Tag Archives: Modulation

Theoretical BER of M-QAM in Rayleigh Fading

We have previously discussed the Bit Error Rate of M-QAM in Rayleigh Fading using Monte Carlo Simulation. We now turn our attention to calculation of Bit Error Rate (BER) of M-QAM in Rayleigh fading using analytical techniques. In particular we look at the method used in MATLAB function berfading.m. In this function the BER of 4-QAM, 16-QAM and 64-QAM is calculated from series expressions having 1, 3 and 5 terms respectively. These are given below (M is the constellation size and must be a power of 2).

if (M == 4)
     ber = 1/2 * ( 1 - sqrt(gamma_c/k./(1+gamma_c/k)) );
elseif (M == 16)
     ber = 3/8 * ( 1 - sqrt(2/5*gamma_c/k./(1+2/5*gamma_c/k)) ) ...
     + 1/4 * ( 1 - sqrt(18/5*gamma_c/k./(1+18/5*gamma_c/k)) ) ...
     - 1/8 * ( 1 - sqrt(10*gamma_c/k./(1+10*gamma_c/k)) );
elseif (M == 64)
     ber = 7/24 * ( 1 - sqrt(1/7*gamma_c/k./(1+1/7*gamma_c/k)) ) ...
     + 1/4 * ( 1 - sqrt(9/7*gamma_c/k./(1+9/7*gamma_c/k)) ) ...
     - 1/24 * ( 1 - sqrt(25/7*gamma_c/k./(1+25/7*gamma_c/k)) ) ...
     + 1/24 * ( 1 - sqrt(81/7*gamma_c/k./(1+81/7*gamma_c/k)) ) ...
     - 1/24 * ( 1 - sqrt(169/7*gamma_c/k./(1+169/7*gamma_c/k)) );

Although using these expressions we get very accurate BER but it is not that simple to calculate (the expressions become even more complicated for higher constellation sizes such as 256-QAM). Therefore we try to simplify these expressions by using only the first term in each expression. To our surprise the results match quite well with the results using the exact formulae. There is very minor difference at low signal to noise ratios but that can be easily bargained for the ease of calculation.

BER Equations

So here is our program for calculating the BER using the approximate method.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FUNCTION TO CALCULATE THE BER OF M-QAM IN RAYLEIGH FADING
% M: Input, Constellation Size
% EbNo: Input, Energy Per Bit to Noise Power Spectral Density
% ber: Output, Bit Error Rate
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [ber]= BER_QAM_fading (M, EbNo)

k=log2(M);
EbNoLin=10.^(EbNo/10);
gamma_c=EbNoLin*k;

if M==4
    %4-QAM
    ber = 1/2 * ( 1 - sqrt(gamma_c/k./(1+gamma_c/k)) );
elseif M==16
    %16-QAM
    ber = 3/8 * ( 1 - sqrt(2/5*gamma_c/k./(1+2/5*gamma_c/k)) );
elseif M==64
    %64-QAM
    ber = 7/24 * ( 1 - sqrt(1/7*gamma_c/k./(1+1/7*gamma_c/k)) );
else 
    %Warning
    warning('M=4,16,64')
    ber=zeros(1,length(EbNo));
end

semilogy(EbNo,ber,'o-')
xlabel('EbNo(dB)')
ylabel('BER')
axis([0 24 0.001 1])
grid on

return
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

QAM BER Approximate

So we see that the results match quite well with the results previously obtained through simulation. We will next tackle the problem of simplifying the expression for higher order modulations such as 256-QAM in both Rayleigh and Ricean channels.

M-QAM Bit Error Rate in Rayleigh Fading

We have previously discussed the bit error rate (BER) performance of M-QAM in AWGN. We now discuss the BER performance of M-QAM in Rayleigh fading. The one-tap Rayleigh fading channel is generated from two orthogonal Gaussian random variables with variance of 0.5 each. The complex random channel coefficient so generated has an amplitude which is Rayleigh distributed and a phase which is uniformly distributed. As usual the fading channel introduces a multiplicative effect whereas the AWGN is additive.

