Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
Tutorials 9-10: Psychophysics and Physiology
Questions:
Question 1:
In a psychophysical experiment of adaptation to a spatial frequency, a square wave
lattice was presented in a 10 cpd frequency, with high contrast.
For the sake of this question, we assume the visual system of the examinee includes 5
channels sensitive to the following ranges, given in cpd: 2  4, 4  8,16  32,32  64 .
This question deals with the 1D case.
a. Determine which of the channels will respond to the displayed lattice.
b. The channel's sensitivity has decreased to zero due to long projection of the
aforementioned signal. In this point the projection of the first lattice is stopped
and the examinee is presented for a short time with a second square lattice in a 5
cpd frequency.
1) Which of the channels will react now?
2) How will the second signal appear to the examinee? Which signal would
have given an identical sensation prior to the adaptation?
Question 2:
The reaction R  x  to the illuminance E  x  in an AGC model of the visual system is
given by:

 E  x  1  k  x   R  x  
R  x  

0
,
 0.5  x  0.5
,
otherwise
R  x  represents the viewer's brightness sensation.
The functions R  x  , E  x  can be taken as the sum of constant components (DC)
represented by CR , CE respectively, with varying components represented by r  x  ,
0.5
 0.5

e  x  respectively   e  x  dx   r  x  dx  0  .
0.5
 0.5

E  x   CE  e  x 
R  x   CR  r  x 
k  x  is a function that characterizes the model.
a. Given k  x   1,  1  x  1 :
1) Given CE , can CR be calculated? If so, calculate it. If not, explain why.
2) What is the connection between r  x  and e  x  ?
b. Assuming the possible variability range of R  x  without reaching a blinding or a
darkening sensation is 0.01  R  x   0.99 , what are the permitted  values in the
next excitation so it will be possible to notice the details of the presented signal?
1
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
c. Given the following k  x  and R  x  , what is the projected E  x  ? Sketch it.
d. The signal E  x  received in section c is presented to a system that acts according
to k  x   1,  1  x  1 . Is it possible to notice the presented signal without
reaching blinding or darkening?

Solve using the fact:
 E  x  1

Question 3:
a. Calculate the Fourier transform of the Gaussian g  x   e  ax .
2
b. Find the connection between the effective width of the Gaussian g1  x   Ke
and the standard deviation  .
c. Find the Fourier transform of a Gaussian modulated with a sine function:
g2  x   Ke
 x  x0 2
2 2
sin 0 x 
2
 x  x0 2
2 2
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
Question 4:
A cell C1 has the following receptive field:
Figure 1 - The receptive field of cell C1
The cell is stimulated by a signal E  x, y  in the shape of a step (figure 2):
Figure 2 - The stimulus signal E  x, y 
a. Sketch a graph describing the cell's reaction to illuminance as a function of the
angle  . Show the cell is sensitive to directionality. State in which angle the cell
response is maximal and in which angle the cell response is zero. Is it possible to
determine the angle  according to the cell response measurement?
b. There's another cell in the system, C2, with the same receptive field, only rotated
in 90 (Figure 3):
Figure 3 – The receptive field of cell C2
1. Calculate the cell response to the signal E  x, y  as a function of  .
2. A certain  generates a C1 cell response of 2 and a C2 cell response of -1.
Is it possible to determine in which angle  the step was presented? If
yes, state the angle.
3
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
Question 5:
We shall examine the activity of cells in the visual cortex according to the channels
model, which have the following receptive field:
c

 g n  x   cos  2 nx 
, 0n3
 s
g
x

sin
2

nx





n

Calculate the cell response to the following signals:
f1  x   p  x 
f2  x   p  2x 
f 3  x   p  x  sin  4 x 
1 ,  0.5  x  0.5
p  x  
0 , otherwise
Question 6:
In light of the findings regarding the localization of cell activity, a more suitable
model is suggested:
c

 g mn  x   p  x  md  cos  2 nx  , 0  n  3
; D 1
 s
g
x

p
x

md
sin
2

nx
,
0

m

3







mn

Calculate the cell response to the following signals:
f1  x   cos  2 x 
f 2  x   sin  6 x 
f 3  x   3  const 
f 4  x   p  x  sin  4 x 
4
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
Solutions:
Question 1:
a. A Fourier Decomposition of a square wave includes all the uneven harmonies.
Therefore, a square wave in a 10 cpd frequency contains components in
frequencies 10 cpd, 30 cpd, 50cpd… The channels will react when one of the
harmonies falls in their range, meaning the channels are sensitive to the
ranges 8  16,16  32,32  64[cpd ] .
b.
1
Similarly, a square wave in a 5 cpd frequency contains only components
in frequencies 5 cpd, 15 cpd, 25cpd, 35cpd… Since the sensitivity of the 3
high channels is zero due to adaptation, the only channel to react is in the
range 4-8 cpd.
 2 The signal will appear to the examinee as a sine wave in an identical
frequency, meaning a sine wave in a 5 cpd frequency would have created an
identical sensation prior to adaptation.
Question 2:
a.
1

 E  x  1  k  x   R  x  
R  x  
0

1
2
otherwise
x
,
,
The feedback in R  x  constitute the AGC (Automatic Gain ontrol)
E  x   CE  e  x 
R  x   CR  r  x 
0.5
0.5
0.5
0.5
k  x  R  x 
 k  x     R    d    R    C
  I  R  x   E  x 1  CR 
R
 II  CR  CE 1  CR 
  III  CR 
CE
1  CE
 2
Subtraction of equations  I    II  will result in:

