What is the Fourier transform of cos(2πf0x - π/4)?

[thatguy14]

Hi, I need help with some basic Fourier transform properties stuff - its fairly simple though I think I am doing something wrong.

So we know from the shifting property
if h(x) has the Fourier transform H(f)
then h(x-a) has the Fourier transform H(f)e^i*2*π*f*a

so I have the function

cos(2πf₀x - π/4)

I know (from a previous question) that the Fourier transform of cos(2πf₀x) is

½[δ(f+f₀) + δ(f-f₀)]

where δ indicates the delta function.

so then if we factor above

cos(2πf₀x - π/4)
cos(2πf₀(x - 1/(8f₀)))

so then shouldn't the answer be that the Fourier transform is

{½[δ(f+f₀) + δ(f-f₀)]} * exp(i*pi*f/(4f₀)

I don't see if I did anything wrong here - and further, can this be simplified more?

Thanks

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[mathman]

part of statement starting at "so if we factor the above" is very unclear.