Problem 2.1915corresponding) pixels fromaS1+bS2. Then we can writeH(aS1+bS2)=Xp12S1andp22S2ap1+bp2=Xp12S1ap1+Xp22S2bp2=aXp12S1p1+bXp22S2p2=aH(S1) +bH(S2)which, according to Eq. (2.6-1), indicates thatHis a linear operator.Problem 2.19The median,³, of a set of numbers is such that half the values in the set are below³andthe other half are above it. A simple example will suffice to show that Eq. (2.6-1) is vi-olated by the median operator. LetS1=f1;¡2;3g,S2=f4;5;6g, anda=b= 1.In this caseHis the median operator. We then haveH(S1+S2) =medianf5;3;9g=5, where it is understood thatS1+S2is the element-by-corresponding-element sumofS1andS2.Next, we computeH(S1) =medianf1;¡2;3g= 1andH(S2) =medianf4;5;6g= 5.Then, sinceH(aS1+bS2)6=aH(S1) +bH(S2), it followsthat Eq. (2.6-1) is violated and the median is a nonlinear operator.Problem 2.20The geometry of the chips is shown in Fig.P2.20(a).From Fig.P2.20(b) and thegeometry in Fig. 2.3, we know that¢x=¸£80¸¡zwhere¢xis the side dimension of the image (assumed square since the viewing screenis square) impinging on the image plane, and the 80 mm refers to the size of the viewingscreen, as described in the problem statement. The most inexpensive solution will resultfrom using a camera of resolution512£512. Based on the information in Fig. P2.20(a),a CCD chip with this resolution will be of size(16¹)£(512) = 8mm on each side.Substituting¢x= 8mm in the above equation givesz= 9¸as the relationship betweenthe distancezand the focal length of the lens, where a minus sign was ignored becauseit is just a coordinate inversion. If a 25 mm lens is used, the front of the lens will haveto be located at approximately 225 mm from the viewing screen so that the size of theFull file at

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