% Copyright (c) 2000 by Paul G. Flikkema. All Rights Reserved.
\documentclass{article}
\usepackage{techexpl}
\begin{document}
%\input{dissmacs.tex}%
\nocaret\bibfile{wdcbib.tex}
\aboveTopic{http://jan.ucc.nau.edu/\%7Epgf/ee599/class/index.html}
\previousTopic{wchaneffects.tex}
\nextTopic{divss.tex}
%\tocPage%

\baselinestretch{1.7}

\vspace*{-0.5in}

\hspace*{1.5in}
\includegraphics{http://jan.ucc.nau.edu/~pgf/ee599/interface/wdclogo.gif}

\vspace*{-0.5in}
%--------------------------------------------------------------------------
\section{1.8 Fading and Frequency Selective Channels}\label{sec-fafschan}
%----------------------------------------------------------------------

\subsection{Learning Objectives}

\begin{itemize}
\item Understand the fading and frequency selectivity characteristics
of wireless channels
\item Understand Rayleigh and Rician fading channel models
\item Understand the following terms:
\begin{itemize}
\item Delay spread and coherence bandwidth
\item Doppler spread and coherence time
\item Mean excess delay and RMS delay spread
\end{itemize}
\end{itemize}

\subsection{Preview}

The characteristics of wireless channels vary with both time and
frequency, and these characteristics can dramatically effect the performance
of wireless communication systems.  The two fundamental effects
are based on
\begin{enumerate}
\item the nature of the channel, i.e., the fact that multipath propagation
can cause fading or frequency selectivity (or both), and 
\item  the fact that
its characteristics can vary with time.
\end{enumerate}

\subsection{Two-Path Channel}

Suppose that a wireless channel consists of two specular paths between the
transmitter and receiver.  For simplicity, we assume the signal
arrives via the first path at time $t=0$, and via the second path at
time $t=\tau_1 > 0$.  The channel impulse response can then be modeled as
$$
h(\tau) = \alpha_0 e^{j \theta_0} \delta(t)
 + \alpha_1 e^{j \theta_1} \delta(t - \tau_1),
$$
Since both paths are specular, two replicas of the received signal
will appear (at times 0 and $\tau_1$), changed only in magnitude and
phase.  If we transmit a pulse $p_T(t)$ of duration $T$, then, ignoring
noise, the receiver input (channel output) will be (show this!)
$$
y(t) = \alpha_0 e^{j \theta_0} p_T(t)
 + \alpha_1 e^{j \theta_1} p_T(t - \tau_1)
$$
In many situations of interest, the pulse duration $T$ is much greater
than the \textit{delay spread} or \textit{multipath spread} $\tau_M$ of the
channel.  In crude terms, the delay spread is the difference between the
arrival times at the receiver of the first and last replicas of the
transmitted pulse (we will refine this later).  In this example,
$\tau_M = \tau_1$.

For the case $\tau_M \ll T$, the pulses almost overlap perfectly.  Thus
we can say $\tau_1 \approx 0$ and write
$$
y(t) \approx (\alpha_0 e^{j \theta_0}  + \alpha_1 e^{j \theta_1})p_T(t)
$$
If $\alpha_0 \approx \alpha_1$, then the two pulses will add
constructively if their phases are about the same, but they can also
add destructively if they are more than $\pi/2$ out of phase---with
the worst case being a phase difference of $\pi$.  This is the essence
of multipath \textit{flat-fading} channels: as time evolves, the phases
and amplitudes of the pulse replicas vary, causing the received
signal to fade in and out.

It is very important to see that the \textit{shape} of the pulse has
changed very little---in spite of changes in magnitude and phase, the
pulse waveform is still intact.  This can be clearly seen by viewing
the problem from the frequency domain: if the transmitted pulse has
Fourier transform (FT) $P_T(f)$, then the received pulse will
have FT
$$
Y(f) = (\alpha_0 e^{j \theta_0}  + \alpha_1 e^{j \theta_1}) P_T(f)=
\alpha e^{j \theta} P_T(f).
$$
Thus the frequency spectra of the transmitted and received pulses
differ only by an attenuation and phase change.

