discrete time Fourier transform in relation with continuous time Fourier transform

Fourier Transforms of Sampled SignalsFernando Diego Sanz Gamiz

Introduction

Computers are able to handle only finite number of data. Hence, ifwe are to study and treat real world signals (i.e. functions $ℝ\to ℂ$) in a computer, a way to characterize signals by a finite numberof data has to be found.

If we sample the values of the signal $x(t)$ at periodictimes $t=nT,n\in \mathbb{Z}$ we form the sequence $x_d[n]=x(nT)$. Does thissequence contain all the information relative to $x(t)$?

We know from the sampling theorem that if the signal $x(t)$is bandlimited, the sampled sequence allows us to recover theoriginal continuous time signal provided the sampling frequency$1/T$ is at least twice the maximum frequency of the signal. However,real signals are not of finite bandwidth as this would imply thesignal to be of infinite time duration. Therefore, a problem ariseof how well can we approximate the original signal $x(t)$ by thesampled sequence $x_d[n]$. In fact, we are interested in studyingthe spectrum of the original signal based upon the samples $x_d[n]$.

While the relation between $x_d[n]$ and the spectrum of $x(t)$is widely used in communication and electronic engineering books, itis difficult to find a rigorous proof. We cover here the gap betweenengineering daily knowledge and rigorous mathematical proof of thenamed relations establishing under what assumptions those relationsare valid.

Relation between Discrete Time Fourier Transform and Continuous time Fourier transform

Theorem 1.

Let $x\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{C}$ be a bounded variation (it can be, in particular,piecewise smooth) $L^{\mathrm{1}}$ function and let $X\mathit{}\mathrm{(}\omega \mathrm{)}$ be its Fouriertransform . Let $T\mathrm{\in }\mathrm{R}$. If $g_n\mathit{}\mathrm{(}\omega \mathrm{)}\mathrm{=}{\mathrm{\sum }}_{k\mathrm{=}\mathrm{-}n}^{\mathrm{+}n}X\mathit{}\mathrm{\left(}\frac{\omega \mathrm{-}\mathrm{2}\mathit{}\pi \mathrm{\cdot }k}{T}\mathrm{\right)}$ converges a.e as $n\mathrm{\to }\mathrm{+}\mathrm{\infty }$ and is thereis $M$ such that a.e. then

$$\frac{1}{T}\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\left(\frac{\omega -2\pi \cdot k}{T}\right)=\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x(k{T}^{+})+x(k{T}^{-})}{2}{e}^{-j\omega k}\mathit{\text{ in }}{L}^2([-\pi ,\pi ])$$(1)

If additionally ${\mathrm{\sum }}_{k\mathrm{=}\mathrm{-}\mathrm{\infty }}^{\mathrm{+}\mathrm{\infty }}X\mathit{}\mathrm{\left(}\frac{\omega \mathrm{-}\mathrm{2}\mathit{}\pi \mathrm{\cdot }k}{T}\mathrm{\right)}$ is continuous, theright hand side of (1) converges uniformly to the left handside in any closed interval $\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{\subset }\mathrm{[}\mathrm{-}\pi \mathrm{,}\pi \mathrm{]}$.

Proof.

By hypothesis we can form the function

$$g(\omega )=\underset{n\to +\mathrm{\infty }}{lim}{g}_n(\omega )=\underset{n\to +\mathrm{\infty }}{lim}\sum _{k=-n}^{+n}X\left(\frac{\omega -2\pi \cdot k}{T}\right)$$

This function is obviously periodic of period $2\pi$ and bounded,hence it can be expanded in its Fourier series which converge in$L^2([-\pi ,\pi ])$; the Fourier theory shows that the convergence isuniformly in $[a,b]\subset [-\pi ,\pi ]$ if $g(\omega )$ iscontinuous.

$$g(\omega )=\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\left(\frac{\omega -2\pi \cdot k}{T}\right)=\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}{c}_k{e}^{j\omega k}$$

where the coefficients$c_k$ are given by

$$c_k=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }g(\omega )e^{-j\omega k}𝑑\omega =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\underset{n\to +\mathrm{\infty }}{lim}{g}_n(\omega )e^{-j\omega k}d\omega$$

As we can appeal the dominated convergencetheorem to write

${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{-\pi }^{\pi }}\underset{n\to +\mathrm{\infty }}{lim}{g}_n(\omega )e^{-j\omega k}d\omega$$=$$\underset{n\to +\mathrm{\infty }}{lim}{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{-\pi }^{\pi }}{g}_n(\omega )e^{-j\omega k}𝑑\omega$

$=$${\displaystyle \frac{1}{2\pi }}{\displaystyle \sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}}{\displaystyle {\int }_{-\pi -2k\pi }^{\pi +2k\pi }}X\left({\displaystyle \frac{\omega }{T}}\right)e^{-j\omega k}𝑑\omega$

$=$${\displaystyle \frac{T}{2\pi }}P.V.{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}X(\omega )e^{j\omega (-kT)}𝑑\omega$

Now, as $x(t)$ is of bounded variation, the Jordan theorem onFourier transform inversion says that

$$\frac{1}{2\pi }P.V.{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}X(\omega )e^{j\omega (-kT)}𝑑\omega =\frac{x(-k{T}^{+})+x(-k{T}^{-})}{2}$$

and the result follows.

