Question

Fourier Transform Help

Answer 1

No this is a Signals course in Electrical engineering

I need help with Part C
What I got is :

( -jw / (1-jw) ) * e^(-jw)

Answer 2

@satellite73

Answer 3

I can show you how to derive the answer, but there are different commonly used notations for the Fourier transform. Is this the definition you're using? Just tell me what's different and then we can get to part C like you want.

\[\large F(\omega) = \int_{-\infty}^\infty f(t) e^{-j2 \pi \omega t} dt\]

Answer 4

What I want to know is the operations that are being applied to g(t) are in the correct order .

Does this order or signal operation seem to be correct .

1. g(t)
2. g( 1 - t ) = g (-t + 1)
3. g(-1* (t -1))

From 3. we can see that the fourier transform is shifted by e^ (-jw (1))
and then inverted ??

Answer 5

Right. I'm saying we can derive this quite quickly from the definition of the Fourier transform in general with a cute little u-substitution in the integral.

Answer 6

yes that is the same form we are using

Answer 7

True you can do it that way but you can skip all that work just by using properties

Answer 8

Specifically , i would like to know what you guys think what the answer is using properties

Answer 9

So you're given this, \(G(t) = \frac{j\omega}{1+j\omega}\) or something, anything. In general it's gonna look like this. I forget the properties in general like + or - signs and figuring it out is really a breeze so I'll just do it. \[\large G(\omega) = \int_{-\infty}^\infty g(t) e^{-j2 \pi \omega t} dt\]

You want to know this though,
\[\large ? = \int_{-\infty}^\infty g(1-t) e^{-j2 \pi \omega t} dt\]
So we substitute in
\[1-t=u\]\[dt=-du\]
\[\int_{\infty}^{-\infty}g(u) e^{-j2 \pi \omega (1-u)} (-du)\]
Notice the limits flipped and we have a negative sign on du coming in so we can also just combine them to get it into a form we know and then use exponent rules to pull out a constant phase factor:

\[ e^{-j2 \pi \omega } \int_{-\infty}^{\infty}g(u) e^{-j2 \pi (-\omega)u} du\]
The integral is what we already have solved, except instead of it being \(G(\omega)\) it has a negative sign on \(\omega\) so we have \(G(-\omega)\) for the integral multiplying the phase factor, so the answer is for the FT of g(1-t):
\[\large e^{-j2 \pi \omega } G(-\omega)= \int_{-\infty}^\infty g(1-t) e^{-j2 \pi \omega t} dt\]

Answer 10

The properties are just derived like this anyways so if you have trouble remembering it's probably best to just derive them it takes pretty much no time at all. Here's a good site for just looking up the properties real fast though they use a slightly different notation it's short sweet to the point unlike me

http://www.thefouriertransform.com/transform/properties.php

Answer 11

Ahh , ok !

Answer 12

Sorry for rushing you .

Answer 13

I'm sorta looking for jobs and got distracted halfway through typing that up lol, no rush for me I just thought you were rushing yourself for some reason, you gotta do what you gotta do haha