Please check my solution for this Fourier transform (time diff. Prop.)

[YehiaMedhat]

The main question is to transform the following expression: $\frac{it}{t^2 + 4}$
1. Using the symmetry property first: $2\pi \frac{i}{4} e^{-2|-\omega|}$
2. Then applying the differentiation in time property: $i\frac{d}{d\omega} [\frac{i\pi}{2} e^{-2|\omega|}]$
3. Differentiating: $\frac{\pi}{4} e^{-2|\omega|}$
Is that right, because the solution in the TA's sheet also shows that it's different answer.

 

[mario99]

The main question is to transform the following expression: $\frac{it}{t^2 + 4}$
1. Using the symmetry property first: $2\pi \frac{i}{4} e^{-2|-\omega|}$
2. Then applying the differentiation in time property: $i\frac{d}{d\omega} [\frac{i\pi}{2} e^{-2|\omega|}]$
3. Differentiating: $\frac{\pi}{4} e^{-2|\omega|}$
Is that right, because the solution in the TA's sheet also shows that it's different answer.

What is the answer in the TA's sheet?

 

[YehiaMedhat]

What is the answer in the TA's sheet?

$$2\pi e^{-2|\omega|}u(-\omega)$$

 

[mario99]

$$2\pi e^{-2|\omega|}u(-\omega)$$

$u(-\omega)$ is there because $f(t)$ is an odd function and we want just negative frequencies.

The main question is to transform the following expression: $\frac{it}{t^2 + 4}$
1. Using the symmetry property first: $2\pi \frac{i}{4} e^{-2|-\omega|}$
2. Then applying the differentiation in time property: $i\frac{d}{d\omega} [\frac{i\pi}{2} e^{-2|\omega|}]$
3. Differentiating: $\frac{\pi}{4} e^{-2|\omega|}$
Is that right, because the solution in the TA's sheet also shows that it's different answer.

How did you get $\frac{\pi}{4}$ in step three?

 

[mario99]

The main question is to transform the following expression: $\frac{it}{t^2 + 4}$
1. Using the symmetry property first: $2\pi \frac{i}{4} e^{-2|-\omega|}$
2. Then applying the differentiation in time property: $i\frac{d}{d\omega} [\frac{i\pi}{2} e^{-2|\omega|}]$
3. Differentiating: $\frac{\pi}{4} e^{-2|\omega|}$
Is that right, because the solution in the TA's sheet also shows that it's different answer.

If I do the transformation by integration, I get $\displaystyle F(\omega) = \frac{\pi}{2}e^{-2|\omega|} \text{sgn}(\omega)$

And because $f(t)$ is odd, we should include a negative sign to the final answer.

$\displaystyle F(\omega) = -\frac{\pi}{2}e^{-2|\omega|} \text{sgn}(\omega)$

And if you will use symmetry, the answer $2\pi\frac{i}{4}e^{-2|\omega|}$ is for $\frac{i}{t^2 + 4}$

For $\frac{it}{t^2 + 4}$, the answer should be $2\pi\frac{i\omega}{4}e^{-2|\omega|}$

 

[mario99]

$\displaystyle F(\omega) = -\frac{\pi}{2}e^{-2|\omega|} \text{sgn}(\omega)$

I did a contour integration, and I got a slightly different result from my first attempt.

Now, I will solve the problem by the Symmetry Property and Differentiation Property.

The main question is to transform the following expression: $\frac{it}{t^2 + 4}$
1. Using the symmetry property first: $2\pi \frac{i}{4} e^{-2|-\omega|}$
2. Then applying the differentiation in time property: $i\frac{d}{d\omega} [\frac{i\pi}{2} e^{-2|\omega|}]$
3. Differentiating: $\frac{\pi}{4} e^{-2|\omega|}$
Is that right, because the solution in the TA's sheet also shows that it's different answer.

Let $\displaystyle g(t) = \frac{i}{t^2 + 4}$ and $\displaystyle f(t) = tg(t)$

The Symmetry Property will give you:

$\mathcal{F}\{g(t)\} = G(\omega) = 2\pi \frac{i}{4} e^{-2|\omega|} = \pi \frac{i}{2} e^{-2|\omega|}$

Now we can use the differentiation property:

$\displaystyle \mathcal{F}\{f(t)\} = \mathcal{F}\{tg(t)\} = F(\omega) = i\frac{dG(\omega)}{d\omega} = \frac{\omega}{|\omega|}\pi e^{-2|\omega|}$

Because $f(t)$ is an odd function, you need to multiply its Fourier transform by $-1$.

$\displaystyle F(\omega) = -\frac{\omega}{|\omega|}\pi e^{-2|\omega|}$

You can replace the fraction $\displaystyle \frac{\omega}{|\omega|}$ by $\text{sgn}(\omega) \ \text{or} \ u(\omega)$.

$\displaystyle F(\omega) = -\pi e^{-2|\omega|}u(\omega)$

We know that $\displaystyle -u(\omega) = u(-\omega)$.

$\displaystyle F(\omega) = \pi e^{-2|\omega|}u(-\omega)$

Which is slightly different than TA's sheet answer! TA's sheet answer has a typo, maybe?!

 

[YehiaMedhat]

Thanks, but I think i will need to read again more than once

 

[mario99]

You are welcome.