Convolution theorem; integration

[unscientific]

Homework Statement

Part(a)/(b)/(c): Fourier transform the following:

Part(d): Prove the convolution theorem

Part (e): Find total displacement

Homework Equations

The Attempt at a Solution

Part(a)
$$f = \frac {1}{sqrt{2\pi}} \int_{-\infty}{\infty} F e^{-iωt} dω$$

$$\frac {1}{\sqrt{2\pi}} \int_{-T}{T}$$
$$= \frac{1}{iω\sqrt{2\pi}} [e^{iωT} - e^{-iωT}]$$
$$= \sqrt{\frac{2}{\pi}} \frac {sin (ωt)}{ω}$$

Part (b)
$$F_2 = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0}e^{-at}e^{iωt} dt + \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty}e^{-at}e^{iωt} dt$$
$$= -\frac{2a}{\sqrt{2\pi}(a^2 + ω^2)}$$

Part(c)

Using result from (b):
$$F_2 = -\frac{2a}{\sqrt{2\pi}(a^2 + ω^2)}$$

Let a = 1,

$$e^{-|t|} = -\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac {2}{1+ω^2} e^{-iωt} dω$$

$$t→-t,$$
$$e^{-|-t|} = \frac {1}{\pi} \int_{-\infty}^{\infty} \frac {1}{1+ω^2} e^{iωt} dω$$

$$t→ω,$$
$$e^{-|ω|} = \frac {1}{\pi} \int_{-\infty}^{\infty} \frac {1}{1+t^2} e^{iωt} dt$$

Part(e)

$$V_{(ω)} = \sqrt{2\pi} F_{(ω)}G_{(ω)} = \frac {2}{\pi} \frac {a sin (ωT)}{a^2 + ω^2}$$

$$sin (ωT) = \frac {1}{2i} (e^{iωT} - e^{-iωT}) ,$$

Taking inverse Fourier transform of V_ω:

$$V_{(t)} = \frac {1}{i\pi}(\frac{a}{\sqrt{2\pi}}) \int_{-\infty}^{\infty} \frac {1}{a^2 + ω^2} e^{iω(T-t)} dω + \frac {1}{i\pi}(\frac{a}{\sqrt{2\pi}}) \int_{-\infty}^{\infty} \frac {1}{a^2 + ω^2} e^{-iω(t+T)} dω$$

My velocity is complex! Strange...

This is related to inverse Fourier transform of F₂, f₂ but as a function of (t-T) and (t+T):

$$i(\frac{1}{\sqrt{2\pi}}) f_2(t-T) - i(\frac{1}{\sqrt{2\pi}}) f_2(t+T)$$
$$= i\frac{1}{\sqrt{2\pi}}[e^{-a|t-T|} - e^{-a|t+T|}]$$

To find the displacement, simply integrate the above with respect to t from -∞ to ∞.

$$s = i\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-a|t-T|} dt - i\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-a|t+T|} dt$$
$$= i\sqrt{\frac{2}{\pi a}}(e^{-aT} - e^{aT})$$

Again complex displacement.. (and negative)

 

[BruceW]

That's some nice work on parts a to c. at the end of part a, it should be $\tau$ (not $t$), but I'm guessing that is just a miss-type. For part e, there is a mistake in this equation:
$$V_{(ω)} = \sqrt{2\pi} F_{(ω)}G_{(ω)} = \frac {2}{\pi} \frac {a sin (ωT)}{a^2 + ω^2}$$
there should be another $\omega$ from $f_1$ (which you have got in part a). And it looks like you've missed out the $\sqrt{2\pi}$ which is on the left hand side of the equation.

Also, there is a simpler way to do part e. You don't need to do the inverse Fourier transform. hint: you have got $\tilde{V}(\omega )$ and you want
$$\int_{-\infty}^{\infty}V(t) dt$$
So, thinking about the definition of the Fourier transform, what is the easiest way to get this?

 

[unscientific]

That's some nice work on parts a to c. at the end of part a, it should be $\tau$ (not $t$), but I'm guessing that is just a miss-type. For part e, there is a mistake in this equation:
$$V_{(ω)} = \sqrt{2\pi} F_{(ω)}G_{(ω)} = \frac {2}{\pi} \frac {a sin (ωT)}{a^2 + ω^2}$$
there should be another $\omega$ from $f_1$ (which you have got in part a). And it looks like you've missed out the $\sqrt{2\pi}$ which is on the left hand side of the equation.

Also, there is a simpler way to do part e. You don't need to do the inverse Fourier transform. hint: you have got $\tilde{V}(\omega )$ and you want
$$\int_{-\infty}^{\infty}V(t) dt$$
So, thinking about the definition of the Fourier transform, what is the easiest way to get this?

$$\tilde{V}(ω) = \int_{-\infty}^{\infty} V_{(t)} e^{iωt} dt$$

I've been thinking of a way to 'remove' the exp[iωt] from the integral and everything would be perfect. The only way is if ω = 0. But if ω = 0, LHS = 0..

 

[BruceW]

yes, use $\omega=0$. The left hand side is not zero, it is $\tilde{V}(0)$ this isn't zero. Even if it was zero, that would still be the correct answer, since the right hand side is exactly what you are looking for.

 

[unscientific]

yes, use $\omega=0$. The left hand side is not zero, it is $\tilde{V}(0)$ this isn't zero. Even if it was zero, that would still be the correct answer, since the right hand side is exactly what you are looking for.

The 0 comes about from the sin (ωT) function and part (a) looks right to me..

 

[BruceW]

you have got the correct answer to part (a), but part (a) evaluated at $\omega=0$ is not equal to zero.

edit: you have
$$\sqrt{\frac{2}{\pi}} \frac {sin (ω\tau)}{ω}$$
but this is not zero for $\omega=0$

 

[unscientific]

you have got the correct answer to part (a), but part (a) evaluated at $\omega=0$ is not equal to zero.

edit: you have
$$\sqrt{\frac{2}{\pi}} \frac {sin (ω\tau)}{ω}$$
but this is not zero for $\omega=0$

Ah ok, I get it, thanks so much!

 

[BruceW]

hehe, no problem!