Home/Chapter 5

The Fourier Transformation

All four Fourier transforms: DDFT, CDFT, DCFT, and CCFT. Fourier series, DFT properties, FFT algorithm, Plancherel/Parseval theorems, transforms of distributions, and the sampling/bandwidth tradeoff.

The Big Idea: What Fourier Transforms Actually Do

The central insight behind all Fourier transforms is surprisingly simple: every signal can be built by adding up sine waves at different frequencies. A musical chord is the sum of its individual notes. A square wave is the sum of infinitely many harmonics. Even a sharp spike can be constructed from sinusoids, if you use enough of them.

The Fourier transform answers the question: "How much of each frequency is present in my signal?"

There are two complementary operations:

  • Forward transform (Analysis / Decomposition): Takes a signal and produces its frequency recipe -- a function (or sequence) that tells you the amplitude and phase of each frequency component. This is like hearing a chord and identifying the individual notes.
  • Inverse transform (Synthesis / Reconstruction): Takes the frequency recipe and rebuilds the original signal by adding up weighted sinusoids. This is like reading sheet music and playing the chord.

How the forward transform works, conceptually:

To detect "how much" of frequency ff is in a signal xx, the transform multiplies xx by the complex sinusoid ei2πfte^{-i2\pi ft} and then sums (or integrates) the result. The exponential ei2πfte^{-i2\pi ft} acts like a tuning fork tuned to frequency ff:

  • If the signal contains energy at frequency ff, the product x(t)ei2πftx(t) \cdot e^{-i2\pi ft} will have a steady, non-cancelling component, and the sum/integral will be large.
  • If the signal does not contain frequency ff, the product oscillates symmetrically around zero, and the sum/integral cancels out to zero (or near zero).

The result is a complex number x^(f)\hat{x}(f) whose magnitude x^(f)|\hat{x}(f)| tells you the amplitude (strength) of frequency ff, and whose angle x^(f)\angle \hat{x}(f) tells you the phase (timing offset) of that component.

A concrete example: If x(t)=cos(2π5t)x(t) = \cos(2\pi \cdot 5 \cdot t), a pure 5 Hz tone, then the Fourier transform will show a spike at f=5f = 5 and f=5f = -5, and zero everywhere else. The negative frequency appears because cos(2π5t)=12ei2π5t+12ei2π5t\cos(2\pi \cdot 5 \cdot t) = \frac{1}{2}e^{i2\pi \cdot 5 \cdot t} + \frac{1}{2}e^{-i2\pi \cdot 5 \cdot t} -- cosine is the sum of two complex exponentials rotating in opposite directions.

Why four transforms? The four Fourier transforms (DDFT, CDFT, DCFT, CCFT) all implement this same idea, but differ in whether the input signal and the output spectrum are discrete or continuous. These choices depend on the mathematical setting -- whether your signal is finite-length, periodic, or defined over all of R\mathbb{R}.

We have seen earlier in Theorem that in a Hilbert space HH, a complete orthonormal sequence {en,nN}\{e_n, n \in \mathbb{N}\} can be used to express any xHx \in H via the relation

x=limNk=1Nx,ekekx = \lim_{N \to \infty} \sum_{k=1}^{N} \langle x, e_k \rangle e_k

We have also seen in Chapters 2 and 3 that complex exponentials provide such a complete orthonormal sequence, and therefore can be used to approximate any square integrable signal arbitrarily well. We saw in Chapter 4 that complex exponentials possess the eigenfunction property for linear time-invariant systems.

The above motivate the use of the representation of signals in terms of complex exponentials and the spectral properties of frequency response / transfer functions. This study is achieved by Fourier transforms. Accordingly, Fourier transforms play a significant role in systems theory and applied mathematics at large. There are four types of Fourier transformations: Discrete-to-Discrete (DDFT), Continuous-to-Discrete (CDFT), Discrete-to-Continuous (DCFT), Continuous-to-Continuous (CCFT).

We will start with the first two of the above. Before we proceed, recall that a bijective transformation T\mathcal{T} is a map from a signal set XX to another one X^\hat{X}, such that T\mathcal{T} is onto and one-to-one; the Fourier transforms will constitute examples of such transformations with further very useful structural and regularity properties to be studied in this chapter and beyond.

Discrete-to-Discrete (DDFT) and Continuous-to-Discrete (CDFT) Fourier Transforms

Fourier Series Expansions

Discrete Time

The NN-dimensional complex vector space l2({0,1,2,,N1};C)l_2(\{0, 1, 2, \ldots, N-1\}; \mathbb{C}) is a Hilbert space with the inner product:

h1,h2=n=0N1h1(n)h2(n),\langle h_1, h_2 \rangle = \sum_{n=0}^{N-1} h_1(n) \overline{h_2(n)},

where, as earlier, the bar notation ()\overline{(\cdot)} denotes the complex conjugate of its argument.

The set of complex harmonic signals:

{1Nei2πkn,n=0,1,2,,N1,k{0,1N,,N1N}},\left\{ \frac{1}{\sqrt{N}} e^{i2\pi k n}, \quad n = 0, 1, 2, \ldots, N-1, \quad k \in \left\{0, \frac{1}{N}, \ldots, \frac{N-1}{N}\right\} \right\},

provides a complete orthonormal sequence, hence, provides a basis for l2({0,1,2,,N1};C)l_2(\{0, 1, 2, \ldots, N-1\}; \mathbb{C}). The Fourier series expansion is given by:

x(n)=1Nk=0N1x^ ⁣(kN)ei2πkNn,n{0,1,,N1}x(n) = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \hat{x}\!\left(\frac{k}{N}\right) e^{i2\pi \frac{k}{N} n}, \quad n \in \{0, 1, \ldots, N-1\}

where

x^ ⁣(kN)=1Nn=0N1x(n)ei2πkNn,k{0,1,,N1}\hat{x}\!\left(\frac{k}{N}\right) = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x(n) e^{-i2\pi \frac{k}{N} n}, \quad k \in \{0, 1, \ldots, N-1\}
Remark.

Observe that x^(kN)=x,ekN\hat{x}(\frac{k}{N}) = \langle x, e_{\frac{k}{N}} \rangle with ekN(n)=1Nei2πkNne_{\frac{k}{N}}(n) = \frac{1}{\sqrt{N}} e^{i2\pi \frac{k}{N} n}, for n=0,1,2,,N1n = 0, 1, 2, \ldots, N-1.

Continuous Time

The complex vector space L2([0,P];C)L_2([0, P]; \mathbb{C}) is a Hilbert space with the inner product:

h1,h2=0Ph1(t)h2(t)dt.\langle h_1, h_2 \rangle = \int_0^P h_1(t) \overline{h_2(t)}\, dt.

The countably infinite sequence of complex harmonic signals:

1Pei2πft,f{kP,kZ},\frac{1}{\sqrt{P}} e^{i2\pi f t}, \quad f \in \left\{\frac{k}{P}, k \in \mathbb{Z}\right\},

provides a complete orthogonal sequence, hence, provides a basis for L2([0,P];C)L_2([0, P]; \mathbb{C}).

The completeness of Fourier series in L2[0,P]L_2[0, P] is based on the argument that trigonometric polynomials are dense in L2([0,P];C)L_2([0, P]; \mathbb{C}) (see Theorem and the discussion in the relevant section).

The Fourier series expansion is given by:

x(t)=1Pkx^ ⁣(kP)ei2πkPt,t[0,P]x(t) = \frac{1}{\sqrt{P}} \sum_k \hat{x}\!\left(\frac{k}{P}\right) e^{i2\pi \frac{k}{P} t}, \quad t \in [0, P]

where

x^(f)=1P0Px(t)ei2πftdt,f{kP,kZ}\hat{x}(f) = \frac{1}{\sqrt{P}} \int_0^P x(t) e^{-i2\pi f t}\, dt, \quad f \in \left\{\frac{k}{P}, k \in \mathbb{Z}\right\}
Remark.

Observe that x^(f)=x,ef\hat{x}(f) = \langle x, e_f \rangle with ef(t)=1Pei2πfte_f(t) = \frac{1}{\sqrt{P}} e^{i2\pi f t}, for t[0,P]t \in [0, P] with f{kP,kZ}f \in \{\frac{k}{P}, k \in \mathbb{Z}\}.

Thus, in the context of the relevant section, a Fourier series expansion is the representation of a square integrable signal in terms of a collection of a complete orthonormal harmonic signal sequence.

Discrete-to-Discrete (DDFT) and Continuous-to-Discrete (CDFT) Fourier Transforms

In view of the above, we define Discrete-to-Discrete (DDFT), Continuous-to-Discrete (CDFT) as follows:

DefinitionDDFT
FDD:l2({0,1,,N1};C)l2 ⁣({0N,1N,,N1N};C)\mathcal{F}_{DD} : l_2(\{0, 1, \ldots, N-1\}; \mathbb{C}) \to l_2\!\left(\left\{\frac{0}{N}, \frac{1}{N}, \ldots, \frac{N-1}{N}\right\}; \mathbb{C}\right)x^=FDD(x)\hat{x} = \mathcal{F}_{DD}(x)

and

x^(f)=n=0N11Nx(n)ei2πfn,f{0N,1N,,N1N}\hat{x}(f) = \sum_{n=0}^{N-1} \frac{1}{\sqrt{N}} x(n) e^{-i2\pi f n}, \quad f \in \left\{\frac{0}{N}, \frac{1}{N}, \ldots, \frac{N-1}{N}\right\}
Remark.

