1.

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

Answer»

Right CHOICE is (C) \(\FRAC{1-acosω}{1-2acosω+a^2}\)

The explanation is: 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

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


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