6.003 Homework #5 Solutions

Transcription

1 6. Homework #5 Soluios Problems. DT covoluio Le y represe he DT sigal ha resuls whe f is covolved wih g, i.e., y[] = (f g)[] which is someimes wrie as y[] = f[] g[]. Deermie closed-form expressios for each of he followig: y[] f[] g[] y a [] u[] u[] y b [] u[] u[] y c [] u[] u[] y d [] u[] u[] Eer expressios for y i [] for ad for < i he boxes below. y a [] for : + y a [] for < : y a [] = k= u[k]u[ k] u[k] = if k < ad u[ k] = if k <. Therefore y a [] = = + if oherwise k= y a [] = ( + ) u[] Noice ha covolvig ay fucio g[] wih u[] resuls i he ruig sum of g[]: (g u)[] = g[k]u[ k] = g[k] k= k= Thus covolvig u[] wih u[] yields u[k] = ( + )u[].

2 6. Homework #5 Soluios / Fall y b [] for : - y b [] for < : y b [] = y b [] = ( k= u[k] ( ) ) u[] ( ) k u[ k] = k= Noice ha his is equivale o he ruig sum of ( ) k = m= ( ) u[]. ( ) m = ( ) + y c [] for : y c [] for < : - - y c [] = y c [] = ( k= ( ) ( ) k u[k] ( ) ) u[] ( ) k u[ k] = ( ) k= ( ) k = ( ) ( ) + Noice ha his is he same as he ui-sample respose of a sysem wih poles a z = ad z = (ad a double zero a z = ). y d [] for : - ( + ) y d [] for < : y d [] = k= - k - u[k] k u[ k] = - k= - () k = ( + ) - y d [] = ( + ) u[] Noice ha his is he same as he ui-sample respose of a sysem wih a double pole a z = (ad a double zero a z = ).

3 6. Homework #5 Soluios / Fall. CT covoluio Le y represe he CT sigal ha resuls whe f is covolved wih g, i.e., y() = (f g)() which is someimes wrie as y() = f() g(). Deermie closed-form expressios for each of he followig: y() f() g() y a () u() u() y b () u() e u() y c () e u() e u() y d () e u() e u() Eer expressios for he o-zero pars of y i () ad he rage of for which he expressio apply i he boxes below. y a () for : y a () for < : y a () = y a () = u() u(τ )u( τ)dτ = dτ = u() Noice ha he covoluio of ay fucio g() wih u() is equal o he idefiie iegral of g(): (g u)() = Thus (u u)() = u(). g(τ)u( τ)dτ = g(τ )dτ y b () for : e y b () for < : y b () = u(τ)e ( τ) u( τ)dτ = e e τ dτ = e (e ) = ( e )u() y b () = ( e )u() Noice ha his is he idefiie iegral of e u().

4 6. Homework #5 Soluios / Fall 4 y c () for : e e y c () for < : y c () = y c () = ( e e ) u() e τ u(τ)e ( τ) u( τ)dτ = e e τ dτ = e (e ) = ( e e ) u() Noice ha his is equal o he impulse respose of a sysem wih poles a s = ad s =. y d () for : e y d () for < : y d () = e τ u(τ)e ( τ) u( τ)dτ = e dτ = e = e u() y d () = e u() Noice ha his is equal o he impulse respose of a sysem wih a double pole a s =.

5 6. Homework #5 Soluios / Fall. Covoluios Cosider he covoluio of wo of he followig sigals. 5 a() b() c() Deermie if each of he followig sigals ca be cosruced by covolvig (a or b or c) wih (a or b or c). If i ca, idicae which sigals should be covolved. If i cao, pu a X i boh boxes. Noice ha here are e possible aswers: (a a), (a b), (a c), (b a), (b b), (b c), (c a), (c b), (c c), or (X,X). Noice also ha he aswer may o be uique. b = b 4 Mus be asymmeric wih large oupu a early imes ad smaller oupu a laer imes. a b = b a 4 The oupu is symmeric, which could happe if oe of he ipus is a flipped-i-ime versio of he oher ipu. There are oly a few such opios. Oe is c c bu ha would resul i a riagle-shaped oupu. Aoher symmeric opio is a b (or equivalely b a), which fis wih he firs ierval beig cocave up. X X = 4 The oupu is symmeric, which could happe if oe of he ipus is a flipped-i-ime versio of he oher ipu. There are oly a few such opios. Oe is c c bu ha would resul i a riagle-shaped oupu. Aoher symmeric opio is a b (or equivalely b a) bu ha would be cocave up i firs ierval. Noe of he provided fucios could resul i his oupu. a a = 4

6 6. Homework #5 Soluios / Fall Mus be asymmeric wih small oupu a early imes ad larger oupu a laer imes. 6 a c = c a 4 The firs ierval looks like he iegral of a. The secod ierval looks like he firs ierval, bu shifed ad flipped i ime. Thus he aswer is a c or c a. X X = 4 The sep discoiuiy i his resul could oly resul if oe of he covolved fucios coais a impulse.

