What is the Inverse Fourier Transform of cos(4ω)?

[helderdias]

Hi everyone,

I'm trying to solve an exercise in which I need to find x(t) considering that X(ω) = cos(4ω). So, I need to find the Inverse Fourier Transform of cos(4ω), but I don't have the inverse Fourier transform table.

So, I thought about applying the duality property. If x(t) <--> X(ω), then X(t) <--> 2π*x(-ω)

cos(4ω) <--> π[δ(ω-4) + δ(ω+4)]

Applying the duality property

π[δ(t-4) + δ(t+4)] <--> 2π.cos(-4ω)

Since cos(x) = cos(-x)

1/2*[δ(t-4) + δ(t+4)] <-->cos(4ω)

Therefore

x(t) = 1/2*[δ(t-4) + δ(t+4)]

Is that correct? WolframAlpha is giving me a different answer :(

 

[jvicens]

Hi there,

It looks like you are on the right track! Your application of the duality property is correct, but there is a small mistake in your final answer. The correct inverse Fourier transform of cos(4ω) should be x(t) = 1/2*[δ(t-4) + δ(t+4)].

WolframAlpha may be giving you a different answer because it simplifies the expression further to just δ(t-4), since δ functions are considered to be 0 everywhere except at the point of interest. However, your answer is also correct and both expressions are equivalent.

Keep up the good work and don't hesitate to ask for further clarification if needed. Good luck with your exercise!