# Derivation of equation for Phase-modulation fluorescence

1. May 12, 2010

### McKendrigo

Hello there,

I'm going through the derivation for the equations relating to frequency-domain lifetime measurements for fluorescent samples. This is a technique whereby a sinusoidally-modulated light source excites a fluorescent sample, and the fluorescence lifetime of the sample, $$\tau$$ is found by relating $$\tau$$ to the phase shift and/or amplitude de-modulation of the fluorescence relative to the excitation source.

I'll go through the derivation up to the point I'm a little stuck!

Firstly, the excitation is given by:

$$L(t)= a + b sin \omega t$$,

where b/a = Ml is the modulation of the excitation light.

The fluorescence emission will respond to this excitation with the same frequency, but with a different phase shift and modulation. We assume that the excited state population of the fluorescent sample is:

$$N(t)= A + B sin(\omega t - \phi)$$,

The intensity, I(t), at any time is proportional to the number of molecules in the excited state N(t).

We suppose that the intensity decay following a delta-function excitation is a single exponential:

$$I(t) = I_{o}*exp(-t/\tau)$$

Now - here comes where I come unstuck! - For a single exponential decay, the differential equation describing the time-dependent excited state population is:

$${dI(t)}/{dt} = -{1/\tau}*I(t) + L(t)$$

....
The textbook then goes on to substitute N(t) for I(t) in that equation, and find relationships between $$\tau$$ and the phase shift / de-modulation. The subsequent steps I'm fine with, I'm just not sure where this last equation comes from. I'm particularly unsure as to why you add L(t), rather than adding its derivative. If anyone could shed some light on how this equation must have been derived, I would be very grateful!

2. May 12, 2010

### LCKurtz

I'm way out of my area as far as your particular application is but here is my take, FWIW.

With your delta impulse the intensity I decreases exponentially according to your equation

$$\frac {dI(t)}{dt} +\frac 1 \tau I(t) =0$$

giving the solution

$$I(t) = I_{o}*e^{-\frac t \tau}$$

Now under the excitation L(t) you have a forcing function on the right side, much like the equation of a damped spring with a forcing function:

$$\frac {dI(t)}{dt} +\frac 1 \tau I(t) =L(t)$$

Just my best guess. Probably I'd have to know the derivation of the DE in the first place to be sure.

3. May 13, 2010

### McKendrigo

That sounds entirely sensible to me. Thanks for your help!