Edition manual third trial win

Then, by substituting this array into the last line of the previous equa- tion we have the 1-D transform along the columns of T x , v. In other words, when a kernel is separable, we can compute the 1-D transform along the rows of the image. Then we compute the 1-D transform along the columns of this in- termediate result to obtain the final 2-D transform, T u , v. We obtain the same result by computing the 1-D transform along the columns of f x , y followed by the 1-D transform along the rows of the intermediate result.

From Eq. From Fig. Based on the information in Fig. This value of z is reasonable, but any other given lens sizes would be also; the camera would just have to be positioned further away. It is given that the defects are circular, with the smallest defect having a diameter of 0. So, all that needs to be done is to determine if the image of a circle of diameter 0. This can be determined by using the same model as in Fig. In other words, a circular defect of diameter 0. If, in order for a CCD receptor to be activated, its area has to be excited in its entirety, then, it can be seen from Fig.

Chapter 3 Problem Solutions Problem 3. Problem 3. The question in the problem statement is to find the smallest value of E that will make the threshold behave as in the equation above. In this truth table, the values of the 8th bit are 0 for byte values 0 to , and 1 for byte values to , thus giving the transformation mentioned in the problem statement. Note that the given transformed values of either 0 or simply indicate a binary image for the 8th bit plane.

Any other two values would have been equally valid, though less conventional. Continuing with the truth table concept, the transformation required to pro- duce an image of the 7th bit plane outputs a 0 for byte values in the range [0, 63], a 1 for byte values in the range [64, ], a 0 for byte values in the range [, ], and a 1 for byte values in the range [, ].

Similarly, the trans- formation for the 6th bit plane alternates between eight ranges of byte values, the transformation for the 5th bit plane alternates between 16 ranges, and so on.

Finally, the output of the transformation for the lowest-order bit plane al- ternates between 0 and depending on whether the byte values are even or odd. Because the number of pixels would not change, this would cause the height of some of the remaining histogram peaks to increase in general. Typically, less variability in intensity level values will reduce contrast.

Because the number of pixels would remain constant, the height of some of the histogram peaks would increase. The general shape of the his- togram would now be taller and narrower, with no histogram components being located past The histogram equalization method has no provisions for this type of artificial intensity redis- tribution process.

We have assumed negligible round-off errors. First, this equation assumes only positive values for r. Recognition of this fact is important. Once recognized, the student can approach this diffi- culty in several ways.

One good answer is to make some assumption, such as the standard deviation being small enough so that the area of the curve under p r r for negative values of r is negligible.

Another is to scale up the values until the area under the negative part of the curve is negligible. This is the cumulative distribution function of the Gaussian density, which is either integrated numerically, or its values are looked up in a table. A third, less important point, that the student should address is the high-end values of r. One possibility here is to make the same assumption as above regarding the standard deviation.

Another is to divide by a large enough value so that the area under the positive part of the PDF past that point is negligible this scaling reduces the standard deviation. Another approach the student can take is to work with histograms, in which case the transformation function would be in the form of a summation. The is- sue of negative and high positive values must still be addressed, and the possible answers suggested above regarding these issues still apply.

The student needs to indicate that the histogram is obtained by sampling the continuous function, so some mention should be made regarding the number of samples bits used.

The most likely answer is 8 bits, in which case the student needs to address the scaling of the function so that the range is [0, ]. Consider the probability density function in Fig. Because p r r is a probability density function we know from the discussion in Section 3. However, we see from Fig.

This implies a one-to-one mapping both ways, meaning that both forward and inverse transformations will be single-valued. Suppose that the neighborhood is moved one pixel to the right we are assuming rectangular neighborhoods.

This deletes the left- most column and introduces a new column on the right. Thus, the only time that the histogram of the images formed by the operations shown in the problem statement can be de- termined in terms of the original histograms is when one both of the images is are constant.

In d we have the additional requirement that none of the pixels of g x , y can be 0. Assume for convenience that the histograms are not normalized, so that, for example, h f rk is the number of pixels in f x , y having intensity level rk. Assume also that all the pixels in g x , y have constant value c. The pixels of both images are assumed to be positive. Finally, let u k denote the intensity levels of the pixels of the images formed by any of the arithmetic oper- ations given in the problem statement.

In other words, the values height of the compo- nents of h sum are the same as the components of h f , but their locations on the intensity axis are shifted right by an amount c. Note that while the spacing between components of the resulting histograms in a and b was not affected, the spacing between components of h prod u k will be spread out by an amount c.

The preceding solutions are applicable if image f x , y is constant also. Their location would be affected as described a through d. When the images are blurred, the boundary points will give rise to a larger number of different values for the image on the right, so the histograms of the two blurred images will be different.

