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Dig Sig Proc&Analys

by: Ophelia Ritchie

Dig Sig Proc&Analys EECS 451

Ophelia Ritchie
GPA 3.8

Andrew Yagle

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Andrew Yagle
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This 16 page Class Notes was uploaded by Ophelia Ritchie on Thursday October 29, 2015. The Class Notes belongs to EECS 451 at University of Michigan taught by Andrew Yagle in Fall. Since its upload, it has received 12 views. For similar materials see /class/231528/eecs-451-university-of-michigan in Engineering Computer Science at University of Michigan.

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Date Created: 10/29/15
EECS 451 IMAGE PROCESSING BASICS Image 2 D function xij Where 0 g Lj g M I for an M X M image 0578 ADI IWJ gt DAI gt 110578 Continuous System space Rs SAMPLER FILTER INTERP7 OLATOR 0 gt0 2 sinB iiT sinB si39T DAi 10573 42100 Zjooyh7 WW 2D 51m Sample t 2T 8 jT S 2777 S gt 2B Where 2Birnage bandwidth Assume Xw1w2 fxts 0 for w gt B and w gt B LTI Ifxn aj gt LTII gt ynija then Zenxn aj gt LTII gt EleniNJ hi jirnpu1se responsepoint spread function Stable BIBO stable ltgt 00 00 hij lt oo 1700 j7ltgto 2D Z Zhij Z Zhijzf1z j Hz1 22transfer function 19077 had MM Y31 Z2 H31 Z2X31 Z2 DSFT Hej 1ejw2 ZZhmne jltwlmw2n HZ1Z2 Zlejw17z2ejw2 Inverse hmn fir fir Hej 1ejw2ejlt 1mw2ndw1 dwg Note Hej 1ejw2 is periodic in M1 and tag with periods 27F Odd 4157 l M x Z j DSFT lfREIX j 17em l Even 0077 DSFT71jIMXejW17ejW2l Any xij xeij xoij DSFTreal and even is real and even Freq cosw1iw2j gt hij gt Hej 1ejw2 cosw1iw2jARGHejwl6W2 Resp Hej 1ejw2 Gain ARGHej 1ejw2Phase shift f f N171 N271 427T 1 1 2k2 0ltk ltN 71 2D DFT liC2 Z Zn20 xn17n2e N1 N2 7 7 1 1 7110 OSkQSN271 i 1 N171 N271 j27r v 1 ka2 ognlgNlii Inverse 011712 7 N1N2 Zk10 Zk20 Xk17k2e 1 2 03n2SN271 CyCliC Mid ltZgt Yk17k2 Hk17k2Xk17k239 Opad 2 D cyclic convolution can be 0 padded to 2 D linear convolution 2D FFT N1N2 point 1 D FFTs to compute 011162 from 011712 AND N2N1 point 1 D FFTs to compute X161762 from 011162 gt N1 log N2 N2 log N1 1ogN1N2 rnults like 1 D EECS 451 IMAGE PROCESSING FILTERING Example 1 WJ 1 1 sinl5w1gt sinl5w2gt sin05w1 sin05w2 i l l l 1 Hejwijejw2 1 1 l 2cosw1l 2cosw2 Note Point So Separable hij h1ih2j gt Hej 1ej 2 H1ej 1H2ejw2 Separable system image gt2 D problem decouples to 1 D problems 2 D lovvpass hij hLihLj Where hLi is 1 D lowpass lter Radial Note Recall l for xww ltB 0 for xww gtB I I OS WIL WZ SW HeW1eW2 is circularly symmetric pill box function Hej 1ej 2 is periodic in M1 and tag with periods 27F Hejw1 7 ejw2 Then Where hij 2 Where R xi2 j2 Derivation Lim p57 J1Bessel function of l kind of order 1 ATM N sinc function Note If But Hej 1ej 2 Hxw w3 for Mm w lt 7r gt hij hi2 39 Hej 1ej 2constant in rest of the region 0 3 tall w 3 7T hij hi2 j2 does NOT gt Hej 1ej 2 Hw MEDIAN FILTERING For While 1D Why 2D Impulsive salt and pepper shot noise arises from bit errors Preserving image edges NOTE Example of nonlinear ltering yn MEDIANxn 2xn lxnxn lxn Preserves edges Note this has no effect on step function Apply 1 D median lter to each row then to each column EDGE DETECTION For WNH Enhancing edges