Now, recall that the received signal spectrum $Y(f)$ can be computed as the
product of the transmit signal spectrum $X(f)$ and the frequency response of
the channel.  That is, for any transmitted signal $x(t)$ and channel
impulse response $h(t)$,
$$
Y(f) = H(f) X(f).
$$

Since the received pulse was not distorted by the channel, we would
expect the channel's magnitude response $|H(f)|$ to be flat (i.e.,
constant).  This is indeed the case, as you can confirm by taking the
Fourier transform of $h(t)$.  In fact, what is necessary is for it to
be flat (and the phase response to be linear) over the range of
frequencies occupied by the transmitted signal.  Technically, we want
the \textit{coherence bandwidth} of the channel---the range of
frequencies over which the magnitude response varies little---to
exceed the bandwidth of the transmitted signal.  The coherence
bandwidth, denoted $(\Delta f)_c$, is a measure of the variation of
the channel with frequency.  In essence, it is a measure of the range
of frequencies over which the channel can be considered flat. It
can also be defined as the range of frequencies over which the
amplitudes of two different frequency components are strongly
correlated over time or space.

Thus we can say that if the channel's coherence bandwidth exceeds the
signal spectrum, the channel is \textit{frequency-flat}, or
\textit{frequency non-selective}.  Below, we will discuss the other
case, that of frequency selectivity.  But first we need to look at two
more realistic models for frequency-flat channels: the
\textit{Rayleigh} and \textit{Ricean} models.

\subsection{Many-Path Flat Fading Channel and Rayleigh Fading}

We can now extend our two-path model to many paths.  Let the total number of
paths be $\Lambda$.  Then the channel impulse response will be
$$
h(t) = \sum_{\lambda=0}^{\Lambda} \alpha_\lambda e^{j \theta_\lambda}
\delta (t - \tau_\lambda)
$$ If we continue to assume flat fading, this means that $\tau_\lambda
- \tau_0$ is very small compared to the pulse duration, and we can 
write (again assuming $\tau_0 = 0$)
$$
h(t) \approx \sum_{\lambda=0}^{\Lambda} \alpha_\lambda e^{j \theta_\lambda}
\delta (t)
$$
Recall that each path $\lambda$ contributes a complex gain
$\alpha_\lambda e^{j \theta_\lambda}$.  A powerful statistical model
that has been corroborated in practice is based on the following assumptions:
\begin{itemize}
\item The real and imaginary parts of every path gain are statistically
independent, and are independent of the real and imaginary parts of
other path gains.
\item They have means of zero.
\item $\Lambda$ is reasonably large ($>10$); that is, there are many paths
from the transmitter to the receiver.
\end{itemize}
Under these assumptions, $|y_I(t)|$ and $|y_Q(t)|$ both become
zero-mean Gaussian-distributed random variables at any time $t$ by the
central limit theorem of probability.  Then it can be shown that the
envelope of the received signal envelope,
$$
\rho(t) = \sqrt{y_I^2(t) + y_Q^2(t)},
$$
is Rayleigh-distributed, and we have the important \textit{Rayleigh-fading}
channel model.

Let us be specific.  First define $\bar{p}$ as the variance, or average power,
of each of the two quadrature components $y_I(t)$ and $y_Q(t)$.  Then
the probability density function (pdf) for the envelope $\rho$
(dropping the dependence on time for clarity) is
$$
f(\rho | \bar{p}) =  \frac{\rho}{\bar{p}}
\exp \left[ - \frac{\rho^2}{2 \bar{p}}\right], \quad \rho \ge 0.
$$
The notation on the left of the equal sign is designed to indicate that
the density of the envelope is conditioned on the power of the quadrature
components $\bar{p}$.

The cumulative density function (cdf) for $\rho$ (again conditioned
on $\bar{p}$) is
$$
F_{\rho}(R) = \text{Prob}[\rho \le R] = \int_0^R f(\rho | \bar{p}) d \rho
= 1 - \exp \left[ - \frac{R^2}{2 \bar{p}}\right], \quad \rho \ge 0.
$$

\subsection{Rician Fading}

In the Rayleigh fading model, each contributing path is random with
mean zero.  This implies that there is no path between the receiver
that provides a strong, non-varying replica of the signal.  Indeed,
the Rayleigh model implies that all components are scattered.
Obviously, we need to be able to handle cases where a strong,
non-varying, \textit{direct} component exists, such as
from a line-of-sight (LOS) path or very strong reflected path.
This can be handled by a simple extension to the Rayleigh model.

