Homec4science
Diffusion epfl (master) Browse

R2653/writing/sp4comm.multipub/40-fourier/20-fa-DTFT.tex
911b2a53aa82master

Last Change
Raw File

20-fa-DTFT.tex
View Options

\section{The Discrete-\!Time Fourier Transform}
\label{sec:fa:dtft}
The DFT provides us with a frequency-domain representation of signals that contain a finite amount of data; periodic signals, which are also uniquely described by a finite number of data samples, use a formally identical frequency representation that, for clarity, we call the DFS. In both cases, the transform is a straightforward change of basis in $\mathbb{C}^N$ and, as such, it can be described algorithmically as a matrix-vector multiplication involving a finite number of operations.
We now consider the problem of obtaining a frequency-domain representation for aperiodic, infinite-length sequences. Such signals are an idealized mathematical concept, of course, but they are indispensable if we want to prove theoretical results that hold independently of the size of the data set and, most importantly, that are still true in the limit case of an infinite number of samples. Obviously we would very much like for this new transform to be a simple extension of the DFT, that is, we would like it to retain its intuitive interpretation as a measure of similarity (that is, an inner product) between an input signal and a set of oscillatory components. While this is ultimately the case, from the mathematical point of view unfortunately something ``breaks down'' and, as we move from finite to infinite lengths, we won't be able to obtain a Fourier transform that acts as an endomorphism on the space of discrete-time sequences; namely, while the frequency representation of a time-domain signal in $\mathbb{C}^N$ is itself a vector in $\mathbb{C}^N$, the Fourier transform of a sequence will not be a sequence but a function of a real variable. Although the exact reasons for this are quite technical and beyond the scope of this book, we will try to provide some intuition in the next pages.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Definition}
The frequency-domain representation of an infinite-length, discrete-time signal $\mathbf{x}$ is called the Discrete-Time Fourier Transform (DTFT)\index{DTFT|mie} and it is defined as
\begin{equation}\label{eq:fa:DTFT}
X(\omega) = \sum_{n=-\infty}^{\infty} x[n] \, e^{-j\omega n}.
\end{equation}
Formally, the DTFT can be seen as an operator that maps discrete-time sequences to complex-valued functions of the frequency variable $\omega \in \mathbb{R}$; since the argument $\omega$ only appears in the phase of the complex exponentials in the sum, a DTFT is always $2\pi$-periodic. To stress out this property\footnote{
And also to provide a convenient notational framework which unifies the DTFT and the $z$-transform, as we will see in Chapter~\ref{ch:zt}.
}
many authors use the notation $X(e^{j\omega})$ to indicate a DTFT, and we may do so too if the context calls for it. Finally, when no confusion can arise, we will also use the vector notation $\mathbf{X}$ to indicate the DTFT of a sequence $\mathbf{x}$.
\itempar{Existence.} The DTFT is guaranteed to exist if the sum in~(\ref{eq:fa:DTFT}) converges, that is, if the partial sum
\begin{equation}\label{eq:fa:DTFTpartial}
X_N(\omega) = \sum_{n=-N}^{N} x[n]\,e^{-j\omega n}
\end{equation}
is finite for all values of $\omega$ as $N$ grows to infinity. Convergence is very easy to prove for \emph{absolutely summable} sequences, that is for sequences satisfying
\begin{equation}
\lim_{N \rightarrow \infty} = \sum_{n=-N}^{N} \bigl|x[n] \bigr| < \infty
\end{equation}
since, according to the triangle inequality,
\begin{equation}
\bigl| X_N(\omega) \bigr| \leq \sum_{n=-N}^{N} \bigl| x[n] \, e^{-j\omega n} \bigr| = \sum_{n=-N}^{N} \bigl|x[n] \bigr|.
\end{equation}
For this class of sequences it can also be proved that the convergence of $X_N(\omega)$ to $X(\omega)$ is uniform and that $X(\omega)$ is continuous. While absolute summability is a sufficient condition for the existence of the DTFT, it can be shown that the sum in~(\ref{eq:fa:DTFTpartial}) exists also for all sequences in $\ell_2(\mathbb{Z})$, the space of \emph{square-summable} (i.e., finite-energy) sequences. For square-summable signals that are not absolutely summable, however, the convergence of~(\ref{eq:fa:DTFTpartial}) is no longer uniform but takes place only in the mean-square sense, i.e.{}
\begin{equation}
\lim_{N\rightarrow\infty} \int_{-\pi}^{\pi} \bigl|X_N(\omega) - X(\omega) \bigr|^2 \, d\omega = 0.
\end{equation}
This type of convergence implies that, while the total energy of the difference between functions goes to zero, the functions may differ in value over a countable set of points.\footnote{
A particular manifestation of this behavior is called the \emph{Gibbs phenomenon}, which has important consequences in the problem of filter design, as we will study in Chapter~\ref{ch:fd}.}
This also implies that, in this case, $X(\omega)$ is no longer guaranteed to be continuous.
\itempar{Inversion.} The DTFT, when it exists, can be inverted via the integration
\begin{equation}\label{eq:fa:DTFTrec}
x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\omega) \,e^{j\omega n} d\omega;
\end{equation}
this can be easily verified by substituting~(\ref{eq:fa:DTFT}) into~(\ref{eq:fa:DTFTrec}) and recalling that
\[
\int_{-\pi}^{\pi} e^{j\omega (n-k)} = 2\pi\, \delta[n-k].
\]
In fact, due to the $2\pi$-periodicity of the DTFT, the integral in~(\ref{eq:fa:DTFTrec}) can be computed over \emph{any} $2\pi$-wide interval on the real line; by convention, though, the DTFT is generally represented over the $[-\pi, \pi]$ interval.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Intuitive interpretations}
As we said, we would like the DTFT to emerge as a natural extension of the DFT, whose simplicity and whose properties we are now familiar and happy with. There are at least three rather intuitive paths that lead from the DFT to the DTFT, although (be forewarned) all of them will rely at some point on a certain amount of willful carelessness with mathematical precision.
\itempar{The DTFT as the limit of a DFT.}
The simplest approach to the DTFT considers an infinite sequence as the limit of a finite-length signal whose length grows larger and larger. In the analysis formula for an $N$-point DFT,
\[
X[k] = \sum_{n=0}^{N-1} x[n] e^{-j\frac{2\pi}{N}nk}, \qquad \quad k = 0, \ldots, N-1,
\]
the fundamental frequency for the space is $\omega_N = 2\pi/N$. As $N$ grows large, $\omega_N$ becomes infinitesimally small and so its multiples $k\omega_N$ will become infinitely dense in the $[0, 2\pi]$ interval, much like a real-valued variable would. By replacing $2k\pi/N$ by $\omega$ and by shifting the reference interval to $[-\pi, \pi]$, the metamorphosis is complete. The same limiting argument can be carried out for the synthesis formula, as in Exercise~\ref{fa:ex:riemann}.
\itempar{The DTFT as the limit of a DFS.}
An aperiodic sequence can also be interpreted as the limit of a periodic signal whose period grows to infinity. Pick an absolutely summable sequence $\mathbf{x}$ and build the $N$-periodic signal\index{periodization}
\begin{equation}\label{eq:fa:periodization}
\tilde{x}_N[n] = \sum_{p = -\infty}^{\infty} x[n + pN]
\end{equation}
with $N>0$; since $\mathbf{x}$ is absolutely summable, the above expression will be finite for all $n$ (see Example~\ref{exa:dt:periodization}). As $N$ grows larger, the copies that make up the periodic signal will be spaced further and further apart and, in the limit,
\[
\lim_{N \rightarrow \infty} \tilde{\mathbf{x}}_N = \mathbf{x}.
\]
For all finite values of $N$, the natural frequency representation for $\tilde{\mathbf{x}}_N$ is its DFS:
\begin{equation}
\tilde{X}_N[k] = \sum_{n=0}^{N-1} \tilde{x}[n] \, e^{-j\frac{2\pi}{N}nk} = \sum_{p = -\infty}^{\infty} \left( \sum_{n=0}^{N-1} x[n + pN] \, e^{-j\frac{2\pi}{N}(n + pN)k} \right);
\end{equation}
in the derivation, we have used the definition of $\tilde{\mathbf{x}}_N$ and exploited the fact that $e^{-j(2\pi /N)nk} = e^{-j(2\pi /N)(n+pN)k}$. Now, for every value of $p$ in the outer sum, the value $(n+pN)$ in the inner sum ranges from $pN$ to $pN + N - 1$ so that the double sum can be simplified to
\begin{equation}
\tilde{X}_N[k] = \sum_{n = -\infty}^{\infty} x[n] \, e^{-j\frac{2\pi}{N}kn}.
