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\basiclecturehead{Wireless Networks}{1}

\begin{center}
\Large{\bf Transmission Fundamentals}
\end{center}

\medskip

\begin{itemize}
\item
Signals
\item
Data rate and bandwidth
\begin{itemize}
\item
 Nyquist sampling theorem
\item
 Shannon capacity theorem
\end{itemize}
\end{itemize}

\section{Signals}
\label{sec:signals}
Any electromagnetic signal (actually, any function $f(t)$ that is
square-integrable) can be expressed as a superposition of sinusoidal
waves.  A general sinusoidal signal is given by the function $s(t) = A
\sin(2\pi ft + \phi)$, where $A$ is the amplitude (usually in volts),
$f$ is the frequency, and $\phi$ is the phase.  A more accurate way of
stating this is using Fourier analysis and Fourier transforms, as
follows.
\begin{eqnarray*}
s(t) & = & \int_{-\infty}^{+\infty} S(f) e^{i2\pi f t} df\\
S(f) & = & \int_{-\infty}^{+\infty} s(t) e^{-i2\pi f t} dt.
\end{eqnarray*}
The function $S(f)$ is referred to as the Fourier transform of $s(t)$,
and captures how the energy of the signal $s(t)$ is distributed within
the frequency spectrum.

Studying signals in the frequency domain greatly faciliates our
understanding of how signals are transmitted.  A channel takes as
input an input signal $s(t)$ and outputs a signal $r(t)$.  A channel
can be typically described by its impulse response function $h(t)$
which is the channel output corresponding to an infinitesimally narrow
pulse of unit area at time $t$.  Given an input $s(t)$ to a channel
with impulse response $h(t)$, the output $r(t)$ is given by
\[
r(t) = \int_{-\infty}^{+\infty} s(\tau) h(t - \tau) d\tau.
\]
Using the Fourier transform equations, we obtain:
\begin{eqnarray*}
r(t) & = & \int_{-\infty}^{+\infty} \left( h(t - \tau)
\int_{-\infty}^{+\infty} S(f) e^{i2\pi f \tau} df \right) d\tau\\
& = & \int_{-\infty}^{+\infty} \left( S(f) \int_{-\infty}^{+\infty}
h(t - \tau) e^{i2\pi f \tau} d\tau\right) df\\
& = & \int_{-\infty}^{+\infty} S(f) e^{i2\pi f t}
\int_{-\infty}^{+\infty} \left( h(t - \tau)
e^{-i2\pi f (t-\tau)} d\tau\right) df\\
& = & \int_{-\infty}^{+\infty} S(f) e^{i2\pi f t}
\int_{-\infty}^{+\infty} \left( h(\tau)
e^{-i2\pi f \tau} d\tau\right) df\\
& = & \int_{-\infty}^{+\infty} S(f) H(f) e^{i2\pi f t} df.
\end{eqnarray*}
If $R(f)$ is the Fourier transform of $r(t)$, then we obtain the
following simple input-output relation in the frequency domain
\[
R(f) = S(f)H(f).
\]

The absolute bandwidth of the signal $s(t)$ (resp., a channel with
impulse response $h(t)$) is the range of frequencies $f$ in which
$S(f) \neq 0$ (resp., $H(f) \neq 0$).  The effective bandwidth, or
simply, the bandwidth of the signal is the range within which most of
the energy is concentrated.  The energy of a signal $s(t)$ (per unit
resistance) is given by
\[
\int_{-\infty}^{+\infty} s(t)^2 dt.
\]

It can be shown that any periodic signal $s(t)$ with frequency $f$ has
non-zero frequency components only at frequencies that are multiples
of $f$.  That is, a periodic signal $s(t)$ with frequency $f$ can be
expressed as:
\[
s(t) = \sum_{n = -\infty}^{\infty} S_n e^{2\pi i nft}.
\]

A periodic square wave with amplitude $A$ and $-A$ can be expressed as
follows.
\[
s(t) = A \times \frac{4}{\pi} \sum_{{\scriptstyle \mbox{odd }} k}
\frac{sin(2\pi kft)}{k}.
\]

\section{Data rate and bandwidth}
\label{sec:rate}
Suppose we superpose sinusoids of frequencies $f$, $3f$, and $5f$,
then we get a wave that resembles a square wave with 3 peaks per half
cycle.  The bandwidth of the signal is $5f - f = 4f$.  In a time
interval of $1/f$ seconds, two bits have been transmitted, achieving a
bit rate of $2f$.  If we double the frequency $f$, this doubles the
bandwidth and hence also doubles the bit rate.  If we reduce the
bandwidth, say by using only sinusoids $f$ and $3f$, then we can
achieve the same bit rate, but the signal has a higher chance of being
misinterpreted since it is ``farther'' from the desired square wave
(and hence can cause more errors).