Now, recall that the Rayleigh model leads to a complex envelope that
can be modeled as a complex Gaussian random variable, with a magnitude
(envelope) that is distributed according to the Rayleigh pdf.  Let the
mean power of this scattered signal be $\bar{q}_s$.  Also, let the
power of the direct component be $\bar{q}_d$.  (Note that this is not
a mean power since it is not random.)  Thus the total mean power is
$\bar{p} = \bar{q}_s + \bar{q}_d$, and the pdf for the
\textit{Rician fading channel} is
$$
f(\rho | \bar{q}_s, \bar{q}_d) =  \frac{\rho}{\bar{q}_s}
\exp \left( - \frac{\rho^2 + 2 \bar{q}_d}{2 \bar{q}_s}\right)
 I_0 \left( \frac{\rho \sqrt{2 \bar{q}_d}}{\bar{q}_s} \right).
$$
Here, $I_0(\cdot)$ is defined as the zeroth-order modified Bessel function
of the first kind.

So far we have shown the probabilistic descriptions for two of the
most important fading models (note that there are more!).  Now we will
conclude this section by reformulating the Ricean model to show more directly
the relative contributions of the direct and scattered components.  We define
the \textit{Ricean K factor} as
$$
K = \frac{\text{direct power}}{\text{mean scattered power}} =
\frac{\bar{q}_d}{\bar{q}_s}
$$
Thus $K$ summarizes the severity of channel fading.  For example, it
is clear that $K = \infty$ implies that the channel is benign---it
does not fade at all, and becomes an ideal channel (cf. Topic YYY).
On the other hand, $K=0$ corresponds to the Rayleigh fading case.
Clearly, the larger $K$ is, the better, since the signal fluctuations
become smaller relative to the non-varying component. In most
wireless environments, $K$ ranges from about 4 to about 16.

By making two substitutions, we can re-write the Ricean pdf using $K$:
$$
f(\rho | \bar{p}, K) =  \frac{\rho}{\bar{p}} (1+K)e^{-K} 
\exp \left( - \frac{1+K}{2 \bar{p}} \rho^2 \right)
I_0 \left( \rho \sqrt{\frac{2 K(1+K)}{\bar{p}}} \right).
$$

Another fading model that you may see in the literature is the
\textit{Nakagami} $m$-\textit{distribution}, where $m$ is a
model parameter.  This model is more complex than the Rayleigh
model. However, it has been a better fit to empirical data in a number
of cases, and simplifies to the Rayleight model when $m=1$.  We will
not cover it here, preferring to focus on the physical insight
provided by the Rayleigh and Ricean models.

\subsection{ISI and Frequency Selectivity}

Up to this time we have considered the case where the delay spread of
the channel is a small fraction of the symbol pulse duration $T$.
Now, if the data rates increases, one option is to simply shorten the
duration of the signaling pulse to send more bits/second through the
channel.  However, this makes the delay spread larger relative to the
pulse duration, causing the received pulse to be smeared version of
the transmitted pulse. For a delay spread of $\tau_M$, the smearing
causes the received pulse to be of duration $T + \tau_M$.

If one could just design a receiver to detect the received pulse, then
all would be OK.  But something more problematic occurs when you
consider transmitting not just an isolated pulse, but a series of
pulses, one after the other. For example, if you want to send a block
of $N$ symbols, the modulator output would be
$$
\sum_{n=0}^{N-1} a_n p_T(t-nT).
$$
Then the receiver input would be 
$$
\sum_{n=0}^{N-1} a_n \tilde{p_T}(t-nT),
$$
where $\tilde{p_T}$ is the received pulse.
However, since the pulse start times arrive at the receiver separated by
$T$ and the received pulses are of duration $T + \tau_M$, they must overlap.
This is \textit{intersymbol interference (ISI)}.

