Question

ri+my» +e-2-wii+m) – 2e-f[mni-r))do་ = - ( ༡༧ +8 , -28)

How do you get from one step to the other in the picture above? I don't understand how to calculate the deltas!

v2

I don't get how you suddenly get a new function?

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Answer #1

Using the concept of inverse fourier transform,

integration of F(w)*exp(jwt)dw = f(t), where f(t) in inverse fourier transform of F(w)

Comparing your equation with LHS of inverse fourier transform formula, F(w) = 1, output of integration wil be f(t) which is inverse fourier transform of 1.

Inverse fourier tranform of 1 is 2*pi*dellta(w)

To be very exact, your equation is shifted in w axis, so delta function corresponding to that particular shifted value.

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