\end{equation}
If we define the following \textit{function} of a real-valued variable $\omega$
\begin{equation} \label{eq:fa:dtftAnteLitteram}
X(\omega) = \sum_{n = -\infty}^{\infty} x[n] \, e^{-j\omega n}
\end{equation}
then we can express the DFS of $\tilde{\mathbf{x}}_N$ as the set of $N$ regularly spaced samples\footnote{
Once again, it is instructive to notice the time-frequency duality pattern: in Section~\ref{sec:fa:dft_elementary} we showed how a maximally compact time-domain signal possesses a maximally wide spectrum, and vice-versa; here we see that increasing the period spacing in time decreases the spacing of samples in frequency.
}
\[
\tilde{X}_N[k] = X\left(\frac{2\pi}{N}k\right);
\]
as shown in Figure~\ref{fig:fa:dsf2dtft} for different values of $N$. As $N$ grows large, the set of samples will grow denser in the $[0, 2\pi]$ interval; since, in the limit, $\tilde{\mathbf{x}}_N$ tends to $\mathbf{x}$, it appears that the frequency-domain representation for $\mathbf{x}$ is indeed the $2\pi$-periodic function $X(\omega)$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[p]
%
%% absolutely summable signal
\def\sig#1{dup \dspPorkpie{#1}{10} 0.2 mul exch #1 sub 2 div dup mul 1 add 1 exch div sub -1 mul }
%% widening periodization
\def\perSigPlot#1{%
\hspace{-3em}
\begin{dspPlot}[width=0.5\dspWidth,height=0.5\dspHeight,xticks=50,xout=true,yticks=none,sidegap=0,ylabel={$\mathbf{\tilde{x}}_{#1}$}]{-50,50}{-.2, 1.2}
\dspSignal[linecolor=ColorDT]{0 -80 #1.0 80 {/i exch def x \sig{i} add } for }
%\dspText(30,0.5){$N=#1$}
\end{dspPlot}
&
\hspace{-5em}
\begin{dspPlot}[width=0.5\dspWidth,height=0.5\dspHeight,xtype=freq,xticks=1,yticks=none,ylabel={$\mathbf{\tilde{X}}_{#1}$}]{0,2}{0, 1.4}
\dspFunc[linecolor=ColorDF!30]{x \dspPeriodize \dspPorkpie{0}{.8}}
\dspSignal[plotpoints=#1,linecolor=ColorDF]{x \dspPeriodize \dspPorkpie{0}{.8}}
\end{dspPlot}}
%
\center
\small
\psset{unit=5mm}
\begin{tabular}{cc}
\hspace{-3em}
\begin{dspPlot}[width=0.5\dspWidth,height=0.5\dspHeight,xticks=50,xout=true,yticks=none,sidegap=0,ylabel={$\mathbf{x}$}]{-50,50}{-.2, 1.2}
\dspSignal[linecolor=ColorDT]{x \sig{0}}
\end{dspPlot}
&
\hspace{-5em}
\begin{dspPlot}[width=0.5\dspWidth,height=0.5\dspHeight,xtype=freq,xticks=1,yticks=none,ylabel={$X(\omega)$}]{0,2}{0, 1.4}
\dspFunc[linecolor=ColorDF]{x \dspPeriodize \dspPorkpie{0}{.8}}
\end{dspPlot}
\\ \\
\perSigPlot{10}
\\
\perSigPlot{20}
\\
\perSigPlot{40}
\\
\perSigPlot{80}
\end{tabular}
\caption{top row: original infinite-length, absolutely summable signal (left) and the function $X(\omega)$ defined in~(\ref{eq:fa:dtftAnteLitteram}); rows 2-5: periodized signal $\mathbf{\tilde{x}}_N$ and its DFS for increasing values of $N$; all DFS values are samples of $X(\omega)$ at multiples of $2\pi/N$.}\label{fig:fa:dsf2dtft}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\itempar{The DTFT as a formal change of basis.}
Finally, we could simply read the analysis formula in~(\ref{eq:fa:DTFT}) as a formal inner product in $\ell_2(\mathbb{Z})$ between a sequence $\mathbf{x}$ and a set of ``basis vectors'' $\left\{ e^{j\omega n} \right\}$ indexed by $\omega\in\mathbb{R}$:
\begin{equation} \label{eq:fa:dtftIP}
X(\omega) = \langle e^{j\omega n}, \mathbf{x} \rangle.
\end{equation}
According to our usual interpretation of the inner product, every point of the DTFT thus captures the similarity between the input sequence $\mathbf{x}$ and an infinitely long oscillation at frequency $\omega$. This is certainly intuitively consistent with our description of the DFT as a change of basis, but mathematically there are two glaring problems: first, these putative basis vector do not form a countable set, since they are indexed by a real-valued variable; secondly, they are not even elements of $\ell_2(\mathbb{Z})$, since their energy is infinite. Nevertheless, the DTFT \textit{is}\/ in fact a change of basis --- the only trick is that we have to look at it \textit{backwards}, as shown in the next section.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The DTFT as a basis expansion}
In Section~\ref{sec:vs:functionSpace} we showed how the elements of a vector space can be standard functions (rather than Euclidean tuples) and the set of square-integrable functions over the $[-\pi, \pi]$ interval is a Hilbert space known as $L_2([-\pi, \pi])$. A vector $\mathbf{F} \in L_2([-\pi, \pi])$ is thus a complex-valued function for which the integral
\[
\int_{-\pi}^{\pi} |F(\omega)|^2\, d\omega < \infty.
\]
The usual definition for the inner product in this space is
\begin{equation}
\langle \mathbf{F, G} \rangle = \frac{1}{2\pi} \int_{-\pi}^{\pi} F^*(\omega)G(\omega)d\omega
\end{equation}
and with this it can be shown that the set
\begin{equation} \label{eq:fa:L2basis}
\left\{ e^{-j\omega n} \right\}_{n\in\mathbb{Z}}
\end{equation}
forms an orthogonal basis\footnote{
While orthogonality is elementary to show, the really tricky part is proving completeness. Please refer to~\cite{Foundations} for details.}.
Note that the set of basis functions is countable (it is indexed by an integer) and that each element is obviously square-integrable over the finite interval $[-\pi, \pi]$.
Although elements of $L_2([-\pi, \pi])$ are defined only over the $[-\pi, \pi]$ interval, they can be transformed into $2\pi$-periodic functions via a simple mapping of the real line\footnote{
For instance, using the mapping $i : \mathbb{R} \rightarrow [-\pi, \pi]$ defined by $i(\omega) = ((\omega + \pi) \mod 2\pi) - \pi$, $X(i(\omega))$ is $2\pi$-periodic for all $X$.}
and therefore every one of them can be considered a bona-fide DTFT. With this in mind, the DTFT inversion formula in~(\ref{eq:fa:DTFTrec}) can be interpreted as the inner product between an element of $L_2([-\pi, \pi])$ and the $n$-th basis vector for the space. In other words, any square-integrable DTFT can be decomposed into a linear combination of vectors from~(\ref{eq:fa:L2basis}) and the coefficients of the expansion form the discrete-time sequence
\begin{equation} \label{eq:fa:dtftInverseIP}
x[n] = \langle e^{-j\omega n}, X(\omega) \rangle.
\end{equation}
While this inverted viewpoint may seem just a bit of mathematical sophistry, it actually allows us to legitimately invoke the properties of Hilbert space for the DTFT and, in particular, the conservation of energy guaranteed by Parseval's theorem:
\begin{equation}
\sum_{n=-\infty}^{\infty} |x[n]|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |X(\omega)|^2 d\omega.
