ICCBN Conference aims to provide a premier forum for researchers, industry practitioners and educators to present and discuss the most recent ideas, innovations, trends, experiences, and concerns in the emerging areas of Communication, Convergence, and Broadband Networking.

Important links

1. Program Schedule

As I see there are three major components for ICCBN 2008 : Technical paper presentations, Industry watch and the Tutorial sessions.

2. Industry watch

Industry leaders present their perspective and share their views on design, development, deployment, direction and regulation of existing and emerging communications and networking technologies that facilitate access, move and storage of network information and content.

3. Tutorials

Around 6 tutorials (some happening in parallel) spread over July19th and 20th are planned. The sessions are on Embedded Linux, Mobile phone OS, Nano technology, WiMax, Intellectual Property management, and Programming.

4. Registration

5. Accommodation

I plan to attend the only the full day tutorial session on Insights Into IEEE 802.16e (WiMAX) PHY and MAC layers Design Concepts and Road Ahead, by Prasad VTSV, Samsung.

Date: Sunday, 20th July 2008

Time: 9:00am to 5:00pm

If you plan to be around, for sure we can meet up.

is called a Rayleigh random variable. Further, the phase is uniformly distributed from . In this post we will try to derive the expression for probability density function (PDF) for and .

The text provided in Section 5.4.5 of [ELECTRONIC-COMMUNICATION:PRADIP] is used as reference.

Joint probability

The probability density function of is

.

Similarly probability density function of is

.

As and are independent random variables, the joint probability is the product of the individual probability, i.e,

.

The joint probability that the random variable lies between and and the random variable lies between and is,

.

Conversion to polar co-ordinate

Given that is in the Cartesian co-ordinate form, we can convert that into the polar co-ordinate where,
and
.

Figure: Cartesian co-ordinate to polar co-ordinate

The area is Cartesian co-ordinate form is equal to the area in the polar co-ordinate form.

.

Simplifying,
.

Summarizing the joint probability density function,

.

Since and are independent, the individual probability density functions are,
,

.

Simulation Model

Simple Matlab/Octave simulation model is provided for plotting the probability density of and . The script performs the following:

(a) Generate two independent zero mean, unit variance Gaussian random variables

(b) Using the hist() function compute the simulated probability density for both and

(c) Using the knowledge of the equation (which we just derived), compute the theoretical probability
density function (PDF)

(d) Plot the simulated and theoretical probability density functions (PDF) and show that they are in good agreement.

Click here to download Matlab/Octave script for simulating the probability density function (PDF) of Rayleigh random variable

Figure: Simulated/theoretical PDF of Rayleigh random variable

Figure: Simulated/theoretical PDF of uniformly distributed theta random variable

Reference

[ELECTRONIC-COMMUNICATION:PRADIP] Principles of Electronic Communications Analog - Digital, by Pradip Kumar Ghosh

Multipath environment

In a multipath environment, it is reasonably intuitive to visualize that an impulse transmitted from transmitter will reach the receiver as a train of impulses.

Figure: Impulse response of a multipath channel

Let the transmit bandpass signal be,
, where

is the baseband signal,
is the carrier frequency and
is the time.

As shown above, the transmit signal reaches the receiver through multiple paths where the path has an attenuation and delay . The received signal is,

.

Plugging in the equation for transmit baseband signal from the above equation,

.

The baseband equivalent of the received signal is,

,

where is the phase of the path.

The impulse response is,

.

Rayleigh fading model

The phase of each path can change by radian when the delay changes by . If is large, relative small motions in the medium can cause change of radians. Since the distance between the devices are much larger than the wavelength of the carrier frequency, it is reasonable to assume that the phase is uniformly distributed between 0 and radians and the phases of each path are independent (Sec 2.4.2 [WIRELESS-COMMUNICATION: TSE, VISWANATH]).

When there are large number of paths, applying Central Limit Theorem, each path can be modelled as circularly symmetric complex Gaussian random variable with time as the variable. This model is called Rayleigh fading channel model.