Figure P3. The values are summarized in Table P3. It is easily verified that the sum of the numbers on the left column of the table is N 2. A histogram is easily constructed from the entries in this table.

A similar tedious procedure yields the results in Table P3. Table P3. Initially, it takes 8 additions to produce the response of the mask. However, when the mask moves one pixel location to the right, it picks up only one new column.

This is the basic box-filter or moving-average equation. To this we add one subtraction and one addition to get R new. Thus, a total of 4 arithmetic operations are needed to update the response after one move. This is a recursive procedure for moving from left to right along one row of the image. When we get to the end of a row, we move down one pixel the nature of the computation is the same and continue the scan in the opposite direction. Because the coefficients of the mask sum to zero, this means that the sum of the products of the coefficients with the same pixel also sum to zero.

Carrying out this argument for every pixel in the image leads to the conclusion that the sum of the elements of the convolution array also sum to zero. This does not affect the conclusions reached in a , so cor- relating an image with a mask whose coefficients sum to zero will produce a correlation image whose elements also sum to zero.

Let f x , y and h x , y denote the image and the filter function, respectively. If h x , y is now applied to this image, the resulting image will be as shown in Fig. Note that the sum of the nonzero pixels in both Figs. Since the sum remains constant, the values of the nonzero elements will become smaller and smaller, as the number of applications of the filter increases. In the limit, the values would get infinitely small, but, because the average value remains constant, this would require an image of infinite spatial proportions.

It is at this junction that border conditions become important. Although it is not required in the problem statement, it is instructive to discuss in class the effect of successive applications of h x , y to an image of finite proportions.

The net effect is that, because the values cannot diffuse out- ward past the boundary of the image, the denominator in the successive appli- cations of averaging eventually overpowers the pixel values, driving the image to zero in the limit.

A simple example of this is given in Fig. We see that, as long as the values of the blurred 1 can diffuse out, the sum, S, of the resulting pixels is 1. Here, we used the commonly made assumption that pixel value imme- diately past the boundary are 0. The mask operation does not go beyond the boundary, however. In this example, we see that the sum of the pixel values be- gins to decrease with successive applications of the mask.

Thus, even in the extreme case when all cluster points are encom- passed by the filter mask, there are not enough points in the cluster for any of them to be equal to the value of the median remember, we are assuming that all cluster points are lighter or darker than the background points.

This conclusion obviously applies to the less extreme case when the number of cluster points encompassed by the mask is less than the maximum size of the cluster.

Thus, two or more dif- ferent clusters cannot be in close enough proximity for the filter mask to encom- pass points from more than one cluster at any mask position. It then follows that no two points from different clusters can be closer than the diagonal dimension of the mask minus one cell which can be occupied by a point from one of the clusters.

Since this is known to be the largest gap, the next odd mask size up is guaranteed to encompass some of the pixels in the segment. This average value is a gray-scale value, not bi- nary, like the rest of the segment pixels. Denote the smallest average value by A min , and the binary values of pixels in the thin segment by B. Clearly, A min is less than B. Then, setting the binarizing threshold slightly smaller than A min will create one binary pixel of value B in the center of the mask.

The phenomenon in question is related to the horizontal separation between bars, so we can simplify the problem by consid- ering a single scan line through the bars in the image. The key to answering this question lies in the fact that the distance in pixels between the onset of one bar and the onset of the next one say, to its right is 25 pixels. Consider the scan line shown in Fig.

The response of the mask is the average of the pixels that it encompasses. In fact, the number of pixels belonging to the vertical bars and contained within the mask does not change, regardless of where the mask is located as long as it is contained within the bars, and not near the edges of the set of bars. The fact that the number of bar pixels under the mask does not change is due to the peculiar separation between bars and the width of the lines in relation to the pixel width of the mask This constant response is the reason why no white gaps are seen in the image shown in the problem statement.

The averaging mask has n 2 points of which we are assuming that q 2 points are from the object and the rest from the background. Note that this assumption implies separation be- tween objects that, at a minimum, is equal to the area of the mask all around each object. The problem becomes intractable unless this assumption is made. This condition was not given in the problem statement on purpose in order to force the student to arrive at that conclusion.

If the instructor wishes to simplify the problem, this should then be mentioned when the problem is assigned. A further simplification is to tell the students that the intensity level of the back- ground is 0. Let B represent the intensity level of background pixels, let a i denote the in- tensity levels of points inside the mask and o i the levels of the objects.

In addi- tion, let S a denote the set of points in the averaging mask, So the set of points in the object, and S b the set of points in the mask that are not object points. Let the maximum expected average value of object points be denoted by Q max.

If this was a fact specified by the instructor, or the student made this assumption from the beginning, then this answer follows almost by inspection. We want to show that the right sides of the first two equations are equal. All other elements are 0. This mask will perform differentiation in only one direction, and will ignore intensity transitions in the orthogonal direc- tion. An image processed with such a mask will exhibit sharpening in only one direction.

