1.

What is the value of XI(ω) given \(\frac{1}{1-ae^{-jω}}\), |a|

Answer»

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

EASIEST explanation: Given, X(ω)=\(\frac{1}{1-ae^{-jω}}\), |a|<1

By multiplying both the numerator and DENOMINATOR of the above equation by the complex conjugate of the denominator, we obtain

X(ω)=\(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)

This expression can be SUBDIVIDED into REAL and imaginary parts, THUS we obtain

XI(ω)=\(\frac{-asinω}{1-2acosω+a^2}\).


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