Consider the situation in the frequency domain.  The smearing in time
implies that the shape of the received pulses is now substantially
different from the transmitted pulses.  The distortion must be from a
filtering effect caused by the channel, and implies that the channel
is not flat (if it were, the pulse shape would be unchanged).
Therefore, the channel has significant variations in its response
within the bandwidth of the signal, or is \textit{frequency-selective}.
Thus we can say that a frequency-selective channel causes ISI.

The frequency selectivity can be assessed by comparing the signal
bandwidth--- call it $W$---with the coherence bandwidth $(\Delta f)_c$
of the channel. If we have $W < (\Delta f)_c$, the channel is flat.
On the other hand, if $W > (\Delta f)_c$, there are significant variations
channel response variations within the signal bandwidth, and the channel
is frequency-selective.

Note that this relationship is the frequency-domain dual of the
relationship between the pulse duration and the delay spread.  In this
sense, \textit{time dispersion is the dual of frequency selectivity}.
In fact, the delay spread and the coherence bandwidth are inversely
proportional.  This can be seen by comparing the frequency responses
of a channels with short (small delay spread) and long (large delay spread)
impulse responses.

A practical time-domain rule of thumb is that the
channel is dispersive (or, equivalently, frequency-selective) if
$\tau_M > \frac{1}{10} T$.  If the channel dispersion exceeds this
threshold, equalization (a form of compensation) is necessary to
maintain reliable communications.  Equalization will be discussed
later in the course.

\subsection{Time Variation and Doppler Spread}

Now suppose that the transmitted signal is a pure carrier.  Then any
motion in the environment---whether the transmitter, receiver, or
other objects--- can lead to a Doppler shift of the carrier.  Since
this shift will vary with time as the environment changes, the overall
effect is termed \textit{Doppler spread}.  Just as delay spread smears
the pulse shape of a transmitted signal, Doppler spread will smear the
carrier frequency of a signal.

A small Dopper spread implies that the channel varies slowly, and is
relatively constant for segments of time.  Such as channel is termed a
\textit{slow-fading channel.} Conversely, a \textit{fast-fading
channel} is associated with a large Doppler spread.  Thus we can
associate the Doppler spread with its time-domain dual, the
\textit{coherence time}, which is the time interval for which the
channel is relatively constant.  In this way, we will have a
relationship analogous to the delay spread/coherence bandwidth
relationship for frequency selectivity.  If $B_d$ is the Doppler
spread and $(\Delta t)_c$ is the coherence time, then we have that
$B_d$ is inversely proportional to $(\Delta t)_c$.


\subsection{Quantitative Measures for Wireless Channels}

The \textit{multipath intensity profile (MIP)}, or power delay
profile, is defined as the received power level as a function of
excess delay (with respect to a fixed delay reference), and is given
for the specular time-invariant channel by
\begin{equation}
P(\tau) = \sum_{\lambda=0}^{\Lambda} P(\tau_{\lambda}) \delta(\tau-\tau_\lambda),
\end{equation}
where $P(\tau_{\lambda}) = |\alpha_\lambda|^2$.  Since the parameters vary due to the
changing multipath environment, the average small-scale power delay
profile can be found experimentally by averaging power delay profile
measurements over a local area.

Once the MIP is found, the delay spread can be characterized several ways.
One of the most useful is the RMS (root mean square) delay spread, which
can be thought of as the standard deviation of the spread.

To find the RMS delays spread we first computer the mean excess delay:
\begin{equation}
\overline\tau = \frac{\sum_{\lambda}P(\tau_{\lambda})\tau_{\lambda}}{\sum_{\lambda}P(\tau_{\lambda})} 
                \label{eq:e211}.
\end{equation}
The RMS (root mean square) delay spread is defined as follows:
\begin{equation}
\sigma_{\tau} = \sqrt{\overline{\tau^{2}} - (\overline\tau)^{2}},
\label{eq:e212}
\end{equation}
where 
\begin{equation}
\overline{\tau^{2}} = \frac{\sum_{\lambda}P(\tau_{\lambda})\tau_{\lambda}^{2}}
                          {\sum_{\lambda}P(\tau_{\lambda})} 
                        \label{eq:e213}.
\end{equation}

Sometimes the maximum excess delay $\tau_{\bf X} -
\tau_{0}$ is used, where $\tau_{0}$ is the first arriving signal and
$\tau_{\bf X}$ is the maximum delay at which a multipath component is
within {\bf X} dB of the strongest arriving multipath signal.

