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q3sols18085f05

# q3sols18085f05 - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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18.085 Quiz 3 December 9, 2005 Professor Strang Your PRINTED name is: SOLUTIONS Grading 1 2 3 1) (30 pts.) (a) Solvc this cyclic convolution cyuation for tllc vcctor d. (I would trans- form corlvolutiorl to multiplication.) Notice that c = (5,0,0,0) - (1,1,1,1). The cyuation is like dcconvolution. (h) IVhy is thcrc no solution d if I change c to C = ( 3 , 1 , 1 , 1 ) ? Try it. Can you find a nonzcro D so that C 8 D = (0,0,0,0) ? Solution (a) Hcrc n = 4. Tllc transform F c is 5(1,1,1,1)-(4,0,0,0) = (1,5,5,5). Tllc right sidc has transform (1,1,1,1). hIultiplication (or division!) gives (1, .2, .2, .2) = .8(1,0,0,0) + .2(l, l , l , 1) which corncs from .8(,, 1 1 , , 1 1 ,) + .2(1,0,0,0) = (.4, .2, .2, .2) = d. (h) Tllc trarlsforrrl FC is 4(1,1,1,1) - (4,0,0,0) = (0,4,4,4). LVc can't dividc by zero! Tllc vcctor D = ( l , l , l , l ) solvcs C * D = (0,0,0,0). Note for the future. Exyrcss tllc same problcrn with circulant matrices:
(h) No solutiorl m~llc~i thc matrix is sirigular a zero cigcn~~aluc! (The cigcrivalucs arc thc discrctc transforms !!)

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