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An LTI system is characterized by its impulse response h(n)=(1/2)^nu(n). What is the spectrum of the output signal when the system is excited by the signal x(n)=(1/4)^nu(n)?(a) \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)(b) \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)(c) \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)(d) \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)This question was addressed to me in quiz.The above asked question is from Frequency Domain Characteristics of LTI System in section Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer»

Correct answer is (b) \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)

The best I can explain: The FREQUENCY response FUNCTION of the system is

H(ω) = \(\sum_{n=0}^∞ (\frac{1}{2})^n e^{-jωn}\)

=\(\frac{1}{1-\frac{1}{2} e^{-jω}}\)

SIMILARLY, the input sequence x(n) has a Fourier transform

X(ω)=\(\sum_{n=0}^∞ (\frac{1}{4})^n e^{-jωn}\)

=\(\frac{1}{1-\frac{1}{4} e^{-jω}}\)

Hence the spectrum of the signal at the output of the system is

Y(ω)=X(ω)H(ω)

=\(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\).


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