A circularly symmetric complex Gaussian random variable is of the form,

,

where real and imaginary parts are zero mean independent and identically distributed (iid) Gaussian random variables. For a circularly symmetric complex random variable ,

.

The statistics of a circularly symmetric complex Gaussian random variable is completely specified by the variance,

.

The magnitude which has a probability density,

is called a Rayleigh random variable.

This model, called Rayleigh fading channel model, is reasonable for an environment where there are large number of reflectors.

Reference

[WIRELESS-COMMUNICATION: TSE, VISWANATH] Fundamentals of Wireless Communication, David Tse, Pramod Viswanath

For the first two posts in the series are:

1. Understanding Shannon’s capacity equation

2. Bounds on Communication based on Shannon’s capacity

The article summarizes the symbol error rate derivations in AWGN for modulation schemes like BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK.

Based on the knowledge of bandwidth requirements for each type of modulation scheme, the capacity in bits/seconds/Hz is listed. Further, using the knowledge that the symbol to noise ratio is times the bit to noise ratio , the symbol error rate vs Eb/No curves are plotted. Using symbol error rate versus Eb/No plots, the Eb/No required for achieving symbol error rate of is identified. Upon having the capacity and Eb/No requirement, the requirements for BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK are mapped on to the Shannon’s capacity vs Eb/No curve.

Further, assuming Gray coded modulation mapping, each symbol error causes one bit out of bits to be in error. So, the relation between symbol error and bit error is,

.

Using this assumption, the Bit Error Rate (BER) for BPSK, QPSK, 4PAM, 16QAM, 16PSK, 64QAM and 32PSK are listed and the BER vs Eb/No curve plotted.

Hope this article serves as a nice overview of the various digital modulation schemes. Click here to read the full article in DSPDesignline.com

Further, the presentation in the IEEE TGN, PAPR in HT-LTF (11-06/1595r1), mentions that in 40MHz mode where a 128pt FFT is used, PAPR of HT-LTF (High Throughput Long Training Field) can be reduced by multiplying the upper 20MHz subcarriers by j. Using quick Matlab simulations, we will try to validate that claim for HT-LTF and further check the PAPR for a general random BPSK and QPSK modulation.

PAPR in HT-LTF

The peak to average power ratio for a signal is defined as
, where
corresponds to the conjugate operator.

Expressing in deciBels,
.

Matlab/Octave script for computing PAPR in HT-LTF
close all; clear all;
nFFT = 128; nDSC = 114;
% The 128pt HT-LTF sequence for 40MHz as defined in 11-06/1595r4
htLTF = [ zeros(1,6) 1 1 -1 -1 1 1 -1 1 -1 1 1 1 1 1 1 -1 -1 1 1 -1 1 -1 1 1 1 1 1 ...
1 -1 -1 1 1 -1 1 -1 1 -1 -1 -1 -1 -1 1 1 -1 -1 1 -1 1 -1 1 1 1 1 -1 -1 -1 1 0 ...
0 0 -1 1 1 -1 1 1 -1 -1 1 1 -1 1 -1 1 1 1 1 1 1 -1 -1 1 1 -1 1 -1 1 1 1 1 1 ...
1 -1 -1 1 1 -1 1 -1 1 -1 -1 -1 -1 -1 1 1 -1 -1 1 -1 1 -1 1 1 1 1 zeros(1,5)];



x1F = htLTF; % no rotation
x2F = [htLTF(1:64) j*htLTF(65:128)]; % with 90degree rotation



% Taking iFFT, time domain
x1t = (nFFT/sqrt(nDSC))*ifft(fftshift(x1F.')).'; % no rotation
x2t = (nFFT/sqrt(nDSC))*ifft(fftshift(x2F.')).'; % with rotation