7 6. Homework #5 Soluios / Fall Egieerig Desig Problems 4. The L operaor The R (righ-shif) operaor ca be used o express fucioal forms for ay block diagram ha is composed of adders, gais, ad delays. Sice he delay operaio is causal (i.e., is oupu does o deped o fuure values of is ipu) fucioals ha build o R ca oly represe causal sysems. Defie he L (lef-shif) operaor o represe aicipaors," which are he ai-causal couerpars o delay elemes. If X is a ipu sequece, he Y = LX meas ha y[] = x[ + ] for all. Le H represe he sysem described by he followig fucioal: Y H = =. X L a. Deermie a differece equaio for his sysem. y[] y[ + ] = x[] b. Fid he ui-sample respose of his sysem. Assume ha he sysem is ai-causal ad sars a res so ha y[] = for >. The we ca solve he differece equaio ieraively: y[] = x[] + y[ + ] y[] = y[ ] = y[ ] = 9 y[ ] = 7 y[ 4] = 8... y[ ] = y[] = u[ ] c. Deermie he pole(s) of his sysem. The ui-sample respose has he form z =. Thus he pole is a z =. d. If a sysem is represeed as he raio of polyomials i R, he he poles of he sysem are he roos of he deomiaor polyomial afer R is replaced by z. Deermie a aalogous rule for sysems represeed by a raio of polyomials i L. Subsiue L z i he sysem fucioal. Reduce o a raio of polyomials i z. The he poles are he roos of he deomiaor. Cosider a sysem ha is described by he followig ui-sample respose: ( ) h [] = for all. 7

8 6. Homework #5 Soluios / Fall e. Deermie a closed-form fucioal represeaio for he sysem described by h []. We ca express he fucioal for h [] as a polyomial i R ad L: H = + ( ) ( ) R + R + R + + ( ) ( ) L + L + L + We ca close he form by recogizig wo power series ( ) ( ) H = R + L = R + L = 4 RL ( R)( L) = f. Deermie he poles of he sysem i par e. Subsiue R z H (z) = = ad L z i he sysem fucioal: 4 RL ( R)( L) = 4 ( z )( The poles are he roos of he deomiaor: z = z) = ad. 4 z (z )( z) g. Do a series expasio of he sysem fucio i par e (i may be helpful o use parial fracios). Explai he relaio bewee he series ad he ui-sample respose h []. Each erm i he series expasio H = + ( ) ( ) R + R + R + + L + correspods o oe poi i he ui-sample respose. ( ) ( ) L + L + 8

9 6. Homework #5 Soluios / Fall 5. Upsamplig Oe way o elarge a image is o upsample by liear ierpolaio, as follows. 9 a. Cosider a oe-dimesioal sigal x[] whose o-zero samples are idexed from o N (show here for N = ). x[] 4 5 Firs, make a ew sigal w[] defied by { w[] = x[ ] =,, 6,..., N oherwise. w[] Nex, covolve w[] wih f[] o produce y[] = (w f)[]. y[] = (w f)[] Deermie f[] f[] b. Exed he liear ierpolaio mehod o wo dimesios. Wrie a program ha uses his mehod o elarge he followig image from is curre dimesios ( 6) o hree imes hose dimesios. Compare he resul of liear ierpolaio wih he resul of replicaig each pixel value imes (see appedix).

10 6. Homework #5 Soluios / Fall The digial form of his image is available o he 6. websie. The appedix coais sample Pyho code. Program o elarge image usig biliear ierpolaio. impor Image as im xx = im.ope( zebra.jpg ) # origial image xpix = xx.load() aa = im.ew( L,[xx.size[]+,xx.size[]+],) # zero-pad edges aa.pase(xx,(,)) apix = aa.load() bb = im.ew( L,[*aa.size[]-,aa.size[]],) # horiz srech bpix = bb.load() cc = im.ew( L,[*aa.size[]-,aa.size[]],) # horiz ierpolaed cpix = cc.load() dd = im.ew( L,[*aa.size[]-,*aa.size[]-],) # verical srech dpix = dd.load() yy = im.ew( L,[*aa.size[]-,*aa.size[]-],) # fial aswer ypix = yy.load() for j i rage(aa.size[]): for i i rage(aa.size[]): # srech image horizoally bpix[*i,j] = apix[i,j] for i i rage(,bb.size[]-): # covolve wih riagle cpix[i,j] =.*bpix[i-,j]+.666*bpix[i-,j]+bpix[i,j] +.666*bpix[i+,j]+.*bpix[i+,j] for i i rage(cc.size[]): for j i rage(cc.size[]): # srech image verically dpix[i,*j] = cpix[i,j] for j i rage(,dd.size[]-): # covolve wih riagle ypix[i,j] =.*dpix[i,j-]+.666*dpix[i,j-]+dpix[i,j] +.666*dpix[i,j+]+.*dpix[i,j+] yy.crop([,,yy.size[]-,yy.size[]-]).save( zebrabiliear.jpg )

11 6. Homework #5 Soluios / Fall Image produced by code provided i appedix.

