Is the Fourier transform possible ?

[prathamesh29]

Can the function :
X^f (w) = cos (w/2) , w= R
be the fourier transform of a discrete signal x[n] ?

Please help with this question.
Regards,
Prathamesh

 

[albbg]

As far as I understood you have the Fourier transform X(ω)=cos(ω/2) of a discrete time signal x and you want to find x. It's not clear to me what you mean with "w=R". However I'll try to answer.

The inverse Fourier transform is:

\[x=\frac{ 1}{2\pi}*\int {X(\omega)}*{2}^{j\omega *n}*d\omega\]

we know that \[cos(\omega/2)=\frac{1}{2}*{(e}^{j\omega /2} +{e}^{-j\omega /2})*d\omega \]

then

\[x=\frac{ 1}{4\pi}*\int {(e}^{j\omega /2} +{e}^{-j\omega /2})*{e}^{j\omega *n}*d\omega\]

that is

\[x=\frac{ 1}{4\pi}*\int {(e}^{j(n+1)\omega /2} +{e}^{j(n-1)\omega /2})*d\omega\]

remembering that \[\delta=\frac{ 1}{\sqrt{4\pi}}*\int {e}^{j\omega}*d\omega\]

Thus:

\[x=sqrt{2\pi}*[\delta{[(n+1)/2]}+\delta{[(n-1)/2]}]\]

Hope it's correct.

 

[prathamesh29]

Hi albbg,
The solution for this says as : X^f(w) must be 2pi periodic if X^f(w) is dtft of x, but cos(0.5pi n) is not periodic,so dtft is not possible. Can you clarify this statement.
PS:w belongs to R i.e Real value
Thanks and regards,
Prathamesh

 

[albbg]

I corrected a bit the final results in my previous calculation, however from your last post it seems your function is X(ω)=cos(ω*n/2) instead of X(ω)=cos(ω/2) of the original post #1. That is also depends from "n". Is it correct ?