\end{equation}
This means that the square magnitude of the DTFT encodes the signal's distribution of energy in the frequency domain, just as in the case of the DFT; and this also guarantees that, given a DTFT in $L_2([-\pi, \pi])$, the associated time-domain sequence will necessarily be in $\ell_2(\mathbb{Z})$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Plotting the DTFT}
\label{fa:sec:DTFTplot}
The DTFT of a discrete-time signal is a complex-valued $2\pi$-periodic function of a real-valued frequency variable. Just as we discussed in Section~\ref{fa:sec:DFTplot}, DTFT plots display separately the magnitude and the phase whose physical significance remains unchanged with respect to the DFT. In this case however the range of choice for the frequency variable is the zero-centered interval $[-\pi, \pi]$, which clearly shows the contribution of positive (counterclockwise) and negative (clockwise) frequencies as shown in Figure~\ref{fig:fa:dtftread}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[t]
\centering
\begin{dspPlot}[xtype=freq,yticks=none,xticks=2,xout=true,height=\dspHeightCol,ylabel={$|X(\omega)|$}]{-1.05,1.05}{0, 2}
\dspFunc[linecolor=ColorDF]{x \dspPorkpie{0}{.3}}
\psset{braceWidthInner=3pt,braceWidthOuter=3pt,braceWidth=1pt}
\def\y{1.3}
\pnode(-1,\y){CC1} \pnode(0,\y){CC2}
\pnode(0,\y){C1} \pnode(1,\y){C2}
\psbrace[linecolor=blue,ref=C,nodesep=-1.5ex,rot=-90](CC2)(CC1){frequencies $< 0$ (clockwise)}
\psbrace[linecolor=blue,ref=C,nodesep=-1.5ex,rot=-90](C2)(C1){frequencies $> 0$ (counterclockwise)}
\def\y{-.55}
\pnode(-1,\y){L1}\pnode(-.8,\y){L2}
\pnode(-.25,\y){L3}\pnode(.25,\y){L4}
\pnode(.8,\y){H1}\pnode(1,\y){H2}
\psbrace[linecolor=green,ref=C,nodesep=2ex,rot=90](L1)(L2){high frequencies (fast)}
\psbrace[linecolor=green,ref=C,nodesep=2ex,rot=90](L3)(L4){low frequencies (slow)}
\psbrace[linecolor=red,ref=C,nodesep=2ex,rot=90](H1)(H2){high frequencies (fast)}
\end{dspPlot}
\vspace{3em}
\caption{Reading a DTFT plot.}\label{fig:fa:dtftread}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
It is important to remember that the DTFT is always $2\pi$-periodic and therefore, if the plot encompasses a larger range of frequencies, both magnitude and phase repeat \textit{ad infinitum}, as shown in Figure~\ref{fig:fa:dtftperiodic}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[t]
\centering
\begin{dspPlot}[xtype=freq,yticks=none,xticks=1,xout=true,height=\dspHeightCol,ylabel={$|X(\omega)|$}]{-5,5}{0, 1.2}
\dspFunc[linecolor=ColorDF]{x \dspPeriodize \dspPorkpie{0}{.3}}
\end{dspPlot}
\caption{Implicit periodicity of the DTFT.}\label{fig:fa:dtftread}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The DTFT of elementary signals}
\itempar{Impulse.} The DTFT of a discrete-time delta is a complex exponential function:
\[
\DTFT{\boldsymbol{\delta}_m} = \sum_{n=-\infty}^{\infty}\delta[n-m]e^{-j\omega n} = e^{-j\omega m};
\]
the magnitude of the transform is constant and equal to one, whereas the phase is linear and proportional to $m$. Again, as we saw for the DFT, the Fourier transform the most ``concentrated'' signal in time is nonzero over its entire freqency support.
\itempar{Rectangular signal.} Consider again the step signal defined in~(\ref{eq:fa:step}), now considered as an infinite-support sequence; it is
\begin{align*} \label{eq:fa:DTDF_step}
X(\omega) &= \sum_{n=0}^{M-1}e^{-j\omega n} \\
&= \frac{1-e^{-j\omega M}}{1-e^{-j\omega}} \\
&= \frac{e^{-j\omega M/2}\,(e^{j\omega M/2}) - e^{-j\omega M/2})}{e^{j\omega/2} \, (e^{j\omega/2} - e^{-j\omega/2})} \\
&= \frac{\sin \left(\omega M/2\right) }{\sin( \omega/2)} e^{-j \omega (M-1)/2}.
\end{align*}
Again, we have manipulated the expression for the transform into a product of a real-valued term and a pure phase term; this allows us to easily plot the DFT as in Figure~\ref{fig:fa:dtftex2}; note how the magnitude is zero at all nonzero multiples of $2\pi/M$ and how the sign changes in the magnitude at those points determine an additional shift of $\pm\pi$ in the phase.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[tb]
\def\M{8.0 }
\center
\begin{dspPlot}[height=0.3\dspWidth,xtype=freq,xticks=4,yticks=2,ylabel={$|X(\omega)|$}]{-1,1}{0, 9}
\dspFunc[linecolor=ColorDF]{x 90 mul dup dup 0 eq {pop pop 1} {\M mul sin exch sin div} ifelse abs}
\end{dspPlot}
\begin{dspPlot}[height=0.3\dspWidth,xtype=freq,xout=true,yticks=custom,ylabel={$\measuredangle X(\omega)$}]{-1, 1}{-1.1,1.1}
\dspFunc[linecolor=ColorDF]{\M 1 sub -2 div x mul %
dup truncate abs 1 add cvi 2 idiv
x 0 gt {-2} {2} ifelse mul sub}
\dspCustomTicks[axis=y]{-1 $-\pi$ 1 $\pi$}
\end{dspPlot}
\caption{DTFT of an \M-point rectangular signal (see~(\ref{eq:fa:step})).}\label{fig:fa:dtftex2}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\itempar{Exponential decay.} Consider the exponential decay defined in section~\ref{sec:dt:elementary},
\[
x[n] = a^n u[n];
\]
for $|a| < 1$ the sequence is absolutely summable and its DTFT is easily computed as
\begin{equation}
X(\omega) = \frac{1}{1 - a e^{-j\omega}}.
\end{equation}
Magnitude and phase are, respectively,
\begin{align}
\bigl|X(\omega) \bigr|^2 & = \frac{(1}{1 + |a|^2 - 2a\cos(\omega)} \\[2em]
\measuredangle X(\omega) & = \arctan \!\left[\frac{-a\sin(\omega)}{1-a\cos(\omega)} \right]
\end{align}
and they are plotted in Figure~\ref{fig:fa:dtftex3}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[tb]
\def\a{0.9 }
\center
\begin{dspPlot}[height=0.3\dspWidth,xtype=freq,yticks=5,ylabel={$|X(\omega)|$}]{-1,1}{0, 11}
\dspFunc[linecolor=ColorDF]{x 180 mul cos \a -2 mul mul \a \a mul 1 add add sqrt 1 exch div }
\end{dspPlot}
\begin{dspPlot}[height=0.3\dspWidth,xtype=freq,xout=true,yticks=custom,ylabel={$\measuredangle X(\omega)$}]{-1, 1}{-1.1,1.1}
\dspFunc[linecolor=ColorDF,xmax=0.99]{x 180 mul sin -\a mul 1 x 180 mul cos \a mul sub atan x 0 gt {360} {0} ifelse sub 180 div}
\dspCustomTicks[axis=y]{-1 $-\pi$ 1 $\pi$}
\end{dspPlot}
\caption{DTFT of the exponential decay for $a=\a$.}\label{fig:fa:dtftex3}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Properties of the DTFT}
\label{fa:sec:DTFTprops}
In this section we will list, without formal proof, a series of elementary properties for the DTFT, which you are encouraged to verify using the definition of the transform.
\itempar{Linearity.} The DTFT is a linear operator:
\begin{equation}
\DTFT{a\mathbf{x} + b\mathbf{y}} = a\,\DTFT{\mathbf{x}} + b\,\DTFT{\mathbf{y}} = a\mathbf{X} + b\mathbf{Y} \quad \forall a, b \in \mathbb{C}.
\end{equation}
\itempar{Time shift.} The shift of a signal in time correspond to multiplication by a linear phase factor in frequency:
\begin{equation}
\DTFT{\mathcal{D}^m\mathbf{x}}(\omega) = e^{-j\omega m}\,X(\omega)
\end{equation}
that is,
\[
x[n - m] \stackrel{\scriptstyle\text{DTFT}}{\longleftrightarrow} e^{-j\omega m}\,X(\omega)
\]
\itempar{Frequency shift.} Multiplication in the time domain by a linear phase term corresponds to a shift in frequency:
\begin{equation}
x[n] \, e^{j\omega_0 n} \stackrel{\scriptstyle\text{DTFT}}{\longleftrightarrow} X(\omega - \omega_0)
\end{equation}
and, by linearity,
\[
x[n] \, \cos(\omega_0 n) \stackrel{\scriptstyle\text{DTFT}}{\longleftrightarrow} \frac{1}{2}\left[ X(\omega - \omega_0) + X(\omega + \omega_0)\right].
\]
\itempar{Time and frequency reversal.} Reversing a signal in time reverses its transform:
\begin{equation}
\DTFT{\mathcal{R}\mathbf{x}} = \mathcal{R}\mathbf{X}.
\end{equation}
Conjugating the time-domain signal results in both conjugation and reversal of the DTFT:
\begin{equation}
\DTFT{\mathbf{x}^*} = \mathcal{R}\mathbf{X}^*.
\end{equation}
\itempar{Symmetries.} In the case of infinite-length sequences and periodic functions, symmetries (and antisymmetries) are intended around zero and have a straightforward definition. With this,
\begin{itemize}
\item the DTFT of a symmetric signal is symmetric
\item the DTFT of a real-valued signal is Hermitian-symmetric
\item the magnitude DTFT of a real-valued signal is symmetric.
\item the DTFT of a real-valued symmetric signal is real-valued and symmetric.
