Finding the Fourier Transform of a function with finite duration

Question

Finding the Fourier Transform of a function with finite duration

Intro_________

In the following question I analyze a brick sliding in an horizontal plane after an initial push (under Coulomb's dry friction), which after some simplifications Newton's laws derives in this equation for its dynamics ($k>0$ and $g>0$):

$$x'' = -k\cdot g\cdot \text{sgn}(x')$$

From the answer to the mentioned question looks I mess it up with some integration constants, but I believe the a solution to the differential equation is given by:

$$x(t) = \frac{k\cdot g}{2}\cdot\left(T-t\right)^2\cdot\theta(T-t) $$ where $T<\infty$ is the finite extinction time determined by initial conditions, and $\theta(t)$ is the Heaviside step function.

Question_________

I am interested into find the Fourier Transform of $x(t)$ in the interval $[0,\ T]$: $$X(iw) = \int\limits_{0}^T x(t)e^{-iwt}\ dt$$

Does $X(iw)$ fulfill the the Plancherel theorem?

I have done attempts but I have not been able to make a transform which energy fits the energy of the signal (as I believe it should happen due Plancherel theorem): I don't know if I just making a mistake, if Plancherel theorem maybe don't apply to signal of finite duration, but I am also suspicious that at the start the Fourier Transform is sees a jump discontinuity which is adding energy, if it is indeed the problem, please explain how to get rid of it.

Motivation__________

As quantum systems have the Heisenberg's uncertainty principle, classic systems have an analogous situation among it time extension and their bandwidth: Gabor's limit, and using this example I want to see how the inequality will behaves as I made $T\to 0$, since after the system stops moving, position and velocity are determined, all of them zero, energy also zero, my intuition tell me I should see a point where the inequality broke, or instead get undefined (this because the spectrum should be an entire function since $x(t)$ is compacted-supported, as states the Paley–Wiener theorem).

Answer 1

I will disregard the constants for this analysis. You can just multiply at the end if needed.

Let $x(t) = (T-t)^2 \theta(T-t)$

Let $y(t) = x(t) \operatorname{Rect}(\frac{t}{T}-\frac12)$, where Rect is UnitBox function.

We are interested in the FourierTransform of $y(t)$ and see if Plancherel Theorem holds.

$$Y(\omega) = Y[y(t)](\omega) = \int\limits_{t=-\infty}^{\infty} y(t) e^{-i 2\pi t \omega} dt$$.

Using Mathematica (I can do it manually and show steps if you want): $$Y(\omega) = \frac{i}{4\pi^3 \omega^3} - \frac{i e^{-2i\pi T \omega}}{4\pi^3 \omega^3}+\frac{T}{2\pi^2\omega^2} - \frac{i T^2}{2\pi\omega}$$

Checking Plancherel Theorem, $\int\limits_{-\infty}^{\infty} |y(t)|^2 dt = \int\limits_{-\infty}^{\infty} |Y(\omega)|^2 d\omega = \frac{T^5}{5}$

Edit: Adding some manual calculations as per OP's request.

The exponential $e^{-i 2\pi t \omega}$ can be written as $\cos(2 \pi \omega t) - i \sin(2 \pi \omega t)$.

$Y(\omega) = I_{1} -i I_{2}$

Where $I_{1} = \int\limits_{0}^{T} (T-t)^2 \cos(2 \pi \omega t) dt$ and $I_{2} = \int\limits_{0}^{T} (T-t)^2 \sin(2 \pi \omega t) dt$

These can be integrated by parts. I am just showing it for $I_{1}$.

$$ \begin{align} I_{1} & = \int\limits_{0}^{T} (T-t)^2 \cos(2 \pi \omega t) dt \\ &= \left. (T-t)^2 \frac{\sin(2\pi \omega t)}{2 \pi \omega} \right\rvert_{0}^{T} + \frac{1}{2\pi \omega}\int\limits_{0}^{T} 2(T-t) \sin(2 \pi \omega t) dt\\ &= \frac{1}{\pi \omega} \left(\left. (T-t) \frac{-\cos(2\pi \omega t)}{2\pi \omega} \right\rvert_{0}^{T} -\int\limits_{0}^{T}\frac{\cos(2\pi \omega t)}{2\pi \omega} dt \right)\\ &= \frac{1}{\pi \omega} \left( \frac{T}{2\pi \omega} - \frac{\sin(2\pi \omega T)}{(2\pi \omega)^2}\right)\\ &= \frac{T}{2 \pi^2 \omega^2} - \frac{\sin(2 \pi \omega T)}{4 \pi^3 \omega^3}\\ \end{align} $$