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Equalization in Digital Communications

This example shows how to apply adaptive filters to channel equalization in digital communications.

Author(s): Scott C. Douglas


Channel equalization is a simple way of mitigating the detrimental effects caused by a frequency-selective and/or dispersive communication link between sender and receiver. For this example, all signals are assumed to have a digital baseband representation. During the training phase of channel equalization, a digital signal s[n] that is known to both the transmitter and receiver is sent by the transmitter to the receiver. The received signal x[n] contains two signals: the signal s[n] filtered by the channel impulse response, and an unknown broadband noise signal v[n]. The goal is to filter x[n] to remove the inter-symbol interference (ISI) caused by the dispersive channel and to minimize the effect of the additive noise v[n]. Ideally, the output signal would closely follow a delayed version of the transmitted signal s[n].

Transmitted Input Signal

A digital signal carries information through its discrete structure. There are several common baseband signaling methods. We shall use a 16-QAM complex-valued symbol set, in which the input signal takes one of sixteen different values given by all possible combinations of {-3, -1, 1, 3} + j*{-3, -1, 1, 3}, where j = sqrt(-1). Let's generate a sequence of 5000 such symbols, where each one is equiprobable.

ntr = 5000;
j = sqrt(-1);
s = sign(randn(ntr,1)).*(2+sign(randn(ntr,1)))+...
Hs = dsp.SignalSource(s,'SamplesPerFrame',1000);
axis([-4 4 -4 4]);
title('Input signal constellation');

Transmission Channel

The transmission channel is defined by the channel impulse response and the noise characteristics. We shall choose a particular channel that exhibits both frequency selectivity and dispersion. The noise variance is chosen so that the received signal-to-noise ratio is 30 dB.

b = exp(j*pi/5)*[0.2 0.7 0.9];
a = [1 -0.7 0.4];
% Transmission channel filter
channel = dsp.BiquadFilter('SOSMatrix',[b,a]);
% Impulse response
hFV = fvtool(channel,'Analysis','impulse','Color','White');
legend(hFV, 'Transmission channel');

% Frequency response
set(hFV, 'Analysis', 'freq')

Received Signal

The received signal x[n] is generated by the transmitted signal s[n] filtered by the channel impulse response with additive noise v[n]. We shall assume a complex Gaussian noise signal for the additive noise.

sig = sqrt(1/16*(4*18+8*10+4*2))/sqrt(1000)*norm(impz(channel));
v = sig*(randn(ntr,1) + j*randn(ntr,1))/sqrt(2);
x = step(channel,s) + v;
Hx = dsp.SignalSource(x,'SamplesPerFrame',Hs.SamplesPerFrame);
axis([-40 40 -40 40]);
title('Received signal x[n]');
set(gcf, 'Color', [1 1 1])

Training Signal

The training signal is a shifted version of the original transmitted signal s[n]. This signal would be known to both the transmitter and receiver.

D = 10;
Hd = dsp.Delay(D);

Trained Equalization

To obtain the fastest convergence, we shall use the conventional version of a recursive least-squares estimator. Only the first 2000 samples are used for training. The output signal constellation shows clusters of values centered on the sixteen different symbol values--an indication that equalization has been achieved.

P0 = 100*eye(20);
lam = 0.99;
h = dsp.RLSFilter(20,'ForgettingFactor',lam,'InitialInverseCovariance',P0);
hlog = dsp.SignalSink('BufferLength',2); % Store 2 frames (2000 samples)
for k = 1:2
    d = step(Hd,step(Hs));
    Rx = step(Hx);
    [y,e] = step(h,Rx,d);
    hold on
    axis([-5 5 -5 5]);
    title('Equalized signal y[n]');
    set(gcf, 'Color', [1 1 1])

Training Error

Plotting the squared magnitude of the error signal e[n], we see that convergence with the RLS algorithm is fast. It occurs in about 60 samples with the equalizer settings chosen.

hold off
ntrain = 1:1000;
xlabel('Number of iterations');
title('Squared magnitude of the training errors');
set(gcf, 'Color', [1 1 1])

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