Question

What is the auto correlation of the sequence x(n)=a^nu(n), 0<a<l?

(a) \(\frac{1}{1-a^2}\) a^l (l≥0)

(b) \(\frac{1}{1-a^2}\) a^-l (l<0)

(c) \(\frac{1}{1-a^2}\) a^|l|(-∞<l<∞)

(d) All of the mentioned

The question was posed to me in my homework.

Query is from Correlation of Discrete Time Signals topic in division Discrete Time Signals and Systems of Digital Signal Processing

Answer 1

Right choice is (d) All of the mentioned

Explanation: rxx(l)=\(\sum_{n=-\infty}^{\infty} x(n)x(n-l)\)

For l≥0, rxx(l)=\(\sum_{n=l}^{\infty} x(n)x(n-l)\)

=\(\sum_{n=l}^{\infty} a^n a^{n-l}\)

=\(a^{-l}\sum_{n=l}^{\infty} a^{2n}\)

=\(\frac{1}{1-a^2}a^l\)(l≥0)

For l<0, rxx(l)=\(\sum_{n=0}^{\infty} x(n)x(n-l)\)

=\(\sum_{n=0}^\infty a^n a^{n-l}\)

=\(a^{-l}\sum_{n=0}^{\infty} a^{2n}\)

=\(\frac{1}{1-a^2}a^{-l}\)

So, rxx(l)=\(\frac{1}{1-a^2}a^{|l|}\) (-∞<l<∞)