Intuition: The DDFT takes a finite-length discrete signal (N samples) and produces N frequency coefficients. Both the input and output are finite sequences, making this the version of the Fourier transform that computers actually compute. It decomposes a finite signal into N harmonic components equally spaced in frequency.

Remark.

What each part of the does (DDFT forward):

x^(f)=n=0N11Nx(n)ei2πfn,f{0N,1N,,N1N}\hat{x}(f) = \sum_{n=0}^{N-1} \frac{1}{\sqrt{N}} x(n) \, e^{-i2\pi f n}, \quad f \in \left\{\frac{0}{N}, \frac{1}{N}, \ldots, \frac{N-1}{N}\right\}

  • x(n)x(n) -- the original signal: a sequence of NN numbers (samples), indexed by n=0,1,,N1n = 0, 1, \ldots, N-1
  • ei2πfne^{-i2\pi f n} -- a complex sinusoid at frequency ff, evaluated at discrete time nn. By Euler's formula, ei2πfn=cos(2πfn)isin(2πfn)e^{-i2\pi f n} = \cos(2\pi f n) - i\sin(2\pi f n). This is the "detector" for frequency ff
  • x(n)ei2πfnx(n) \cdot e^{-i2\pi f n} -- multiplying the signal by the detector at each sample. If the signal oscillates at frequency ff, this product will tend to align (not cancel out)
  • 1N\frac{1}{\sqrt{N}} -- a normalization factor that ensures the transform is unitary (energy-preserving). Some conventions put 1/N1/N on the forward transform and 11 on the inverse, but the symmetric 1/N1/\sqrt{N} keeps the transform an isometry
  • n=0N1\sum_{n=0}^{N-1} -- sum over all NN samples. This is the discrete analogue of integration. If frequency ff is present, the terms reinforce and the sum is large; if not, the terms cancel and the sum is small or zero
  • x^(f)\hat{x}(f) -- the result: a complex number for each frequency ff. Its magnitude x^(f)|\hat{x}(f)| is the amplitude of that frequency component, and its angle x^(f)\angle \hat{x}(f) is the phase shift
DefinitionCDFT
FCD:L2([0,P];C)l2 ⁣(Z1P;C)\mathcal{F}_{CD} : L_2([0, P]; \mathbb{C}) \to l_2\!\left(\mathbb{Z}\frac{1}{P}; \mathbb{C}\right)x^=FCD(x)\hat{x} = \mathcal{F}_{CD}(x)

and

x^(f)=τ=0P1Px(τ)ei2πfτdτ,f{kP,kZ}\hat{x}(f) = \int_{\tau=0}^{P} \frac{1}{\sqrt{P}} x(\tau) e^{-i2\pi f \tau}\, d\tau, \quad f \in \left\{\frac{k}{P}, k \in \mathbb{Z}\right\}
Remark.

Intuition: The CDFT takes a continuous-time signal defined on a finite interval [0,P][0,P] and produces a discrete (countable) set of Fourier coefficients at frequencies k/Pk/P. This is the classical Fourier series: a periodic continuous signal is decomposed into a countable set of harmonics. The spacing 1/P1/P between frequencies is inversely proportional to the period -- longer signals yield finer frequency resolution.

Remark.

What each part of the does (CDFT forward):

x^(f)=τ=0P1Px(τ)ei2πfτdτ,f{kP,kZ}\hat{x}(f) = \int_{\tau=0}^{P} \frac{1}{\sqrt{P}} x(\tau) \, e^{-i2\pi f \tau} \, d\tau, \quad f \in \left\{\frac{k}{P},\, k \in \mathbb{Z}\right\}

  • x(τ)x(\tau) -- the original continuous-time signal defined on the interval [0,P][0, P]
  • ei2πfτe^{-i2\pi f \tau} -- a complex sinusoid at frequency ff, acting as the detector. By Euler's formula, ei2πfτ=cos(2πfτ)isin(2πfτ)e^{-i2\pi f \tau} = \cos(2\pi f \tau) - i\sin(2\pi f \tau)
  • x(τ)ei2πfτx(\tau) \cdot e^{-i2\pi f \tau} -- the product of the signal with the detector. When the signal has energy at frequency ff, this product has a non-zero average over the interval
  • 1P\frac{1}{\sqrt{P}} -- normalization factor. Dividing by P\sqrt{P} makes the transform unitary, so energy is preserved between the time and frequency representations
  • 0Pdτ\int_0^P d\tau -- integration over one full period. This is the continuous analogue of summation: it accumulates the correlation between the signal and the detector sinusoid. If frequency ff is present, the integral is large; if not, positive and negative contributions cancel out
  • x^(f)\hat{x}(f) -- the Fourier coefficient at frequency f=k/Pf = k/P. Note that the output is discrete: you only get coefficients at the harmonic frequencies 0,±1/P,±2/P,0, \pm 1/P, \pm 2/P, \ldots The magnitude x^(f)|\hat{x}(f)| gives the amplitude of harmonic kk, and the angle gives its phase

The inverses of these can also be defined:

DefinitionInverse DDFT
FDD1:l2({0/N,1/N,,(N1)/N};C)l2({0,1,,N1};C)\mathcal{F}_{DD}^{-1} : l_2(\{0/N, 1/N, \ldots, (N-1)/N\}; \mathbb{C}) \to l_2(\{0, 1, \ldots, N-1\}; \mathbb{C})x=FDD1(x^)x = \mathcal{F}_{DD}^{-1}(\hat{x})

and

x(n)=k=0N11Nx^ ⁣(kN)ei2πkNn,n{0,1,,N1}x(n) = \sum_{k=0}^{N-1} \frac{1}{\sqrt{N}} \hat{x}\!\left(\frac{k}{N}\right) e^{i2\pi \frac{k}{N} n}, \quad n \in \{0, 1, \ldots, N-1\}
Remark.

Intuition: The inverse DDFT reconstructs the original time-domain samples from their frequency coefficients. The fact that this inverse exists and perfectly recovers the signal means no information is lost when transforming to the frequency domain -- the two representations are equivalent.

Remark.

What each part of the does (Inverse DDFT -- Synthesis):

x(n)=k=0N11Nx^ ⁣(kN)ei2πkNn,n{0,1,,N1}x(n) = \sum_{k=0}^{N-1} \frac{1}{\sqrt{N}} \hat{x}\!\left(\frac{k}{N}\right) e^{i2\pi \frac{k}{N} n}, \quad n \in \{0, 1, \ldots, N-1\}

  • x^(k/N)\hat{x}(k/N) -- the Fourier coefficient at frequency k/Nk/N. This complex number encodes the amplitude and phase of the kk-th harmonic component. It controls how much of that frequency to add back
  • ei2πkNne^{i2\pi \frac{k}{N} n} -- a complex sinusoid at frequency k/Nk/N, evaluated at time nn. Note the positive sign in the exponent (compare with ei2πfne^{-i2\pi f n} in the forward transform). This generates the actual oscillation at frequency k/Nk/N
  • x^(k/N)ei2πkNn\hat{x}(k/N) \cdot e^{i2\pi \frac{k}{N} n} -- one frequency component, weighted by its coefficient. Each term in the sum adds one sinusoidal "ingredient" back into the signal
  • 1N\frac{1}{\sqrt{N}} -- the same normalization as the forward transform, ensuring unitarity
  • k=0N1\sum_{k=0}^{N-1} -- sum over all NN frequency components. By adding all the weighted oscillations together, you reconstruct the original signal sample x(n)x(n) exactly. This is the synthesis step: the frequency recipe is turned back into a signal
DefinitionInverse CDFT
FCD1:l2 ⁣(Z1P;C)L2([0,P];C)\mathcal{F}_{CD}^{-1} : l_2\!\left(\mathbb{Z}\frac{1}{P}; \mathbb{C}\right) \to L_2([0, P]; \mathbb{C})x=FCD1(x^)x = \mathcal{F}_{CD}^{-1}(\hat{x})

and

x(t)=kZ1Px^ ⁣(kP)ei2πkPtdt,t[0,P]x(t) = \sum_{k \in \mathbb{Z}} \frac{1}{\sqrt{P}} \hat{x}\!\left(\frac{k}{P}\right) e^{i2\pi \frac{k}{P} t}\, dt, \quad t \in [0, P]
Remark.

Intuition: The inverse CDFT synthesizes a continuous-time signal from its countably many Fourier coefficients. Each coefficient x^(k/P)\hat{x}(k/P) weights a complex exponential at frequency k/Pk/P, and the sum of all these weighted harmonics reconstructs the original signal. This is the synthesis formula of the Fourier series.

