)\right \vert ^{2}d\omega \\ \\ & \geq &\left \vert \int _{-\infty }^{\infty }\left \{tf(t)\,\overline{f^{{\prime}}(t)}\right \}dt\right \vert ^{2},\quad \text{ by Schwarz's inequality}\\ \\ & \geq &\left \vert \int _{-\infty }^{\infty }t \cdot \frac{1} {2}\left \{f^{{\prime}}(t)\,\overline{f(t)} + \overline{f^{{\prime}}(t)}f(t)\right \}dt\right \vert ^{2} \\ \\ & =&\frac{1} {4}\left [\int _{-\infty }^{\infty }t\left ( \frac{d} {dt}\,\vert f\vert ^{2}\right )dt\right ]^{2} = \frac{1} {4}\left \{\left [t\big\vert f(t)\big\vert ^{2}\right ]_{ -\infty }^{\infty }-\int _{ -\infty }^{\infty }\vert f\vert ^{2}dt\right \}^{2} \\ \\ & =&\frac{1} {4}\,\big\|f\big\|_{2}^{4}, \end{array} }$$

in which \(\sqrt{ t}\,f(t) \rightarrow 0\) as | t | → was used to eliminate the integrated term.

This completes the proof of inequality (1.7.3).

If we assume f (t) is proportional to tf(t), that is, f (t) = atf(t), where a is a constant of proportionality, this leads to the Gaussian signal \(f(t) = C\,e^{-bt^{2} },\) where C is a constant of integration and \(b = -\dfrac{a} {2}> 0\).

Remarks.

 
  1. 1.

    In a time-frequency analysis of signals, the measure of the resolution of a signal f in the time or frequency domain is given by σ t and σ ω . Then, the joint resolution is given by the product (σ t )(σ ω ) which is governed by the Heisenberg uncertainty principle. In other words, the product (σ t )(σ ω ) cannot be arbitrarily small and is always greater than the minimum value \(\dfrac{1} {2}\) which is attained only for the Gaussian signal.

     
  2. 2.

    In many applications in science and engineering, signals with a high concentration of energy in the time and frequency domains are of special interest. The uncertainty principle can also be interpreted as a measure of this concentration of the second moment of f2(t) and its energy spectrum \(\hat{f}^{2}(\omega )\).

     

1.8 Exercises

  1. 1.
    Find the Fourier transforms of each of the following functions:
    1. (a)

      \(f(t) = t\,e^{-a\vert t\vert },\quad a> 0,\qquad \text{(b)}\quad f(t) = t\,e^{-at^{2} },\quad a> 0,\)

       
    2. (c)

      \(f(t) = t^{2}\,e^{-t^{2}2 },\ \qquad \quad \quad \quad \ \text{(d)}\quad f(t) = e^{-at^{2}+bt },\)

       
    3. (e)

      \(f(t) = \vert t\vert ^{a-1},\ \quad \qquad \quad \qquad \text{(f)}\quad f(t) = \dfrac{\sin ^{2}at} {\pi at^{2}}.\)

       
     
  2. 2.
    If \(f(t) = \left (1 - \dfrac{t} {a}\right )\chi _{[-a,a]}(t),\) where a > 0, show that
    $$\displaystyle{\hat{f}(\omega ) = \dfrac{\sin ^{2}(a\pi \omega )} {a(\pi \omega )^{2}}.}$$
     
  3. 3.
    Show that the Fourier transform of
    $$\displaystyle{f(t) = \dfrac{\cos bt} {a^{2} + t^{2}},\quad a> 0,}$$
    for any constant b is
    $$\displaystyle{\hat{f}(\omega ) = \dfrac{\pi } {2a}\Big[e^{-a\vert \omega -b\vert } + e^{-a\vert \omega +b\vert }\Big].}$$
     
  4. 4.
    If f(t) has a finite discontinuity at a point t = a, prove that
    $$\displaystyle{\mathcal{F}\left \{f^{{\prime}}(t)\right \} = (i\omega )\hat{f}(\omega ) - e^{-ia\omega }[f]_{ a},}$$
    where [f] a = f(a + 0) − f(a − 0). Generalize this result for \(\mathcal{F}\left \{f^{(n)}(t)\right \}\).
     
  5. 5.