In general, the question relating bit rate and channel bandwidth is
not about what bit rate can a given channel bandwidth support since
over any finite length of time, there is a non-zero probability that
every bit transmitted is in error.  Consequently, any discussion of
capacity of a channel has to be done together with the notion of an
error rate.  Shannon framed the capacity question as: given a
particular channel bandwidth, what is the maximum rate at which
information can be sent, for a given error rate?  Shannon's channel
coding theorem gives a surprising answer to this question, which
implies that non-zero information rate is achievable even for an
arbitrarily low required error rate.

\begin{Theorem}[Nyquist Sampling Theorem]{~\cite[Section 6A]{stallings:wirelessBook}}
Any signal $s(t)$ that contains no frequency component more than $W$
Hz is completely determined by its values at times $j/2W$, where $j$
is an integer in the range $[-\infty, +\infty]$.
\end{Theorem}
\begin{Proof}
Consider a periodic pulse function $p(t)$ that consists of a unit
pulse (Dirac delta function) every $T = 1/2W$ seconds.  Since any
periodic function can be expressed as a sum of sinusoidals with
frequencies that are multiples of the function, we have
\[
p(t) = \sum_{n = -\infty}^{\infty} P_n e^{i(2\pi n/T)t}
\]
for some appropriate values of $P_n$.  Let the input signal be $s(t)$
with Fourier transform $S(f)$.  The sampled signal $x(t)$ can be
written as
\[
x(t) = s(t)p(t) = \sum_{n = -\infty}^{\infty} P_n s(t) e^{i(2\pi
n/T)t}.
\]
Therefore, the Fourier transform of the sampled signal is
\begin{eqnarray*}
X(f) & = & \int_{-\infty}^{+\infty} \sum_{n = -\infty}^{\infty} P_n x(t)
e^{i(2\pi n/T)t} e^{-i2\pi f t} dt\\
& = & \int_{-\infty}^{+\infty} \sum_{n = -\infty}^{\infty}
P_n x(t) e^{-i2\pi (f - n/T)} dt.
\end{eqnarray*}
We also have
\[
S(f - n/T) = \int_{-\infty}^{+\infty} x(t) e^{-i2\pi (f - n/T)} dt.
\]
Therefore, we obtain 
\[
X(f) = \sum_{-\infty}^{+\infty} P_n S(f - n/T) 
\]
We know that $S(f) = 0$ for $|f| \ge W$.  Since $1/T = 2W$, we obtain
that for $-W \le f \le W$, $X(f) = P_0 S(f)$.  Thus the frequency
spectrum of $s$ can be entirely determined by the frequency spectrum
of the sampled signal $x$.
\end{Proof}

There are several different proofs of the Nyquist sampling theorem.
One equality that directly captures the equivalence of the
band-limited signal $s$ and the sample $x$ is the following.
\begin{eqnarray}
s(t) = \sum_{n = -\infty}^{\infty} x_n \frac{\sin(\pi(2Wt -
n))}{\pi(2Wt - n)} \label{eqn:nyquist}
\end{eqnarray}

\begin{Theorem}[Shannon Capacity Theorem]{~\cite{shannon:noise}}
Let $P$ be the average transmitted power, and suppose the noise of
white thermal noise of power $N$ in the frequency band $W$.  Then,
there exists an encoding scheme that allows to transmit bits at a rate
\[
C = W \log_2 \frac{P + N}{N}
\]
with as small an error probability as desired.  It is not possible by
any encoding method to send at a higher rate and have an arbitrarily
low frequency of errors.
\end{Theorem}
\begin{Sketch}
We consider a signal $s(t)$ over an interval of period $T$.  Using
Nyquist sampling, we specify the signal using $2TW$ samples $x_1$
through $x_{2TW}$.  Using Equation~\ref{eqn:nyquist} and the energy
formula, one can show that the energy $E$ (into unit resistance) of
the signal $s(t)$ equals
\[
\int_{-\infty}^{\infty} s(t)^2 dt = \frac{1}{2W} \sum_{n = 1}^{2TW}
x_n^2.
\]
So if $d$ is the distance of the point $(x_1, x_2, \ldots, x_n)$ from
the origin, then we have 
\[
d^2 = 2WE = 2WPT,
\]
where $P$ is the average signal power during the period of length $T$.
Thus any signal over time period $T$ with average power at most $P$
can be represented by a point in a ball of radius $\sqrt{2WPT}$ in
$2WT$-dimensional space.

We now consider the effect of additive white Gaussian noise (AWGN) on
this signal.  An AWGN signal has the property that each sample is
perturbed independently of all others and the distribution of the
perturbation is Gaussian with standard deviation $\sqrt{N}$ where $N$
is the average noise power.