Remark.

What each part of the does (Inverse CDFT -- Synthesis):

x(t)=kZ1Px^ ⁣(kP)ei2πkPt,t[0,P]x(t) = \sum_{k \in \mathbb{Z}} \frac{1}{\sqrt{P}} \hat{x}\!\left(\frac{k}{P}\right) e^{i2\pi \frac{k}{P} t}, \quad t \in [0, P]

  • x^(k/P)\hat{x}(k/P) -- the Fourier coefficient at frequency k/Pk/P. This tells you the amplitude and phase of the kk-th harmonic. It controls how much of that frequency to include in the reconstruction
  • ei2πkPte^{i2\pi \frac{k}{P} t} -- a complex sinusoid at frequency k/Pk/P, now as a continuous function of time tt. The positive sign in the exponent (vs. negative in the forward CDFT) indicates this is synthesis, not analysis
  • x^(k/P)ei2πkPt\hat{x}(k/P) \cdot e^{i2\pi \frac{k}{P} t} -- one harmonic component, weighted by its Fourier coefficient. Each term in the sum contributes one "pure tone" to the reconstructed signal
  • 1P\frac{1}{\sqrt{P}} -- normalization matching the forward transform
  • kZ\sum_{k \in \mathbb{Z}} -- sum over all integer harmonics kk, from -\infty to ++\infty. Unlike the finite DDFT inverse, this sum is infinite -- you need infinitely many harmonics to perfectly reconstruct a continuous-time signal from its Fourier series. In practice, the partial sums converge in the L2L_2 sense

Properties of the Discrete Fourier Transforms

TheoremParseval's Equality

The transformations FDD\mathcal{F}_{DD} and FCD\mathcal{F}_{CD} are unitary, that is:

x,x=x^,x^\langle x, x \rangle = \langle \hat{x}, \hat{x} \rangle
Remark.

Intuition: Parseval's equality says that energy is conserved when you transform between time and frequency domains. The total energy of a signal, computed by summing/integrating its squared amplitude over time, equals the total energy computed from its frequency coefficients. This is why the Fourier transform is called a "unitary" or "isometric" transformation -- it preserves distances and energies.

Remark.

Time-shift, periodicity and differentiation will also be discussed. An important property is with regard to convolution. Suppose that in the following, we write f(t)=f(tmodP)f(t) = f(t \mod P), in which case the convolution operation is also termed cyclical convolution.

TheoremConvolution Theorem for CDFT

Let f,gL2([0,P])f, g \in L_2([0, P]). Then,

FCD(fg)(k)=PFCD(f)(k)FCD(g)(k),kZ ⁣(1P).\mathcal{F}_{CD}(f * g)(k) = \sqrt{P}\, \mathcal{F}_{CD}(f)(k)\, \mathcal{F}_{CD}(g)(k), \quad k \in \mathbb{Z}\!\left(\frac{1}{P}\right).

That is, convolution is equivalent to point-wise multiplication in the frequency domain.

Remark.

Intuition: The convolution theorem is one of the most practically important results in signal processing. Computing a convolution directly requires integrating a product over all time shifts, but in the frequency domain it reduces to simple pointwise multiplication. This is why filtering is often done by transforming to the frequency domain, multiplying, and transforming back -- especially with the FFT, this is far more efficient.

Worked Exercise

ExampleFourier Series of Indicator Functions

Exercise.

a) For some NNN \in \mathbb{N}, let xl2((N,(N1),,N1,N);C)x \in l_2\big((-N, -(N-1), \ldots, N-1, N); \mathbb{C}\big) with x(n)=1{nN1}x(n) = 1_{\{|n| \le N_1\}}. Find the Fourier series expansion of xx. Study the case with N1=NN_1 = N, and the case with N1=0N_1 = 0.

b) For some TR+T \in \mathbb{R}_+, let xL2([T2,T2];C)x \in L_2([-\frac{T}{2}, \frac{T}{2}]; \mathbb{C}) with x(t)=1{tT1}x(t) = 1_{\{|t| \le T_1\}}. Find the Fourier series expansion. Study the cases with (i) T1=T2T_1 = \frac{T}{2}, and (ii) the case where, with T1=12nT_1 = \frac{1}{2n}, we have xn(t)=n1{t12n}x_n(t) = n \cdot 1_{\{|t| \le \frac{1}{2n}\}}, as nn \to \infty.

Solution. a) We have for k=0,1,,2N+1k = 0, 1, \ldots, 2N+1:

x^ ⁣(k2N+1)=n=N1N112N+1ei2πk2N+1n=12N+1ei2πk2N+1N1n=02N1ei2πk2N+1n\hat{x}\!\left(\frac{k}{2N+1}\right) = \sum_{n=-N_1}^{N_1} \frac{1}{2N+1} e^{-i\frac{2\pi k}{2N+1}n} = \frac{1}{2N+1} e^{-i\frac{2\pi k}{2N+1}N_1} \sum_{n=0}^{2N_1} e^{-i\frac{2\pi k}{2N+1}n}

For k0k \neq 0, we have:

x^ ⁣(k2N+1)=12N+1ei2πkN12N+11ei2π(2N1+1)2N+11ei2πk2N+1\hat{x}\!\left(\frac{k}{2N+1}\right) = \frac{1}{2N+1} e^{-i\frac{2\pi k N_1}{2N+1}} \frac{1 - e^{-i\frac{2\pi(2N_1+1)}{2N+1}}}{1 - e^{-i\frac{2\pi k}{2N+1}}}

For k=0k = 0, we have:

x^ ⁣(02N+1)=2N1+12N+1\hat{x}\!\left(\frac{0}{2N+1}\right) = \frac{2N_1 + 1}{\sqrt{2N+1}}

If N1=NN_1 = N: For k0k \neq 0, x^ ⁣(k2N+1)=0\hat{x}\!\left(\frac{k}{2N+1}\right) = 0, and for k=0k = 0, x^ ⁣(02N+1)=2N+1\hat{x}\!\left(\frac{0}{2N+1}\right) = \sqrt{2N+1}.

For the other extreme, if N1=0N_1 = 0, we have for all k{0,,2N+1}k \in \{0, \ldots, 2N+1\}: x^ ⁣(k2N+1)=12N+1\hat{x}\!\left(\frac{k}{2N+1}\right) = \frac{1}{\sqrt{2N+1}}.

b) We have the expansion (in the L2L_2-sense):

x(t)=kZx^ ⁣(kT)1Tei2πkTtx(t) = \sum_{k \in \mathbb{Z}} \hat{x}\!\left(\frac{k}{T}\right) \frac{1}{\sqrt{T}} e^{i2\pi \frac{k}{T} t}

with

x^ ⁣(kT)=T/2T/2x(t)1Tei2πkTtdt\hat{x}\!\left(\frac{k}{T}\right) = \int_{-T/2}^{T/2} x(t) \frac{1}{\sqrt{T}} e^{-i\frac{2\pi k}{T} t}\, dt

Thus, with xx as given, we have:

x^ ⁣(kT)=T1T11Tei2πkTtdt=ei2πkT1Tei2πkT1Ti2πk/T=sin(2πkT1T)2πk1T2T1T\hat{x}\!\left(\frac{k}{T}\right) = \int_{-T_1}^{T_1} \frac{1}{\sqrt{T}} e^{-i\frac{2\pi k}{T} t}\, dt = \frac{e^{i\frac{2\pi k T_1}{T}} - e^{-i\frac{2\pi k T_1}{T}}}{i2\pi k / \sqrt{T}} = \frac{\sin(2\pi k \frac{T_1}{T})}{2\pi k \frac{1}{T}} \frac{2T_1}{\sqrt{T}}

If T1=T2T_1 = \frac{T}{2}, we have that x^(kT)=0\hat{x}(\frac{k}{T}) = 0 for all k0k \neq 0 and x^(0T)=T\hat{x}(\frac{0}{T}) = \sqrt{T}.

If T1=12nT_1 = \frac{1}{2n}, with xn(t)=n1{tT1}x_n(t) = n \cdot 1_{\{|t| \le T_1\}}, then we observe that x^n(kT)1T\hat{x}_n(\frac{k}{T}) \to \frac{1}{\sqrt{T}} for all kZk \in \mathbb{Z}.

We observe from these examples that, as a general insight (which can be made more rigorous under additional conditions), if we expand the signal in time domain, we shrink it in the frequency domain; and if we shrink it in time domain, we expand it in the frequency domain.

Computational Aspects: The FFT Algorithm

The Fast Fourier Transform (FFT) is a very important algorithm to implement DTFT (FDD\mathcal{F}_{DD}) in a computationally efficient fashion in practice. The fft command in Matlab generates the transform.