A Laplacian mask with a -4 in the center and 1s in the vertical and horizontal directions will obviously produce an image with sharpening in both directions and in general will appear sharper than with the previous mask. In other words, the number of coefficients and thus size of the mask is a direct result of the definition of the second derivative.

In fact, as explained in part b , just the opposite occurs. To see why this is so, consider an image consisting of two vertical bands, a black band on the left and a white band on the right, with the transition be- tween the bands occurring through the center of the image.

That is, the image has a sharp vertical edge through its center. As the center of the mask moves more than two pixels on either side of the edge the entire mask will en- compass a constant area and its response would be zero, as it should.

However, suppose that the mask is much larger. As its center moves through, say, the black 0 area, one half of the mask will be totally contained in that area. However, de- pending on its size, part of the mask will be contained in the white area. The sum of products will therefore be different from 0. This means that there will be a response in an area where the response should have been 0 because the mask is centered on a constant area. The progressively increasing blurring as a result of mask size is evident in these results.

Convolv- ing f x , y with the mask in Fig. Then, because these operations are linear, we can use superposition, and we see from the preceding equation that using two masks of the form in Fig.

Convolving this mask with f x , y produces g x , y , the unsharp result. The right side of this equation is recognized within the just-mentioned propor- tionality factors to be of the same form as the definition of unsharp masking given in Eqs.

Thus, it has been demonstrated that subtract- ing the Laplacian from an image is proportional to unsharp masking. The fact that images stay in the linear range implies that images will not be saturated at the high end or be driven in the low end to such an extent that the camera will not be able to respond, thus losing image information irretrievably.

The only way to establish a benchmark value for illumination is when the variable daylight illumination is not present. Let f 0 x , y denote an image taken under artificial illumination only, with no moving objects e. This be- comes the standard by which all other images will be normalized. There are nu- merous ways to solve this problem, but the student must show awareness that areas in the image likely to change due to moving objects should be excluded from the illumination-correction approach.

One way is to select various representative subareas of f 0 x , y not likely to be obscured by moving objects and compute their average intensities.

We then select the minimum and maximum of all the individual average values, denoted by, f min and f max. The objective then is to process any input image, f x , y , so that its minimum and maximum will be equal to f min and f max , respectively. Another implicit assumption is that moving objects com- prise a relatively small area in the field of view of the camera, otherwise these objects would overpower the scene and the values obtained from f 0 x , y would not make sense.

If the student selects another automated approach e. We support this conclusion with an example. Consider a one-pixel-thick straight black line running vertically through a white image.

As the size of the neighborhood increases, we would have to be further and further from the line before the center point ceases to be called a boundary point. That is, the thickness of the boundary detected increases as the size of the neighbor- hood increases. If the intensity is smaller than the intensity of all its neighbors, then increase it.

Else, do not nothing. Note: In rule 1, all positive differences mean that the intensity of the noise pulse z 5 is less than that of all its 4-neighbors. The converse is true when all the differences are negative. A mixture of positive and negative differences calls for no action because the center pixel is not a clear spike.

In this case the correction should be zero keep in mind that zero is a fuzzy set too. Membership function ZR is also a triangle. It is centered on 0 and overlaps the other two slightly. This diagram is similar to Fig.

This rule is nothing more that computing 1 minus the minimum value of the outputs of step 2, and using the result to clip the ZR membership function. It is important to understand that the output of the fuzzy system is the center of gravity of the result of aggregation step 4 in Fig. This would produce the complete ZR membership function in the implication step step 3 in Fig. The other two results would be zero, so the result of aggregation would be the ZR function.

This is as it should be because the differences are all positive, indicating that the value of z 5 is less than the value of its 4-neighbors. Fuzzify inputs. Apply fuzzy logical 3. Apply d2 d4 d6 d8 aggregation method max. Defuzzify center of v gravity. It is a phase term that accounts for the shift in the function.

The magnitude of the Fourier transform is the same in both cases, as expected. The last step follows from Eq. Problem 4. The continuous Fourier trans- form of the given sine wave looks as in Fig. In terms of Fig. For some values of sampling, the sum of the two sines combine to form a single sine wave and a plot of the samples would appear as in Fig.

Other values would result in functions whose samples can describe any shape obtainable by sampling the sum of two sines. But, we know from the translation property Table 4. This proves that multiplication in the frequency domain is equal to convolution in the spatial domain. The proof that multiplication in the spatial domain is equal to convolution in the spatial domain is done in a similar way.

Because, by the convolution theorem, the Fourier transform of the spatial convolution of two functions is the product their transforms, it follows that the Fourier transform of a tent function is a sinc func- tion squared. Substitut- ing Eq. Substituting Eq. We do this by direct substitution into Eq.

