Inverse Fourier Transform of sin(t)(sin(2t))^2 question

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SUMMARY

The discussion centers on the inverse Fourier transform of the function X(w) = (sin(w/2) * exp(-j2w)) / (jw + 2). Participants analyze the transformation properties and the implications of the Fourier transform on the time domain representation. Key insights include the realization that the multiplication of two Fourier functions corresponds to convolution in the time domain, and the importance of recognizing sinc functions in the context of Fourier transforms. The final expression for x(t) is confirmed as x(t) = j/2[exp(-2(t-2.5))u(t-2.5) - exp(-2(t-1.5))u(t-1.5)].

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Homework Statement



Inverse Fourier transform of X(w) = (sin(w/2).exp(-j2w))/(jw + 2)

Homework Equations


From table:

exp(-bt).u(t) → 1/(jw+b)

multiplication by sin: x(t)sin(w0t → j/2[X(w+w0)-X(w-w0]

w0 being 0.5 here.

shifted left or right in time: x(t-c) → X(W)exp(-jwc)

The Attempt at a Solution



I got x(t) = j/2[exp(-2(t-1.5)).u(t-1.5) - exp(-2(t-2.5).u(t-2.5)]
but I think it should be: x(t) = j/2[exp(-2(t-2.5).u(t-2.5) - exp(-2(t-1.5)).u(t-1.5)]

Any opinions?

Thanks
 
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Well x(t) is real. I know this because |X(w)| is an even function and ∠X(w) is an odd function, so spotting j in x(t) means either your result is not simplified to make it real or it's not quite right.

I'm not quite sure how you did the inverse transform with the transform properties you listed. Keep in mind two Fourier functions multiplied together is the same as convolution in the time domain, not the product of two time domain functions. Though I'm not sure if that is what you did.

When I see sin(w/2) in the Fourier transform I am immediately thinking sinc as in sinc(w/2) = sin(w/2)/(w/2). The inverse transform of a sinc function is a rectangle in the time domain. I would look at whether X(w) could be written in terms of sinc.

I only looked at this quickly so keep that in mind! I'm about to sign off but wanted to give some feedback since you asked in that other thread.
 
Last edited:
aralbrec said:
I only looked at this quickly so keep that in mind! I'm about to sign off but wanted to give some feedback since you asked in that other thread.

Fairenough, I appreciate that, hopefully another time.

Well I took the whole thing to be 1/(jw+b)

which was multiplied by sin and shifted right by 2 (from the exp(-j2w)).

I was given as the answer:
x(t) = j/2[exp(-2(t-2.5).u(t-2.5) - exp(-2(t-1.5)).u(t-1.5)]

but when I did the above to try and get it from X(w) the things in the square brackets were the other way around. So maybe it can be simplified to be real but it probably doesn't matter.

When you say that "|X(w)| is an even function and ∠X(w) is an odd function" how can you tell that?

P.S
So did I take your meaning correctly, like "...Likewise integrating: sin(t).(sin(2t))^2
would be zero?"

Thanks again!
 

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