Lecture 12 Convolutions and summary of course

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Lecture 12Convolutions and summary of course

• Today
• Convolutions
• Some examples
• Summary of course up to this point

Remember Phils Problems and your notes
everything
http//www.hep.shef.ac.uk/Phil/PHY226.htm
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Convolutions
Imagine that we try to measure a delta function
in some way
Despite the fact that the true signal is a spike,
our measuring system will always render a signal
that is instrumentally limited by something
often called the resolution function.

Detected signal
True signal

If the resolution function g(t) is similar to the
true signal f(t), the output function c(t) can
effectively mask the true signal.
Resolution function
Convolved signal
True signal


http//www.jhu.edu/signals/convolve/index.html
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Convolutions




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Deconvolutions
We have a problem! We can measure the resolution
function (by studying what we believe to be a
point source or a sharp line. We can measure the
convolution. What we want to know is the true
signal!
This happens so often that there is a word for it
we want to deconvolve our signal.

There is however an important result called the
Convolution Theorem which allows us to gain an
insight into the convolution process. The
convolution theorem states that-

i.e. the FT of a convolution is the product of
the FTs of the original functions.
We therefore find the FT of the observed signal,
c(x), and of the resolution function, g(x), and
use the result that
in order to find f(x).


If
then taking the inverse transform,
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Deconvolutions
Of course the Convolution theorem is valid for
any other pair of Fourier transforms so not only
does ..
and therefore

allowing f(x) to be determined from the FT

but also
and therefore

allowing f(t) to be determined from the FT

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Example of convolution
I have a true signal
between 0 lt x lt 8 which I detect using a device
with a Gaussian resolution function given by

What is the frequency distribution of the
detected signal function C(?) given that
?
Lets find F(?) first for the true signal
Lets find G(?) now for the resolution signal
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Example of convolution
What is the frequency distribution of the
detected signal function C(?) given that
?
Lets find G(?) now for the resolution signal
so
We solved this in lecture 10 so lets go straight
to the answer
So if
then ..
and ..
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Example of convolution
Detection of Dark Matter using NaI(Tl) crystal
Sodium iodide crystal
PMT
Caesium iodide coating (turns tiny light signals
into single electrons)
Dynode stages (amplify single electrons)
The PMT therefore turns a tiny light pulse into
an electrical signal which we view on an
oscilloscope
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Example of convolution
Detection of Dark Matter using NaI(Tl) crystal

• Single photoelectron pulse
• 10ns pulse duration
• 8mV pulse height

The rate of light production is convolved with
the single electron pulse to yield a net pulse
shape

• Net pulse produced by the crystal following
  particle interaction
• 400ns pulse duration
• 1200mV pulse height

By deconvolving the pulse shape we can
discriminate between events since neutrons and
gammas for example give off light at different
rates
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Dirac Delta Function
The delta function d(x) has the value zero
everywhere except at x 0 where its value is
infinitely large in such a way that its total
integral is 1.
Formally, for any function f(x)
d(x x0) is a spike centred at x x0
d(x) is a spike centred at x 0
For example
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Examples of Fourier Transforms Diffraction
Consider a small slit width a illuminated by
light of wavelength ?.
Minima in diffraction pattern occur at-
where m is an integer.
Taken from PHY102 Waves and Quanta
Intensity
We now can rewrite as
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Examples of Fourier Transforms Diffraction
At the slit, if the light amplitude is f(x), then
the light intensity f(x)2, will be similar to
the top hat function from example 1.
X domain
K domain
q
The diffraction pattern on a distant screen, the
intensity at any point has a sinc2 distribution
and is given by the Fraunhofer diffraction
equation which is related to the Fourier
transform squared F(k)2.
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Convolutions
Example in Optics Diffraction
Weve said that the Fraunhofer diffraction
pattern from a single slit is the modulus squared
of the Fourier transform of the scattering
function.

For double slit diffraction, the scattering
function at the screen is a convolution of two
functions.

Here we convolve a top hat function with 2 delta
functions so as to yield the characteristic
equation representative of light intensity
produced from infinitely narrow slits.

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Convolutions
Example in Optics Diffraction
Consider two delta functions convolved with the
single slit top hat function
This example is specially useful because the
convolution of a true signal with a delta
function is the only time that you can simply
express the convolution


single top hat function two delta functions
convolution

Let us find the Fourier transforms of f and g,
and their moduli squared. From earlier, we know
that the FT of a top hat function is a sinc
function


so
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Convolutions
Example in Optics Diffraction

For the two delta functions we have-

So
These are the familiar cos2 fringes we expect
for two infinitely narrow slits.


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Convolutions
Example in Optics Diffraction
The diffraction pattern observed for the double
slits will be the modulus squared of the Fourier
transform of the whole diffracting aperture, i.e.
the convolution of the delta functions with the
top hat function.

Hence

We see cos2 fringes modulated by a sinc2 envelope
as expected


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Convolutions
We see cos2 fringes modulated by a sinc2 envelope
as expected
(a) Double slits, separation 2d,
infinitely narrow
(b) A single slit of finite width 2a
(c) Double slits of separation 2d, width 2a
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Double-slit Interference Slits of finite width
Fringe patterns for double slits of different
widths
(a) d 50l, a l
Slits are 2d apart Slit width is 2a
(b) d 50l, a 5l
(c) d 50l, a 10l
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Parsevals Theorem
At the end of the Fourier series lectures, we
very very briefly met Parsevals theorem which
states that the total energy in a wavefunction is
equal to the sum of the energy in each harmonic
mode.
It can be shown that it is also true to say ..
Relating f(x) and F(k),
or F(w) and f(t)
For example, when are considering Fraunhofer
diffraction, for f(x) and F(k) this formula means
the total amount of light forming a diffraction
pattern on the screen is equal to the total
amount of light passing through the apertures or
for F(w) and f(t), the total amount of light that
is recorded by the spectrometer (dispersed
according to frequency) is equal to the total
amount of light that entered the detector in that
time interval.
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• Convolving Two Gaussians

Very often we must convolve a true signal and
resolution function both of which are Gaussians
..
Remember when we calculated the FT of a Gaussian ?
has FT
Similarly
has FT
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• Convolving Two Gaussians

So their convolution c(x) has FT
So
so
where
Therefore
This is just the FT of a single Gaussian c(x)
where ..
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• Convolving Two Gaussians

This is an important result we have shown that
the convolution of two Gaussians characterised by
a and b is also a Gaussian and is characterised
by D.
The value of D is dominated by whichever is the
smaller of a or b, and is always smaller then
either of them. Since we found that the full
width of the Gaussian f(x) is in
lecture 9, it is the widest Gaussian that
dominates, and the convolution of two Gaussians
is always wider than either of the two starting
Gaussians.
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• Convolving Two Gaussians

Depending on the experiment, physicists and
especially astronomers sometimes assume that both
the detected signal and the resolution functions
are Gaussians and use this relation in order to
estimate the true width of their signal.