Why Image segmentation image understanding V2xij xi 1j xi 1j xij 1 xij 1 4xij Find zerocrossings of Laplacian V2xij ltgt gradient extrema If cam 23 n im n2 gt 77 declare El edge H WM Use gradients for directional edge detection lj l lj 1 m central differences If V930 gt 77 declare El edge in vertical direction Use only local maxima of Vwij edge thinning EECS 451 IMAGE PROCESSING TOMOGRAPHY Given Goal Note Radon transform pt 9 f f fx y6t xcos 9 ysin 0dx dy Reconstruct image fxy from its projections pt0 No longer using discrete pixel representation xi j of image Why Example Then Basic tomography problem Image reconstruction from projections CAT Computed Axial Tomography scan x ray tomography pt 9absorption of probing x ray along the line t x cos 9 y sin 0 f x ylocal absorption coef cient at x 3 this is what is imaged SOLUTION FILTERED BACKPROJECTION FBP FBP Where fx 3 i f fHpt 96t xcosQ ysin 9dt d0 39HHilbert 71905 filflklf tl and W jk jSGlel Why Next Compute ftmk of Radon transform gtthe projectionslice theorem Pllt 9 f f fx ye jkw COS de e jkysmedy Fllt cos 9 k sin 0 Inverse 2 D Fourier transform of Fllt907 kg in polar coordinates fx 3 g f f Fllt cos 91 sin mew 9kysin9gtk dk d9 i f Hpt x cos 9 ysin 9d0 QED thwH Rotate around objectlt gtvary 9 Must interpolate between angles 0139 Direct Fourier method Interpolate Fllt cos 9 k sin 9 on polar raster Algebraic Reconstruction Technique ART System of equations Synthetic Aperture Radar SAR ltgt tomography David Munson VITAL MATLAB COMMANDS NH ooumme39 w K3 To load mandrillmat gtgtload mandrillmatAxx12561256 To load mandrilltif gtgtAimread mandrilltif tif Can handle the following formats jpg tif gif bmp png pcx Display matrix A as imagegtgtimagescAcolormapgrayaxis off 39 gtgtimsh0WA automatically gives you a grayscale image fft2AMNifft2AMNc0nv2HX lter2ABXfreqz2H Vertical edge detection with threshold ngtgtedgeA sobel n vertica Laplacian of Gaussian edge detectiongtgtedgeA log 5 X 5 2 D median lteringgtgtmed lt2A55 default 3 X 3 Resize imagegtgtYimresizeXF Y is F times the size of X Default method nearest neighbor interpolation Radon transform commandsgtgtrad0nirad0nphant0m Yrad0nX default each of 180 columns is projection at an angle iradon uses the ltered backprojection method phantom default 256 X 256 modi ed Shepp Logan head Hremez20 01521111001 edgex sobelquot Y2Xfoorrand201201O1 medfi112Y2 EECS 451 CONTINUOUS AND DISCRETE TlME FOURlER TRANSFORMS TYPE APPLICATION CONTINUOUS 7 TIME DISCRETE 7 TIME 2 7 sided Atlanta Airport xt ff te 5tdt 227 nk 7 Example Inverse Rational eatut ls 7 a Za un 7 a EX ROC Noncausal Stability s Res gt Rea z gt lol 1 7 sided lnitial Condition iot fooo te 5tdt 220 nk 7 Formula Differ Equations 7f sXs 7 x0 Zxn 7 1 2 1Xz M71 Relation Fourier 7 Laplace 7 Z s jw z ej Fourier Frequency Response ft ff xte j tdt DTFTn 227 nle jwn Example Hw ljw 7 a Hej ll 7 ae j lnverse Filter6Response x05 ff Xwej tdw fir Xej ej dw Example 3 2 LiLeiv ge ht sinwot7rt sinwon7rn Series Periodic functions x05 2700 Xke rk iT DTFS 201 Xke mkN Coef cients Line Spectrum Xk fOT xte j2quotktTdt Xk Egg xne j2quot kN DFT Discrete Spectrum Xk Eff xne j2quot kN 2201 Xke MkN Relation DFT 7 DTFT 7 Z Compute using FFT to 