\end{itemize}
\clearpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The DTFT of power signals} \label{sec:fa:dirac}\label{fa:sec:functionals}\index{functional}
The DTFT is mathematically well defined for all finite-energy sequences so that every signal in $\ell_2(\mathbb{Z})$ is uniquely associated to a square-integrable spectrum in $L_2([-\pi, \pi])$. However, as we discussed in Section~\ref{powersig}, many interesting signals are \textit{power} signals, that is, signals that are not square summable but whose energy per unit of time remains finite; examples include a constant sequence, the unit step, and sinusoidal oscillations. With some care, the DTFT formalism can be extended to power signals as well to obtain a meaningful frequency representation that we can interpret as a change of basis; this can be achieved via the use of \textit{functionals.}\index{functional}\index{distribution|Fourier transform}
\itempar{Functionals.} For our purposes, a functional $f$ is a linear operator that maps the elements of a given vector space $V$ to the set of complex scalars. For example, let's place ourselves in $\ell_2(\mathbb{Z})$ and define a functional $\delta$ that extracts the value of a sequence $\mathbf{x}$ at $n=0$; the functional is described in terms of its \textit{action}\/ on a vector, and we will write
\begin{equation}
\delta\{\mathbf{x}\} = x[0];
\end{equation}
functionals can be parametrized so that, for instance, we can generalize the value-extracting functional as $\delta_{n_0}$, with $n_0 \in \mathbb{Z}$, whose action on elements of $\ell_2(\mathbb{Z})$ is
\begin{equation}
\delta_{n_0}\{\mathbf{x}\} = x[n_0].
\end{equation}
Another example is given by the integrating functional, defined as
\begin{equation}
1\{\mathbf{x}\} = \sum_{n=-\infty}^{\infty}x[n].
\end{equation}
Interestingly, all the examples so far can be expressed in terms of an inner product:
\begin{align*}
\delta_{n_0}\{\mathbf{x}\} &= \langle \boldsymbol{\delta}_{n_0}, \mathbf{x} \rangle \\
1\{\mathbf{x}\} &= \langle \mathbf{1}, \mathbf{x} \rangle;
\end{align*}
and, in fact, any sequence $\mathbf{f}$ can be used to \textit{induce}\/ a functional $f$ over $\ell_2(\mathbb{Z})$ as
\begin{equation}
f\{\mathbf{x}\} = \langle \mathbf{f}, \mathbf{x} \rangle.
\end{equation}
Note that $\mathbf{f}$ does not need to be in $\ell_2(\mathbb{Z})$ and, indeed, the integrating functional is induced by the infinite-energy sequence $\mathbf{1}$.
If we now move to the space $L_2([-\pi, \pi])$, we can again define a parametrized value-extracting functional
\begin{equation}
\delta_{\omega_0}\{\mathbf{X}\} = X(\omega_0)
\end{equation}
(which we will indicate simply as $\delta$ when $\omega_0 = 0$), as well as an integrating functional
%, parametrized in order to select an interval around the origin
\begin{equation}
1\{\mathbf{X}\} = \frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)d\omega.
\end{equation}
Interestingly, while the last functional can be defined in terms of an inner product in $L_2([-\pi, \pi])$, the first one can not; indeed we have
\[
1\{\mathbf{X}\} = \langle \mathbf{1}, \mathbf{X} \rangle
\]
(where $\mathbf{1}$ is now a function of $\omega$ identically equal to one over $[-\pi, \pi]$) but it is impossible to define a standard function $\delta(\omega)$ for which the following holds for all $\mathbf{X}$:
\begin{equation}
\frac{1}{2\pi} \int_{-\pi}^{\pi}\delta(\omega)X(\omega)d\omega = X(0).
\end{equation}
In general, while any reasonably well-behaved function $F(\omega)$ can induce a functional $F$ in $L_2([-\pi, \pi])$ via the inner product
\[
F\{\mathbf{X}\} = \langle \mathbf{F}, \mathbf{X} \rangle = \frac{1}{2\pi}\int_{-\pi}^{\pi}F(\omega)X(\omega)d\omega,
\]
the converse is not true and there are functionals that are not induced by regular functions.
\itempar{Fourier transform of a functional.} As we said, a functional is defined by its action on the elements of a vector space; functionals that operate on elements of $\ell_2(\mathbb{Z})$ can be regarded as time-domain functionals, while those that operate on elements of $L_2([-\pi, \pi])$ can be considered frequency-domain functionals.
Given a time-domain functional $f$, its Fourier transform is defined as the frequency-domain functional $F$ for which the equality
\begin{equation}\label{fa:eq:FTfunc}
f\{\mathbf{x}\} = F\{\mathbf{X}\}.
\end{equation}
is satisfied for \textit{any}\/ regular DTFT pair $\mathbf{x} \leftrightarrow \mathbf{X}$. The rationale behind this definition is that the action of a functional and that of its Fourier transform should be equivalent across domains when applied to a given signal. The definition is also an extension of the standard DTFT: indeed, if a time-domain functional $f$ is induced by a sequence $\mathbf{f} \in \ell_2(\mathbb{Z})$, then the corresponding frequency-domain functional will be induced by $\mathbf{F} = \DTFT{\mathbf{f}} \in L_2([-\pi, \pi])$:
\begin{align}
F\{\mathbf{x}\} &= \langle \mathbf{F}, \mathbf{X} \rangle = \frac{1}{2\pi} \int_{-\pi}^{\pi}F^*(\omega)X(\omega)d\omega \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{n=-\infty}^{\infty} f^*[n] e^{j\omega n}\ X(\omega)d\omega \\
&= \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} f^*[n] \int_{-\pi}^{\pi} X(\omega) e^{j\omega n} d\omega \\
&= \sum_{n=-\infty}^{\infty} f^*[n] x[n] \\
&= \langle \mathbf{f}, \mathbf{x} \rangle = f\{\mathbf{x}\}.
\end{align}
For finite-energy sequences, therefore, the DTFT and the Fourier transform in the sense of functionals coincide; but consider now a sequence such as the constant $\mathbf{1}$, which has infinite energy. Given any DTFT pair $\mathbf{x} \leftrightarrow \mathbf{X}$, we have
\begin{align*}
\mbox{in $\ell_2(\mathbb{Z})$:} & \qquad 1\{\mathbf{x}\} = \langle \mathbf{1}, \mathbf{x} \rangle = \sum_{n=-\infty}^{\infty}x[n] \\
\mbox{in $L_2([-\pi, \pi])$:} & \qquad \delta\{\mathbf{X}\} = X(0) = \sum_{n=-\infty}^{\infty}x[n]
\end{align*}
so that, in the functional sense,
\begin{equation}\label{fa:eq:deltaFunctional}
\DTFT{1} = \delta.
\end{equation}
Similarly, the prototypical oscillation $e^{j\omega_0 n}$ has infinite energy but
\[
\langle e^{j\omega_0 n}, \mathbf{x} \rangle = \sum_{n=-\infty}^{\infty} e^{-j\omega_0 n}x[n] = X(\omega_0) = \delta_{\omega_0}\{\mathbf{X}\}
\]
so that, in the functional sense,
\begin{equation}
\DTFT{e^{j\omega_0 n}} = \delta_{\omega_0}.
\end{equation}
With the same method, and exploiting the linearity of functionals, we obtain
\begin{align}
\DTFT{\cos(\omega_0 n)} &= \frac{1}{2}(\delta_{\omega_0} + \delta_{-\omega_0}) \\
\DTFT{\sin(\omega_0 n)} &= -\frac{j}{2}(\delta_{\omega_0} - \delta_{-\omega_0}).
\end{align}
\itempar{Physical interpretation.}
While the theory of functionals (whose surface we ever so slightly just scratched) provides an elegant and consistent mathematical framework for the Fourier transform of power signals, what is the underlying physical phenomenon that we are trying to model? To gain some intuition, consider the delta functional, that is, the generalized Fourier transform of the constant unit signal. Although there is no regular function that can induce the delta functional on $L_2([-\pi, \pi])$, if there \textit{were}\/ such a function what would it look like?
To explore the issue let's move to the time domain and consider a double-sided decaying exponential $\mathbf{d}_a$, with $d_a[n] = a^{|n|}$ and $0 < a < 1$. The sequence belongs to $\ell_2(\mathbb{Z})$ since
\[
\sum_{n=-\infty}^{\infty} |a^{|n|}|^2 = \frac{1 + a^2}{1 - a^2}
\]
and its DTFT is (see Exercise~\ref{fa:ex:doublesided})
\begin{equation}
D_a(\omega) = \frac{1 - a^2}{1 - 2a\cos\omega + a^2};
\end{equation}
note that the area under the DTFT is independent of $a$ and equal to $2\pi$ since
\[
\int_{-\pi}^{\pi} D_a(\omega)d\omega = 2\pi\, d_a[0] = 2\pi.
