Fourier Transform of cos(2*pi*t)

[BriWel]

I have a practice question, which is to find the Fourier Transform of cos(2^pi^t)

By substitution into the FT formula, and use of eulers formula,I have managed to reduced to:

INTEGRALOF ( (cos(2*pi*t) * ( cos(2*pi*F*t) - j*sin(2*pi*F*t) ) )

By plotting the frequency graph of the original function, I know that the answer I am looking for is: delta(1) + delta(-1)

I have also been told that the integral of two trig functions multiplied together equals 0 if the functions have different frequencies. This indicates that the above formula is only non-zero where F = 1.

My problem is that I don't know how to get from the above formula to delta(1) and delta (-1). Can anybody help?

Also, I'm relatively new to Fourier Transforms, so as much detail as possible in answers will be appreciated!

thanks in advance for any help

 

[tiny-tim]

Welcome to PF!

Hi BriWel! Welcome to PF!

(have a pi: π and a delta: δ and an integral: ∫ and try using the X² tag just above the Reply box )

I have also been told that the integral of two trig functions multiplied together equals 0 if the functions have different frequencies. This indicates that the above formula is only non-zero where F = 1.

My problem is that I don't know how to get from the above formula to delta(1) and delta (-1). Can anybody help?

You need to learn your https://www.physicsforums.com/library.php?do=view_item&itemid=18" …

in this case, 2cosxcosy = cos(x+y) + cos(x-y) and 2sinxcosy = sin(x+y) + sin(x-y) …

and then you should be able to prove that the ∫₀^2π is 0 unless x = ±y