27rkN z e WkN BlBO Stability Causal systems Polesin LHP Repn lt 0 Polesinside unit circle lt l DTFT Xej of is periodic in to With period27r DFT Xk of is single period of periodic in k With periodN lDFT of Xk is single period of periodic in n With periodN lf is real all Fourier transforms are conjugate symmetric Xej Xej2 and x XNk Xej 220 ak coskwj 221 bk sinkw lXej l and are real non negative and even in k Real Xej and RealXk are real and even in k ArgXej and ArgXk are real and imaginary in k lmagXej and lmagXk are real and odd in k EECS 451 PRONY S METHOD COMPUTING LINE SPECTRA Given Want Point DOF A1 cosw1n 01 Ap coswpn 99 for n I3p To compute parameters tubIi 9m 1 from Now are given that is sum of p sinusoids unlike nonparametric So problem is only to compute unknown parameters m A 0 3p nonlinear equations in 339 unknowns Solvable but looks ugly Idea Props Rewrite f1IBiejwin Bfeijwin Where Bi Ai2ej92 This satis es difference equation of AR order 239 with AR coef cients 14239 a0 alz a2pz2P Hf1z ejwixz eijwz zi gt Az ZZPA1Z gt a a219 Also note a0 agp I AR Toeplitz system 2120 aixn 0 for 2 I g n 3 4p 2p initial conditions Write as the Toeplz39tz system of equations constant along diagonals M21 219 1 x0 11 x2p 1 x2p I x2p x2 a2 x2p 2 seep 1 a4 2 ma a p at But Matrix a a219 a0 agp I gtonly needp equations gtneed ToeplitzHankel constant along diagonalsantidiagonals Ex Matrix Here Poly Modes cosn 2cosnn 012345 can use these as well x3 x2 x1 x0 a1 x4 x4 x3 x2 x1 a2 x5 which becomes x5 x4 x3 x2 a3 x6 using a a219 was x5 x3 a4 1 3 x7 a1 a1 2l Iagl Iw al a agl I 2 l a0a1a2a3a4 1 2 1 using 13 a1 and a4 a0 1 Roots of polynomial z ejEj7T2eij774 gt m 7T2w2 7r4 Get A Solve Note Cl cosn 1 sinn Cg cosn 2 sinn 1 0 1 0 C1 x0 3 C1 1 0 1 g g 51 x1 i 51 i 0 1 0 0 1 02 x2 1 02 i 2 0 1 Q g 52 x3 52 0 2 Nonlinear problem gtnonlinear problem roots we know how to solve EECS 451 DISCRETE TIME FOURIER TRANSFORM DTFT Xej 2730700 xne j m XZlzem Z Xform on unit circle Inverse fjWXej ej mdw 7T7T gt p 7Tp if for any 19 Period Xej is periodic with period 27F Highest frequency w 7T Dual Fourier series Expand X ejw as a Fourier series with period 27F xnFourier coef cients computed using DTFT 1 formula above Uniform Z lt 00 absolutely summable gt uniform convergence converge m max lXNej Xejwl 0 where XNej 257Nxnejwn Naoo Mean 2 lt 00 nite energy gt mean square convergence square Nlinoo 7 lXNej Xej l2dw 0 Weaker than uniform sinBn jw 1 0 S lwl lt B Slnc 7m gtXe 0 KB lt w S W Finite 0 0 3 1 4 2 5 0 0 4 nite length 5 gt length Xej 3ej2 1e 4 2e j 5e j2 XZ Zejw signal Xej 4 3cosw 8cos2w jsinw 2sin2w Expo a un gt Xej 200 ane j m 11 ae j nential a un b u n 1 gt Xe W provided lal lt 1 lt lbl stable ltgtROC must include the unit circle 1 DISCRETE TIME FOURIER SERIES DTFS DTFS Xk 2715 xne j27mkN 5701 XkejgmkN Discreteperiodic in time domainltgtDiscreteperiodic in frequency N71 j27rnkN N if N divides k 2710 e 0 otherwise Periodic xnXk ejgmkN are all periodic in n and k with periods N Parseval 2715 2531 le l2power in the periodic 1 if0ltnltL1 ideividesk else gtXk Basis Orthogonal function Square 0 if L S n S N 1 sprzr7flfIIfeij7rkLilN Continuous L ltgt J zes I sjw I fZxn6t n Discrete Z ltgt DTFT