 IV  r  x   e  x 1  CR   e  x  1 

 r  x 
e  x
1  CE
5
CE
1  CE
 e  x

 1  CE
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
b. The average of the given signal is:
CE  0.75
Addition of equations  III  +  IV  will result in:
R  x 
E  x
1  CE
 max
  max  3.84
1  0.75 max
0.5 min
Min : 0.01 
  min  0.02
1  0.75 min
Therefore:
0.02    3.84
 Max : 0.99 
c. E  x  
R  x
1 k  x  R  x
We shall sketch k  x   R  x  first:
6
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty

 0.5

d. CE    E  x  dx   E  x  dx  1  CE  1

 0.5

E  x
According to section b: R  x  
 0.5E  x 
1  CE
According to section c:
Emax  4  R  x   2  blinding
Emin  0.5  R  x   0.25  No darkening
Question 3:
a.

e
 ax2  xy

y2
4
e a

dx   e
2
y2

 
2a  4a
 a  x  y

dx
y2
4
e a


e
 az 2
tz a
dz 


e
 a  x  y


2 a 
2
z  x
dx

y
2a

y2
1 4a
e
a

e
t 2
dt 


a
y2
4
e a

The Fourier transform of the Gaussian g  x   e  ax is:
2
G   

 g  x e

 j x

dx 
e
 ax2  j x
e

dx 

e

 ax2  j x
y  j
dx  

a
  j 2
e 4a


a
 2
e 4a
Conclusions:
1. The Fourier transform of a Gaussian is a Gaussian.
2. The width of the Gaussian in the frequency domain is in opposite ratio to
the one in the time domain.
b. The effective width of a function is defined as the distance between the points
where the function value decreases by 2 (in 3dB) from the maximum.
The maximal value of the Gaussian g1  x  is K, received for x  x0 .
7
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
 x  x0 2

2 2
Therefore we have to find x1,2 for which Ke
1
K
2
By reducing K and operating ln on both sides we receive:
 x  x0 
2
 ln
2 2
 2
  x  x0    2 ln 2
2
 x  x0   ln 2
The effective width of the Gaussian is therefore:
x  2 ln 2  1.66
Meaning the effective width of the Gaussian depends on its standard deviation  .
c. Multiplication in the space domain equals convolution in the frequency domain:
F
sin 0 x  
  j    0      0  
 x  x0 2
Ke
2 2

e
F
 jx0
K

1


e
2
1
2 2
 K
2 e
2
 jx0
 2 2
e
2
2 2
A convolution with  is equal to a shift, therefore:
2

 0 2  2 
  jx     0   2


jx







 j x
2
2
2

G2     g  x  e
dx   j K 2 e 0 0 e
e 0 0 e




Sketching the absolute value of G2   results in the following image:
8
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
Question 4:
a. First we shall examine

4
 
5
:
4
r  E  x, y  , g  x, y     E  x, y  , g  x, y  dxdy



R


    E0 R 2 
3 


r
 E0 










  
 2 


2
4
4  
2 
2 




 The sector of the circle with +1 The sector of the circle with -1
3

  :
Now we shall examine 
4
4
2
r
 
E R2 
 R2

 

 
 E0                0  2  
2
2 
2
 
4
 
 4
9
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
A sketch of r as a function of  :
Since r is not an injective function,  cannot be determined with certainty.


b. 1 r2    r1    
2

The result is the red dashed line in the sketch shown in section b.
 2
r1    2
r2    1
2  3
r1  
2
 2 

r2  
2 
 3
2
2




13
 195
12
Notice this is the area where r1 is positive and r2 is negative!
Question 5:
c

 g n  x   cos  2 nx 
, 0n3
 s

 g n  x   sin  2 nx 
According to the receptive field's output:
r  f  x, g
c
n
c
n
 x


 f  x  g  x  dx
c
n

rns  f  x  , g ns  x  

 f  x  g  x  dx
s
n

10
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
 f1  x   p  x 
rnc 
1
2
 cos  2 nx  dx

rns 
1
2
1
2
 sin  2 nx  dx

1
2
r0s  0
r0c 
1
2
 1 dx  1

1
2
For g ns all the responses rns are zero (because sin is an uneven function)
 f2  x   p  2 x 
r0s  0
r0c 
1
4

r1c 
1
 cos  2 x   dx  
1
4
1
4
 cos  4 x   dx  0

r3c 
1
4
1
4

r2c 
1
 1 dx  2
1
4
1
4
1
 cos  6 x   dx   3

1
4
 f 3  x   p  x  sin  4 x 
rnc 
1
2
 sin  4 x  cos  2 nx   dx

rns 
1
2
1
2
 sin  4 x  sin  2 nx   dx

1
2
11
Visual and Auditory Systems
Tutorial 9-10
The Technion - Israel Institute of Technology
Electrical engineering faculty
Since this is a case of trigonometric functions in a section including an integer number
of periods, for n  2 we receive orthogonality. For the same reason, all the elements
related to cosine will be zero.
For n  2 :
r2s 
1
2
1
 sin  4 x  dx  2
2

1
2
Question 6:
c

 g mn  x   p  x  md  cos  2 nx  , 0  n  3
; D 1
 s
g
x

p
x

md
sin
2

nx
,
0

m

3







mn

 f1  x   cos  2 x 
rmc1  x  
1
m
2

cos 2  2 x   dx 
1
 m
2
1
, m  0,1, 2,3
2
All the rest of the responses become equal to zero (orthogonality)!
Following the same principle:
 f 2  x   sin  6 x 
1
2
All the rest of the responses become equal to zero.
rmc 2  x  
 f3  x   3
1
m
2
rmc 0  x  

3dx  3
1
 m
2
All the rest of the responses become equal to zero.
 f 4  x   p  x  sin  4 x 
r02c  x  
1
2
1
 sin  4 x  dx  2
2

1
2
All the rest of the responses become equal to zero.
12