\]
Figure~\ref{fa:fig:dsexp} shows a few DTFT pairs $\mathbf{d}_a \leftrightarrow \mathbf{D}_a$ for different values of $a$ and it is obvious that
\[
\lim_{a \rightarrow 1} \mathbf{d}_a = \mathbf{1}
\]
that is, the finite-energy sequences $\mathbf{d}_a$ tend to the infinite energy sequence $\mathbf{1}$ as $a$ gets closer to one. At the same time, the corresponding Fourier transforms $\mathbf{D}_a$, whose area remains constant, become more and more concentrated around the origin while their maximum value $D_a(0)$ grows to infinity:
\[
\lim_{a \rightarrow 1} D_a(\omega) = \begin{cases}
\displaystyle \lim_{a \rightarrow 1} \frac{1+a}{1-a} & \omega = 0 \\ \\
0 & \omega \neq 0. \end{cases}
\]
Being finite-energy, every $\mathbf{d}_a$ induces a functional in $\ell_2(\mathbb{Z})$ and, according to~(\ref{fa:eq:FTfunc}),
\[
\langle \mathbf{d}_a, \mathbf{x} \rangle = \langle \mathbf{D}_a, \mathbf{X} \rangle
\]
for any DTFT pair $\mathbf{x} \leftrightarrow \mathbf{X}$; we also know from~(\ref{fa:eq:deltaFunctional}) that $1\{\mathbf{x}\} = \delta\{\mathbf{X}\} = X(0)$. But now:
\[
1\{\mathbf{x}\} = \lim_{a \rightarrow 1} \langle \mathbf{d}_a, \mathbf{x} \rangle =
\lim_{a \rightarrow 1} \langle \mathbf{D}_a, \mathbf{X} \rangle = \lim_{a \rightarrow 1} \frac{1}{2\pi} \int_{-\pi}^{\pi}D_a(\omega)X(\omega)d\omega = X(0).
\]
Moving the limit inside the integral and calling
\[
\delta(\omega) = \lim_{a \rightarrow 1} D_a(\omega)
\]
we can finally (if daringly) write
\[
\frac{1}{2\pi} \int_{-\pi}^{\pi}\delta(\omega)X(\omega)d\omega = X(0);
\]
that is, the ``function'' that induces the functional $\delta$ is something that looks more or less like
\begin{equation}
\delta(\omega) = \begin{cases} \infty & \omega = 0 \\ 0 & \omega \neq 0 \end{cases}
\end{equation}
and whose area is equal to $2\pi$. This odd animal is known in engineering circles as the Dirac delta.
TODO: show that this can be obtained very generally using other signal families: rect(n/2N), sinc(n/N), etc.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[t]
\def\plotRow#1{%
\hspace{1em}
\begin{dspPlot}[width=0.5\dspWidth,height=0.5\dspHeight,xticks=50,xout=true,yticks=1,sidegap=0]{-50,50}{0, 1.2}
\dspFunc[linecolor=ColorDT]{x abs #1 exch exp}
\dspText(-34,0.2){$\mathbf{d}_{#1}$}
\end{dspPlot}
&
\hspace{-1em}
\FPupn\dmax{#1 1 - #1 1 + / 0 trunc}
\begin{dspPlot}[width=0.5\dspWidth,height=0.5\dspHeight,xtype=freq,xticks=1,yticks=custom]{-1,1}{0,1.2}
\dspFunc[linecolor=ColorDF!30]{x 180 mul cos #1 -2 mul mul 1 add #1 dup mul add 1 #1 sub dup mul exch div }
\dspText(-0.7,0.21){$\mathbf{D}_{#1}$}
\dspCustomTicks[axis=y]{1, \dmax}
\end{dspPlot}}
%
\center
\small
\psset{unit=5mm}
\begin{tabular}{cc}
\plotRow{0.5}
\\
\plotRow{0.9}
\\
\plotRow{0.99}
\\
\plotRow{0.999}
\end{tabular}
\caption{Double-sided complex exponential sequences $d_a[n] = a^{|n|}$ and their DTFTs for increasing values of $a$.}\label{fa:fig:dsexp}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\itempar{The Dirac delta.} Talk about the use of the Dirac delta in engineering etc.
%This will require us to make the acquaintance with an oddball notational entity called the Dirac delta function, that is very popular with physicists and engineers.
\clearpage
\subsection{The DTFT of the unit step sequence}
The unit step $\mathbf{u}$ is a signal that is most commonly used as a multiplicative mask in order to impose causality to the mathematical expression for a sequence; we have often written the one-sided decaying exponential, for example, as
\begin{equation}\label{eq:fa:expdecay}
h_a[n] = a^n\,u[n], \qquad |a| < 1.
\end{equation}
But what is the DTFT of the unit step on its own? Obviously $\mathbf{u}$ is a power signal so its frequency representation will only exist in the sense of a functional. We could naively begin by remarking that the exponential decay in~(\ref{eq:fa:expdecay}) converges in the limit to the unit step as the parameter $a$ gets closer to one. Since
\[
H_a(\omega) = \frac{1}{1 - a e^{-j\omega}}
\]
we could consequently assume that
\[
U(\omega) = \lim_{a \rightarrow 1} H_a(\omega) = \frac{1}{1 - e^{-j\omega}}.
\]
Let's call
\[
S(\omega) = \frac{1}{1 - e^{-j\omega}}
\]
and notice right away that, by collecting $e^{-j\omega/2}$ at numerator and denominator we can write
\begin{equation}
S(\omega) = \frac{1}{2} - j\frac{1}{2}\cot\frac{\omega}{2}.
\end{equation}
Although divergent in $\omega = 0$, the function $\mathbf{S}$ seems sufficiently well-behaved to induce a functional over $L_2([-\pi, \pi])$, so we can try to verify if indeed it corresponds to the Fourier transform of the unit step. For this to happen, as detailed in Section~\ref{fa:sec:functionals}, for every valid DTFT pair $\mathbf{x} \leftrightarrow \mathbf{X}$, it should be
\[
u\{\mathbf{x}\} = S\{\mathbf{X}\}
\]
that is,
\[
\sum_{n = 0}^{\infty} x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S^{*}(\omega)X(\omega) d\omega.
\]
Now let's pick $\mathbf{x}$ real and symmetric, so that its DTFT will also be real and symmetric; the right-hand side becomes
\[
% \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(\frac{1}{2} + j\frac{1}{2}\cot\frac{\omega}{2}\right)X(\omega) d\omega &=
% \frac{1}{4\pi} \int_{-\pi}^{\pi} X(\omega) d\omega + \frac{j}{4\pi} \int_{-\pi}^{\pi} \cot\frac{\omega}{2}X(\omega) d\omega
\frac{1}{2\pi} \int_{-\pi}^{\pi} S^{*}(\omega)X(\omega) d\omega = \frac{1}{2} x[0] + \frac{j}{4\pi} \int_{-\pi}^{\pi} \cot\frac{\omega}{2}X(\omega).
\]
In order to compute the last integral, given that the cotangent is divergent in zero, we could use Cauchy's principal value theorem; but since our chosen $\mathbf{X}$ is real and symmetric, while $\cot(\omega/2)$ is real and antisymmetric, it is geometrically obvious that, if the integral is finite, its value will be zero. Since we can easily find real symmetric sequences for which
\[
\sum_{n = 0}^{\infty} x[n] \neq \frac{1}{2} x[0]
\]
$S$ is \textit{not}\/ the Fourier transform of the unit step.
Disappointed but not yet discouraged, we may try another approach; let's take the difference of two unit step sequences shifted by one:
\begin{equation}\label{fa:eq:unitstepdiff}
u[n] - u[n-1] = \delta[n].
\end{equation}
In the frequency domain this becomes
\[
U(\omega) - e^{-j\omega}U(\omega) = 1
\]
and, once again, we obtain $U(\omega) = S(\omega)$ --- which we know to be wrong! Clearly there is something going on here. The key observation is that the first-order difference in~(\ref{fa:eq:unitstepdiff}) remains valid even if we add an offset to the unit step\footnote{
Just as differentiation ``hides'' a constant offset from a function...}; call
\[
u_c[n] = u[n] + c, \qquad c \in \mathbb{R}
\]
and indeed
\[
u_c[n] - u_c[n-1] = \delta[n] \quad \forall c.
\]
This implies that there is an infinite number of sequences whose DTFT is $\mathbf{S}$, which contradicts the unicity of the Fourier transform. So now the question is: is there a value for the offset $c$ so that $\mathbf{u}_c$ and $\mathbf{S}$ form a true DTFT pair? The answer comes again from symmetry considerations: since $S(\omega) - 1/2$ is purely imaginary, the associated time-domain signal $u_c[n] - (1/2)\delta[n]$ must be antisymmetric; this implies
\[
u[n] + c - (1/2)\delta[n] = -(u[-n] + c - (1/2)\delta[n]);
\]
and, by considering $n$ first positive, then negative, then equal to zero, we can see that the condition is satisfied only by $c=-1/2$. The signal $\mathbf{s}$, defined as
\begin{equation}
s[n] = u[n] - \frac{1}{2} = \begin{cases} +\frac{1}{2} & n \ge 0 \\ -\frac{1}{2} & n < 0 \end{cases}
\end{equation}
is called the \textit{signum}\/ sequence and its DTFT (in the sense of functionals) is indeed $\mathbf{S}$. Since $u[n] = s[n] + (1/2)$, we finally have
\begin{equation}
U(\omega) = \frac{1}{1 - e^{-j\omega}} + \frac{1}{2}\tilde{\delta}(\omega).