Note that this holds for positive and negative values of k. We prove the validity of Eq. The other half of the discrete convolution theorem is proved in a similar manner. To avoid aliasing we have to sample at a rate that exceeds twice this frequency or 2 0.

So, each square has to correspond to slightly more than one pixel in the imaging system. This is not the case in zooming, which introduces additional samples.

Although no new detail in introduced by zooming, it certainly does not reduce the sampling rate, so zooming cannot result in aliasing. The linearity of the inverse transforms is proved in exactly the same way. There are various ways of proving this. The vector is cen- tered at the origin and its direction depends on the value of the argument.

This means that the vector makes an integer num- ber of revolutions about the origin in equal increments. This produces a zero sum for the real part of the exponent. Similar comments apply the imaginary part. Proofs of the other properties are given in Chapter 4. Recall that when we refer to a function as imaginary, its real part is zero.

We use the term complex to denote a function whose real and imaginary parts are not zero. We prove only the forward part the Fourier transform pairs. Similar techniques are used to prove the inverse part. Proof: Because f x , y is imaginary, we can express it as j g x , y , where g x , y is a real function. And conversely.

Proof: From Example 4. Proof: Because f x , y is real, we know that the real part of F u , v is even and its imaginary part is odd. If we can show that F is purely imaginary, then we will have completed the proof.

Proof: We know that when f x , y is imaginary, then the real part of F u , v is odd and its imaginary part is even. If we can show that the real part is 0, then we will have proved this property.

Because f x , y is imagi- nary, we can express it as j g x , y , where g is a real function. Proof: If f x , y is imaginary, we know that the real part of F u , v is odd and its imaginary part is even. Proof: Here, we have to prove that both the real and imaginary parts of F u , v are even. Recall that if f x , y is an even function, both its real and imaginary parts are even.

The second term is the DFT of a purely imaginary even function, which we know is imaginary and even. Thus, we see that the the transform of a complex, even function, has an even real part and an even imaginary part, and is thus a complex even function.

This concludes the proof. Proof: The proof parallels the proof in h. The second term is the DFT of purely imaginary odd function, which we know is real and odd.

Thus, the sum of the two is a complex, odd function, as we wanted to prove. Imagine the image on the left being duplicated in- finitely many times to cover the x y -plane. The result would be a checkerboard, with each square being in the checkerboard being the image and the black ex- tensions.

Now imagine doing the same thing to the image on the right. The results would be identical. Thus, either form of padding accomplishes the same separation between images, as desired.

These can be strong horizontal and vertical edges. These sharp transitions in the spatial domain introduce high-frequency components along the vertical and horizontal axes of the spectrum.

This is as expected; padding an image with zeros decreases its average value. The last step follows from the fact that k 1 x and k 2 y are integers, which makes the two rightmost exponentials equal to 1. The other part of the convolution theorem is done in a similar manner. Consider next the second derivative. Thus, we see that the amplitude of the filter decreases as a function of distance from the origin of the centered filter, which is the characteristic of a lowpass filter.

A similar argument is easily carried out when considering both variables simultaneously. From property 3 in Table 4. The negative limiting value is due to the order in which the derivatives are taken. The important point here is that the dc term is eliminated and higher frequencies are passed, which is the characteristic of a highpass filter. As in Problem 4. For val- ues away from the center, H u , v decreases as in Problem 4.

The important point is the the dc term is eliminated and the higher frequencies are passed, which is the characteristic of a highpass filter. The Fourier transform is a linear process, while the square and square roots involved in computing the gradient are nonlinear operations.

The Fourier transform could be used to compute the derivatives as differences as in Problem 4. The explanation will be clearer if we start with one variable. This result is for continuous functions.

To use them with discrete variables we simply sample the function into its desired dimensions. The inverse Fourier transform of 1 gives an impulse at the origin in the highpass spatial filters.

However, the dark center area is averaged out by the lowpass filter. The reason the final result looks so bright is that the discontinuity edge on boundaries of the ring are much higher than anywhere else in the image, thus dominating the display of the result.

The order does not matter. We know that this term is equal to the average value of the image. So, there is a value of K after which the result of repeated lowpass filtering will simply produce a constant image.

Note that the answer applies even as K approaches infinity. In this case the filter will ap- proach an impulse at the origin, and this would still give us F 0, 0 as the result of filtering. We want all values of the filter to be zero for all values of the distance from the origin that are greater than 0 i. However, the filter is a Gaussian function, so its value is always greater than 0 for all finite values of D u , v. But, we are dealing with digital numbers, which will be designated as zero whenever the value of the filter is less than one-half the smallest positive number representable in the computer being used.

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