Observe that for the operations described in the DDFT definition:

FDD:l2({0,1,,N1};C)l2 ⁣({0N,1N,,N1N};C)\mathcal{F}_{DD} : l_2(\{0, 1, \ldots, N-1\}; \mathbb{C}) \to l_2\!\left(\left\{\frac{0}{N}, \frac{1}{N}, \ldots, \frac{N-1}{N}\right\}; \mathbb{C}\right)

so that x^=FDD(x)\hat{x} = \mathcal{F}_{DD}(x), with

x^(f)=n=0N11Nx(n)ei2πfn,f{0N,1N,,N1N},\hat{x}(f) = \sum_{n=0}^{N-1} \frac{1}{\sqrt{N}} x(n) e^{-i2\pi f n}, \quad f \in \left\{\frac{0}{N}, \frac{1}{N}, \ldots, \frac{N-1}{N}\right\},

there are NN complex multiplications and NN complex additions for each ff (and thus there will be NN such computations). Thus, the FFT algorithm in the form above has the computational complexity of N2N^2 complex operations (with one complex operation being equivalent to one complex addition and one complex multiplication).

If NN is even, we can write

x^ ⁣(kN)=n=0N11Nx(n)ei2πkNn\hat{x}\!\left(\frac{k}{N}\right) = \sum_{n=0}^{N-1} \frac{1}{\sqrt{N}} x(n) e^{-i2\pi \frac{k}{N} n} =m=0N/211Nx(2m)ei2πkN2m+m=0N/211Nx(2m+1)ei2πkN(2m+1)= \sum_{m=0}^{N/2 - 1} \frac{1}{\sqrt{N}} x(2m) e^{-i2\pi \frac{k}{N} 2m} + \sum_{m=0}^{N/2 - 1} \frac{1}{\sqrt{N}} x(2m+1) e^{-i2\pi \frac{k}{N}(2m+1)} =m=0N/211Nx(2m)ei2πkN/2m+(m=0N/211Nx(2m+1)ei2πkN/2m)ei2πkN= \sum_{m=0}^{N/2 - 1} \frac{1}{\sqrt{N}} x(2m) e^{-i2\pi \frac{k}{N/2} m} + \left(\sum_{m=0}^{N/2 - 1} \frac{1}{\sqrt{N}} x(2m+1) e^{-i2\pi \frac{k}{N/2} m}\right) e^{-i2\pi \frac{k}{N}}

Define x0(m)=x(2m)x_0(m) = x(2m) and x1(m)=x(2m+1)x_1(m) = x(2m+1), for m=0,1,,N21m = 0, 1, \ldots, \frac{N}{2} - 1. Then, the above leads to

x^ ⁣(kN)=12x^0 ⁣(kN/2)+ei2πkN12x^1 ⁣(kN/2)\hat{x}\!\left(\frac{k}{N}\right) = \frac{1}{\sqrt{2}} \hat{x}_0\!\left(\frac{k}{N/2}\right) + e^{-i2\pi \frac{k}{N}} \frac{1}{\sqrt{2}} \hat{x}_1\!\left(\frac{k}{N/2}\right)

Thus, as we see, the above leads to a parallel processing of two smaller length transforms. If NN is a power of 2, we can continue with this approach to a building block of length N=2N = 2. By inductively splitting the summations in the expansions as above, the FFT algorithm then reduces the computational complexity for the FDD\mathcal{F}_{DD} from N2N^2 complex operations to Nlog2(N)N \log_2(N) (with NN being a power of 2) such operations.

Remark.

The FFT algorithm achieves a dramatic speedup: for N=1024N = 1024, the naive DFT requires about 10610^6 operations, while the FFT requires only about 10410^4. This makes real-time signal processing practical.

The Discrete-to-Continuous Fourier Transform (DCFT): FDC\mathcal{F}_{DC}

The Discrete-to-Continuous Fourier Transform can be viewed as the inverse of FCD\mathcal{F}_{CD} (with PP taken to be 1 and a negation in the index):

DefinitionDCFT

A signal xl2(Z;R)x \in l_2(\mathbb{Z}; \mathbb{R}) can be expanded as:

x(n)=01x^(f)ei2πfndf,nZx(n) = \int_0^1 \hat{x}(f) e^{i2\pi f n}\, df, \quad n \in \mathbb{Z}

with

x^(f)=nZx(n)ei2πfn,f[0,1)\hat{x}(f) = \sum_{n \in \mathbb{Z}} x(n) e^{-i2\pi f n}, \quad f \in [0, 1)

We write x^=FDC(x)\hat{x} = \mathcal{F}_{DC}(x).

Remark.

The DCFT maps a discrete-time signal to a continuous function of frequency. The output x^(f)\hat{x}(f) is periodic with period 1. This transform is the workhorse of discrete-time signal processing -- it takes a sequence of numbers and produces a continuous frequency spectrum.

Remark.

What each part of the does (DCFT forward):

x^(f)=nZx(n)ei2πfn,f[0,1)\hat{x}(f) = \sum_{n \in \mathbb{Z}} x(n) \, e^{-i2\pi f n}, \quad f \in [0, 1)

  • x(n)x(n) -- the original discrete-time signal: an infinite sequence of numbers indexed by all integers nZn \in \mathbb{Z}
  • ei2πfne^{-i2\pi f n} -- a complex sinusoid at frequency ff, evaluated at discrete time nn. By Euler's formula, ei2πfn=cos(2πfn)isin(2πfn)e^{-i2\pi f n} = \cos(2\pi f n) - i\sin(2\pi f n). This is the frequency detector, just as in the other transforms
  • x(n)ei2πfnx(n) \cdot e^{-i2\pi f n} -- the product of the signal sample with the detector at that sample. If xx oscillates at frequency ff, these products reinforce rather than cancel
  • nZ\sum_{n \in \mathbb{Z}} -- sum over all integer time indices, from -\infty to ++\infty. This replaces integration (since the signal is discrete) and replaces the finite sum of the DDFT (since the signal is now infinite-length)
  • x^(f)\hat{x}(f) -- the result is a continuous function of frequency f[0,1)f \in [0,1), and it is periodic with period 1. The output lives on a continuous interval because the input is an infinite-length sequence -- more input data yields a denser (continuous) frequency representation. The magnitude x^(f)|\hat{x}(f)| gives the spectral amplitude at frequency ff, and the angle gives the phase

Note that there is no explicit 1/P1/\sqrt{P} normalization here because the convention uses P=1P = 1 and the normalization is absorbed into the inverse transform's integration.

Remark.

What each part of the does (DCFT inverse -- Synthesis):

x(n)=01x^(f)ei2πfndf,nZx(n) = \int_0^1 \hat{x}(f) \, e^{i2\pi f n} \, df, \quad n \in \mathbb{Z}

  • x^(f)\hat{x}(f) -- the continuous frequency spectrum. For each frequency ff, this complex-valued function tells you the amplitude and phase of that component
  • ei2πfne^{i2\pi f n} -- a complex sinusoid at frequency ff, evaluated at discrete time nn. The positive sign (vs. negative in the forward transform) signals that this is synthesis
  • x^(f)ei2πfn\hat{x}(f) \cdot e^{i2\pi f n} -- one infinitesimal frequency component, weighted by the spectrum. Each value of ff contributes a tiny sinusoidal piece to the reconstructed signal
  • 01df\int_0^1 df -- integration over all frequencies in one period [0,1)[0,1). Since the spectrum is continuous, reconstruction requires integrating (not summing) over all frequency components. This integral adds up a continuum of weighted oscillations to recover x(n)x(n)
  • x(n)x(n) -- the reconstructed signal sample at time nn, perfectly recovered from the continuous spectrum

The CCFT: FCC\mathcal{F}_{CC} on S\mathcal{S} and its Extension to L2(R)L_2(\mathbb{R})

We will define the CCFT by the relation

DefinitionCCFT
x(t)=f=x^(f)ei2πftdf,fRx(t) = \int_{f=-\infty}^{\infty} \hat{x}(f) e^{i2\pi f t}\, df, \quad f \in \mathbb{R}

with

x^(f)=τ=x(t)ei2πftdt,fR\hat{x}(f) = \int_{\tau=-\infty}^{\infty} x(t) e^{-i2\pi f t}\, dt, \quad f \in \mathbb{R}

We write x^=FCC(x)\hat{x} = \mathcal{F}_{CC}(x).

Remark.

Intuition: The CCFT is the "fully continuous" Fourier transform: it maps a continuous-time signal defined on all of R\mathbb{R} to a continuous frequency spectrum on all of R\mathbb{R}. Unlike the Fourier series (which produces discrete frequencies), the CCFT produces a continuous frequency density. This is the appropriate transform for non-periodic signals of finite energy, and it is the transform most commonly meant when engineers simply say "the Fourier transform."

Remark.

