# Deriving the transfer function

Suppose we have a free particle in one dimension with position $$x$$ and momentum $$p$$, and some damping $$\Gamma$$: \begin{equation} \begin{aligned} \dot{x} &= p/m, \\ \dot{p} &= -\Gamma p+F(t). \end{aligned} \end{equation} I want to find the impulse response if some force $$F(t)$$ is applied to the system. Since the equations are linear we may Fourier transform both sides: \begin{equation} \begin{aligned} i\omega x &= p/m, \\ i\omega p &= -\Gamma p +F, \end{aligned} \end{equation} where all variables are now understood to be in Fourier space. We may then re-arrange to obtain the transfer function $$\begin{equation} p=\frac{1}{\Gamma+i\omega}F \end{equation}$$ $$\begin{equation} x=\frac{1}{im\omega(\Gamma+i\omega)}F \end{equation}$$ So it seems like the impulse response should be $$\chi_x(\omega)=\frac{1}{im\omega(\Gamma+i\omega)}.$$ However, when you inverse transform this into the time domain this doesn't seem to behave how it should.

# Taking the inverse Fourier transform

Now let's compute the inverse Fourier transform, to understand $$\chi_x$$ in the time domain: $$\chi_x(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{e^{i\omega t}}{im\omega(\Gamma+i\omega)}d\omega.$$ We may approach this problem using contour integration. There are two simple poles, at $$\omega=0$$ and $$\omega=i\Gamma$$. Their residues are $$\mathrm{Res}_{\omega=0}\left(\frac{e^{i\omega t}}{im\omega(\Gamma+i\omega)}\right)=\lim_{\omega\rightarrow 0}(\omega-0)\frac{e^{i\omega t}}{im\omega(\Gamma+i\omega)}=\frac{1}{im\Gamma},$$ $$\mathrm{Res}_{\omega=i\Gamma}\left(\frac{e^{i\omega t}}{im\omega(\Gamma+i\omega)}\right)=\lim_{\omega\rightarrow i\Gamma}(\omega-i\Gamma)\frac{e^{i\omega t}}{im\omega(\Gamma+i\omega)}=\frac{ie^{-\Gamma t}}{m\Gamma}.$$ Depending on if $$t>0$$ or $$t<0$$, we use different contours in the complex plane.

## t>0 For positive $$t$$ we take a contour in the upper half-plane. The integral over the entire contour should be $$\oint=2\pi i\left(\frac{ie^{-\Gamma t}}{m\Gamma}\right)=\frac{-2\pi e^{-\Gamma t}}{m\Gamma}.$$ The integral over $$C_{\gamma}$$ vanishes, the integral over the real axis gives us $$\sqrt{2\pi}$$ times the inverse Fourier transform, while the integral over $$C_{\epsilon}$$ gives $$-\pi i$$ times the residue (by the fractional Residue theorem, and we are going clockwise so a minus sign). We therefore have: $$\frac{-2\pi e^{-\Gamma t}}{m\Gamma}=\sqrt{2\pi}\chi_x(t)-\frac{\pi}{m\Gamma},$$ $$\chi_x(t)=\frac{\pi}{m\Gamma\sqrt{2\pi}}\left(1-2e^{-\Gamma t}\right).$$ In particular, $$\chi_x(0)=-\frac{\pi}{m\Gamma\sqrt{2\pi}}$$, rather than zero.

## t<0 For negative $$t$$ we take a contour in the lower half-plane. Now the integral over the entire contour should be $$\oint=0.$$ The integral over $$C_{\gamma}$$ vanishes, the integral over the real axis gives us $$\sqrt{2\pi}$$ times the inverse Fourier transform, while the integral over $$C_{\epsilon}$$ now gives $$\pi i$$ times the residue (positive as we are going clockwise). Putting this all together gives: $$0=\sqrt{2\pi}\chi_x(t)+\frac{\pi}{m\Gamma},$$ $$\chi_x(t)=-\frac{\pi}{m\Gamma\sqrt{2\pi}}.$$

# Analysing the inverse Fourier transform

We have $$\begin{equation} \chi_x(t)= \begin{cases} \frac{\pi}{m\Gamma\sqrt{2\pi}}\left(1-2e^{-\Gamma t}\right) & t>0 \\ -\frac{\pi}{m\Gamma\sqrt{2\pi}} & t<0 \end{cases} \end{equation}$$ Let's plot $$\chi_x(t)$$ for three different values of damping $$\Gamma$$: This doesn't seem correct, if we interpret the transfer function as the response to a unit impulse at $$t=0$$. We would expect the response to be zero for $$t=0$$, and for the system to increase in $$x$$ and then slow down and approach a constant value, where this constant value is smaller for greater damping values.

We can get the behaviour we expect by adding an offset of $$+\frac{\pi}{\Gamma\sqrt{2\pi}}$$ to the transfer function: This is exactly what we would expect! I think this is what the transfer function must be. However, this would mean there must be an extra term proportional to $$\delta(\omega)$$ in $$\chi_x(\omega)$$, and I don't see where this would come from. Have I made a mistake with my inverse Fourier transform, perhaps due to the pole at $$\omega=0$$? Or is there some other subtlety I am missing?