Download Chapter 9. Computer vision

Chapter 9. Computer vision
Sample problems
9-S1
Figure P9.1 Represents an optical system, a point of the object space is imaged on the image
space.
1) On the basis of the developments presented in sections 9.2 to 9.4 write the relationship
between input and output.
2) Is the above relationship valid for monochromatic light as well as for polychromatic
light? What is the difference?
3) How you would generalize the above relationship from a single point to the points that
make up a full object. What operation is involved in this relationship?


Utilize the reduced coordinates,  r  , r 
M
M
Solution to 9-S1
The graph represents the relationship between object points and image points assuming that the
lens system is a linear system,
(  0 ,  0 )  h( u  u 0 , v  v0 )
The object is represented as an aggregate of luminous sources mathematically represented by the
two dimensional Diract’s δ. Assuming that the optical system is linear implies that the
relationship between the input and the output of the system is a linear map. This means that all
the properties listed in section 9.2 are satisfied. The fundamental result in system theory is that
any LSI system can be characterized entirely by a single function called the system's impulse
response in this case h(u).
The impulse response of a lens system is characterized by the Airy’s ring, of radius
R
(see section 8.4.2.)
r  0.610
a
It is possible to see that the radius is a function of the wavelength hence it will change with λ. In
the case of white light then one can assume that one will have a distribution of intensities caused
by the different components of the white light. The concept is still valid for incoherent light. The
radius of the Airys’ ring is computed assuming a value of λ in the green region λ=540 nm. For a
full object the image will be the result of the addition of the impulse response of the lens (9.8)
E( u, v )   h( u, v ) E M ( r , r ) d r dr
A
The image is the result of convolving the geometrical image of the object with the impulse
response of the lens according to the shift invariance of the lens.
There are several points to be taken into account that are important. The linearity of the response
in the case of coherent illumination corresponds to the electrical field expressed in V/m. In the
case of incoherent illumination the linearity applies to the irradiance expressed in W/m 2. This is
an important aspect because the coherent imaging will cause phenomena of interference that will
not be present in the incoherent illumination causing changes in the final irradiance of the image.
The validity of the convolution integral depends on satisfying the shift invariance property, real
lenses are shifting invariant for paraxial beams and on a limited region of the lens depending on
the lens quality that are called isoplanatic regions the lens.
There is a final consideration to make, when there is a sensor to capture the response of the
system depends on the sensor. If the sensor is non linear the convolution relationship will not be
valid.
9-S2
Equation E( u, v )   h( u, v ) E M ( r ,r ) d r dr represents the field in the image plane. Since as it
A
was observed in the preceding problem h ( u, v ) depends on λ (because it represents the Airy’s
ring consequence of the diffraction effect of the pupil controlling the formation of the image),
comment the consequences of this fact and the effect of the wave length of light in the image
formation.
Solution to 9-S2
Taking the FT of the preceding equation,


FTE( u, v ) FT   h( u, v ) E M ( r ,r ) d r dr 
A

Recalling that the convolution on the object space leads to a product in the frequency space and
taking into consideration (9.8) and (9.9) one gets
G i (f u , f v )  H(f u , f v ) G 0 (f u , f v )
In this equation G i (f u , f v ) denotes the image spectrum and H(f u , f v ) is the spectrum of the
impulse response at the lens output, and is also called the lens system transfer function and
relates the object and the image multiplicatively. The above relationship means that the spectrum
of spatial frequencies of the image is a function of the wavelength of light since the impulse
response function is a function of the wave length of the light forming the image.
Also the normalization coordinates in the formation of the image are function of the wavelength