\end{equation}
\clearpage
\begin{comment}
\itempar{Harmonic sinusoids.} The fundamental frequency for $\mathbb{C}^{N}$ is $2\pi/N$ and a complex exponential at a multiple of this frequency will coincide with a Fourier basis vector. Given the signal $\mathbf{x}$ defined by
\[
x[n] = e^{j\frac{2\pi}{N}mn}
\]
it is clearly $\mathbf{x} = \mathbf{w}_m$ and its DFT coefficients can be easily computed using~(\ref{eq:fa:orthogonalityCompact}) as
\[
X[k] = \langle \mathbf{w}_m, \mathbf{w}_k \rangle = N\delta[m-k]
\]
so that
\begin{equation}
\DFT{\mathbf{w}_m} = N\boldsymbol{\delta}_m;
\end{equation}
up to a normalization factor, this is the dual of~(\ref{eq:fa:DFTdelta}). Note that the result in the previous section, for the constant signal $\mathbf{x=1}$, is just a particular case of the above relationship when $m=0$.
We can now easily compute the DFT of standard trigonometric functions in which the frequency is harmonically related to the fundamental frequency for the space. Consider for instance the signal $\mathbf{x}$ defined by
\[
x[n] = \cos \left(\frac{\pi}{8} n \right), \qquad n=0,1,2,\ldots, 63.
\]
With a simple manipulation we can write:
\begin{align*}
\cos \left(4\frac{2\pi}{64} n \right) &= \frac{1}{2}\left( e^{j4\frac{2\pi}{64}n} + e^{-j4\frac{2\pi}{64}n}\right) \\
&= \frac{1}{2}\left( e^{j4\frac{2\pi}{64}n} + e^{j60\frac{2\pi}{64}n}\right)
\end{align*}
where we have used the fact that we can take all frequency indexes modulo 64 because of the aliasing property of complex exponentials; we can therefore express $\mathbf{x}$ as
\[
\mathbf{x} = \frac{1}{2}\mathbf{w}_4 + \frac{1}{2}\mathbf{w}_{60}.
\]
Since the signal is the sum of two Fourier basis vectors, orthogonality implies that only two of the inner products in~(\ref{eq:fa:fouriercoefficient}) will be nonzero:
\begin{align*}
X[4] &= \langle \mathbf{w}_4, \mathbf{x} \rangle = \langle \frac{1}{2}\mathbf{w}_4, \mathbf{w}_4 \rangle = 32 \\
X[60] &= \langle \mathbf{w}_{60}, \mathbf{x} \rangle = \langle \frac{1}{2}\mathbf{w}_{60}, \mathbf{w}_{60} \rangle = 32.
\end{align*}
The DFT of the signal is plotted in Figure~\ref{fig:fa:ex1}; the spectrum shows how the entire frequency content of the signal is concentrated over two single frequencies. Since the original signal is real-valued, the DFT component at $k=60$ ensures that the imaginary parts in the reconstruction formula cancel out; this symmetry is a general property of the Fourier transform that we will examine in more detail later.
Consider now a slight variation of the previous signal obtained by introducing a phase offset:
\[
x[n] = \cos \left(\frac{\pi}{8} n + \frac{2\pi}{3} \right), \qquad n=0,1,2,\ldots, 63.
\]
Again, we can easily manipulate the signal to obtain
\[
\mathbf{x} = \frac{e^{j2\pi/3}}{2}\mathbf{w}_4 + \frac{e^{-j2\pi/3}}{2}\mathbf{w}_{60}
\]
so that the resulting DFT coefficients are all zero except for
\begin{align*}
X[4] &= 32\,e^{j2\pi/3} \\
X[60] &= 32\,e^{-j2\pi/3}.
\end{align*}
The resulting DFT is plotted in Figure~\ref{fig:fa:ex2}; the magnitude does not change but the phase offset is reflected in nonzero values for phase at $k=4, 60$.
\subsection{Reading a DTFT Plot}
The DFT, being a change of basis in $\mathbb{C}^N$, maps a length-$N$ vector $\mathbf{x}$ onto another length-$N$ vector $\mathbf{X}$. The graphical representation of $\mathbf{x}$ is intuitively clear as a succession of values in time; for the DFT vector, as we said, we choose a graphical representation that splits the information into magnitude and phase. The (discrete) horizontal axis for the plot corresponds to the DFT coefficients' index $k$ which uniquely identifies the underlying frequency of the Fourier vector $\omega_k = (2\pi/N)k$; as we move along the horizontal axis from left to right, therefore, we will encounter coefficients that correspond to increasingly faster frequencies up to the midpoint $\lfloor k/2 \rfloor$, after which the frequencies will start to decrease again as shown in Figure~\ref{fig:fa:dftread}. The first half of the Fourier coefficients correspond to counterclockwise rotations, while the second half to clockwise rotations, as we saw in Section~\ref{sec:fa:cexp}.
As an example, consider the sequence:
\begin{equation}\label{DTFTexeq}
x[n] = \left\{ \! \begin{array}{ll}
1 & \mbox{ for } -N \leq n \leq N \\
0 & \mbox{ otherwise}
\end{array}
\right.
\end{equation}
Its DTFT can be computed as the sum\footnote{Remember that
$ \displaystyle %\[
\sum_{n=0}^{N}x^n = \frac{1 - x^{N+1}}{1-x} \, $.}
%\]}
\begin{align*}
X(\omega) &= \sum_{n=-N}^{N}e^{-j\omega n}
\\ &
= \sum_{n=1}^{N} e^{j\omega n} + \sum_{n=0}^{N} e^{-j\omega n}
\\
&= \frac{1-e^{-j\omega (N+1)}}{1-e^{-j\omega}} + \frac{1-e^{j\omega (N+1)}}{1-e^{j\omega}} - 1
\\
&= e^{j\omega/2}\,\frac{1-e^{-j\omega (N+1)}}{e^{j\omega/2} - e^{-j\omega/2}} + e^{-j\omega/2}\,\frac{1-e^{j\omega (N+1)}}{e^{-j\omega/2} - e^{j\omega/2}} - 1
\\
&= \frac{e^{j\omega (N+\frac{1}{2})} - e^{-j\omega
(N+\frac{1}{2})}}{e^{j\omega/2} - e^{-j\omega/2}}
\\ &
=
\frac{\sin \left(\omega \left(N+\frac{1}{2} \right)\right) }{\sin( \omega/2)}
\end{align*}
\begin{comment}
%%% 4.10 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\def\specShapeEx{x RadtoDeg %
dup 10.5 mul sin exch %
0.5 mul sin div}
\begin{figure}[h]
\center\small
\begin{psDFplot}[dy=5,pifrac=2]{-7}{23}
\psDFplotFunction[linewidth=1.2pt]{\specShapeEx}
\end{psDFplot}
\vskip-5mm
\caption{The DTFT of the signal in~(\ref{DTFTexeq}).}\label{DTFTexampleFigA}
\end{figure}
The DTFT of this particular signal turns out to be real (we
will see later that this is a consequence of the signal's
symmetry) and it is plotted in Figure~\ref{DTFTexampleFigA}.
When, as is very often the case, the DTFT is complex-valued,
the usual way to represent it graphically takes the
magnitude and the phase separately into account. The DTFT is
always a $2\pi$-periodic\index{spectrum!periodicity|mie}
function and the standard convention is to plot the interval
from $-\pi$ to $\pi$. Larger intervals can be considered if
the periodicity needs to be made explicit;
Figure~\ref{DTFTexampleFigB}, for instance, shows five full
periods of the same function.
%% 4.11
\begin{figure}[h]
\center\small
\begin{psDFplot}[dy=5,reps=5]{-7}{23}
\psDFplotFunction[linewidth=1.2pt,linecolor=gray]{\specShapeEx}
\psDFplotFunction[linewidth=1.2pt,fmin=-3.14,fmax=3.14]{\specShapeEx}
\end{psDFplot}
\vskip-5mm
\caption{The DTFT of the signal in~(\ref{DTFTexeq}), with
explicit periodicity.}\label{DTFTexampleFigB}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The DTFT as the Limit of a DFS}
A way to gain some intuition about the structure of the DTFT
formulas is to consider the DFS of periodic sequences with
larger and larger periods.
%increasingly longs periods.
Intuitively, as we look at the
structure of the Fourier basis for the DFS, we can see that
the number of basis vectors in~(\ref{DFTsyn}) grows with the
length $N$ of the period and,
\pagebreak%
consequently, the frequencies
of the underlying
complex exponentials become ``denser''
between $0$ and $2\pi$. We want to show that, in the limit,
we end up with the reconstruction formula of the DTFT.