Time 2 ej EECS 451 FIR FILTER DESIGN AND EXAMPLES FIR MA Why Finite Impulse Response Mn has nite duration All poles at origin Moving Average lter ynb0xnb1xn l bNxn N orderN Always stable add delay to make causal easy to design see below LINEAR PHASE TYPE FORM OF hn APPLICATION RESTRICTION I 97 a7 07 a7 b Low or highpass None II 97 a7 a7 b Lowpass only Hej 0 III 97 a7 07 a7 b Differentiator Hej0 Hej 0 IV 97 a7 a7 b Differentiator Hej0 0 Why Zeros No phase distortion just time delay affects only Fourier magnitudes Complex conjugate reciprocal quadruples 20727 1 1 Coincide 207 z THREE FIR FILTER DESIGN TECHNIQUES WINDOW DESIGN Example hIDEAL 71 Solution Matlab hn hIDEALnwn for some data Window eg7 Hamming wn Design a 5 point digital di erentiator using a rectangular window 7 7T 39 jam L71quotL i IMAG REAL 7 fiwjwe dwi n i 271797 17 2 ODD gt EVEN l 1 Mn 27 797 I7 gt yn7xn2xnlixn lxn 2 hfir1N 1 W ftype window Wvector of passband cutoffs FREQUENCY SAMPLING Example Constraints Solution Answer Alternate Matlab N71 Solve 2710 hne Wk HDESIREDQW for some frequencies wk Design a 5 point digital lowpass lter using frequency sampling Type I lter Hej0 I Hej 2 34 Hej7r 0 Lowpass hna7 b7 07 b7 u Hejwc2bcosw2acos2wDTFThn w 0 c2b2al w 7T ci2b2a0 w 7T2 ci2a34 Solving gives hna7b7g7b7a 1 167i7g7z7 LLil ifft1 34 0 3458 14 18 14aliasedhn 16 1678 hfir2 N l FMwindow where Mmagnitudes at frequenciesF EQUli RIPPLE Huh Solution Matlab Where 4w 4w 4w in N 74w ll MfXE6j IWW MfXW6j IHDW Znolhlnl6 j quotI Minimize largest error No frequency far off Worst case minimized Iterative algorithm using RemeZ exchange and alternation theorems hfirpm N l FMweightsftype Mmagnitudes at frequenciesF Fvector of Maximum frequency w7r gt FI F includes 071 EECS 451 TRANSFER OR SYSTEM FUNCTIONS DEF Poles Zeros as de ned below roots of Dz 0 ARMA Zaiyn j roots of Nz 0 Impulse response 6n gt gt Transfer functions associated With zero initial conditions ZSR IO ARMA EX 1 Soln Input gt gtoutput yn S Hz S hn YZXZ Zhn ZaU Mn ibjxltn j g M a M plot BltzgtAltzgt 0H 1 2 un gt yn Find X02 2239YZ 11101 Wt First term of ynforced response Second terrnnatura1 response EX 2 Soln Zil Difference eqn Zfl 174 gt yn yn l Find difference equation implementing H Write Hw and crossmultiply Yzl z 1 2z 2 3z 3 Xzl 7z 1 gtdifference equation yn yn 1 2yn 2 3yn 3 7x01 1 EX 3 Soln Find step response of system with zero at I pole at 3 and H 0 I UZ zfl YZ 32g 101 3 23 7 H z 3 333211 4 yn 3yn 1 3xn 3xn 1 Modes7 Poles yn 3yn I 2yn 2 xn 1 Modes 12 g Poles 2 poles C modes ZIR ZSR yn 012711101 021711101 depending on initial conditions yn 032711101 natural responseforced responserv If no polezero cancellation then poles modes BIBO Stable LTI system BIBO stable iff 2730700 lt 00 is nite Same as unit circle I C ROC of Causal Causal LTI BIBO stable iff poles inside unit circle Anticausal BIBO stable iff poles outside unit circle EECS 451 FREQUENCY RESPONSE Recall Z gt gt Eigenfunction of LTl Now 2 BMW gt gt Hejw0ejw0quot and 2 e jwon gt gt He jw0e jw0quot cosw0n gt gt Hej 0 cosw0n argHejw0 Gain Amplitude increases by factor of Hejw0 Phase Shift by argHejw0 tan 1 W DTFT Hej 0 Zhne jw0 DTFThn Zero has a zero at eijwquot gt yn 0 in steady state