# Cross-Correlation of Two Exponential Sequences

Compute and plot the cross-correlation of two 16-sample exponential sequences, ${x}_{a}=0.8{4}^{n}$ and ${x}_{b}=0.9{2}^{n}$, with $n\ge 0$.

N = 16;
n = 0:N-1;

a = 0.84;
b = 0.92;

xa = a.^n;
xb = b.^n;

r = xcorr(xa,xb);

stem(-(N-1):(N-1),r)

Determine $c$ analytically to check the correctness of the result. Use a larger sample rate to simulate a continuous situation. The cross-correlation function of the sequences ${x}_{a}\left(n\right)={a}^{n}$ and ${x}_{b}\left(n\right)={b}^{n}$ for $n\ge 0$, with $0, is

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

cn = (1 - (a*b).^(N-abs(nn)))/(1 - a*b) .* ...
(a.^nn.*(nn>0) + (nn==0) + b.^-(nn).*(nn<0));

Plot the sequences on the same figure.

hold on
plot(nn,cn)

xlabel('Lag')
legend('xcorr','Analytic')

Verify that switching the order of the operands reverses the sequence.

figure

stem(-(N-1):(N-1),xcorr(xb,xa))

hold on
stem(-(N-1):(N-1),fliplr(r),'--*')

xlabel('Lag')
legend('xcorr(x_b,x_a)','fliplr(xcorr(x_a,x_b))')

Generate the 20-sample exponential sequence ${x}_{c}=0.7{7}^{n}$. Compute and plot its cross-correlations with ${x}_{a}$ and ${x}_{b}$. Output the lags to make the plotting easier. xcorr appends zeros at the end of the shorter sequence to match the length of the longer one.

xc = 0.77.^(0:20-1);

[xca,la] = xcorr(xa,xc);
[xcb,lb] = xcorr(xb,xc);

figure

subplot(2,1,1)
stem(la,xca)
subplot(2,1,2)
stem(lb,xcb)
xlabel('Lag')