2ax  2ay
x
, y
. In Section 9.3 it was shown that from red to violet there is a large change of
z i
z i
this factor that reflects in the image spatial frequency. If one has incoherent light of different
frequencies, the different colors will have different normalization coordinates.
Problem 9-S3
Comment the consequences of (9.9) in the formation of the image of a sinusoidal object in the
object space with monochromatic illumination and a circular pupil. The sinusoid has a visibility
one: What will be the visibility of the image?
Solution to 9-S3
The response of the system is characterized by the pupil function represented in the normalized
a
space by a low pass filter of cut-off frequency f co 
. It the frequency of the sinusoid is
 zi
smaller than the cut-off it will pass without change since the filter has a normalized value either
one or zero. Furthermore the ideal lens will not introduce changes of phase as a function of the
frequency, then the sinusoid will continue to be a sinusoid and the contras will continue to be
one.
The ideal mathematical model ignores the effect of noise in the signal. Noise will be an
important factor in coherent imaging. The presence of foreign particulates in the optical system
will cause light emission and interference effects. The introduced normalization provides values
of the transmittance either one or zero, ignoring losses by Fresnel reflections inside the optical
system or losses of energy due to absorption and light scattering inside the lenses. Also phase
changes due to the difference between and an ideal lens and a real lens are ignored. G 0 (f u , f v )
will be a complex quantity wan amplitude component and a phase component.
9-S4
Compare the formation of the image of an object with coherent and incoherent illumination.
Which is better? Comment your conclusions. Utilize Figure 9.12 for this comparison.
Solution to 9-S4
Considering the amplitude of the signal we know that the amplitude of the image depends on the
modulus of the OTF or the so called modulation transfer function MTF. As it shown in Figure
9.9 that represents the MTF of a lens system with a circular aperture the amplitude of the signal
is reduced with the frequency. This is illustrated in Figure 9.12 that shows the reduction of the
visibility of the first harmonic of a bar system that is the bar target to analyze the properties of a
lens system, in the case of the Figure 9.12 a microscope objective. Hence we can anticipate that
the image of the object will experience a distortion since the amplitude of the harmonics that
make up the object has been reduced as a function of the frequency. There is another aspect to be
considered, the fact that the coherent imaging reduces the amount of harmonics that can be
captured by a lens that arises from the comparison of the cut-off frequencies of the transmittance
of coherent optics as compared to the OTF of an incoherent system. Hence depending of what
we want to see in an image is difficult to make an absolute judgment concerning the type of
illumination.
9-S5
Relate the MTF of a lens system to the numerical aperture of the lens, to the diameter of the
Airy’s ring and to the Raleigh and Sparrow criteria of resolution. See Figure 9.13.
Solution to 9-S5
In (9.48) the numerical aperture of a lens is defined as,
N  f/# 
f
D
D is the diameter of the entrance pupil of the lens and f the focal distance. This definition applies
for an object at infinity. (As described in section 6.8) Calling s1 the distance of the object to the
principal plane of the optical system and s2 the distance of the image to the principal plane helps
define (f/#)object space=s1/D and (f/#)image space=s2/D. Consequently the autocorrelation function can
be defined as a function of the numerical aperture utilizing these different definitions. In section
9.4 the cut-off frequency for the circular section is twice the cut-off frequency for the coherent
2a
illumination is 0  2f ct 
for the case of the objet to infinity zi=f and a=D/2 replacing in
 zi
2D / 2 
1
 /# .
the expression of the cut-off frequency 0  2f ct 
f
f
For finite distances this expression is valid replacing zi by si. Looking at the definition of the
R
radius of the Airy’s ring r  0.610
in the nomenclature utilized to derive the above equation
a
R=f then for the diameter of the ring d ay  2.4 f /# . Considering the normalized frequency
r
1
 0.826 /# this results corresponds to the radius of the Airy’s ring and is the
/#
1.21f
f
Raleigh criterion of resolution. In the sparrow criterion the separation between adjacent Airy’s
ring is taken as 0.47 in place of 0.826 resulting in a frequency 0.94 of the ideal cut-off frequency
The results of this analysis are shown in Figure 9.13
r 
9-S6
Analyze the effect of the sensor array in the MTF function.
Hint. In the analysis it is necessary to introduce the effect of the quantification of the space and
the intensity in the derivations introduced in Chapter 9.