    Use result (1.2.17) to find

    (a) \(\mathcal{F}\left \{t^{n}\,e^{-t^{2}/2 }\right \},\qquad \quad \quad \quad \text{(b)}\quad \mathcal{F}\left \{t^{n}\,e^{-at^{2} }\right \}.\)

     
  6. 6.
    Prove the following convolution properties:
    1. (a)

      h(t + a + b) = f(t + a) ∗ g(t + b), h = fg

       
    2. (b)

      \(h(at + b + c) = \vert a\vert \big[f(at + b) {\ast} g(at + c)\big],\quad h = f {\ast} g\)

       
    3. (c)

      h(tt0) = f(tt0) ∗ g(t), h = fg

       
    4. (d)

      \(\overline{\big(f(t) {\ast} g(t)\big)} =\big (\overline{f} {\ast}\overline{g}\big)(t),\)

       
    5. (d)

      \(t\big[f(t) {\ast} g(t)\big] =\big [tf(t) {\ast} g(t)\big] +\big [f(t) {\ast} tg(t)\big]\)

       
     
  7. 7.
    Show that
    $$\displaystyle{\mathcal{F}\left \{e^{-a^{2}t^{2} }\right \} {\ast}\mathcal{F}\left \{e^{-b^{2}t^{2} }\right \} = \dfrac{\pi \,e^{-\omega ^{2}/4c^{2} }} {ab},\quad \text{where}\quad \dfrac{1} {c^{2}} = \left ( \dfrac{1} {a^{2}} + \dfrac{1} {b^{2}}\right ).}$$
     
  8. 8.
    If \(G(t) = \dfrac{1} {\pi } \,e^{-t^{2} }\), then show that the family of functions G λ (t) = λG(λt), λ > 0 forms a summability kernel on \(\mathbb{R}\). Moreover, show that
    $$\displaystyle{G_{\lambda }(t) {\ast} f(t) =\int _{ -\infty }^{\infty }e^{-\omega ^{2}/4\lambda ^{2} }\,\hat{f}(\omega )\,e^{i\omega t}d\omega.}$$
     
  9. 9.
    Consider \(f: \mathbb{Z}_{N} \rightarrow \mathbb{C}\), such that f(n) = ean , for some constant a. Show that the discrete Fourier transform of f is
    $$\displaystyle{\hat{f}(n) = \dfrac{1 - e^{-aN}} {1 - e^{-a-2\pi in/N}}.}$$
     
  10. 10.

    For positive integer N, let w = e2πiN . Show that

    (a) \(w\,\overline{w} = 1,\quad \quad \quad \quad \quad \text{(b)}\quad w^{k} = \overline{w^{-k}} = w^{N+k} = \overline{w^{N+k}},\,\text{for any integer}\ k\),

    (c) 1 + w + w2 + + w N−1 = 0, N > 1. 

     
  11. 11.
    Show that the fast Fourier transform of \(f(n) =\big (1,0,i,2,-i,1,0,i\big) \in l^{2}(\mathbb{Z}_{8})\) is
    $$\displaystyle\begin{array}{rcl} \hat{f}(n)& =& \Big(4 + i,2 - 2\sqrt{2} + i,-i,\sqrt{2} + i(1 -\sqrt{2}),-2 - i,2 + 2\sqrt{2} + i,2 - 3i, {}\\ & & \quad \;\;\; -\sqrt{2} + i(1 + \sqrt{2})\Big). {}\\ \end{array}$$
     
  12. 12.

    Verify the equality of the uncertainty principle for the Gaussian function \(f(t) = e^{-t^{2}/2 }\) and its Fourier transform \(\hat{f}(\omega ) = \sqrt{2\pi }\,e^{-\omega ^{2}/2 }\).

     
  13. 13.

    Repeat the calculation of the previous exercise using the function \(f(t) = \dfrac{1} {2\sqrt{\pi a}}\,e^{-t^{2}/4a }\) and its Fourier transform \(\hat{f}(\omega ) = e^{-a\omega ^{2} }\). Show that equality of the Heisenberg inequality is preserved for all values of a.

     

1

Footnotes

  1. 1.

    The following list includes books and research papers that have been useful for the preparation of these notes as well as some which may be of interest for further study.

Bibliography

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Copyright information

© Springer International Publishing AG 2017

Authors and Affiliations

  • Lokenath Debnath
    • 1
  • Firdous A. Shah
    • 2
  1. 1.School of Mathematical and Statistical SciencesUniversity of Texas – Rio Grande ValleyEdinburgUSA
  2. 2.Department of MathematicsUniversity of KashmirAnantnagIndia

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