# Fourier transform of Langevin equation (integral cancellation problem)

1. Apr 5, 2013

### stephenx_86

Hi,

(To cut a long story short, can I cancel the integrals in Eq. 6 to leave me with Eq. 7?)

I am trying to follow the method for modelling the motion of a tethered bead from a couple of papers ("Te Velthuis, A. J. W. et al. (2010) Biophys. J. 99 1292–1302" and "Lansdorp, B. M., & Saleh, O. A. (2012) Rev. Sci. Instrum. 83 025115"), but am getting stuck going from the Langevin equation (where γ is the friction coefficient and k is the tether stiffness):

[1] $\gamma \dot{x} (t)+kx(t)=F_{therm}$

to the power spectrum:

[2] $P(\omega )\equiv \left |x^{2}(\omega )\right| =\frac{2\gamma k_{B}T}{\gamma ^{2}\omega ^{2}+k^{2}}$

I've followed a few lecture handouts I've found online and think I understand the majority of the problem. From what I understand, I need to take the three components of the first equation and individually apply the Fourier transforms (all in the range ∞→-∞):

[3] $\dot{x}(t)=\frac{d}{dt}\left|\frac{1}{2\pi }\int x(\omega )e^{i\omega t}d\omega \right|=\frac{1}{2\pi }\int x(\omega ) i\omega e^{i\omega t}d\omega$

[4] $x(t)=\frac{1}{2\pi }\int x(\omega )e^{i\omega t}d\omega$

[5] $F(t)=\frac{1}{2\pi }\int F(\omega )e^{i\omega t}d\omega$

If I substitute all these back into the original equation I get:

[6] $\gamma \frac{1}{2\pi }\int x(\omega ) i\omega e^{i\omega t}d\omega +k\frac{1}{2\pi }\int x(\omega )e^{i\omega t}d\omega = \frac{1}{2\pi }\int F(\omega )e^{i\omega t}d\omega$

From the lecture handouts I've found, I should be able to rearrange this to get the following (or something similar):

[7] $\gamma i\omega x(\omega) +kx(\omega) = F(\omega)$

Now, here's my question. Is it possible for me to cancel the parts of the integral that appear in all three components of in Eq. 6 (this is $\frac{1}{2\pi}\int e^{i\omega t} d\omega$) to leave me with Eq. 7?

Any help on this would be greatly appreciated, since I've been stuck on this for a couple of days now!

Thanks
Stephen

p.s. I also seem to end up with a rogue i, which will become -1 when I square P(x). Have I done my Fourier transform equations wrong?

2. Apr 5, 2013

### MisterX

Yes, but it is not the way I would do it.

Instead, I would just take the Fourier transform of the original equation.

use this
$x(\omega )=\int x(t )e^{-i\omega t}dt$
$i\omega x(\omega )=\int x'(t )e^{-i\omega t}dt$
$F_{therm}(\omega )=\int F_{therm}(t )e^{-i\omega t}dt$

the FT is linear:
$y(\omega ) + z(\omega )=\int (y(t) + z(t))e^{-i\omega t}dt$

Just apply the FT to both sides original equation.

3. Apr 5, 2013

### MisterX

As for the question about power spectrum, I think the equation is wrong. It should be
$P(\omega) = \left|x(\omega) \right|^2 = (x(\omega))^* x(\omega)$

4. Apr 5, 2013

### stephenx_86

That's brilliant, thanks for your help.