What each part of the does (CCFT forward):

x^(f)=τ=x(t)ei2πftdt,fR\hat{x}(f) = \int_{\tau=-\infty}^{\infty} x(t) \, e^{-i2\pi f t} \, dt, \quad f \in \mathbb{R}

  • x(t)x(t) -- the original signal in continuous time, defined for all tRt \in \mathbb{R}
  • ei2πfte^{-i2\pi ft} -- a complex sinusoid at frequency ff (the "detector"). By Euler's formula, this equals cos(2πft)isin(2πft)\cos(2\pi ft) - i\sin(2\pi ft). It probes the signal for content at frequency ff
  • x(t)ei2πftx(t) \cdot e^{-i2\pi ft} -- multiplying the signal by the detector. If the signal contains frequency ff, this product will have a steady (non-cancelling) component
  • dt\int_{-\infty}^{\infty} dt -- integrating (summing up) over all time. If frequency ff is present, the integral is large; if not, positive and negative parts cancel and the integral is small or zero
  • x^(f)\hat{x}(f) -- the result: a complex number for each frequency fRf \in \mathbb{R}, forming a continuous spectrum. The magnitude x^(f)|\hat{x}(f)| tells you the amplitude at frequency ff, and the angle x^(f)\angle \hat{x}(f) tells you the phase shift of that component

This is the "fully continuous" version: continuous time in, continuous frequency out. Both the signal and its transform live on R\mathbb{R}, and neither is assumed periodic.

Remark.

What each part of the does (CCFT inverse -- Synthesis):

x(t)=f=x^(f)ei2πftdf,tRx(t) = \int_{f=-\infty}^{\infty} \hat{x}(f) \, e^{i2\pi f t} \, df, \quad t \in \mathbb{R}

  • x^(f)\hat{x}(f) -- the continuous frequency spectrum (the "frequency recipe"). For each frequency ff, this complex number encodes the amplitude and phase of that component
  • ei2πfte^{i2\pi ft} -- a complex sinusoid at frequency ff. Note the positive sign in the exponent (vs. ei2πfte^{-i2\pi ft} in the forward transform). This generates the actual oscillation at frequency ff
  • x^(f)ei2πft\hat{x}(f) \cdot e^{i2\pi ft} -- one infinitesimal frequency component, weighted by the spectrum. The coefficient x^(f)\hat{x}(f) controls how much of frequency ff to add back
  • df\int_{-\infty}^{\infty} df -- integrating over all frequencies. Since the spectrum is continuous, we sum a continuum of weighted oscillations. Each infinitesimal frequency band [f,f+df][f, f+df] contributes x^(f)ei2πftdf\hat{x}(f) e^{i2\pi ft} df to the total signal
  • x(t)x(t) -- the reconstructed signal at time tt, perfectly recovered by summing all frequency contributions

The synthesis integral assembles the original signal from its frequency ingredients, just as the forward transform decomposed it. The two operations are perfect inverses of each other.

Two important properties are given in the following.

TheoremDifferentiation Properties of CCFT

For ϕS\phi \in \mathcal{S} and mZ+m \in \mathbb{Z}_+:

a) FCC ⁣(dmdtmϕ) ⁣(f)=(i2πf)mFCC(ϕ)(f)\displaystyle \mathcal{F}_{CC}\!\left(\frac{d^m}{dt^m} \phi\right)\!(f) = (i2\pi f)^m \mathcal{F}_{CC}(\phi)(f).

b) dmdfmFCC(ϕ)(f)=FCC ⁣((i2πt)mϕ) ⁣(f)\displaystyle \frac{d^m}{df^m} \mathcal{F}_{CC}(\phi)(f) = \mathcal{F}_{CC}\!\left((-i2\pi t)^m \phi\right)\!(f)

Remark.

Theorem is one of the most powerful properties of the Fourier transform. Part (a) says that differentiation in time corresponds to multiplication by i2πfi2\pi f in frequency. Part (b) says that multiplication by tt in time corresponds to differentiation in frequency (up to a constant). These properties are what make the Fourier transform so useful for solving differential equations.

TheoremCCFT maps Schwartz space to itself

FCC\mathcal{F}_{CC} is a continuous linear map on S\mathcal{S} to S\mathcal{S}.

Remark.

Intuition: Schwartz functions are "maximally well-behaved" signals -- they are infinitely smooth and decay faster than any polynomial. This theorem says the Fourier transform preserves this niceness: transforming a Schwartz function gives another Schwartz function. This makes the Schwartz space a natural starting point for Fourier analysis, from which results can then be extended to larger spaces like L2L_2.

We will see in the following that the CCFT is a transformation also from L2(R;C)L2(R;C)L_2(\mathbb{R}; \mathbb{C}) \to L_2(\mathbb{R}; \mathbb{C}).

The Inverse Transform

The following is known as the Riemann-Lebesgue lemma.

TheoremRiemann-Lebesgue Lemma

Let gL1(R;R)g \in L_1(\mathbb{R}; \mathbb{R}). Then,

limfg(t)ei2πftdt=0\lim_{f \to \infty} \int g(t) e^{-i2\pi f t}\, dt = 0
Remark.

Intuition: The Riemann-Lebesgue lemma says that the Fourier transform of any integrable signal decays to zero at extreme frequencies. Physically, this means an L1L_1 signal cannot have unbounded high-frequency content -- eventually, the frequency components must die out. This is a fundamental regularity property of the Fourier transform.

Lemma

The following holds for all ϕS\phi \in \mathcal{S}:

limnsin(nx)πxϕ(x)dxδ~(ϕ)=ϕ(0)\lim_{n \to \infty} \int \frac{\sin(nx)}{\pi x}\, \phi(x)\, dx \to \tilde{\delta}(\phi) = \phi(0)
Remark.

Intuition: This lemma shows that the "sinc-like" function sin(nx)πx\frac{\sin(nx)}{\pi x} behaves like a Dirac delta as nn grows large -- it concentrates all its mass at the origin and "picks out" the value ϕ(0)\phi(0). This is the key technical fact that makes the Fourier inversion formula work: the integral of the inverse transform converges to the original signal value.

With this discussion, we can show that the inverse FCC1\mathcal{F}_{CC}^{-1} is defined on S\mathcal{S} as:

FCC1(ϕ^):=ϕ^(f)ei2πftdf.\mathcal{F}_{CC}^{-1}(\hat{\phi}) := \int \hat{\phi}(f) e^{i2\pi f t}\, df.

Observe that

ϕ^(f)ei2πftdf\int_{-\infty}^{\infty} \hat{\phi}(f) e^{i2\pi f t}\, df =limAf=AAϕ^(f)ei2πftdf= \lim_{A \to \infty} \int_{f=-A}^{A} \hat{\phi}(f) e^{i2\pi f t}\, df =limAf=AA(ϕ(τ)ei2πfτdτ)ei2πftdf= \lim_{A \to \infty} \int_{f=-A}^{A} \left(\int \phi(\tau) e^{-i2\pi f \tau}\, d\tau\right) e^{i2\pi f t}\, df =limAϕ(τ)f=AAei2πfτei2πftdfdτ= \lim_{A \to \infty} \int \phi(\tau) \int_{f=-A}^{A} e^{-i2\pi f \tau} e^{i2\pi f t}\, df\, d\tau =limAϕ(τ)(f=AAei2πf(τt)df)dτ= \lim_{A \to \infty} \int \phi(\tau) \left(\int_{f=-A}^{A} e^{-i2\pi f(\tau - t)}\, df\right) d\tau

For every fixed AA, we can justify the above by (Fubini's) Theorem A.3.1. By Lemma, the fact that

f=AAei2πf(τt)df=1π(tτ)sin(2Aπ(tτ)),\int_{f=-A}^{A} e^{-i2\pi f(\tau - t)}\, df = \frac{1}{\pi(t - \tau)} \sin(2A\pi(t - \tau)),

represents a distribution which converges to the δ~t\tilde{\delta}_t distribution (which then satisfies ϕ(τ)δ~t(τ)dτϕ(τ)δ(tτ)dτ=ϕ(t)\int \phi(\tau)\tilde{\delta}_t(\tau)\, d\tau \equiv \int \phi(\tau)\delta(t - \tau)\, d\tau = \phi(t)) leads to the desired result.

Plancherel's Identity / Parseval's Theorem

TheoremPlancherel's Identity / Parseval's Theorem

For every h,gSh, g \in \mathcal{S} (as well as L2(R;C)L_2(\mathbb{R}; \mathbb{C})), the Fourier transforms f^,g^\hat{f}, \hat{g} satisfy:

h^,g^=h,g\langle \hat{h}, \hat{g} \rangle = \langle h, g \rangle
Remark.

When h=gh = g, Plancherel's identity reduces to Parseval's theorem: h^2=h2\|\hat{h}\|^2 = \|h\|^2. This says that the Fourier transform preserves energy -- the total energy of a signal computed in the time domain equals the total energy computed in the frequency domain. In engineering terms, the Fourier transform is an isometry (distance-preserving map).

Extension of FCC\mathcal{F}_{CC} on L2(R;C)L_2(\mathbb{R}; \mathbb{C}) and Plancherel's Theorem

The above discussion applies for h,gSh, g \in \mathcal{S} (where we also allow for C\mathbb{C}-valued functions in S\mathcal{S}). We will show that one can extend the Fourier transform for signals in L2(R;C)L_2(\mathbb{R}; \mathbb{C}). Furthermore, as not every function in L2(R)L_2(\mathbb{R}) is also in L1(R)L_1(\mathbb{R}) (for example: f(t)=11+t2f(t) = \frac{1}{\sqrt{1+t^2}} is in L2L_2 but not in L1L_1), one cannot define a Fourier transform of a general function in L2(R)L_2(\mathbb{R}) pointwise directly by an integral.

