Fourier transform of cos(100t)

[rolls]

Homework Statement

Find the Fourier transform of cos(100t)

The Attempt at a Solution

now I know just from looking at a Fourier transform table that if the equation is in the form cos(2Pi*k*t) then the answer is just 1/2(delta(f+k) + delta(f-k))

So in this case is the answer 1/2(delta(f+100/2pi) + delta(f - 100/2pi)) ?

I'm not that good at integrals so I haven't attempted to do this problem the traditional long way.

Thanks :)

 

[dx]

Write out the Fourier transform integral, and write cos(100t) as a sum of exponentials using Euler's formula. Then use the following fact:

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ixy} dy = \delta(x)$$

 

[rolls]

Ok starting with S e^(-2Pi*i*t) * (e^(i100/2pi*t) + e^(-i100/2pi*t))/2

I got it down to 1/2PI S (e ^(-2Pi*i(t + 100/2Pi) + e ^ (-2Pi*i(t - 100/2pi))

Which then equals delta(f + 100/2Pi) + delta (f - 100/2pi)

I'm not sure if I did it correctly though or just worked my way backwards from the answer (I attempted this originally and then got stuck). Does that look like I am going in the right direction?

 

[dx]

cos(100t) = (e^i100t + e^-i100t)/2

I don't understand where you got the 2pi's in it.

 

[rolls]

cos(100t) = (e^i100t + e^-i100t)/2

I don't understand where you got the 2pi's in it.

I did that originally so I could use the transform table, its ok I understand what to do now. I will have a shot at doing it from scratch and report back.

Thanks guys.