[SOLVED] Fourier transform of delta(9-5n)

[blade88]

Hello,

I solved an exercise, asking for the Fourier transform of the signal \[\delta(9-5n)\].

First, I changed the signal form:

Because delta function is symmetric, I rewrote it as \[\delta(5n-9)\]
and then I used the scale property: \[\delta(5(n-\frac{9 }{5 }))=\frac{1 }{5 }\delta(n-\frac{9 }{5 })\]

Then, the fourier transform is: \[\frac{1 }{5 }exp(-j\frac{ 9}{ 5}\Omega)\].

However, the solution given by the professor is : \[exp(-j\frac{ 9}{ 5}\Omega)\].

Could someone please tell me whether I am doing something wrong in my solution?

Thank you

 

[andre_luis]

Seems like you did not scaled correctly.
Check if this is what should be done:

\[\delta(5(n-\frac{9 }{5 }))=5\delta(n-\frac{9 }{5 })\]

 

[blade88]

I am applying this property https://en.wikipedia.org/wiki/Dirac_delta_function#Scaling_and_symmetry

where it seems that the number 5 should go into the denominator.

Also, I computed the Fourier transofrm in Wolfram Alpha and got this result: https://www.wolframalpha.com/input/?i=Fourier+transform+delta%289-5n%29

Basically, the whole problem is the number 5 in the denominator, that it doesn't exist in my professor's solution.

 

[blade88]

It seems like the problem is solved:

The scaling property does not hold for discrete sequences, as I have in my exercises.

Therefore, the correct solution for the discrete case is just with the exponential.

Thank you for your help.