# Fourier trasnform problem

1. Mar 14, 2014

### asdf12312

1. The problem statement, all variables and given/known data

2. Relevant equations
here is list of fourier transforms:
http://uspas.fnal.gov/materials/11ODU/FourierTransformPairs.pdf

3. The attempt at a solution
so I know the solution but I don't know how to get it. Here is what I think to do: the ramp function r(t) and the rect pτ(t). I know that r(t) is the integral of step u(t) which is integral of d(t), and I know the transform of u(t)..would that help? I guess I'm not realy sure what to do. In the solution they use integral of pτ(t) minus delta function d(t)..not sure why.

Last edited: Mar 14, 2014
2. Mar 14, 2014

### vela

Staff Emeritus
Well, you haven't really made much of an attempt here. All you've done is throw out some random facts that may or may not have to do with the problem.

You're given a graphical representation of x(t), and you want to find the Fourier transform
$$X(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x(t) e^{-i\omega t}\,dt.$$ To evaluate the integral, you're going to need to figure out what to plug in for x(t). Start there.

Last edited: Mar 15, 2014
3. Mar 15, 2014

### rude man

I think you r integrand is in error ... the exponential?

4. Mar 15, 2014

### vela

Staff Emeritus
Fixed it.

5. Mar 15, 2014

### rude man

The OP should also be forewarned that there are two versions of the Fourier transform pair.

vela gave one in his post #2. I gave the other in the duplicate post, to wit,
$$X(f) = \int_{-\infty}^\infty x(t) e^{-i\omega t}\,dt$$
where ω = 2πf.

I personally prefer mine since the annoying 1/2π factor is absent. In vela's way that fracton has to go in front of the transform but left out in the inverse transform. On the other hand, my way introduces f as the transform parameter while keeping ω in the exponential. Of course, you can get around this by substititing ω = 2πf in the exponential, but I never do. Pretty easy to remember ω = 2πf.

6. Mar 15, 2014

### asdf12312

well i could say x(t) is X+1 from -1<x<0 and 1 from 0<x<1 but i dont think thats doable with transforms. I am trying to follow how they got the solution, since x(t) has r(t) and u(t), could use integral ∫ of (u(t) - d(t)). but i think that, since the r(t) is only from -1<x<0 they used the rect function instead of u(t) (which is kinda same thing I guess). so I understand this much:

∫p1(λ+1/2) - d(λ-1) dλ

and transform of rect pτ(t) is τsinc(τω/2pi), transform of d(t-c) is e-jωc.

so they should get something like ∫sinc(ω/2pi)ejω/2 - ∫e-jω which would be...

2/jω (sinc(ω/2pi)ejω/2) + 1/jω (e-jω ) ?

7. Mar 15, 2014

### rude man

No you couldn't. Look again at what x(t) is from t = -1 to t=0.
Then use the integral vela or I gave you (depends on how your teach defines the Fourier transform) with suitable limits of integration.

8. Mar 15, 2014

### LCKurtz

You are making it way too hard. Your function is zero except on $(-1,1)$. Just plug in the formulas for $x(t)$ and do the integration.

9. Mar 15, 2014

### asdf12312

i am getting confused.. why is x(t) not t+1 from t=-1 to t=0 though? I guess the problem is I have no way to find transform of r(t+1), so instead I have to use integral of u(t) (or the rect p(t) which is easier). Is there another way to do it? I am just doing it the way they did it, to get the answer which is:

1/jω [ sinc(ω/2pi)ejω/2 - e-jω ]

just trying to figure out how they got that (how they could factor out 1/jω like that).

Last edited: Mar 15, 2014
10. Mar 15, 2014

### rude man

There is also the integration property: for the left side. the ramp is the integral of a constant from -1 to 0
The right side is the integral of the delta function delta(t) from 0 to +1
etc. There are various possibilities depending on where you want to start.

11. Mar 15, 2014

### LCKurtz

It is $t+1$ as you said, except you didn't use $t$ as the variable.

12. Mar 15, 2014

### rude man

The answer I got is X(ω) = [1 - exp(jω) + jw exp(-jω)]/ω2.

It sounds like they want you to use the transform of the triangle function for -1<t<0 and the rectangle for 0<t<1. I looked at that approach briefly & didn't see much advantage to it. Quite the opposite, actually. But I'm no Fourier expert & there may well be a trick to it that I don't see.

I suppose you could let sinc(ω/2pi) = sin(ω/2pi)/(ω/2pi)
then sin(ω/2pi) = [exp j(ω/2pi) - exp -j(ω/2pi)]/2j

which should eventually give the answer I got (assuming I got it right to begin with).

13. Mar 15, 2014

### vela

Staff Emeritus
I think you're saying that
$$x(t) = \int_{-\infty}^t [p_1(\lambda+1/2)-\delta(\lambda-1)]\,d\lambda,$$ where $p_1(t)$ denotes a rectangular pulse of unit width and centered at the origin. If so, that's fine.

No, you don't just replace the integrand with its Fourier transform. Use the property in the table you linked to for the transform of
$$\int_{-\infty}^t f(\tau)\,d\tau.$$

14. Mar 15, 2014

### LCKurtz

I agree with that answer (using your version of the FT integral).