Recall from Theorem, and the discussions following it, that continuous functions are dense in integrable functions, and from Corollary one can approximate a continuous function by its convolution with a smooth approximate identity sequence leading to a smooth approximation. These lead to the following.

TheoremDensity of Schwartz Space

S\mathcal{S} is dense in L2(R)L_2(\mathbb{R}).

Remark.

Intuition: This density result is what allows us to extend the Fourier transform from Schwartz functions (where everything converges nicely) to all of L2(R)L_2(\mathbb{R}). Since every L2L_2 signal can be approximated by Schwartz functions, we can define its Fourier transform as the limit of the transforms of the approximating Schwartz functions.

With this theorem, our goal will be to define the CCFT of a signal ff in L2(R;C)L_2(\mathbb{R}; \mathbb{C}) as follows. Let {ϕn}\{\phi_n\} be a sequence of functions in S\mathcal{S} converging to ff (This is possible by the denseness of S\mathcal{S} in L2L_2). We define the CCFT of ff as the L2L_2 limit of a sequence of CCFTs of {ϕn}\{\phi_n\}. Such an extension defines the unique extension of FCC\mathcal{F}_{CC} from S\mathcal{S} to S\mathcal{S} which is continuous on L2L_2. This result builds on the following theorem.

TheoremBounded Linear Extension

Let T:MYT : M \to Y be a linear mapping, YY a Banach space, MM a dense linear subspace of a normed linear space XX. Furthermore, let TT be bounded in the sense that

T=supxM,x0Txx<.\|T\| = \sup_{x \in M, x \neq 0} \frac{\|Tx\|}{\|x\|} < \infty.

Then, there exists a unique extension of TT on XX to YY, denoted by Tˉ:XY\bar{T} : X \to Y, such that Tˉ(x)=T(x)\bar{T}(x) = T(x) for xMx \in M (that is, TT and Tˉ\bar{T} are in agreement on MM). Furthermore, with

Tˉ:=supxX,x0Tˉxx,\|\bar{T}\| := \sup_{x \in X, x \neq 0} \frac{\|\bar{T}x\|}{\|x\|},

we have T=Tˉ\|T\| = \|\bar{T}\|.

Remark.

Intuition: This theorem is the general machine for extending operators from a dense subspace to the entire space. If a linear map is bounded (continuous) on a dense subset of a normed space, it can be uniquely extended to the whole space without increasing its norm. This is precisely how the CCFT is extended from the Schwartz space to all of L2L_2 -- the Fourier transform is bounded on S\mathcal{S} (by Parseval), so it extends uniquely and continuously to L2L_2.

Now, let M=SM = \mathcal{S}, X=Y=L2(R;C)X = Y = L_2(\mathbb{R}; \mathbb{C}), T=FCCT = \mathcal{F}_{CC} and note by Theorem that FCC=1\|\mathcal{F}_{CC}\| = 1 (when viewed as a mapping from a subset of L2(R;C)L_2(\mathbb{R}; \mathbb{C})). In view of this, we define FCC\mathcal{F}_{CC} on L2(R;C)L_2(\mathbb{R}; \mathbb{C}) to be the unique extension of FCC\mathcal{F}_{CC} from SL2(R;C)\mathcal{S} \to L_2(\mathbb{R}; \mathbb{C}). Thus, for hL2(R;C)h \in L_2(\mathbb{R}; \mathbb{C}),

h,h=h^,h^,h^=FCC(h).\langle h, h \rangle = \langle \hat{h}, \hat{h} \rangle, \quad \hat{h} = \mathcal{F}_{CC}(h).

In view of the above, we have that Plancherel's Theorem also applies to signals in L2(R;C)L_2(\mathbb{R}; \mathbb{C}).

g,h=g^,h^,h,gS\langle g, h \rangle = \langle \hat{g}, \hat{h} \rangle, \quad h, g \in \mathcal{S}

that

FCC=1.\|\mathcal{F}_{CC}\| = 1.

By the stated (extension) theorem, it follows that FCC\mathcal{F}_{CC} is also unitary on L2(R;C)L_2(\mathbb{R}; \mathbb{C}) with the same operator norm, that is:

g,h=g^,h^,h,gL2(R;C).\langle g, h \rangle = \langle \hat{g}, \hat{h} \rangle, \quad h, g \in L_2(\mathbb{R}; \mathbb{C}).

To verify this, let gg and hh be in L2(R;C)L_2(\mathbb{R}; \mathbb{C}) and gngg_n \to g, and let hnhh_n \to h with gn,hnSg_n, h_n \in \mathcal{S} being Schwartz signals. By Plancherel's identity:

gn,hn=g^n,h^n\langle g_n, h_n \rangle = \langle \hat{g}_n, \hat{h}_n \rangle

Let us take the limit as nn \to \infty on both sides. For the left hand side the Cauchy-Schwarz inequality implies that

limgn,hn=g,h,\lim \langle g_n, h_n \rangle = \langle g, h \rangle,

because

gn,hng,h=gng,hn+g,hnhgnghn+ghnh0|\langle g_n, h_n \rangle - \langle g, h \rangle| = |\langle g_n - g, h_n \rangle + \langle g, h_n - h \rangle| \leq \|g_n - g\|\|h_n\| + \|g\|\|h_n - h\| \to 0

Likewise

g^n,h^ng^,h^=g^ng^,h^n+g^,h^nh^g^ng^h^n+g^h^nh^0|\langle \hat{g}_n, \hat{h}_n \rangle - \langle \hat{g}, \hat{h} \rangle| = |\langle \hat{g}_n - \hat{g}, \hat{h}_n \rangle + \langle \hat{g}, \hat{h}_n - \hat{h} \rangle| \leq \|\hat{g}_n - \hat{g}\|\|\hat{h}_n\| + \|\hat{g}\|\|\hat{h}_n - \hat{h}\| \to 0

Here, the convergence to zero follows from the fact that g^ng^0\|\hat{g}_n - \hat{g}\| \to 0 (since gng=g^ng^0\|g_n - g\| = \|\hat{g}_n - \hat{g}\| \to 0 by the discussion above) and that h^n\|\hat{h}_n\| is bounded. This generalizes Plancherel's identity for signals in L2(R;C)L_2(\mathbb{R}; \mathbb{C}).

Fourier Transform of Distributions (FCC\mathcal{F}_{CC} on S\mathcal{S}^*)

Recall that S\mathcal{S}^* is the dual space on S\mathcal{S}, that is the space of distributions (linear and continuous functions) on S\mathcal{S}.

DefinitionFourier Transform of a Distribution

The Fourier transform of a distribution is defined by the following relation: Let TST \in \mathcal{S}^*. Then, with T^=FCC(T)\hat{T} = \mathcal{F}_{CC}(T), we have

T^(ϕ)=T(ϕ^)ϕS\hat{T}(\phi) = T(\hat{\phi}) \quad \phi \in \mathcal{S}

This definition is consistent with the Fourier transform of a regular distribution (represented by some function ψS\psi \in \mathcal{S}) being a distribution which is represented by the Fourier transform of ψ\psi. That is,

ψ^(f)ϕ(f)df=ψ(t)ϕ^(t)dt.\int \hat{\psi}(f) \phi(f)\, df = \int \psi(t) \hat{\phi}(t)\, dt.

This equality follows from Fubini's theorem by expressing ψ^(f)=ψ(t)ei2πftdt\hat{\psi}(f) = \int \psi(t) e^{-i2\pi f t}\, dt.

Remark.

Intuition: Defining the Fourier transform of a distribution via T^(ϕ)=T(ϕ^)\hat{T}(\phi) = T(\hat{\phi}) is an elegant "trick": instead of trying to transform the distribution directly (which may not be a regular function), we transform the test function and let the distribution act on the result. This allows us to take Fourier transforms of objects like the Dirac delta, step functions, and periodic signals that do not have classical Fourier transforms.

DefinitionInverse Fourier Transform of a Distribution

The inverse FCC1\mathcal{F}_{CC}^{-1} of a distribution is defined with the relation

FCC1(T)(ϕ)=T(FCC1(ϕ))ϕS\mathcal{F}_{CC}^{-1}(T)(\phi) = T(\mathcal{F}_{CC}^{-1}(\phi)) \quad \phi \in \mathcal{S}
Remark.

Intuition: The inverse Fourier transform of a distribution follows the same "dual" strategy: apply the inverse transform to the test function, then let the distribution act on the result. Together with the forward transform, this gives us a complete and invertible Fourier theory on the space of distributions.

With the above, we can conclude that T^\hat{T} itself is a distribution. Just as the CCFT is a map from S\mathcal{S} to itself, the CCFT is also a mapping from S\mathcal{S}^* to itself. Thus, every distribution has a Fourier Transform.