Figure P9.2.Effect of a linear array in the capture of the intensity of in the object space.
Figure P9.2 shows the areas in the object plane that contribute to the different sensors of the
linear array. One can make a first observation that all the intensity coming from an area that falls
in a single sensor will be averaged by the sensor. The second observation is that one has to make
some assumption concerning the sensor itself. The first assumption is that sensor must respond
linearly to the energy received. The second assumption is that the intensity levels will be below
the saturation limit. We can now utilize for a single sensor the same procedure adopted for the
whole image. The detector response can be described utilizing the coordinate system of Figure
9.6. and calling E the field emerging from the object by,
w / 2
u
Ed ( u)   E( u)du  E( u)  rect
w
w / 2
Where as defined in section 8.7.1 the symbol * indicates the convolution integral.
This operation has a similarity with the assumption that the basic unit in an image produced by a
lens is an Airy’s ring. In this case the basic unit is the square pixel sensor. The field collected by
the sensor is the integral of all the components of the field received by the sensor. Taking the FT
of the above expression one gets,
sin f u w
FTE d ( u )  FTE( u )x
f u w
following similar procedures to those used in the formation of the image and introducing the
concept of impulse response one gets,
MTFf u   sin c sin fu w ;
Where the double bar indicates the absolute value. This expression is very important because
relates the effect of the sensor width w in the formation of the quantized image produced by the
sensor that is different from the image formed by a continuum medium receptor assumed in the
analysis of the images presented in Chapter 9. Extending the analysis to two dimensions and
Figure P.9.3.Capture of the full field by the detector array.
utilizing the assumption that rows are independent of each other as it is done in the formation of
images we can extend to two dimensions,
MTFf u, f v   sin c sin fu w sin c sin f v w
Figure P.9.4.MTF of the array detector.
It is interesting to see the effect of the size of the detector in the recovery of the information of
the original signal coming from the object. For that one can look to the plot of MTF as a function
of fu. One can see that the smaller is the detector size the larger is the frequency resolution of the
image. This result has to be understood in context of the size of the detector. The meaning is
clear the smaller is the size of the pixel for a given detector size the better is the obtained spatial
resolution.
Problems to solve
9.1
Assume that s1=50 mm and that s2=100 mm. The lens system has a circular pupil of r=10 mm
and λ=635 nm. Compute the cut-off frequency in the object space and in the image space.
9.2
In the case of sample problem 9-S6 compute at what spatial frequency relative to the cut-off
frequency the MTF will be 80 % of the maximum.
9.3
An object is located at the distance s1=105 mm of the principal plane of a lens system, the focal
length is 50 mm, the aperture has a radius r=10 mm, the wavelength of the illuminating light
λ=545 nm. What is the maximum spatial frequency that the lens can resolve? If the illumination
is incoherent what will be the resolution.
9.4
A microscope illuminated by non coherent light has a field of view of 391 microns, the highest
frequency observed is 0.8 125/μm. What is the angular aperture of the microscope objective.
Take the wavelength λ= 540 nm.
9.5
What will be the highest frequency observed if the illumination is changed to coherent with
λ=635 nm.
9.6
A detector array utilized in a microscope has a field of view of 391 microns and has 1024 square
pixels. What is the maximum spatial frequency that can be captured by the sensor assuming that
the microscope has the necessary optical resolution.
9.7
What happens to the resolution of the image in problem 9.6 if the detector has only 512 square
pixels.
9.8
In the case of problem 9.6 one has to record a sinusoidal fringe pattern that has 300 fringes. Will
the sensor be capable of recording these fringes?
9.9
A camera with a 2/3 “ sensor of size 8.57.1 mm has 24562058 pixels of 3.45 μm  3.45
μm.We want to observe a 60 mm diameter disk and leave some pixels outside of the disk area
surrounding the disk.Calling hx and hy the field of view in the x-direction and y-direction that
includes the disk. Select a hx and hyin such a way that a maximum number of pixels is inside the
the sensor. To process the information we need to utilize FFT, assume that 2024 is the maximum
number of pixels that can be utilized. If we have at our disposal a 50 mm lens determine the
position of the object with respect to the lens. Assume that the lens is a good quality lens that
allows a f/#=2. Compare the the optical resolution and the resolution of the sensor. Which one
limits the spatial resolution? Assume coherent illumination with λ=635 nm.