Pole has a pole at eijwquot gt yn gt oo blows up EX 1 hn and 1707 1707 l cos7T2 quot Hej ll e jw ll 08967j263960 at w yn 089 cos7r 2quot 2660 087 047 087 047 08704 EX 2 hn and l7 l7 l7 17 l cos7rn Hej ll at w 7T gt yn cos7rn l Why 2171 7 WW 26V 1 H 1 Z 1 1 11 EX 33 hn a gt H63 l e jw cos e jw2 Lowpass EX 31 hm 57 gt mew 51 e jw sinej w2 High Notch Hz z ejwoz e jwog z 2cosw0 z l lter Hej 2cosw 2cosw0 hn 17 2coswa7 l 1 M 7L A Comb H z m Zk7M z gt zeros at z eLlt2M1gt for k ll l LM unless L divides k See p 349 Reson BZQKZ rejwo z re jwo BZQZQ 27quot cosw0z W ator On unit circle l7 Hej peaks at w lcos 112 cos we 7quot gt 1 Resonant freq we 3 dB bandwidth 21 7quot See p 342 i D 271 A1Z i w i All Hziz H17 gtHzHlzil gtHej il pass zkz l or ZNAZ lt l gtstable and causal Az Zcausal signal LOWPASS FILTER DESIGN USING POLES AND ZEROS Given such that Sum of two continuous time signals Call this sum x1t x2 Signal x1t is bandlimited to 000 250 HZ low pass signal Signal 205 is bandlimited to 250 500 HZ highpass signal Given and Goal A DSP system having a sampling rate1 kHZ1000 S ggfg A DSP chip having a total of 10 delays storage registers Design a digital lowpass lter to pass x1t and reject x205 Freqs 1000 HZ 500 HZ 250 HZ 27T 7T 7T 2 sampling Nyquist cutoff contin discrete identity 1 for0 lwl lt7T2 0 for 7T2lt lwl 3 7T Ho Ideal Problems Mn sinn 7m sinc function Impulse response of HR lter 1 Noncausal 2 Unstable 3 No difference equation implements Zeros Poles Note eijWQ eij37T4 ej to reject high frequencies zeros on unit circle 6 SeijW4 SeijWZ to pass low frequencies 6 works better here Need all the poles inside the unit circle lt 1 for BIBO stability Hz 1 HMgtltzeejquot2gtltzeej3quot4gtwashWee i 278ejW2zi8e j7 2278e7quot4278e 77 4276 39 using and and z ejfrZXZ e j 2z ej z jzjz1 z3z2z1 z 633774z 6 937T4 22 2cosjT7Tz 1 22 z 1 Performing similar computations for the denominator factors Hz z52414z43414233414z22414z1 Yz 7 257173Z41962371492208427025 7 Cross multiply ylnl xlnl 173yn 1 196yn 2 149yn 3 084yn 4 025yn 5 2414xn 1 3414xn 2 3414xn 3 2414xn 4 xln 5 Z P8 B expjpi2 eXp jpi2 eXp3jpi4 eXp 3jpi4 expjpi expjpi2 8exp jpi2 8expjpi4 8exp jpi4 6 polyZ ApolyP HW freqzBA plot WabsH zplane BA 40 20 gt9 gt2 Imaginary Part C Real Part EECS 45 NUMERICAL ANALYSIS USING THE DT FT Given Goal Plan Sample System A continuoustime signal 33t bandlimited to B Hz day To compute its derivative dt and inde nite integral xudu Sample w xnA digital lter gt gt Sample every A ltgtsa1npling rate i gtgt 25 ltgt 28A ltlt 1 some AAAA gLPF BHZ gt flD a W gt m gt 2 inter palate antialias lter sampler dgeital x Forward Backward Central Computation of Derivative of from m 3312 1A xnAA Compute yt g 3nA 23 e lAgquotA using nite di erences Meg m i A1Al2A Ref Math 116 DTFT Forward Backward Central Plots Lesson Compute DTFTS of each difference operator then scale by A Hpejw 8J0 l gt Hpeij A 63 OJA lA Hgej l gt HBej AA x 1 e39ijA 1106 83quot i e jw2 gt Hcej AA 6WA e quotv7 A2A See plots below of frequency responses Hej vs no for all of these All approximate ideal di erentiator jw for small w zero at origin Plots Show amount of oversampling necessary for good results Series Central Error But IF M m lij 1jw Hi