The exact relationship between delay spreads and coherence bandwidth
is a function of the multipath structure.  A common approximation is
\begin{equation} 
(\Delta f)_{c} \approx \frac{1}{2\pi{\sigma_{\tau}}}\quad, \label{eq:e215}
\end{equation}
where we note that indeed the delay spread and coherence bandwidth
are inversely related.

To summarize, a highly frequency-selective channel (i.e., with a small
coherence bandwidth) will cause severe delay dispersion of a
transmitted pulse.  The degree of this dispersion is captured by the
delay spread.

The delay spread statistics for wireless channels differ significantly
based on the environment.  The table below gives typical RMS delay spread
values for three typical wireless environments.

\vspace*{0.2in}
\begin{table}
\begin{center}
\begin{tabular}{|c||c|c|}
\hline
                  & {\bf RMS Delay Spread}    &            \\
{\bf Environment} & {\bf (typical)}           &  {\bf Comment}          \\
\hline 
{\bf Macrocell} & 0.1--10 $\mu$s    &  Worst case 100 $\mu$s          \\ 
(1--20 km)      &     &                  \\ \hline
{\bf Microcell} & 10--100 ns        &             \\ 
(0.1--1 km)     &     &                  \\ \hline
{\bf Indoor}    & 30--60 ns    &       Worst case 300 ns      \\ 
                &     &                (large buildings)  \\ \hline
\end{tabular}
\end{center}
\caption{\label{ta-bt}Typical values of RMS delay spread.}
\end{table}
\vspace*{0.1in}


\subsection{Summary}

\begin{itemize}
\item The wireless channel may distort the received signal, causing
ISI.  If the delay spread is more than a small fraction of the signal
pulse duration (or, equivalently, if the coherence bandwidth is less
than the pulse bandwidth), then the channel is frequency-selective.
If not, then the channel is called flat, or frequency non-selective.
Note that in both cases, multiple replicas of the transmitted signal
arrive at the receiver.  The key distinction is how closely the
replicas overlap in time.

\item Changes of the multipath structure of the channel with time
cause signal fading.  
\end{itemize}

\subsection{Reading Assignment}

Chapter 2, Sections 2.5--2.9

\subsection{Homework}

1. This problem is concerned with the Rayleigh fading channel.  Show
using simple arguments that the mean-square value of the envelope,
given as $E[\rho^2]$, is $2 \bar{p}$.

2. For the Rayleigh channel, why is $\rho$ always non-negative?
 
3. It turns out that $I_0(0)=1$. Use this result to show that the Ricean
density simplifies to the Rayleigh density in the case where there is
no direct component.

4. Calculate the mean excess delay and the RMS delay spread for the MIP
shown below. Estimate the coherence bandwidth of the channel.

\includegraphics{http://jan.ucc.nau.edu/~pgf/ee599/interface/dBgraph.gif}

5. Suppose that the RMS delay spread of a channel is 5 $\mu$s.  What is
the highest channel symbol rate that can be employed without equalization?
(Hint: recall that the channel symbol rate is the inverse of the pulse
duration.)

6. Assume a channel's impulse response is a square pulse of duration
$\tau$ (thus its maximum delay spread is $\tau$).  By considering the
magnitude of the frequency response (FT) of the channel, describe
how the delay spread and coherence bandwidth are inversely proportional.

\subsection{References}

The classic source for fading channels is:

\textit{Microwave Mobile Communications}, W. C. Jakes (ed.), 1974 (reprinted
by IEEE Press).

\vspace*{0.1in}
\hrule
\vspace*{0.1in}

\textbf{Wireless Digital Communications}
\copyright 2000 by Paul G. Flikkema. All Rights Reserved.

\includegraphics{http://www.nau.edu/image/small-blue.gif}
\end{document}




