Question

What is the energy density spectrum Sxx(ω) of the signal x(n)=a^nu(n), |a|<1?

(a) \(\frac{1}{1+2acosω+a^2}\)

(b) \(\frac{1}{1+2asinω+a^2}\)

(c) \(\frac{1}{1-2asinω+a^2}\)

(d) \(\frac{1}{1-2acosω+a^2}\)

The question was asked during an online interview.

I would like to ask this question from Frequency Analysis of Discrete Time Signal topic in chapter Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer 1

Correct answer is (d) \(\frac{1}{1-2acosω+a^2}\)

Explanation: Since |a|<1, the sequence x(n) is absolutely summable, as can be verified by applying the geometric summation formula.

\(\sum_{n=-∞}^∞|x(n)| = \sum_{n=-∞}^∞ |a|^n = \frac{1}{1-|a|} \lt ∞\)

Hence the Fourier transform of x(n) exists and is obtained as

X(ω) = \(\sum_{n=-∞}^∞ a^n e^{-jωn}=\sum_{n=-∞}^∞ (ae^{-jω})^n\)

Since |ae^-jω|=|a|<1, use of the geometric summation formula again yields

X(ω)=\(\frac{1}{1-ae^{-jω}}\)

The energy density spectrum is given by

Sxx(ω)=|X(ω)|^2= X(ω).X*(ω)=\(\frac{1}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1}{1-2acosω+a^2}\).