Autocorrelation Function of Exponential Sequence

Compute the autocorrelation function of a 28-sample exponential sequence, $$x=0.95^n$$ for $$n\ge 0$$.

a = 0.95;

N = 28;
n = 0:N-1;
lags = -(N-1):(N-1);

x = a.^n;
c = xcorr(x);

Determine $$c$$ analytically to check the correctness of the result. Use a larger sample rate to simulate a continuous situation. The autocorrelation function of the sequence $$x(n)=a^n$$ for $$n\ge 0$$, with $$|a|<1$$, is

fs = 10;
nn = -(N-1):1/fs:(N-1);

dd = (1-a.^(2*(N-abs(nn))))/(1-a^2).*a.^abs(nn);

Plot the sequences on the same figure.

stem(lags,c);
hold on
plot(nn,dd)
xlabel('Lag')
legend('xcorr','Analytic')
hold off

Repeat the calculation, but now find an unbiased estimate of the autocorrelation. Verify that the unbiased estimate is given by $$c_u(n)=c(n)/(N-\left|n\right|)$$.

cu = xcorr(x,'unbiased');

du = dd./(N-abs(nn));

stem(lags,cu);
hold on
plot(nn,du)
xlabel('Lag')
legend('xcorr','Analytic')
hold off

Repeat the calculation, but now find a biased estimate of the autocorrelation. Verify that the biased estimate is given by $$c_b(n)=c(n)/N$$.

cb = xcorr(x,'biased');

db = dd/N;

stem(lags,cb);
hold on
plot(nn,db)
xlabel('Lag')
legend('xcorr','Analytic')
hold off

Find an estimate of the autocorrelation whose value at zero lag is unity.

cz = xcorr(x,'coeff');

dz = dd/max(dd);

stem(lags,cz);
hold on
plot(nn,dz)
xlabel('Lag')
legend('xcorr','Analytic')
hold off

See Also

Functions

• xcorr