∎

There are, however, very which do notsatisfy the conditions required in the theorem above. Such is thecase of the pulse function

whose Fourier transform is $\frac{sin(x/2)}{x/2}$ -with the definition made infootnote 1-. $sinc(x)$ behaves as $\frac{1}{x}$, so${\sum }_{k=-\mathrm{\infty }}^{+\mathrm{\infty }}sinc\left(\frac{\omega -2\pi \cdot k}{T}\right)$ will not converge in general ; it will converge for those $T$ that makethe series an alternating (http://planetmath.org/AlternatingSeries) one, but not for the rest values of $T$.Therefore we need another result which somehow relates both sides ofeq 1.

As we have pointed out, the problem with the pulsefunction is that its Fourier transform does not decay rapid enoughfor the series to converge. So we will try smoothing the signal outso that its Fourier transform will decay faster and, hopefully, theseries converges. We wish the smoothed version of $x(t)$ to resemblethe original signal, so uniform approximation seems reasonable. But,as we will see, for an infinite number of samples $\{nT,n\in \mathbb{Z}\}$each of these might require a different degree of approximation andit could be impossible to find an uniform approximation for all thesamples. So, we will focus on time limited signal, for which we havethe following result.

Notation.

$𝒟(ℝ)$ will denote the set of test functions on $ℝ$ and $𝒮$the set of rapidly decreasing functions on $ℝ$ -the Schwartzspace-. The symbol $\widehat{}$ above a function will denote its Fouriertransform. We know that $𝒟(ℝ)\subset 𝒮$ and that the Fouriertransform is an isomorphism of $𝒮$ onto itself. The product oftwo functions $X$ and $\varphi$ at $\omega$ will be denoted by$X\varphi (\omega )$. Convolution will be denoted by $\ast$.

Theorem 2.

Let $\varphi \mathrm{\in }\mathrm{D}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$ be a test function and $\mathrm{\{}{\varphi }_j\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}j\mathrm{\cdot }\varphi \mathit{}\mathrm{(}j\mathrm{\cdot }t\mathrm{)}\mathrm{,}j\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{\dots }\mathrm{\}}$ be an approximate identity.Let $x\mathit{}\mathrm{(}t\mathrm{)}$ be a time limited bounded variation signal. If${\mathrm{\sum }}_{k\mathrm{=}\mathrm{-}\mathrm{\infty }}^{\mathrm{+}\mathrm{\infty }}X\mathit{}\widehat{{\varphi }_j}\mathit{}\mathrm{\left(}\frac{\omega \mathrm{-}\mathrm{2}\mathit{}\pi \mathrm{\cdot }k}{T}\mathrm{\right)}$ converges for all$j\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{\dots }$ to a continuous function, then

$$\underset{j\to \mathrm{\infty }}{lim}\frac{1}{T}\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\widehat{{\varphi }_j}\left(\frac{\omega -2\pi \cdot k}{T}\right)=\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x(k{T}^{+})+x(k{T}^{-})}{2}{e}^{-j\omega k}$$(2)

Proof.

Take the signal $(x\ast {\varphi }_j)(t)$ whose Fourier transform is$X\widehat{\varphi }(\omega )$. This signal satisfies, by hypothesis, theconditions of Theorem 1, so we can write

$$\frac{1}{T}\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\widehat{{\varphi }_j}\left(\frac{\omega -2\pi \cdot k}{T}\right)=\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{(x\ast {\varphi }_j)(k{T}^{+})+(x\ast {\varphi }_j)(k{T}^{-})}{2}{e}^{-j\omega k}$$

The right hand side is actually a sum of a finite number of termssince the signal $(x\ast {\varphi }_j)(t)$ is time limited -being theconvolution of two compact supported functions-. This makes theFourier series a continuous function, which, together with thehypothesis that $\frac{1}{T}{\sum }_{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\widehat{{\varphi }_j}\left(\frac{\omega -2\pi \cdot k}{T}\right)$ is continuous, shows thatequality in the above equation is pointwise.

Now let $j\to \mathrm{\infty }$ and use the fact that$\{{\varphi }_j(t)\}$ is an approximate identity to obtain equation(2).∎

The rationale behind choosing test functions in the lasttheorem is that, in mostcases,${\sum }_{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\widehat{{\varphi }_j}\left(\frac{\omega -2\pi \cdot k}{T}\right)$ will converge eventhough ${\sum }_{k=-\mathrm{\infty }}^{+\mathrm{\infty }}X\left(\frac{\omega -2\pi \cdot k}{T}\right)$ do not. This is becauserapidly decreasing functions decay faster than $\frac{1}{x^{\alpha }}$for any $\alpha$. So, for example, the Fourier transform of thepulse function has been tamed enough to make the series converge.

Remark 1.

When the signal $x(t)$ is continuous, the right hand side of eqs.(1) or (2) reads

$$\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}{x}_d[k]e^{-j\omega k}$$

where $x_d[k]=x(kT)$. This is defined as the(Discrete Time) Fourier Transform, DTFT of the sequence$x_d[n]=x(nT)$