12 6. Homework #5 Soluios / Fall Image produced by biliear ierpolaio. This image has fewer jaggy edges.

13 6. Homework #5 Soluios / Fall 6. Smiley Cosider he sequece of s ad s show below as x[]. x[] I his x[], here is a sigle occurrece of he paer,,. I occurs sarig a = ad edig a =. Oe mehod o auomaically locae paricular paers of his ype is called mached filerig. Le p[] represe he paer of ieres flipped abou =. The isaces of he paer ca be foud by fidig he imes whe y[] = (p x)[] is maximized. a. Deermie a mached filer p[] ha will fid occurreces of he sequece:,,. Desig p[] so ha (p x)[] has maxima a pois ha are ceered o he desired paer, i.e., a = for he sequece above. The same approach ca be used o fid paers i picures by geeralizig he covoluio operaor o wo dimesios: y[, m] = (x p)[, m] = x[k, l]p[ k, m l]. k= l= A file called fidsmiley.jpg (which ca be dowloaded from he 6. websie) coais a radom paer of whie pixels (coded as 55) ad black pixels () as well as a sigle isace of he followig smiley face: b. Fid he row ad colum of fidsmiley ha correspods o smiley s ose. Noe: mached filerig will work bes if machig whie pixels AND machig black pixels coribue posiively o he aswer. For ha reaso, 55 ad may o be he opimum choices for he values associaed wih whie ad black pixels.

14 6. Homework #5 Soluios / Fall 4 Program o locae ose. I made a image file smiley.jpg o simplify he code. (By reaig he desired arge as a file, i was easy o debug he code wih a differe (simpler) image i a differe file.) Noice ha 7.5 is subraced from each pixel value before applyig he covoluio operaio. This resuls i he followig possible oupus (per pixel): image pixel smiley pixel resul black black ( 7.5) ( 7.5) = +7.5 whie whie (7.5) (7.5) = +7.5 black whie ( 7.5) (7.5) = 7.5 whie black ( 7.5) (7.5) = 7.5 which were chose so ha boh ypes of maches give he same oupu, boh ypes of mismaches give he same oupu, ad maches ad mismaches give opposie sigs. Also, 7.5 was added back o he resul, so ha he resul of he covoluio could be viewed as a image yy. impor Image as im xx = im.ope( fidsmiley.jpg ) xwidh,xheigh = xx.size ww = im.ope( smiley.jpg ) wwidh,wheigh = ww.size yy = im.ew( L,[xwidh-wwidh+,xheigh-wheigh+]) ywidh,yheigh = yy.size xpix = xx.load() wpix = ww.load() ypix = yy.load() xmax = - coords = [,] for j i rage(xheigh-wheigh+): for i i rage(xwidh-wwidh+): ss = for jj i rage(wheigh): for ii i rage(wwidh): ss += (xpix[i+ii,j+jj]-7.5)*(wpix[ii,jj]-7.5)/ ypix[i,j] = ss/wwidh/wheigh if ss>xmax: xmax = ss coords = [i,j] pri coords[]+,coords[]+ yy.save( aswer.jpg ) colum = 765; row = 4. c. Oe advaage of he mached filer mehod is ha i works eve whe he sigal coais some oise. Oe ca make a oisy versio of fidsmiley usig Pyho s gauss fucio: impor radom... pix[i,j] += radom.gauss(,dev)

15 6. Homework #5 Soluios / Fall 5 where dev represes he sadard deviaio of he added oise. Deermie he larges value of dev for which your code sill fids smiley. Program o add oise o image ad he fid ose: impor Image as im impor radom dev = 5 xx = im.ope( fidsmiley.jpg ) xwidh,xheigh = xx.size ww = im.ope( smiley.jpg ) wwidh,wheigh = ww.size yy = im.ew( L,[xwidh-wwidh+,xheigh-wheigh+]) ywidh,yheigh = yy.size xpix = xx.load() wpix = ww.load() ypix = yy.load() for j i rage(xx.size[]): for i i rage(xx.size[]): xpix[i,j] += radom.gauss(,dev) xmax = - coords = [,] for j i rage(xheigh-wheigh+): for i i rage(xwidh-wwidh+): ss = for jj i rage(wheigh): for ii i rage(wwidh): ss += (xpix[i+ii,j+jj]-7.5)*(wpix[ii,jj]-7.5)/ ypix[i,j] = ss/wwidh/wheigh if ss>xmax: xmax = ss coords = [i,j] pri coords[]+,coords[]+ yy.save( aswer.jpg ) This program reliably locaes smiley eve if dev is as high as 5. This oise level is HUGE: he sadard deviaio is greaer ha half he differece bewee whie ad black! Mached filerig works very well i his applicaio. A differe oise sequece is geeraed each ime his program is ru. Therefore, repeaed rus wih he same value of dev may produce differe resuls.

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