% computing the peak to average power ratio for symbol without rotation
meanSquareValue1 = sum(x1t.*conj(x1t),2)/nFFT;
peakValue1 = max(x1t.*conj(x1t),[],2);
paprSymbol1 = peakValue1./meanSquareValue1;
paprSymbol1dB = 10*log10(paprSymbol1)
% computing the peak to average power ratio for symbol with 90 degree rotation
meanSquareValue2 = sum(x2t.*conj(x2t),2)/nFFT;
peakValue2 = max(x2t.*conj(x2t),[],2);
paprSymbol2 = peakValue2./meanSquareValue2;
paprSymbol2dB = 10*log10(paprSymbol2)

As claimed in slide7 of power point presentation PAPR In HT-LTF (11-06/1595r1), the PAPR with

(a) No rotation is 5.6317dB and

(b) With 90 degree rotation is 3.4066dB.

So, around 2.2dB reduction in PAPR is achieved.

PAPR for random BPSK and QPSK

Armed with the above result, I tried to simulate the PAPR for with random BPSK (and QPSK) modulation on a 128pt FFT with and without rotating the upper 20MHz subcarriers by j.

Click here to download the Matlab/Octave script for simulating PAPR for BPSK/QPSK modulation OFDM with j rotation

The results from the simulations is shown below.

Figure: OFDM PAPR for BPSK/QPSK with j rotation

Observations

1. For random BPSK sequence, the multiplication by j resulted in increase of PAPR by around 0.5dB.

2. For random QPSK sequence, the PAPR with and without multiplication by j is almost the same.

Based on the above simulation results, may I conclude that there is no noticable reduction in PAPR by multiplying upper 20MHz subcarriers by j. There might be some special cases (like HT-LTF sequence), where the multiplication by j reduces PAPR. However for random BPSK/QPSK sequence, there is no improvement.

In the first post in this series, we have discussed Shannon’s equation for capacity of band limited additive white Gaussian noise channel with an average transmit power constraint. The capacity is,

bits/second

where

is the capacity in bits per second, is the bandwidth in Hertz, is the signal power and is the noise spectral density.

Capacity with increasing signal power

Increasing the signal power will mean that we can split the signal level into more number of levels even while ensuring low probability of error. Hence increasing signal power will lead to more capacity. However, as the increase in capacity is a logarithmic function of power, the returns are diminishing.


Matlab/Octave script for plotting capacity vs power
B=1;
N0=1;
P= [0:10^4];
C = B.*log2(1+P./(N0*B));
plot(P,C); xlabel(’power, P’); ylabel(’bandwidth,B’); ylabel(’capacity, C bit/sec’); title(’Capacity vs Power’)

Figure: Capacity vs Power, keeping Noise and Bandwidth to unity

Can observe that increase in capacity is diminishing as we keep increase the value of power.

Capacity with increasing bandwidth

The second variable to play with is the bandwidth. Increasing the bandwidth has two effects:

1. More bandwidth means we can have more transmissions per second, hence higher the capacity.

2. However, more bandwidth also means that there is more noise power at the receiver.

The latter reduces the performance.

Let us try to evaluate the capacity equation when bandwidth tends to infinity i.e
.

From the Taylor series expansion, we know that

.

Applying this to the above equation,

.

This means that increasing bandwidth alone will not lead to increase of the capacity.

Matlab/Octave script for plotting capacity vs bandwidth
P = 1;
N0 = 1;
B = [1:10^3];
C = B.*log2(1+P./(N0*B));
plot(B,C)
xlabel(’bandwidth, B Hz’); ylabel(’capacity, C bit/sec’); title(’Capacity vs Bandwidth’)

Figure: Capacity vs Bandwidth, keeping signal power and noise power to unity

Can observe that the maximum achievable capacity by increasing bandwidth is 1.44 times the value.

Capacity (in bit/sec/Hz) vs Bit to noise ratio (Eb/No)

From our discussion till now, we have understood that a practical communication should have a rate which is lower than capacity , i.e.

bits/second.

Dividing both sides of the equation by bandwidth ,

bits/second/Hz.

