def function1D(self, xvals, **optparms):
r""" Fourier transform of the symmetrized stretched exponential integrated
within each energy bin.
The Symmetrized Stretched Exponential:
height * exp( - |t/tau|**beta )
The Fourier Transform:
F(E) \int_{-infty}^{infty} (dt/h) e^{-i2\pi Et/h} f(t)
F(E) is normalized:
\int_{-infty}^{infty} dE F(E) = 1
Let P(E) be the primitive of F(E) from minus infinity to E, then for element i of
xvals we compute:
1. bin_boundaries[i] = (xvals[i]+xvals[i+1])/2
3. P(bin_boundaries[i+1])- P(bin_boundaries[i])
:param xvals: list of values where to evaluate the function
:param optparms: alternate list of function parameters
:return: P(bin_boundaries[i+1])- P(bin_boundaries[i]), the difference of the primitive
"""
rf = 16
parms, de, energies, fourier = function1Dcommon(self, xvals, rf=rf, **optparms)
if parms is None:
return fourier # return zeros if parameters not valid
denergies = (energies[-1] - energies[0]) / (len(energies)-1)
# Find bin boundaries
boundaries = (xvals[1:]+xvals[:-1])/2 # internal bin boundaries
boundaries = np.insert(boundaries, 0, 2*xvals[0]-boundaries[0]) # external lower boundary
boundaries = np.append(boundaries, 2*xvals[-1]-boundaries[-1]) # external upper boundary
primitive = np.cumsum(fourier) * (denergies/(rf*de)) # running Riemann sum
transform = np.interp(boundaries[1:] - parms['Centre'], energies, primitive) - \
np.interp(boundaries[:-1] - parms['Centre'], energies, primitive)
return transform * parms['Height']
def function1D(self, xvals, **optparms):
""" Fourier transform of the Symmetrized Stretched Exponential
The Symmetrized Stretched Exponential:
height * exp( - |t/tau|**beta )
The Fourier Transform:
F(E) \int_{-infty}^{infty} (dt/h) e^{-i2\pi Et/h} f(t)
F(E) is normalized:
\int_{-infty}^{infty} dE F(E) = 1
"""
parms, de, energies, fourier = function1Dcommon(self, xvals, **optparms)
if parms is None:
return fourier #return zeros if parameters not valid
transform = parms['Height'] * np.interp(xvals-parms['Centre'], energies, fourier)
return transform
def function1D(self, xvals, **optparms):
""" Fourier transform of the Symmetrized Stretched Exponential
The Symmetrized Stretched Exponential:
height * exp( - |t/tau|**beta )
The Fourier Transform:
F(E) \int_{-infty}^{infty} (dt/h) e^{-i2\pi Et/h} f(t)
F(E) is normalized:
\int_{-infty}^{infty} dE F(E) = 1
"""
parms, de, energies, fourier = function1Dcommon(self, xvals, **optparms)
if parms is None:
return fourier #return zeros if parameters not valid
transform = parms['Height'] * np.interp(xvals-parms['Centre'], energies, fourier)
return transform
def function1D(self, xvals, **optparms):
""" Fourier transform of the symmetrized stretched exponential integrated
within each energy bin.
The Symmetrized Stretched Exponential:
height * exp( - |t/tau|**beta )
The Fourier Transform:
F(E) \int_{-infty}^{infty} (dt/h) e^{-i2\pi Et/h} f(t)
F(E) is normalized:
\int_{-infty}^{infty} dE F(E) = 1
Let P(E) be the primitive of F(E) from minus infinity to E, then for element i of
xvals we compute:
1. bin_boundaries[i] = (xvals[i]+xvals[i+1])/2
3. P(bin_boundaries[i+1])- P(bin_boundaries[i])
:param xvals: list of values where to evaluate the function
:param optparms: alternate list of function parameters
:return: P(bin_boundaries[i+1])- P(bin_boundaries[i]), the difference of the primitive
"""
rf = 16
parms, de, energies, fourier = function1Dcommon(self,
xvals,
rf=rf,
**optparms)
if parms is None:
return fourier # return zeros if parameters not valid
denergies = (energies[-1] - energies[0]) / (len(energies) - 1)
# Find bin boundaries
boundaries = (xvals[1:] + xvals[:-1]) / 2 # internal bin boundaries
boundaries = np.insert(boundaries, 0, 2 * xvals[0] -
boundaries[0]) # external lower boundary
boundaries = np.append(boundaries, 2 * xvals[-1] -
boundaries[-1]) # external upper boundary
primitive = np.cumsum(fourier) * (denergies /
(rf * de)) # running Riemann sum
transform = np.interp(boundaries[1:] - parms['Centre'], energies, primitive) - \
np.interp(boundaries[:-1] - parms['Centre'], energies, primitive)
return transform * parms['Height']