Furthermore, the map FCC:SS\mathcal{F}_{CC} : \mathcal{S}^* \to \mathcal{S}^* is continuous, linear, and one-to-one. The continuity follows from the definition of the Fourier transform and continuity in S\mathcal{S}^*.

Since any singular distribution can be expressed as a weak* limit of such regular distributions (represented by signals in S\mathcal{S}), the definition above is consistent with the FCC\mathcal{F}_{CC} of regular distributions.

Remark.

The power of defining the Fourier transform on distributions is that we can take Fourier transforms of objects that are not ordinary functions -- like the Dirac delta δ\delta, the unit step function, and periodic signals. This vastly extends the applicability of Fourier analysis.

ExampleFourier Transform of the Dirac Delta

Show that δ^\hat{\delta} has its FCC\mathcal{F}_{CC} as a distribution represented by the function h(f)=1h(f) = 1 for all fRf \in \mathbb{R}.

Accordingly, if we approximate the Dirac Delta distribution δ\delta as a limit in S\mathcal{S}^* of a sequence of regular distributions represented by an approximate identity sequence, the Fourier transform of this sequence converges to the FCC\mathcal{F}_{CC} of δ^\hat{\delta}, which is the distribution represented by h(f)=1h(f) = 1. Often, informally, we write δ^(f)=1\hat{\delta}(f) = 1 for all fRf \in \mathbb{R}.

This is important for applications in systems theory: the impulse response and its Fourier transform, the frequency response, of a linear time-invariant system are critical functions defining such a system. When the input is idealized as a Dirac delta function, viewed as a distribution, the output is the impulse response. And when the Fourier of the Dirac delta function, δ^(f)=1\hat{\delta}(f) = 1 is taken as the input in frequency domain, also to be viewed as a distribution, the output of the system is the frequency response.

ExampleFourier Transform of Cosine

We can compute the Fourier transform of cos(2πf0t)\cos(2\pi f_0 t) by viewing it as a distribution, in the sense that it represents a distribution. Observing that cos(2πf0t)=12ei2πf0t+12ei2πf0t\cos(2\pi f_0 t) = \frac{1}{2}e^{i2\pi f_0 t} + \frac{1}{2}e^{-i2\pi f_0 t}, we consider:

limTTTei2πf0fϕ^(f)df\lim_{T \to \infty} \int_{-T}^{T} e^{i2\pi f_0 f} \hat{\phi}(f)\, df=limTTTei2πf0f(ϕ(t)ei2πftdt)df= \lim_{T \to \infty} \int_{-T}^{T} e^{i2\pi f_0 f} \left(\int \phi(t) e^{-i2\pi f t}\, dt\right) df=limTϕ(t)(TTei2πf0fei2πftdf)dt= \lim_{T \to \infty} \int \phi(t) \left(\int_{-T}^{T} e^{i2\pi f_0 f} e^{-i2\pi f t}\, df\right) dt=limTϕ(t)(TTei2π(f0t)fdf)dt= \lim_{T \to \infty} \int \phi(t) \left(\int_{-T}^{T} e^{i2\pi(f_0 - t)f}\, df\right) dt=limTϕ(t)sin(2π(f0t)T)π(f0t)dt= \lim_{T \to \infty} \int \phi(t) \frac{\sin(2\pi(f_0 - t)T)}{\pi(f_0 - t)}\, dt=δf0(ϕ)(ϕ(t)δf0(t)dt)= \delta_{f_0}(\phi) \left(\equiv \int \phi(t) \delta_{f_0}(t)\, dt\right)

In the analysis above, we use the fact that ϕ^S\hat{\phi} \in \mathcal{S}, use Fubini's Theorem to change the order of the integrations and finally invoke Lemma. Thus, the CCFT of a cosine will be 1/2δf0(t)+1/2δf0(t)1/2\,\delta_{f_0}(t) + 1/2\,\delta_{-f_0}(t). Here, the last in brackets is meant to be in an intuitive sense.

FCC\mathcal{F}_{CC} of Periodic Signals

The CCFT of a periodic signal can also be viewed as a distribution. Let x(t)x(t) be continuous and periodic with period PP and ϕS\phi \in \mathcal{S}. Then,

x^,ϕ=x,ϕ^\langle \hat{x}, \phi \rangle = \langle x, \hat{\phi} \rangle =x(f)ϕ^(f)df= \int x(f) \hat{\phi}(f)\, df =limNk=NNx^ ⁣(kP)1Pei2πkPfϕ^(f)df= \lim_{N \to \infty} \int \sum_{k=-N}^{N} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} e^{i2\pi \frac{k}{P} f}\, \hat{\phi}(f)\, df =limNlimTTTk=NNx^ ⁣(kP)1Pei2πkPfϕ^(f)df= \lim_{N \to \infty} \lim_{T \to \infty} \int_{-T}^{T} \sum_{k=-N}^{N} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} e^{i2\pi \frac{k}{P} f}\, \hat{\phi}(f)\, df =limNlimTϕ(t)(k=NNx^ ⁣(kP)1Pei2πkPf(ϕ(t)ei2πftdt)df)= \lim_{N \to \infty} \lim_{T \to \infty} \int \phi(t) \left(\sum_{k=-N}^{N} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} e^{i2\pi \frac{k}{P} f}\left(\int \phi(t) e^{-i2\pi f t}\, dt\right) df\right) =limNlimTϕ(t)(k=NNx^ ⁣(kP)1PTTei2πkPfei2πftdf)dt= \lim_{N \to \infty} \lim_{T \to \infty} \int \phi(t) \left(\sum_{k=-N}^{N} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} \int_{-T}^{T} e^{i2\pi \frac{k}{P} f} e^{-i2\pi f t}\, df\right) dt =limNlimTϕ(t)(k=NNx^ ⁣(kP)1Psin(2π(kPt)T)π(kPt))dt= \lim_{N \to \infty} \lim_{T \to \infty} \int \phi(t) \left(\sum_{k=-N}^{N} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} \frac{\sin(2\pi(\frac{k}{P} - t)T)}{\pi(\frac{k}{P} - t)}\right) dt =k=x^ ⁣(kP)1PδkP(ϕ)= \sum_{k=-\infty}^{\infty} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} \delta_{\frac{k}{P}}(\phi) (ϕ(t)k=x^ ⁣(kP)1PδkP(t)dt)\left(\equiv \int \phi(t) \sum_{k=-\infty}^{\infty} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} \delta_{\frac{k}{P}}(t)\, dt\right)

In the above, the step involving the limit over NN builds on the dominated convergence theorem (Theorem A.1.5) given that ϕ^S\hat{\phi} \in \mathcal{S}.

Thus, we can essentially first view a periodic signal with its CDFT and then replace the values at kP\frac{k}{P} with δkP\delta_{\frac{k}{P}}. This is in agreement with an engineering insight: If one is to express a periodic signal with its FCD\mathcal{F}_{CD} expression:

x(t)=kZx^ ⁣(kP)1Pei2πkPtx(t) = \sum_{k \in \mathbb{Z}} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} e^{i2\pi \frac{k}{P} t}

one would expect that this would be equivalent to the integral form:

x(t)=kZx^ ⁣(kP)1PδkP(f)ei2πftdf,x(t) = \int \sum_{k \in \mathbb{Z}} \hat{x}\!\left(\frac{k}{P}\right) \frac{1}{\sqrt{P}} \delta_{\frac{k}{P}}(f) e^{i2\pi f t}\, df,

whose equivalence is made precise through a distributional approach presented above.

Remark.

The Fourier transform of a periodic signal is a train of Dirac deltas located at the harmonic frequencies kP\frac{k}{P}, weighted by the Fourier series coefficients. This is the mathematical formalization of the intuitive idea that a periodic signal "lives" only at discrete frequencies.

Band-limited vs Time-limited Functions

Let {x(t),tR}\{x(t), \quad t \in \mathbb{R}\} be a continuous-time signal with x^=FCC(x)\hat{x} = \mathcal{F}_{CC}(x). If this signal has a finite (or bounded) bandwidth BB, that is if

x^(f)=0,f>B,\hat{x}(f) = 0, \quad |f| > B,

(in which case we also say that x^\hat{x} has bounded support), then it is not possible for the signal xx to have a bounded support.

TheoremBand-limited Signals Cannot Be Time-limited

Let gg, with g20\|g\|_2 \neq 0 have a finite bandwidth. Then, gg cannot have a bounded support.

Remark.

This is a fundamental result in signal processing known as the uncertainty principle for the Fourier transform. It says you cannot simultaneously have a signal that is both time-limited (zero outside some interval) and band-limited (zero outside some frequency interval). This is analogous to the Heisenberg uncertainty principle in quantum mechanics -- you cannot simultaneously localize a signal perfectly in both time and frequency. A further interpretation and derivation of such a result will be provided in Chapter 10.

Transform Quick Reference

This section consolidates all four Fourier transforms into a single reference for comparison.