To do so, let us restrict ourselves to the domain of
absolute summable sequences; for these sequences, we know
that the sum in~(\ref{DTFTsyn}) exists. Now, given an
absolutely summable sequence $x[n]$, we can always build an
$N$-periodic sequence $\tilde{x}[n]$ as
\index{periodization}
\begin{equation}\label{periodicx}
\tilde{x}[n] = \sum_{i = -\infty}^{\infty} x[n + iN]
\end{equation}
for any value of $N$ (see
Example~\ref{periodizationExample}); this is guaranteed by
the fact that the above sum converges for all $n \in
\mathbb{Z}$ (because of the absolute summability of $x[n]$)
so that all values of $\tilde{x}[n]$ are finite. Clearly,
there is overlap between successive copies of $x[n]$; the
intuition, however, is the following: since in the end we
will consider very large values for $N$ and since $x[n]$,
because of absolute summability, decays rather fast with
$n$, the resulting overlap of ``tails''{} will be negligible.
%In other words, we have:
This can be expressed as
\[
\lim_{N \rightarrow \infty} \tilde{x}[n] = x[n]
\]
Now consider the DFS of $\tilde{x}[n]$:
\begin{equation}
\tilde{X}[k] = \sum_{n=0}^{N-1} \tilde{x}[n] \,
e^{-j\frac{2\pi}{N}nk} = \sum_{i = -\infty}^{\infty}
\left( \sum_{n=0}^{N-1} x[n +iN] \, e^{-
j\frac{2\pi}{N}(n+iN)k} \right)
\end{equation}
where in the last term we have used~(\ref{periodicx}),
interchanged the order of the summation and exploited the
fact that $e^{-j(2\pi /N)(n+iN)k} = e^{-j(2\pi /N)nk}$. We
can see that, for every value of $i$ in the outer sum, the
argument of the inner sum varies between $iN$ and $iN+N -
1$, i.e.\ non-overlapping intervals, so that the double
summation can be simplified as
\begin{equation}
\tilde{X}[k] = \sum_{m = -\infty}^{\infty} x[m] \, e^{-j\frac{2\pi}{N}mk}
\end{equation}
and therefore
\begin{equation}\label{DFSCoefAsDTFT}
\tilde{X}[k] = X(e^{j\omega}) \bigl|_{\omega = \frac{2\pi}{N}k}
\end{equation}
This already gives us a noteworthy piece of intuition: the
DFS coefficients for the periodized signal are a discrete
set of values of its DTFT (here considered solely as a
formal operator) computed at multiples of $2\pi/N$. As $N$
grows, the spacing between these frequency intervals narrows
more and more so that, in the limit, the DFS converges to
the DTFT.
To check that this assertion is consistent, we can now write
the DFS reconstruction formula using the DFS values given to
us by inserting (\ref{DFSCoefAsDTFT}) in~(\ref{DFTrec}):
\begin{equation}
\tilde{x}[n] =
\frac{1}{N} \sum_{k = 0}^{N-1} X(e^{j\frac{2\pi}{N}k})
\, e^{j\frac{2\pi}{N}nk}
\end{equation}
By defining $\Delta = (2\pi/N)$, we can rewrite the above expression as
\begin{equation}
\tilde{x}[n] = \frac{1}{2\pi} \sum_{k = 0}^{N-1}
X(e^{j(k\Delta)}) \, e^{j(k\Delta) n}\, \Delta
\end{equation}
and the summation is easily recognized as the Riemann sum with step~$\Delta$ approximating the integral of $f(\omega) = X(e^{j\omega})e^{j\omega n}$ between $0$ and $2\pi$. As $N$ goes to infinity (and therefore $\tilde{x}[n] \rightarrow x[n]$), we can therefore write
\begin{equation}\label{finalDFSDTFT}
\tilde{x}[n] \rightarrow \frac{1}{2\pi} \int_{0}^{2\pi} X(e^{j\omega})\,
e^{j\omega n}\, d\omega
\end{equation}
which is indeed the DTFT reconstruction
formula~(\ref{DTFTrec}).\footnote{Clearly~(\ref{finalDFSDTFT})
is equivalent to~(\ref{DTFTrec}) in spite of the different
integration limits since all the quantities under the
integral sign are $2\pi$-periodic and we are integrating
over a period.}
\subsection{The DTFT as a Formal Change of Basis}\label{FourierDirac}
We %will
now show that, if we are willing to sacrifice
mathematical rigor, the DTFT can be cast in the same
conceptual framework we used for the DFT and DFS, namely as
a basis change in a vector space. The following formulas are
to be taken as nothing more than a set of purely symbolic
derivations, since the mathematical hypotheses under which
the results are well defined are far from obvious and are
completely hidden by the formalism. It is only fair to say,
however, that the following expressions represent a very
handy and intuitive toolbox to grasp the essence of the
duality between the discrete-time and the frequency domains
and that they can be put to use very effectively to derive
quick results when manipulating sequences.
One way of interpreting Equation~(\ref{DTFTsyn}) is to see that,
for any given value $\omega_{0}$, the corresponding value of
the DTFT is the inner product in $\ell_2(\mathbb{Z})$ of the
sequence $x[n]$ with the sequence $e^{j\omega_{0}n}$;
formally, at
least, we are still performing a projection in a
vector space akin to $\mathbb{C}^{\infty}$:
\[
X(e^{j\omega}) = \bigl\langle e^{j\omega n}, x[n] \bigr\rangle
\]
Here, however, the set of ``basis vectors''{}
$\{e^{j\omega n}\}_{\omega \in \mathbb{R}}$ is indexed by
the real variable $\omega$ and is therefore uncountable.
This uncountability is mirrored in the inversion
formula~(\ref{DTFTrec}), in which the usual summation is
replaced by an integral; in fact, the DTFT operator maps
$\ell_2(\mathbb{Z})$ onto $L_2 \bigl([-\pi, \pi] \bigr)$ which is a
space of $2\pi$-periodic, square integrable functions. This
interpretation preserves the physical meaning given to the
inner products in~(\ref{DTFTsyn}) as a way to measure the
frequency content of the signal at a given frequency; in
this case the number of oscillators is infinite and their
frequency separation becomes infinitesimally small.
To complete the picture of the DTFT as a change of basis, we
want to show that, at least formally, the set $\{e^{j\omega
n}\}_{\omega \in \mathbb{R}}$ constitutes an orthogonal
``basis''{} for $\ell_2(\mathbb{Z})$.\footnote{You can see here
already why this line of thought is shaky
unsafe: indeed,
$e^{j\omega n} \not\in \ell_2(\mathbb{Z})$!}
In order to do
so, we need to introduce a quirky mathematical entity called
the Dirac delta functional; this is defined in an implicit
way by the following formula\index{Dirac delta|mie}
\begin{equation}\label{deltadef}
\int_{-\infty}^{\infty} \delta(t-\tau) f(t)\, dt = f(\tau)
\end{equation}
where $f(t)$ is an arbitrary integrable function on the real line;
in particular
\begin{equation}
\int_{-\infty}^{\infty} \delta(t) f(t) \,dt = f(0)
\end{equation}
While no ordinary function satisfies the above equation,
$\delta(t)$ can be interpreted as shorthand for a limiting
operation. Consider, for instance, the family of parametric
functions\footnote{The rect function is discussed more
exhaustively in Section~\ref{idealFilters} its definition
is
\[ \mbox{rect}(x) = \left\{\!
\begin{array}{ll}
1 & \mbox{ for } |x| \leq 1/2 \\
0 & \mbox{ otherwise}
\end{array}
\right.
\]
}
\begin{equation}\label{limtodelta}
r_{k}(t) = k\,\mbox{rect}(kt)
\end{equation}
which are plotted in Figure~\ref{fig-delta-limit}. For any
continuous function $f(t)$ we can write
\begin{equation}
\int_{-\infty}^{\infty} r_k(t) f(t) \,dt
=
k \int_{-1/2k}^{1/2k} f(t)\, dt
=
f ( \gamma)\bigr|_{\gamma \in [-1/2k, 1/2k]}
\end{equation}
where we have used the Mean Value theorem. Now, as $k$ goes
to infinity, % we have that
the integral converges to $f(0)$; hence
%and so
we can say that the limit of the series of functions
$r_k(t)$ converges then to the Dirac delta.
