1.

What is the Fourier transform of the signal x(n)=a^|n|, |a|

Answer»

Right answer is (B) \(\frac{1-a^2}{1-2acosω+a^2}\)

Explanation: First we observe x(N) can be expressed as

x(n)=x1(n)+x2(n)

where x1(n)= a^n, n>0

=0, elsewhere

x2(n)=a^-n, n<0

=0, elsewhere

Now applying Fourier transform for the above TWO signals, we get

X1(ω)=\(\frac{1}{1-ae^{-jω}}\) and X2(ω)=\(\frac{ae^{jω}}{1-ae^{jω}}\)

Now, X(ω)=X1(ω)+ X2(ω)=\(\frac{1}{1-ae^{-jω}}+\frac{ae^{jω}}{1-ae^{jω}}=\frac{1-a^2}{1-2acosω+a^2}\).


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