Transcript L12

Lecture 12
Convolutions 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
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
True signal
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.
True signal
Resolution
function
http://www.jhu.edu/~signals/convolve/index.html
Convolved
signal
Convolutions
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:-
C (k ) 
2 F ( k ) G ( k )
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
If
F (k ) 
C (k )
G ( k ) 2
F (k ) 
C (k )
G ( k ) 2
in order to find f(x).
then taking the inverse transform, f ( x ) 
1
2

e

ikx
C (k )
G (k )
dk
Deconvolutions
Of course the Convolution theorem is valid for any other pair of Fourier
transforms so not only does …..
C (k ) 
2 F ( k ) G ( k ) and therefore F ( k ) 
allowing f(x) to be determined from the FT
but also C ( ) 
f ( x) 
C (k )
G ( k ) 2
1
2



2  F ( ) G ( ) and therefore F ( ) 
allowing f(t) to be determined from the FT
f (t ) 
1
2
F (k )e
ikx
dk
C ( )
G ( ) 2 



F ( ) e
i t
d
Example of convolution
f (t )  e
I have a true signal
 at
between 0 < x < ∞ which I detect using a
g (t ) 
device with a Gaussian resolution function given by
  at 2
exp 
2
 2
a




What is the frequency distribution of the detected signal function C(ω) given
that C ( ) 
2  F ( ) G ( ) ?
Let’s find F(ω) first for the true signal …
F ( ) 
1
2


e

 at
e
 i t

F ( ) 
dt 
 t ( a  i )

1 e

 
2  a  i   0
1
2



e
 t ( a  i )
1 
1 
0



a  i 
2 
F ( ) 
dt
1
2

e
 1 


a

i

2 

1
Let’s find G(ω) now for the resolution signal …
2



f (t ) e
 i t
dt
 t ( a  i )


1
 t ( a  i )
dt 


1 e


2  a  i   0
Example of convolution
What is the frequency distribution of the detected signal function C(ω) given
that C ( ) 
2  F ( ) G ( ) ?
Let’s find G(ω) now for the resolution signal …
g (t ) 
  at 2
exp 
2
 2
a




so G ( ) 
1
2



G ( ) 
  at 2
exp 
2
 2
a
1
2



g (t ) e
 i t
dt
  i t
e
dt


  at 2

We solved this in lecture 10 so let’s go


G ( ) 
exp

i

t
dt

 2

straight to the answer
2  


 2 
1

G ( ) 
exp 

2 
 2a 
a

So if C ( ) 
F ( ) 
2  F ( ) G ( ) then …..
 2
1
 1 
exp 

 G ( ) 
2  a  i  
2 
 2a
1
C ( ) 

2 





and …..
 2
 1
exp 

2 a  i   2 
 2a
1
1

1

exp

 2  a  i 
 2 


 2a 


Example of convolution
Detection of Dark Matter using NaI(Tl) crystal
Sodium iodide crystal
Caesium iodide coating
PMT
(turns tiny light signals into single electrons)
The PMT therefore turns a tiny light pulse into an
electrical signal which we view on an oscilloscope
Dynode stages
(amplify single electrons)
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
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)



f ( x )d ( x  x 0 ) dx  f ( x 0 )
d(x – x0) is a spike centred at x = x0
d(x) is a spike centred at x = 0




d ( x ) dx  1
For example





x d ( x  a ) dx  a
2
2
d ( x  x 0 ) dx  1
Examples of Fourier Transforms: Diffraction
Consider a small slit width ‘a’ illuminated by light of wavelength λ.
Minima in diffraction pattern occur at:- a sin
Taken from
PHY102 Waves
and Quanta
  m
Intensity
 sin(  a (sin  ) /  ) 
I  I0 


a
(sin

)
/



where m is an integer.
We now can rewrite as
2
  a sin  
I  I 0 sinc 




2
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 
K domain
2
a
k 
2 sin 


F (k ) 
 ka 
sinc 

2
 2 
a
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.
2   a sin  
I ( )  I 0 sinc 




Convolutions
Example in Optics: Diffraction
We’ve 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.
C (k ) 
2 F ( k ) G ( k )
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
a
a
d
single ‘top hat’ function
f ( x)  1
a  x a
d
d
d
two delta functions
convolution
g ( x)  d ( x  d )  d ( x  d )
c( x)  1
for
x  d  a and
xd a
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:
F (k ) 
2
2
sin ka
k
so
F (k )
2

2 sin

2
k
2
ka
Convolutions
Example in Optics: Diffraction
a
a
d
d
d
d
For the two delta functions we have:G (k ) 
So
1
2


| G (k ) | 
2

[d ( x  d ) d ( x  d )] e
2

2
cos kd
These are the familiar
cos2 fringes we expect
for two ‘infinitely narrow’
slits.
 ikx
dx 
1
2
[e
ikd
e
 ikd
]
2
2
cos kd
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
C (k )
2
2
 2 F ( k ) G ( k )
2

4 sin

2
2
k
2
ka
cos
2
kd
We see cos2 fringes modulated by a sinc2 envelope as expected
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
Double-slit Interference: Slits of finite width
Fringe patterns for double
slits of different widths
(a) d = 50, a = 
Slits are 2d apart
Slit width is 2a
(b) d = 50, a = 5
(c) d = 50, a = 10
Parseval’s Theorem
At the end of the Fourier series lectures, we very very briefly met Parseval’s
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 …..
or F() and f(t)
Relating f(x) and F(k),




dk | F ( k ) | 
2



dx | f ( x ) |
2



d  | F ( ) | 
2

dt | f ( t ) |
2

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.
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 ?
f ( x) 
a
2
e
2
 ax / 2
has FT
F (k ) 
1
2
e
k
2
2a
Similarly
g ( x) 
b
2
e
f ( x)
bx
2
2
has FT
G (k ) 
1
2
e
k
2
2b
g ( x)
Convolving Two Gaussians
So their convolution c(x) has FT
C (k ) 
So
C (k ) 
2 F (k )G (k ) 
1
2

e
Therefore C (k ) 
k
2

2a
e
1
2
k
1
2
2
e
k
2
1
2a
2
2
2

k
2
2
1
C (k ) 
so
2b
e
e
where
2
1


1
a
e
c( x) 
2
e
 x
2
2
2
k 1 1

  
2 a b

1
b
This is just the FT of a single Gaussian c(x) where …..

k
2b
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 .
c( x) 

2
e
 x
2
1
2


1

a
1
b
The value of  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 2 2 a 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.
f (x) 
a
2
x 
2
a
a
2
2
e
 ax / 2
Convolving Two Gaussians
C (k ) 
2 F (k )G (k ) 
2
1
2
e
k
2
2a
1
2
e
k
2
2b
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.
1 1 1
ab
  x 2 2
 

c( x) 
e
2
 a b
ab
f (x) 
a
2
x 
2
a
a
2
2
e
 ax / 2

c( x) 
2
x 
2


2
2
e
 x 2