As already stated, % we said,
the delta cannot be considered as a proper function, so the
expression $\delta(t)$ outside of an integral sign has no
mathematical meaning; it is customary however to associate
an ``idea''{} of function to the delta and we can think of it
as being undefined for $t \neq 0$ and to have a value of
$\infty$ at $t=0$. This interpretation, together
with~(\ref{deltadef}), defines the so-called \emph{sifting
property} of the Dirac delta; this property allows us to
write (outside of the integral sign):
\begin{equation}\label{sifting}
\delta(t-\tau)f(t) = \delta(t-\tau) \, f(\tau)
\end{equation}
%% 4.12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[h]
\center\small
\begin{psDFplot}[dy=2,labelx=false]{0}{7}
\selectPlotFont
\definecolor{mygray}{gray}{0.5}
\psset{linecolor=mygray}
\psDFplotData[linewidth=1.5pt](-1,0)(-0.5, 0)(-0.5,1)(0.5,1)(0.5,0)(1,0)
\definecolor{mygray}{gray}{0.4} \psset{linecolor=mygray}
\psDFplotData[linewidth=1.5pt](-1,0)(-0.25, 0)(-0.25,2)(0.25,2)(0.25,0)(1,0)
\definecolor{mygray}{gray}{0.3} \psset{linecolor=mygray}
\psDFplotData[linewidth=1.5pt](-1,0)(-0.1666, 0)(-0.1666,3)(0.1666,3)(0.1666,0)(1,0)
\definecolor{mygray}{gray}{0.2} \psset{linecolor=mygray}
\psDFplotData[linewidth=1.5pt](-1,0)(-0.125, 0)(-0.125,4)(0.125,4)(0.125,0)(1,0)
\definecolor{mygray}{gray}{0.1} \psset{linecolor=mygray}
\psDFplotData[linewidth=1.5pt](-1,0)(-0.1, 0)(-0.1,5)(0.1,5)(0.1,0)(1,0)
\definecolor{mygray}{gray}{0} \psset{linecolor=mygray}
\psDFplotData[linewidth=1.5pt](-1,0)(-0.0833, 0)(-0.0833,6)(0.0833,6)(0.0833,0)(1,0)
\psset{xunit=3.14}
\uput[-90](-1,0){-1} \uput[-90](1,0){1}
\uput[-90](-.5,0){-0.5} \uput[-90](0.5,0){0.5}
\uput[45](0.5,1){$k=1$}
\uput[45](0.25,2){$k=2$}
\uput[45](0.1666,3){$k=3$}
\uput[45](0.125,4){$k=4$}
\uput[45](0.1,5){$k=5$}
\uput[45](0.0833,6){$k=6$}
\end{psDFplot}
\vskip-5mm
\caption{The Dirac delta as the limit of a family of
rectangular functions.}\label{fig-delta-limit}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The physical interpretation of the Dirac delta is related to
quantities expressed as continuous \emph{distributions} for
which the most familiar example is probably that of a
probability distribution (pdf). These functions represent a
value which makes physical sense only over an interval of
nonzero measure; the punctual value of a distribution is
only an abstraction. The Dirac delta is the operator that
extracts this punctual value from a distribution, in a sense
capturing the essence of considering smaller and smaller
observation intervals.
To see how the Dirac delta applies to our basis expansion,
note that equation~(\ref{deltadef}) is formally identical to
an inner product over the space of functions on the real
line; by using the definition of such an inner product we
can therefore write
\begin{equation}\label{RecFormCont}
f(t) = \int_{-\infty}^{\infty}
\bigl\langle \delta(s-\tau), f(s) \bigr\rangle \, \delta(t-\tau) \, d\tau
\end{equation}
which is, in turn, formally identical to the reconstruction
formula of Section~\ref{InfSeqSpaceDef}. In reality, we are
interested in the space of $2\pi$-periodic functions, since
that is where DTFTs live; this is easily accomplished by
building a $2\pi$-periodic version of the delta
as\index{Dirac delta!pulse train}
\begin{equation}
\tilde{\delta}(\omega) = 2\pi \sum_{k=-\infty}^{\infty} \delta(\omega - 2\pi k)
\end{equation}
where the leading $2\pi$ factor is for later convenience.
The resulting object is called a \emph{pulse train},
similarly to what we built for the case of periodic
sequences in $\tilde{\mathbb{C}}^N$. Using the
%\emph
{pulse train}
and given any $2\pi$-periodic function $ f(\omega)$, the
reconstruction formula~(\ref{RecFormCont}) becomes
\begin{equation}
f(\omega) = \frac{1}{2\pi}
\int_{\sigma}^{\sigma+2\pi}
\bigl\langle \tilde\delta(\theta-\phi), f(\theta) \bigr\rangle
\, \tilde\delta(\omega-\phi) \, d\phi
\end{equation}
for any $\sigma \in \mathbb{R}$.
Now that we have the delta notation in place, we are ready
to start. First of all, we %will
show the formal
orthogonality of the basis functions $\{ e^{j\omega n}
\}_{\omega \in \mathbb{R}}$. We can write
\begin{equation}
\frac{1}{2\pi}\int_{-\pi}^{\pi} \tilde{\delta}
(\omega - \omega_{0})\, e^{j\omega n}d\omega =
e^{j\omega_0 n}
\end{equation}
The left-hand side of this equation has the exact form of
the DTFT reconstruction formula~(\ref{DTFTrec}); %and so
hence we
have found the fundamental relationship\index{Dirac delta!DTFT of} \begin{equation}\label{DTFTofexp}
e^{j\omega_0 n} \;
\stackrel{\scriptstyle\text{DTFT}}{\longleftrightarrow} \;
\tilde{\delta}(\omega - \omega_{0})
\end{equation}
Now, the DTFT of a complex exponential $e^{j\sigma n}$ is,
in our change of basis interpretation, simply the inner
product $\langle e^{j\omega n}, e^{j\sigma n} \rangle$;
because of~(\ref{DTFTofexp}) we can therefore express this as % write:
\begin{equation}
\bigl\langle e^{j\omega n}, e^{j\sigma n} \bigr\rangle =
\tilde{\delta}(\omega - \sigma)
\end{equation}
which is formally equivalent to the orthogonality relation in~(\ref{exportog}).
%\vspace{1em}
We %will
now recall for the last time that the delta notation
subsumes a limiting operation: the DTFT
pair~(\ref{DTFTofexp}) should be interpreted as shorthand
for the limit of the partial sums
\[
s_k(\omega) = \sum_{n=-k}^{k} e^{-j\omega n} \nonumber
\]
(where we have chosen $\omega_0 = 0$ for the sake of
example). Figure~\ref{DTFTconstFig} plots $|s_k(\omega)|$ for
increasing values of $k$ (we show only the $[-\pi,\pi]$
interval, although of course the functions are
$2\pi$-periodic). The family of functions $s_k(\omega)$ is exactly
equivalent to the family of functions $r_k(t)$ we saw
in~(\ref{limtodelta}); they too become
%narrower and narrower
increasingly narrow
while keeping a constant area (which turns out to be
$2\pi$). That is why we can %say
simply state that $s_k(\omega)
\rightarrow \tilde{\delta}(\omega)$.
\eject
%% 4.13 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[H]
\vskip2mm
\center\small
\begin{psDFplot}[dy=20,pifrac=2]{0}{81}
\psDFplotFile[linewidth=1.4pt,plotstyle=line]%,linecolor=gray]%
{data/fdelta5.txt}
\psDFplotFile[linewidth=0.9pt,plotstyle=line]%,linecolor=darkgray]%
{data/fdelta15.txt}
\psDFplotFile[linewidth=0.3pt,plotstyle=line]%,linestyle=dashed]%
{data/fdelta40.txt}
\selectPlotFont
\psset{xunit=3.14}
\uput[45](0.065,8){$k=5$} %0.5
\uput[45](0.02,20){$k=15$} %0.05
\uput[45](0.00,70){$k=40$}
\end{psDFplot}
\vskip-5mm
\caption{The sum
$\bigl|\sum_{n=-k}^{k} \, e^{-j\omega n} \bigr|$ for different values
of $k$.}\label{DTFTconstFig}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip
From~(\ref{DTFTofexp}) we can easily obtain other
interesting results: by setting $\omega_0 = 0$ and by
exploiting the linearity of the DTFT operator, we can derive
the DTFT of a constant sequence:
\begin{equation}
\alpha \; \stackrel{\scriptstyle
\text{DTFT}}{\longleftrightarrow} \;
\alpha \tilde{\delta}(\omega)
\end{equation}
or, using Euler's formulas, the DTFTs of sinusoidal functions:
\begin{align}
&\cos(\omega_0 n + \phi) \; \stackrel{\scriptstyle
\text{DTFT}}{\longleftrightarrow} \;
\frac{1}{2} \bigl[e^{j\phi} \tilde{\delta}(\omega -\omega_0) + e^{-j\phi} \tilde{\delta}(\omega + \omega_0 ) \bigr]
\label{DTFTCosDef} \\
&\sin(\omega_0 n + \phi) \;
\stackrel{\scriptstyle \text{DTFT}}{\longleftrightarrow}
\; \frac{-j}{2} \bigl[e^{j\phi} \tilde{\delta}(\omega -\omega_0)
- e^{-j\phi} \tilde{\delta}(\omega + \omega_0) \bigr] \label{DTFTSinDef}
\end{align}
As we can see from the above examples, we are defining the
DTFT for sequences which are not even square-summable;
again, these transforms are purely a notational formalism
used to capture a behavior, in the limit, as we showed before.\index{DTFT|)}
\end{comment}
c4science · Help