It can be shown that for any desired error probability $\eps$, if $T$
is sufficiently large, then the noise causes the perturbation to lie
within a ball of radius $\sqrt{2WNT}$ from the signal point $(x_1,
\ldots, x_{2WT})$ with probability $1 - \eps$.

When transmitting the information, we need to make sure that two
signals carrying different information should not be decoded as the
same.  So we need to select signal points such that the balls of
radius $\sqrt{2WNT}$ around them do not intersect.  How many such
signal points can we select?  The answer is essentially the ratio of
the volume of the ball of radius $\sqrt{2WT(N+P)}$ to that of
$\sqrt{2WNT}$.  For $2WT$-dimensional space, this ratio equals
\[
M = \left(\frac{\sqrt{2WT(N+P)}}{\sqrt{2WNT}} \right)^{2TW} = \left(\frac{P+N}{N}\right)^{TW}.
\]
It is fairly easy to see, using a ball-packing argument, that the
right-hand-side is an upper bound on $M$.  To see the lower bound,
Shannon gave a much more involved argument.  It relies on random
coding, in which we map each of $M$ pieces of an information to a
random point in the ball of radius $\sqrt{2WNT}$.  Shannon then goes
on to show that the probability that any two balls overlap can be made
arbitrarily close to zero by setting $T$ sufficiently high.

The term $M$ is the number of different pieces of information we can
send in $T$ time units.  So the information rate is given by
\[
C = \frac{\log_2 M}{T} = W \log_2 \frac{P+N}{N}.
\]
\end{Sketch}

\BfPara{Implications of Shannon's Theorem} Shannon's Theorem is
universally applicable.  If we wish to increase the capacity
(information rate), Shannon's Theorem says that we can achieve that by
either increasing the bandwidth $W$ of the underlying channel or
increasing the signal strength $P$.  

Note that neither $W$ nor $P$ can be increased arbitrarily.  There are
limits on $W$ due to technology (semiconductor constraints) and
regulation; furthermore, increase in bandwidth also increases the
associated noise.  In fact, under idealized conditions $N$ increases
linearly with $W$.  In particular, $N = N_0 W$, where $N_0$ is
referred to as the Noise Power Density; $N_0 = kT$, where $k$ is the
Boltzmann constant and $T$ is the temperature of the median (in
Kelvin).

The signal strength $P$ also cannot be increased arbitrarily.  There
are regulatory constraints; furthermore, there are potential health
hazards of very high-powered signals.

\section{Analog and digital data, signals, and transmissions}
\label{sec:transmit}
Simply put, the term {\em analog}\/ refers to ``continuous''
functions, over time, while {\em digital}\/ refers to ``discrete''
functions over time.  Examples of analog data include voice, video,
and graphics, while examples of digital data include text and stored
images.  Analog signals are continuously varying signals; unguided
wireless media transmit analog signals.  A digital signal is a
sequence of voltage pulses; it is generally cheaper and less
susceptible to noise, though suffers more attenuation than analog
signals.

We can have either analog or digital data transmitted using analog or
digital signals.  The process of converting data into signals that can
be carried by a given transmission medium is called {\em modulation}\/
and the reverse process is called {\em demodulation}.  We briefly
discuss the four possibilities here; these will be elaborated in more
detail in the following weeks.

\BfPara{Analog-to-analog} A canonical example of transmitting analog
data using analog signals is the telephone.  In this case, the
frequency spectrum of the underlying data (voice) matches that of the
carrier (telephone wire) -- this is referred to as baseband
transmission.  For several other applications (e.g., cellular
telephony), one needs to modulate the analog data so that the
frequency spectrum of the resulting signal matches that of the
carrier.

\BfPara{Digital-to-analog} When digital data are sent using analog
signals, the discrete data is encoded in terms of the carrier signals
of the underlying transmission medium.  Some transmission media such
as wireless only admit analog signals, so transmitting digital content
through them requires digital-to-analog conversion.  Another example
is digital data over telephone wires.

\BfPara{Analog-to-digital} When analog data are transmitted using
digital signals, the data is essentially digitized to obtain a bit
stream, which is transmitted using a digital signal.  Digitization may
be done even in cases where the final transmission medium admits
analog signals; e.g., digitzation of voice signals prior to
transmission over wireless media, to improve quality.

\BfPara{Digital-to-digital} Finally, when digital data is transmitted
using digital signals, a simple encoding scheme can be used that
essentially represents the two bit values using two different voltage
levels.  There are other encoding schemes that are used when the
signal needs to satisfy some other properties (e.g., synchronization)
which are not satisfied by this simple scheme.

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