l jonn e iHBej A wA ltlt 1 a HCerAA e 1 ij 1 ij2A jw Forward and backward 0w2A Central 0w3A2 better See below Note that Eds has a zero at 2 1 5 a 7r IIR lter DTFT But Truncate Let HUM 33quot M lt 7 HIIRlejwalA 309 lt mm 1 nn 0 Implements a exactly HR lter 77777 rr zneed all data to implement trouble with end effects Truncating hg x n l Kn 1 A oentral difference FIR filter gt gtHzremezN 10101 di 39erentiator freqzH1 35 35 5 a 25 25 L E g 2 e 2 m E g 15 1 a 05 05 quot o 4 o 2 2 Frequency Frequency EECS 451 CIRCULAR CONVOLUTION D f yn 261 ltgt Yk XkUk Where N ltgtN point periodic extension of Cyclic circular Order N point or order N ltgt yn hn all have length N CD HOW to Compute Circular Convolutions Goal Compute 1 234 0 12 3 in three different ways Ref Phillips and Parr Signals Systems and Transforms p576 580 1 Take only one cycle of the two cycles shown for each y0 gai ggg a 00 20 30 40 16 20 i a 00 20 4 30 4 42 18 y2 E i a 02 20 30 43 16 a3 Eaif i a 03 22 4 30 4 40 10 2 Compute the linear convolution and then alias it Linear 1 2 34 gtilt 0 1 2 3 0 1 4 10 16 17 12 rnult Z transforrns Alias gt 0 161 174 1210 16181610 checks 3 Compute 4 point DFTs multiply compute 4 point inverse DFT Hk DFT1 2 3 4 10 2 j2 2 2 j2 Con rm this Uk DFT0 1 2 3 06 2 j2 2 2 j2 Con rm this HkUk 006 24j2 2j2 2 2 2 j2 2 j2 60 78 4J8 yn yn DFT 160 j84j8 16 18 16 10 checks Using Cyclic Convs and DFTS t0 Compute Linear Convs 0pad 1234000 0 123000 01410 16 1712 Note Linear conv of two 4 point gt 4 4 1 7 point sequence Long Often input signal is much longer than lter input Chop up into segments and compute hn gtilt un Use Overlapsave or overlapadd methods see text p430 433 Compute quickly by multiplying 7 point DFTs then inverse DFT EECS 451 DIRECT TWO SIDED Z TRANSFORM DEF ROC May be Xz 2730700 xnz n compare to fxte Stdt z Xz converges has forrn r1 lt lt m for some 701702 0 1zlltm 0ltlzlltr rltlzlltoo 0ltlzlltoo 0 1zlltoo This matters see below Could usually ignore for Laplace Xforrn Finite length signal 0 0 3 1A 2 50 0 4 nitelength 5 gt Xz 3Z2 z4 2z 1 52quot2 324 Z3 4212 22 5z2 ROC 0 lt lt 00 converges except at 0 and 00 Z6n D z D Causal expon entials Za un 200 anz 11 az l 1 ROC 1az 1l lt 1 gt gt 1a for the series to converge EX Z nun ROC 12 gt 1 g1 g Sinu soids Zcosw0nun Zejwonun Ze jwonun linear ROC gt lejw 1 172 1 coswo 1 1 1 1 i 5 1767 W02 1 517e jwoz 1 i Anti causal expon ential 1722 1 coswoz 2 39 707 z Zi anu n 1 214 211371 m 1 ROC 1a 1zl lt 1 gt lt 1a for the series to converge Anticausal but has same Z transforrn as causal a unl POINT Need to know ROC to get from 1 7 az l 39 Two sided Any ROC ZWWW WK n 1 m ROC 1a lt lt lb if 1b lt 1a then ROCernpty never converges NOTE Z Xforrn of sum gtROC ROCs of each term Zxcausal Zxanticausaln E xcausalcn ZZ ZZBiqu n 1 for some poles 39 q 1largest pole of xwusa lt lt 1srnallest pole of xantiwum n Small rnagnitude poleslt gtcausal part large poleslt gtanticausal part ROC is always an annulus ring whose radii are successive poles Also Note 1 201 Z log1 z l ROC gt 1 if you recognize the in nite series else no cigar But see p 167 Convolutionlt gtpolynornial rnultiplication Z gtilt X Eigen funcs of LTI z gt gt z in gt scaled 2 out hn gtilt z