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FFTAmplitudes

< Initial phase in a TF experiment (GPS) | Index | The time shape of the Pulsed Muon Beam >

How to rephase FFT amplitudes from µSR asymmetries

This is work in progress (private notes), do not trust

Assume we have acquired the muon asymmetry on GPS from the UP - DOWN detectors (3 - 4), with with an applied field B, Spin Rotation on, and time t=0 chosen to correspond to the prompt peaks.

We want to obtain a pure absorption spectrum from Fast Fourier Transforms (FFT) of the data. The experimental asymmetry is actually available only for times greater than a given deadtime {$\tau$}, typically 4-5 ns. Hence: {$ (1) \qquad\qquad A(t) = a(t) [1-W(t,\tau)] $} where {$a(t)$} is the asymmetry with no deadtime and {$W(t,\tau)$} is a window function, equal to 1 for {$0 \le t \le \tau$}. This window can be seen as the slit function of elementary optics (whose Fourier transform is the sinc function), of width {$\tau$}, translated by {$\tau/2$} from the origin of times. Let us call {$f(\nu)$} the transform of {$a(t)$}. Whatever its detailed shape, it is typically a function sharply peaked around a frequency {$\nu = \gamma B$} (this is incidentally true also for fields {$B$} of internal origin, in magnetically ordered materials).

For the purpose of a rough estimate we could approximate {$a(t) = \cos(2\pi\nu t-\phi)$}, hence:

{$ (2) \qquad\qquad f(\nu) \approx \frac 1 2 (\cos \phi [ \delta(\nu-\gamma B) + \delta (\nu+\gamma B)] + i \sin\phi \int_{-\infty}^{+\infty} [\frac {\delta (\nu-\gamma B)}{\nu-\nu^'} - \frac {\delta(\nu+\gamma B)}{\nu+\nu^'}] d\nu^' )$}

We can easily compute the transform of {$W(t,\tau)$}:

{$ \int_0^t e^{-i 2\pi\nu t^'} dt^' = \tau \mbox{sinc}(\pi\nu\tau) e^{-i\pi\nu\tau}, $}

The FFT of {$A(t)$} is given by:

{$ F(\nu) = f(\nu) - f(\nu) \otimes [ \tau \mbox{sinc}(\pi \nu \tau) e^{-i \pi\nu\tau} ] $}

where the second term is the convolution of {$f(\nu)$} with the FFT of {$W(t,\tau)$}.

To understand what happens consider the approximation of Eq. 2 for the convolution, and in particular the real term:

{$ (3) \qquad\qquad \Re(F(\nu)) \approx \Re(f(\nu)) - \frac \tau {2\sqrt{2\pi}} \cos\phi[\mbox{sinc}(\pi (\nu-\gamma B) \tau) \cos(\pi(\nu-\gamma B) \tau) + \mbox{sinc}(\pi (\nu+\gamma B) \tau) \cos(\pi(\nu+\gamma B) \tau) ] $}

This amounts to the Fourier Transform {$\Re({f(\nu))$} we are seeking for, plus a smoothly oscillating, background term of small amplitude (since it is weighted by the short time {$\tau$}). In the presence of a natural width {$1/T$} for {$a(t)$} this weight will be of order {$\frac \tau T$}. In practical experimental cases {$\tau, \phi$} and the local field {$B$} may be only roughly known prior to analysis. Therefore one may fit an experimental FFT spectrum by

{$ (4) \qquad \qquad \Re(F(\nu)) \approx \Re(f(\nu)) - A (\mbox{sinc}((\omega-\omega_0)\tau) + \mbox{sinc}((\omega+\omega_0)\tau))$}

with parameters {$ \omega_0, \tau, A=\frac{\tau\cos\phi} {\sqrt{2\pi}} $}. It is easy to see that Eq. (4) defines an even function. Note that the sum of the two sinc is similar to $1.8*\mbox{sinc}(\nu t)$ and that no scaling of the parameters can actually make the two functions coincide.

The imaginary term in Eq. (2) will lead to a similar convolution of a sharp function, {$\Im({f(\nu)})$}, with the smooth sinc, leading to analogous results. The baselines can be fitted with functions like that of Eq. (4), with three free parameters.

{$ (5) \qquad\qquad \Im(F(\nu))\approx \Im(f(\nu))- A(\frac{\cos(2\pi(\nu- \gamma B)\tau)-\mbox{sinc}((\nu-\gamma B)\tau)}{\nu-\gamma B}-\frac{\cos(2\pi(\nu+ \gamma B)\tau)-\mbox{sinc}((\nu+\gamma B)\tau)}{\nu+\gamma B})$}

< Initial phase in a TF experiment (GPS) | Index | The time shape of the Pulsed Muon Beam >