## DSP System Toolbox |

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

Author(s): Scott C. Douglas

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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].

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)))+... j*sign(randn(ntr,1)).*(2+sign(randn(ntr,1))); Hs = dsp.SignalSource(s,'SamplesPerFrame',1000); plot(s,'o'); axis([-4 4 -4 4]); axis square xlabel('Re\{s(n)\}'); ylabel('Im\{s(n)\}'); title('Input signal constellation');

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 hFV.Analysis = 'freq';

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); plot(x,'.'); xlabel('Re\{x[n]\}'); ylabel('Im\{x[n]\}'); axis([-40 40 -40 40]); axis square title('Received signal x[n]'); fig = gcf; fig.Color = [1 1 1];

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);

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); step(hlog,e); plot(y,'.'); hold on xlabel('Re\{y[n]\}'); ylabel('Im\{y[n]\}'); axis([-5 5 -5 5]); axis square title('Equalized signal y[n]'); fig = gcf; fig.Color = [1 1 1]; end

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; semilogy(ntrain,abs(hlog.Buffer(ntrain)).^2); xlabel('Number of iterations'); ylabel('|e[n]|^2') title('Squared magnitude of the training errors'); fig = gcf; fig.Color = [1 1 1];