The function “QAM_fading” has three inputs, ‘n_bits’, ‘M’, ‘EbNodB’ and one output ‘ber’. The inputs are the number of bits to be passed through the channel, the alphabet size and the Energy per Bit to Noise Power Spectral Density in dB respectively whereas the output is the bit error rate (BER).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FUNCTION THAT CALCULATES THE BER OF M-QAM IN RAYLEIGH FADING
% n_bits: Input, number of bits
% M: Input, constellation size
% EbNodB: Input, energy per bit to noise power spectral density
% ber: Output, bit error rate
% Copyright RAYmaps (www.raymaps.com)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function[ber]= QAM_fading(n_bits, M, EbNodB)
% Transmitter
k=log2(M);
EbNo=10^(EbNodB/10);
x=transpose(round(rand(1,n_bits)));
h1=modem.qammod(M);
h1.inputtype='bit';
h1.symbolorder='gray';
y=modulate(h1,x);

% Channel
Eb=mean((abs(y)).^2)/k;
sigma=sqrt(Eb/(2*EbNo));
w=sigma*(randn(n_bits/k,1)+1i*randn(n_bits/k,1));
h=(1/sqrt(2))*(randn(n_bits/k,1)+1i*randn(n_bits/k,1));
r=h.*y+w;

% Receiver
r=r./h;
h2=modem.qamdemod(M);
h2.outputtype='bit';
h2.symbolorder='gray';
h2.decisiontype='hard decision';
z=demodulate(h2,r);
ber=(n_bits-sum(x==z))/n_bits
return
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

64-QAM Constellation

M-QAM Bit Error Rate in Rayleigh Fading
M-QAM Bit Error Rate in Rayleigh Fading

The bit error rates of four modulation schemes 4-QAM, 16-QAM, 64-QAM and 256-QAM are shown in the figure above. All modulation schemes use Gray coding which gives a few dB of margin in the BER performance. As with the AWGN case each additional bit per symbol requires about 1.5-2 dB in signal to ratio to achieve the same BER.

Although not shown here similar behavior is observed for higher order modulation schemes such as 1024-QAM and 4096-QAM (the gap in the signal to noise ratio for the same BER is increased to about 5dB).

Lastly we explain some of the terms used above.

Rayleigh Fading

Rayleigh Fading is a commonly used term in simulation of Digital Communication Systems but it tends to differ in meaning in different contexts. The term Rayleigh Fading as used above means a single tap channel that varies from one symbol to the next. It has an amplitude which is Rayleigh distributed and a phase which is Uniformly distributed. A single tap channel means that it does not introduce any Inter Symbol Interference (ISI). Such a channel is also referred to as a Flat Fading Channel. The channel can also be referred to as a Fast Fading Channel since each symbol experiences a new channel state which is independent of its previous state (also termed as uncorrelated).

Gray Coding

When using QAM modulation, each QAM symbol represents 2,3,4 or higher number of bits. That means that when a symbol error occurs a number of bits are reversed. Now a good way to do the bit-to-symbol assignment is to do it in a way such that no neighboring symbols differ by more than one bit e.g. in 16-QAM, a symbol that represents a binary word 1101 is surrounded by four symbols representing 0101, 1100, 1001 and 1111. So if a symbol error is made, only one bit would be in error. However, one must note that this is true only in good signal conditions. When the SNR is low (noise has a higher magnitude) the symbol might be displaced to a location that is not adjacent and we might get higher number of bits in error.

Hard Decision

The concept of hard decision decoding is important when talking about channel coding, which we have not used in the above simulation. However, we will briefly explain it here. Hard decision is based on what is called “Hamming Distance” whereas soft decision is based on what it called “Euclidean Distance”. Hamming Distance is the distance of a code word in binary form, such as 011 differs from 010 and 001 by 1. Whereas the Euclidean distance is the distance before a decision is made that a bit is zero or one.  So if the received sequence is 0.1 0.6 0.7 we get a Euclidean distance of 0.8124 from 010 and 0.6782 from 001. So we cannot make a hard decision about which sequence was transmitted based on the received sequence of 011. But based on the soft metrics we can make a decision that 001 was the most likely sequence that was transmitted (assuming that 010 and 001 were the only possible transmitted sequences).

Bit Error Rate of QPSK

Simulating a QPSK system is equivalent to simulating two BPSK systems in parallel. So there is no difference in bit error rate(BER). Since the simulation is at baseband we multiply the in-phase and quadrature streams by 1 and j respectively (instead of cos and sin carriers). At the receiver we just use the real and imag functions to separate the two symbol streams. The BER is the average BER of the two parallel streams.