Further, from our discussion on Bit error rate for 16PSK modulation using Gray mapping, we know that symbol to noise ratio is times the bit to noise ratio, i.e.
.

Substituting this into the capacity equation,
bits/second/Hz.

For notational convenience, let us define as the spectral efficiency in bits/second/Hertz.

The above equation can be equivalently represented as,

.

In the above equation, when tends to zero, the bit to noise ratio should be,

.

(Thanks to L’Hospital’s rule).

This means that for reliable communication, we need to have or equivalently expressing in decibels, .


Matlab/Octave script for plotting the capacity in Bits/sec/Hz vs Bit to noise ratio
r = [0:.001:10];
Eb_No_lin = (2.^r -1)./r;
Eb_No_dB = 10*log10(Eb_No_lin);
semilogy(Eb_No_dB,r)
axis([-2 20 0.1 10]); grid on
xlabel(’Bit to noise ratio, Eb/No dB’); ylabel(’Spectral efficiency, R/W bit/sec/Hz’)
title(’Spectral efficiency vs Bit to Noise ratio’)


Figure: Spectral efficiency vs bit to noise ratio

The above plot captures the equation,

.

It divides the area into two regions:

(a) In the region below the curve, reliable communication is possible and

(b) in the region above the curve, reliable communication is not possible.

Closer the performance of a communication system is to the curve, more optimal is the system.

In the next post in this series, we will discuss the performance of various modulation schemes like BPSK, QPSK, QAM etc by mapping them into various points in the above plot.

Reference

[COMM-SYS-PROAKIS-SALEHI] Fundamentals of Communication Systems by John G. Proakis, Masoud Salehi

Further, the following writeup is based on Section 12.5.1 from Fundamentals of Communication Systems by John G. Proakis, Masoud Salehi

Simple example with voltage levels

Let us consider that we have two voltage sources:
(a) Signal source which can generate voltages in the range to volts
(b) Noise source which can generate voltage levels to volts.

Figure: Discrete voltage levels with noise

Let us now try to send information at discrete voltage levels from the source (thick black lines as shown in the above figure). It is intuitive to guess that the receiver will be able to decode the received symbol correctly if the received signal lies within .

So, the number of different discrete voltages levels (information) which can be sent, while ensuring error free communication is the total voltage level divided by the noise voltage level i.e.
.

Extending to Gaussian channel

Let us transmit randomly chosen discrete voltage levels meeting the average power constraint,

, where is the signal power.

The noise signal follows the Gaussian probability distribution function

with mean and variance .

The noise power is,

.

The average total (signal plus noise) voltage over symbols is .

Similiarly, the average noise voltage over symbols is .

Combining the above two equations, the number of different messages which can be ‘reliably transmitted‘ is,

.

Note:

1. The product of the signal and noise accumulated over many symbols average to zero, i.e

.

2. Since the noise is Gaussian distributed, the noise can theoretically go from to . So the above result cannot ensure zero probability of error in receiver, but only arbitrarily small probability of error.

Converting to bits per transmission

With different messages, the number of bits which can be transmitted per transmission is,

bits/transmission.

Bringing bandwidth into the equation

Let us assume that the available bandwidth is .

Noise is of power spectral density spread over the bandwidth . So the noise power in terms of power spectral density and bandwidth is,

.

From our previous post on transmit pulse shaping filter that minimum required bandwidth for transmitting symbols with symbol period isHz. Conversely, if the available bandwidth is , the maximum symbol rate (transmissions per second) is .

Multiplying the equation for bits per transmission with transmission per second of and replacing the noise term , the capacity is

bits/second.

Voila! This is Shannon’s equation for capacity of band limited additive white Gaussian noise channel with an average transmit power constraint.

References

A Mathematical Theory for Communication, by Mr. Claude Shannon

[COMM-SYS-PROAKIS-SALEHI] Fundamentals of Communication Systems by John G. Proakis, Masoud Salehi