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1、 Digital Image ProcessingSecond EditionProblem SolutionsStudent SetRafael C. Gonzalez Richard E. WoodsPrentice HallUpper Saddle River, NJ history10 9 8 7 6 5 4 3 2 1Copyright c 1992-2002 by Rafael C. Gonzalez and Richard E. Woods1PrefaceThis abbreviated manual contains detailed solutions to all pro
2、blems marked with a star in Digital Image Processing, 2nd Edition. These solutions can also bedownloaded from the book web site ().2Solutions (Students)Problem 2.1The diameter, x, of the retinal image corresponding to the dot is obtained from similar triangles, as shown in Fig. P2.1. That is, (d=2)
3、0:2=(x=2) 0:014 which gives x = 0:07d. From the discussion in Section 2.1.1, and taking some liberties ofinterpretation, wecanthinkofthefoveaasasquaresensorarrayhavingontheorderof 337,000 elements, which translates into an array of size 580580 elements. Assuming equal spacing between elements, this
4、gives 580 elements and 579 spaces on a line 1.5 mm long. The size of each element and each space is then s = (1:5mm)=1;159 = 1:3106m. If the size (on the fovea) of the imaged dot is less than the size of a single resolution element, we assume that the dot will be invisible to the eye. In other words
5、, the eye will not detect a dot if its diameter, d, is such that 0:07(d) :0if D(u;v) D0+W 2(b) Butterworth bandpass filter:HBbp(u;v)=1 11 +hD(u;v)W D2(u;v)D20i2n=hD(u;v)W D2(u;v)D20i2n1 +hD(u;v)W D2(u;v)D20i2n:26Chapter 5 Solutions (Students)(c) Gaussian bandpass filter:HGbp(u;v)=1 “1 e12 D2(u;v)D20
6、 D(u;v)W2#=e12 D2(u;v)D20 D(u;v)W2:Problem 5.14We proceed as follows:F(u;v)=ZZ11f(x;y)ej2(ux +vy)dxdy=ZZ11Asin(u0x +v0y)ej2(ux +vy)dxdy:Using the exponential definition of the sine function:sin =1 2jej ejgives usF(u;v)=jA 2ZZ11h ej(u0x+ v0y)ej(u0x+v0y)i ej2(ux+vy)dxdy=jA 2ZZ11ej2(u0x=2 +v0y=2)ej2(ux
7、+ vy)dxdy jA 2ZZ11ej2(u0x=2 + v0y=2)ej2(ux+vy)dxdy :These are the Fourier transforms of the functions1 ej2(u0x=2 +v0y=2)and 1 ej2(u0x=2 +v0y=2) respectively.The Fourier transform of the 1 gives an impulse at the origin, and the exponentials shift the origin of the impulse, as discussed in Section 4.
8、6.1. Thus,F(u;v) =jA 2h u u0 2;v v0 2 u +u0 2;v +v0 2i :Problem 5.16From Eq. (5.5-13),g(x;y) =ZZ11f(;)h(x ;y )dd:Problem 5.1827It is given that f(x;y) = (x a); so f(;) = ( a): Then, using the impulse response given in the problem statement,g(x;y)=ZZ11( a)e(x)2+(y)2dd=ZZ11( a)e(x)2e(y)2dd=Z11( a)e(x)
9、2dZ11e(y)2d=e(xa)2Z11e(y)2dwhere we used the fact that the integral of the impulse is nonzero only when = a: Next, we note thatZ11e(y)2d =Z11e(y)2dwhich is in the form of a constant times a Gaussian density with variance 2= 1=2 or standard deviation = 1=p2. In other words,e(y)2=p2(1=2)“ 1p2(1=2)e(1=
10、2) (y)2 (1=2)#:The integral from minus to plus infinity of the quantity inside the brackets is 1, sog(x;y) =pe(xa)2which is a blurred version of the original image.Problem 5.18Following the procedure in Section 5.6.3,H(u;v)=ZT0ej2ux0(t)dt=ZT0ej2u(1=2)at2dt=ZT0ejuat2dt=ZT0cos(uat2) j sin(uat2)dt=r T2
11、 2uaT2C(puaT) jS(puaT)whereC(x) =r2TZx0cost2dt28Chapter 5 Solutions (Students)andS(x) =r2 Zx0sint2dt:These are Fresnel cosine and sine integrals. They can be found, for example, the Hand- book of Mathematical Functions, by Abramowitz, or other similar reference.Problem 5.20Measure the average value
12、of the background. Set all pixels in the image, except the cross hairs, to that gray level. Denote the Fourier transform of this image by G(u;v). Since the characteristics of the cross hairs are given with a high degree of accuracy, we can construct an image of the background (of the same size) usin
13、g the background gray levels determined previously. We then construct a model of the cross hairs in the correct location (determined from he given image) using the provided dimensions and gray level of the crosshairs. Denote by F(u;v) the Fourier transform of this new image . Theratio G(u;v)=F(u;v)
14、is an estimate of the blurring function H(u;v). In thelikelyevent of vanishing values in F(u;v), we can construct a radially-limited filter using the method discussed in connection with Fig. 5.27. Becausewe know F(u;v) and G(u;v),and an estimate of H(u;v), we can also refine our estimate of the blur
15、ring function by substituting G and H in Eq. (5.8-3) and adjusting K to get as close as possible to a goodresultfor F(u;v) theresultcan beevaluated visually bytaking theinverseFouriertransform. The resulting filter in either case can then be used to deblur the image of the heart, if desired.Problem 5.22This is a simple plugin problem. Its purpose is to gain familiarity with the various termsof the Wiener filter. From Eq. (5.8-3),HW(u;v) =“ 1 H(u;v)jH(u;v)j2 jH(u;v)j2+