As in the case of BPSK we can show that the baseband representation (using 1 and j)  is equivalent to using the passband representation (using cosine and sine). Lets assume the following signal model for QPSK.

s(t)=a(t)*cos(2*pi*f*t)+b(t)*sin(2*pi*f*t)

where a(t) and b(t) contain the information to be transmitted over the channel. Now at the receiver we multiply this signal with cos( ) to recover a(t) and sin( ) to recover b(t).

cos(2*pi*f*t)*s(t)

=cos(2*pi*f*t) [a(t)*cos(2*pi*f*t)+b(t)*sin(2*pi*f*t)]

=a(t)*cos(2*pi*f*t)*cos(2*pi*f*t) +b(t)*sin(2*pi*f*t)*cos(2*pi*f*t)

=(a(t)/2)*(1+cos(4*pi*f*t))+(b(t)/2)*sin(4*pi*f*t)

=(a(t)/2)+(a(t)/2)*cos(4*pi*f*t)+(b(t)/2)*sin(4*pi*f*t)

After low pass filtering (LPF) we get (a(t)/2), which it the required in-phase component scaled by a constant 1/2. Similarly we can find the quadrature component b(t).

sin(2*pi*f*t)*s(t)

=sin(2*pi*f*t) [a(t)*cos(2*pi*f*t)+b(t)*sin(2*pi*f*t)]

=a(t)*sin(2*pi*f*t)*cos(2*pi*f*t)+b(t)*sin(2*pi*f*t)*sin(2*pi*f*t)

=(a(t)/2)*sin(4*pi*f*t)+(b(t)/2)*(1-cos(4*pi*f*t))

=(a(t)/2)*sin(4*pi*f*t)+(b(t)/2)-(b(t)/2)cos(4*pi*f*t)

Again after low pass filtering (LPF) we are left with the quadrature component b(t) scaled by the constant 1/2. So we can conclude that the multiplication by the carrier terms at the transmitter and receiver is not required in simulation and we can simply transmit a(t) and b(t). We just have to make sure that a(t) and b(t) are orthogonal to each other so that they do not interfere.

So the transmitted QPSK signal would have the form.

s(t)=a(t)+j*b(t)

The steps involved in the simulation are

1. Generate a random sequence of symbols for the in-phase and quadrature components (-1 corresponding to binary value of 0 and +1 corresponding to binary value of 1). Add the in-phase and quadrature components in the form a(t)+j*b(t).

2. Generate complex samples of Additive White Gaussian Noise (AWGN) with the required variance (noise power = noise variance OR noise power = square of noise standard deviation OR noise power = noise power spectral density * signal bandwidth).

3. Add AWGN samples to the QPSK signal.

4. Detection is performed at the receiver by determining the sign of the in-phase and quadrature components.

5. And finally the bit error rate (BER) is calculated for the in-phase and quadrature components. Total bit error rate is the mean of the two values.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FUNCTION TO CALCULATE BER OF QPSK IN AWGN
% l - Input, length of the symbol sequence
% EbNo - Input, energy per bit to noise power spectral density
% ber - Output, bit error rate
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function[ber]=err_rate2(l,EbNo)       
si=2*(round(rand(1,l))-0.5);               % In-phase component
sq=2*(round(rand(1,l))-0.5);               % Quadrature component
s=si+j*sq;                              % QPSK symbol  
n=(1/sqrt(2*10^(EbNo/10)))*(randn(1,l)+j*randn(1,l)); % AWGN Noise
r=s+n;                                  % Received signal with noise
si_=sign(real(r));                        % Detected in-phase component 
sq_=sign(imag(r));                        % Detected quadrature component 
ber1=(l-sum(si==si_))/l;                  % Bit error rate in-phase
ber2=(l-sum(sq==sq_))/l;                  % Bit error rate quadrature
ber=mean([ber1 ber2]);                    % Mean bit error rate 
return
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

QPSK Constellation

QPSK BER

One final comment that I want to make is that bit error rate and symbol error rate is not always the same. Taking the example of QPSK a symbol error might occur when there is an error in the in-phase stream or the quadrature stream or both. So it is not a one to one mapping!!!

Note:

1. For a QPSK constellation centered around the origin of the co-ordinate system, the decision boundaries are simply defined by the x and y axes.

2. The reason for the name QPSK is that there are four symbols in the constellation, each having one of four possible phases (45, 135, 225, 315).