Z hiz i z Zhiz i 2 plays same role as 6 in continuous time EECS 451 DISCRETE FOURIER TRANSFORM DFT DFT DTFT DTFS except Text Xk 2715 xne j27mkNxn 5701 XkejgmkN lengthg N Xk XZ Zej27rkN Xejwlw2WkN sampled on unit circle 7 periodic extension of periodic extension of X factoroflNmoved Note k01N 1 andn01N 1 Uses Xllt not Xk I hate thatitoo easy to confuse With Xz What Why Finite length 277k Use DFT to compute DTFT at w N equispaced samples on uc Use for spectral analysis and to recover from DTFT BUT has nite length L gtusing N 2 L gtrecover from Xk But N lt L gtcan only recover 2k xn kN gt aliasedxn EX DFT N24 IDFT Why xn 1 2 34 5 gt Xejw 12e jw3e 2jw4ej3w5ej4w Sample DTFT on unit circle at w 0 g 7T 3777 to get DFT X0 1234515X1 1 2j 34j532j X21 23 4503X312j 3 4j53 2jXf x0 l15 3 2j 3 3 2j 6 incorrect 615 aliased x1 151 3 2jj 3 1 3 2j j 2 correct x2 151 3 2j 1 31 3 2j 1 3 correct x3 1151 34 2j j 3 1 3 2jj 4 correct Undersampled X 69 on unit circle gtaliasing just like before Inter polate DTFT Let have length L g N We re given Xej2 kN k 01 N 1 Then interpolate Xej 5701 Xej27TkNSw wk N V2IUT71tijS1Mn 0 Zero padding EX N gt L gt DFTx0 xL 1 0 0zer0 pad gtabove This smoothes DTFT ner sampling in w BUT no additional info DFTNiZiLgol 501 WETWML DN Window blurs Cyclic convol Ex Check ywmwwwz mmwwmenmn x01 N ltgtN point periodic extension of Cyclic circular 1 2 3 4 5 6 Take one complete cycle of each y0 l j 14263531 051212 15243631 M2 2333353332 16253428 i 4 Hex0 123456 90 313128 YO checks EECS 451 NON PARAMETRIC SPECTRAL ESTIMATION Given Goal Using A cosw1n 81 A2 04an 6g0 g n g L 1 To determine the unknown wi from 0 g n S L 1 DFT FFT7 convolution ltering sine functions windowing Easy But w Compute Y8 2 2201 yne mkN0 g k g N 1 where N gtgt L Look for peaks in 117 at k 2 k1 k2 Then wt gtfreque11cies Consider yn eoswon for 0 n g L l ugt coswonwn YR AVIeglw w39dlW63 Where 1 0 S n g N l and W7ej DTFTwn mainlwmlwwwUf sinqwm previous HO EX 0pad Note Data length is L 2 25 Zero padded N 2048 10 Does NOT increase the resolution smoothes Sidelobes gt leakage from text p 434439 3 A l J Lobe 2 Ill85w has rst zero crossing at w L Well known in optics Need Roughly lwl 92 gt z to resolve two peaks Bigger N gtno help r lur a EX I 00100 m h L so DFT 1v 2048 j 3 31s Freqs 02w0227r067r if 6 0 Lz50 Can t resolve E u l 2 L100 CAN resolve 2 text p435 9 1 0 z 0 f5 7 2 Windows DSP kind not the Microsoft kind Idea Instead of rectangular win ow 1 0 g n g L y 1 use some smeeiher window to avoid sidelobes ripple But Reduce sidelobesaincrease main lobe Width gt lose reeolution gm 83912 was HR Why Lowpass lter Use for various 11121 designgdxwith Blackman Lowpass o 0 0 Figure 810 Low Fl i u 811 L w FR 4 lters g 40 designed with recxiaasgsnlaf i m jagged with 3311 willow 2of W do S g 40 39Yerl m39 LM61gtV In W E a V I 1 60 l 40 de ned 5 60 l 60 I n p 60 E l l P overleaf 3039 30 l I sco m p 629 quot00 0 0 04 05 quot MO 9 02 03 04 95 m 0 0 02 Normalizcd frequency Normamx mummy Normanud


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Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


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Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.