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

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’.
True signal
Detected 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
Convolved signal http://www.jhu.edu/~signals/convolve/index.html

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:-
( )

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 in order to find f(x) .
If

 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 …..
( )

2

F k G k and therefore allowing f(x) to be determined from the FT f ( x )



1
2

 
 
F ( k ) e ikx dk but also C (

)

2

F (

) G (

) and therefore
F (

)

G
C (

)
(

) 2
 allowing f(t) to be determined from the FT f ( t )

1
2





F (

) e i
 t d


Example of convolution f ( t )
 e
 at
I have a true signal between 0 < x <
which I detect using a
( t a
2
 exp
 
 at 2
2


What is the frequency distribution of the detected signal function C( ω) given
C
    
Let’s find F(ω) first for the true signal …
F (

)

1
2




 e
 at e
 i
 t dt

1
2




 e
 t ( a
 i

) dt
F (

)

1
2




 f ( t ) e
 i
 t dt
1
2




 e
 t ( a
 i

) dt


1
2


 e
 t ( a
 i

) a
 i




0
F (

)


1
2


 e
 t ( a
 i

) a
 i




0


1
2

0
 a

1 i


1
2

Let’s find G(ω) now for the resolution signal … a

1 i


Example of convolution
What is the frequency distribution of the detected signal function C( ω) given
C
    
Let’s find G(ω) now for the resolution signal … g ( t )
 a
2
 exp
 
 at
2
2

 so G (

)

1
2

 
 
G (

)
 a
2
 exp
 
 at
2
2

 e
 i
 t dt
1
2




 g ( t ) e
 i
 t dt
G (

)

F (

)

2
 a 


 exp


 at
2
2
 i
 t

 dt
We solved this in lecture 10 so let’s go straight to the answer
G (

)

2
1
 exp
 


2 a
2


C
 
1
2

   a

1 i

C (

)

G (

)

2
1
 exp
 


2 a
2


2


 and …..
1
2
 a
1
 i


 2
1
 exp
 


2 a
2



2


1 a
 i

 exp
 


2 a
2



Example of convolution
Sodium iodide crystal
Caesium iodide coating
(turns tiny light signals into single electrons)
PMT
Dynode stages
The PMT therefore turns a tiny light pulse into an
(amplify single electrons) electrical signal which we view on an oscilloscope

Example of convolution
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

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) is a spike centred at x = 0
 
  d
( x ) dx

1 d
(x – x
0
) is a spike centred at x = x
0
 
  d
( x
 x
0
) dx

1
For example
 
  x
2 d
( x
 a ) dx
 a
2

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
  m
 where m is an integer.
Taken from
PHY102 Waves and Quanta
Intensity
I

I
0

 sin(

 a a (sin
(sin


) /
) /


)


2
We now can rewrite as
I

I
0 sinc
2


  a sin






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 k

2
 a k

 

F ( k )
 a
2
 sinc ka
2
The diffraction pattern on a distant screen, the intensity at any point has a sinc 2 distribution and is given by the Fraunhofer diffraction equation which is related to the Fourier transform squared: | F ( k )| 2 .
I (

)

I
0 sinc
2
 a sin




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.
( )

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 d single ‘top hat’ function two delta functions f ( x )

1
 a
 x
 a g ( x )
 d
( x
 d )
 d
( x
 d ) c ( x )

1
 d d convolution 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 ka k
2

Convolutions
Example in Optics: Diffraction
 a a  d
For the two delta functions we have:-
G ( k )

1
2





[ d
( x
 d )
 d
( x
 d )] e
 ikx dx
 d
 d d
1
2

[ e ikd
 e
 ikd ]

2
2
 cos kd
So | G ( k ) |
2

2
 cos
2 kd
These are the familiar cos 2 fringes we expect for two ‘infinitely narrow’ slits.

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

F ( k )
2
G ( k )
2 

4
2 sin
2 k
2 ka cos
2 kd
We see cos 2 fringes modulated by a sinc 2 envelope as expected

Convolutions
We see cos 2 fringes modulated by a sinc 2 envelope as expected
(a) Double slits, separation 2d, infinitely narrow
(b) A single slit of finite width 2 a
(c) Double slits of separation 2d, width 2a

Fringe patterns for double slits of different widths
d = 50

, a =

Slits are 2d apart
Slit width is 2a
d = 50

, a = 5

d = 50

, a = 10


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 …..
Relating f(x) and F(k),


 dk F k
2  

 dx f x
2 or F(

) and f(t)


 d

| F

2  


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 ?
 a
2
 e
 ax
2
/ 2 has FT
F ( k )

1
2
 e
 k
2
2 a
g ( x )
 b
2
 e
 bx
2
2 has FT
G ( k )

1
2
 e
 k
2
2 b f ( x ) g ( x )

Convolving Two Gaussians
So their convolution c(x) has FT
C ( k )

2

F ( k ) G ( k )

2

1
2
 e
 k
2
2 a
1
2
 e
 k
2
2 b
C ( k )

1
2
 e
 k
2 a
2 k
2 e

2 b
C ( k )

1
2
 e
 k
2
2
1 a
1 b
C ( k )

1
2
 k e

2

2 where
1

1 1 a b
c ( x )


2
 e
  x
2
2

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
2
1

1 1 a 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 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.
a
2

 a
2
 e
 ax
2
/ 2
 x

2 a

Convolving Two Gaussians
C ( k )

2

F ( k ) G ( k )

2

1
2
 e
 k
2
2 a
1
2
 e
 k
2
2 b
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.
c ( x )


2
 e
  x
2
2
1

1 1 a b
  a ab
 b a
2

 a
2
 e
 ax
2
/ 2

2
 c ( x )


2
 e
  x
2
2
 x

2 a
 x

2