Summary of the Four Fourier Transforms

DDFT (Discrete-to-Discrete Fourier Transform) -- FDD\mathcal{F}_{DD}

Input domainDiscrete time: x(n)x(n), n{0,1,,N1}n \in \{0, 1, \ldots, N-1\}
Output domainDiscrete frequency: x^(k/N)\hat{x}(k/N), k{0,1,,N1}k \in \{0, 1, \ldots, N-1\}
Forwardx^(f)=1Nn=0N1x(n)ei2πfn\displaystyle \hat{x}(f) = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x(n)\, e^{-i2\pi f n}
Inversex(n)=1Nk=0N1x^(k/N)ei2πkNn\displaystyle x(n) = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \hat{x}(k/N)\, e^{i2\pi \frac{k}{N} n}
When to useFinite-length discrete signals (NN samples). This is the transform that computers compute (via the FFT).
Key propertiesParseval: $\sum

CDFT (Continuous-to-Discrete Fourier Transform) -- FCD\mathcal{F}_{CD}

Input domainContinuous time on a finite interval: x(t)x(t), t[0,P]t \in [0, P]
Output domainDiscrete frequency: x^(k/P)\hat{x}(k/P), kZk \in \mathbb{Z}
Forwardx^(f)=1P0Px(t)ei2πftdt\displaystyle \hat{x}(f) = \frac{1}{\sqrt{P}} \int_0^P x(t)\, e^{-i2\pi f t}\, dt
Inversex(t)=1PkZx^(k/P)ei2πkPt\displaystyle x(t) = \frac{1}{\sqrt{P}} \sum_{k \in \mathbb{Z}} \hat{x}(k/P)\, e^{i2\pi \frac{k}{P} t}
When to usePeriodic continuous-time signals with period PP. This is the classical Fourier series.
Key propertiesParseval: $\int_0^P

DCFT (Discrete-to-Continuous Fourier Transform) -- FDC\mathcal{F}_{DC}

Input domainDiscrete time (infinite sequence): x(n)x(n), nZn \in \mathbb{Z}
Output domainContinuous frequency on [0,1)[0,1) (periodic with period 1): x^(f)\hat{x}(f)
Forwardx^(f)=nZx(n)ei2πfn\displaystyle \hat{x}(f) = \sum_{n \in \mathbb{Z}} x(n)\, e^{-i2\pi f n}
Inversex(n)=01x^(f)ei2πfndf\displaystyle x(n) = \int_0^1 \hat{x}(f)\, e^{i2\pi f n}\, df
When to useInfinite-length discrete-time signals. The workhorse of discrete-time signal processing and digital filter design.
Key propertiesParseval: $\sum_n

CCFT (Continuous-to-Continuous Fourier Transform) -- FCC\mathcal{F}_{CC}

Input domainContinuous time: x(t)x(t), tRt \in \mathbb{R}
Output domainContinuous frequency: x^(f)\hat{x}(f), fRf \in \mathbb{R}
Forwardx^(f)=x(t)ei2πftdt\displaystyle \hat{x}(f) = \int_{-\infty}^{\infty} x(t)\, e^{-i2\pi f t}\, dt
Inversex(t)=x^(f)ei2πftdf\displaystyle x(t) = \int_{-\infty}^{\infty} \hat{x}(f)\, e^{i2\pi f t}\, df
When to useNon-periodic, continuous-time signals of finite energy. This is "the" Fourier transform in most engineering and physics contexts. Extends to distributions for periodic signals and impulses.
Key propertiesPlancherel/Parseval: $\int

Pattern Summary

All four transforms share the same structure:

  • Forward (analysis): multiply xx by ei2πf(time)e^{-i2\pi f \cdot (\text{time})}, then sum or integrate over time
  • Inverse (synthesis): multiply x^\hat{x} by e+i2πf(time)e^{+i2\pi f \cdot (\text{time})}, then sum or integrate over frequency
  • The sign in the exponent distinguishes analysis (-) from synthesis (++)
  • All four are unitary (energy-preserving): x=x^\|x\| = \|\hat{x}\|
  • All four turn convolution in one domain into pointwise multiplication in the other

The discrete/continuous distinction follows a duality:

  • Discrete in one domain \Leftrightarrow periodic in the other. A finite-length (or periodic) time signal has a discrete frequency spectrum. An infinite-length, non-periodic time signal has a continuous frequency spectrum.
  • Continuous in one domain \Leftrightarrow non-periodic in the other. This duality is why there are exactly four combinations.

Exercises

Problem 5.4.2CCFT of the Unit Step Function

Compute the FCC\mathcal{F}_{CC} of the unit step function, viewed as a distribution.

Problem 5.7.1Fourier Series Convergence

a) Let xL2([0,2];C)x \in L_2([0, 2]; \mathbb{C}) be given by:

x(t)=1{t1}t+1{1t2}(1t)x(t) = 1_{\{t \le 1\}} t + 1_{\{1 \le t \le 2\}} (1 - t)

Let x^(k2)\hat{x}(\frac{k}{2}) denote the Fourier series coefficient corresponding to {12ei2πk2t,t[0,2]}\{\frac{1}{\sqrt{2}} e^{i2\pi \frac{k}{2} t}, t \in [0, 2]\}.

With Matlab, generate the plot of the signal

xN(t)=k=NNx^ ⁣(k2)12ei2πk2t,t[0,2]x_N(t) = \sum_{k=-N}^{N} \hat{x}\!\left(\frac{k}{2}\right) \frac{1}{\sqrt{2}} e^{i2\pi \frac{k}{2} t}, \quad t \in [0, 2]

for N=5,10N = 5, 10 and 1515. Here x^(kP)\hat{x}(\frac{k}{P}) are the Fourier Series expansion coefficients. Observe that, the signal looks more and more like the original signal as NN gets larger.

b) Prove that limN(x(t)xN(t))2dt=0\lim_{N \to \infty} \int (x(t) - x_N(t))^2\, dt = 0.

Hint: Use the properties of Hilbert spaces and the fact that {1Pei2πkPt}\{\frac{1}{\sqrt{P}} e^{i2\pi \frac{k}{P} t}\} forms a complete orthonormal sequence. You could invoke this result directly in your argument.

c) Does for a general xL2([0,2];C)x \in L_2([0, 2]; \mathbb{C}),

supt[0,2]x(t)xN(t)0,\sup_{t \in [0,2]} |x(t) - x_N(t)| \to 0,

as NN \to \infty? Explain your argument.

Problem 5.7.2CCFT of a Gaussian Signal

CCFT is a map from S\mathcal{S} to S\mathcal{S}. One typical example is the Gaussian signal eat2/2e^{-at^2/2} for some a>0a > 0:

Show that for a>0a > 0, the CCFT of ϕ(t)=eat2/2\phi(t) = e^{-at^2/2} is equal to

ϕ^(f)=Ke2π2f2/a,\hat{\phi}(f) = K e^{-2\pi^2 f^2 / a},

for some KK and conclude that ϕ^\hat{\phi} is also a Schwartz signal. Show that KK is independent of ff. Find KK.

Problem 5.7.3Unitarity of the CCFT

Show that CCFT is a unitary transformation from L2(R;C)L_2(\mathbb{R}; \mathbb{C}) to itself. That is, show that Plancherel's Identity holds for functions in L2(R;C)L_2(\mathbb{R}; \mathbb{C}).

Problem 5.7.4CCFT on Distributions

a) Show that CCFT is a continuous map from S\mathcal{S}^* to S\mathcal{S}^*. That is, CCFT maps distributions to distributions and this is continuous map on S\mathcal{S}^*.

b) Show that the CCFT of the δ\delta-distribution is another distribution represented by a function which is equal to 1 for all ff: That is,

FCC(δ~)=H,\mathcal{F}_{CC}(\tilde{\delta}) = H,

with

H(ϕ)=ϕ(t)dtH(\phi) = \int \phi(t)\, dt

Observe that this is in agreement with the general understanding that δ^(f)=1\hat{\delta}(f) = 1 for all ff.

Problem 5.7.5Fourier Transform of an Impulse Train

Consider an impulse train defined by:

wP(t)=nZδ(t+nP)w_P(t) = \sum_{n \in \mathbb{Z}} \delta(t + nP)

so that the distribution that we can associate with this impulse train would be defined by:

wP(ϕ)=nZϕ(nP),\overline{w_P}(\phi) = \sum_{n \in \mathbb{Z}} \phi(nP),

for ϕS\phi \in \mathcal{S}.

a) Show that wP\overline{w_P} is a distribution.

b) Show that

wP^(ϕ)=1Pw1P(t)ϕ(t)dt,\overline{\widehat{w_P}}(\phi) = \int \frac{1}{P} w_{\frac{1}{P}}(t) \phi(t)\, dt,

that is, the FCC\mathcal{F}_{CC} of this train is another impulse train.

Problem 5.7.6Band-limited vs Time-limited

Consider a square-integrable signal with non-zero L2L_2 norm, with bounded support. That is, there exists a compact set, outside of which this signal is identically zero. Can the CCFT of such a signal, with a bounded support in time-domain, also have bounded support in frequency domain?