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# Intro Signals&Syst EECS 216

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GPA 3.8

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This 19 page Class Notes was uploaded by Ophelia Ritchie on Thursday October 29, 2015. The Class Notes belongs to EECS 216 at University of Michigan taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/231548/eecs-216-university-of-michigan in Engineering Computer Science at University of Michigan.

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Date Created: 10/29/15

EECS 216 2nd ORDER CIRCUITS SINUSOIDAL SOURCES NATURAL RESPONSE NO SOURCE PRESENT Given Initial t0 Goal Series RLC circuit resistorinductorcapacitor connected together Capacitor is charged up to 110 and inductor is juiced up to Close or throw the switch at t 0 Recall and vt don7t jump Compute current through inductor and everything else for t gt 0 Devices KVL Note 239 0 gt 1115 5 gemW and v Lg gt 23905 Hidew g ffooz39tdt L 2 R 0 Take gt ii 0 Any circuit variable satis es this differential equation So use the fol lowing procedure to compute inductor current iLt or capacitor volt age v0t usually easiest7 then compute other quantities from those Soln So Where and The trial solution I 6 solves the above differential equation if 82 s Iest 0 gt 82 s 0 characteristic equation Cleslt 02632t for some constants 017027817 82 Math 216 idea 01 and Cg determined by initial conditions 110 and 81 and 82 are the two roots of the characteristic equation Where Note 2gt gt8172 i 2 lt0 2lt gt3172 ij1 2 aijwd Three different cases Overdamped7 underdamped7 critically damped so it always decays 2 the damped natural frequency 2 acts like a time constant7 affecting amplitude of sinusoids and wd R 06 l 04 2 016 Cgte at called critical damping COMPLETE RESPONSE NOW INCLUDE SOURCE Formula Where Note isst Cleslt 02632t fort gt 0 V steadyistate 01 and Cg are determined by the initial conditions 110 and Follows from linearity of differential equation 01 and Cg are deter mined by the initial conditions7 used to determine i0 and 0 transient decays to 0 EECS 216 LECTURE NOTES LOW PASS FILTER DESIGN USING POLES AND ZEROS Given such that Sum of two continuous time signals Call this sum 36115 36215 Signal 36115 is bandlimited to 000 250 HZ low pass signal Signal 36215 is bandlimited to 250 500 HZ highpass signal Given and Goal A DSP system having a sampling ratel kHZ1000 g g A DSP chip having a total of 10 delays storage registers Design a digital lowpass lter to pass 36115 and reject 30215 Freqs 1000 HZ 500 HZ 250 HZ 27T 7T 7T 2 sampling Nyquist cutoff contin discrete identity I for0 w lt7T2 0 for 7T2lt w 37f Hltwgt Ideal Problems Mn singn 7TH sinc function Impulse response of IIR lter I Noncausal 2 Unstable 3 No difference equation implements Zeros Poles Note EMT27 ejEjSW l7 6 to reject high frequencies zeros on unit circle 67 86ij7T47 SeijWQ to pass low frequencies 6 works better here Need all the poles inside the unit circle lt I for BIBO stability Hz T HMgtltzeej 2gtltzeej3 4gtlt27ej3 4gtltzeej gt i 2785j7r22785 j7 22785j 42785 j7 4276 39 using and and z ejf2z e j 2z 6 z jzjz I Z3 212 zl z 633W4z 673 22 2cos377rz I 22 2z 1 Performing similar computations for the denominator factors Hz z52414z434I42334I4z22414z1 Y2 i Z57173241962371492208427025 Xz Cross multiply ylnl a n I73yn I I96yn 2 I49yn 3 084yn 4 025yn 5 24I4n I 34I4an 2 34I4an 3 24I4an 4 n 5 Z P8 w expjpi2exp jpi2exp3jpi4eXp 3jpi4expjpi expjpi28exp jpi28expjpi48exp jpi46 polyZApolyPHWfreqzBAplotWabsHzplaneBA 40 20 Imaginary Part Real Part EEOS 216 LECTURE NOTES BASIC SYSTEM PROPERTIES What Input gt SYSTEM gt yn output Why Design the system to lter input DEF A system is LINEAR if these two properties hold 1 Scaling If gt SYS gt gn7 then CL I L gt SYS gt ayn for any constant a NOT true if a varies with time ie7 an 2 Superposition If m n gt SYS gt yl and 2 gt SYS gt ygn7 Then a1n b2n gt SYS gt ayl 932 for any constants a7 b NOT true if a or b vary with time ie7 an7 EX yn 3n 2 yn n1 nn2n 1 yn NOT 21M 2M 21M sinn 21M Mn 21M n W0 1 NOT yn 1 try it This is called an af ne system HOW If any nonlinear function of mm7 not linear Nonlinear of just n OK DEF A system is TIMEINVARIANT if this property holds If gt SYS gt gn7 then n N gt SYS gt yn N for any integer time delay N NOT true if N varies with time eg7 EX 21M 3n 2 21M sinn 21M n 1 NOT 21M WM 21M OSW 21M 2n 21M HT HOW If n appears anywhere other than in mm7 not time invariant Else OK DEF A system is CAUSAL if it has this form for some function yn n 17n 2 present and past input only Note Physical systems must be causal But DSP lters need not be causal DEF A system is MEMORYLESS if yn present input only DEF A system is BIBO STABLE iff Let gt SYSTEM gt If lt M for some constant M 7 then lt N for some N ie Every bounded input Bl produces a bounded output BO77 HOW BIBO stable ltgt 20700 lt L for some constant L Where Impulse 6n gt SYS gt hnimpulse response EX A timeinvariant system is observed to have these two responses Q7073 gt SYS gt 917072 and 9707071 gt SYS gt 17271 Prove The system is nonlinear Proof By contradiction Suppose the system is linear But then 9707071 gt SYS gt 1727 1 implies Q707 3 gt SYS gt 37673 since we know it is time invariant Then Q7 07 3 produces two outputs EEOS 216 LECTURE NOTES CONVOLUTION AND IMPULSE RESPONSE Note 3717476 ltgt 36n 1 16n 46n 1 66n 2 gtllt 6n sifting property of impulse Delay Fold Both n D is shifted right later if D gt 0 left earlier if D lt 0 n is ipped folded reversed around n 0 MN n is n shifted right if N gt 0 since M0 is now at n N OTlJkpoNH 6 FOR LINEAR TIMEINVARIANT LTI SYSTEMS l 5n gt LTI gt Mn Time invariant delay by i gt LTI gt Linear scale by n gt LTI gt Mn De nition of Impulse response gt Linear superposition gt LTI gt yn Mn gtllt Convolution Input into LTI system with no initial stored energy gtoutput NH PROPERTIES OF DISCRETE CONVOLUTION yn Mn gtllt gtllt Mn ZMn Mn7an both causal 0 for n lt 0 and 0 for n lt 0 gt yn 2210 2210 Mn also causal 00 Note Note Suppose Mn y 0 only for 0 S n S L has length L 1 Suppose y 0 only for 0 S n S M has length M 1 Then yn y 0 only for 0 S n S L M has length L M 1 LengthynLengthMnLengthan 1 y0 M0ac0 yL M MLaM gtllt 5n D n D gt h1n gt h2 gt yn cascade connection Equivalent to gt h1n gtllt hg gt ewnil airmail A gt 63 gt yn parallel connection Equivalent to gt h1n hg gt MA Huh Note yn b0n b1n 1 bqan q Moving Average Present outputweighted average of q most recent inputs Equivalent to yn bn gtllt Where Mk bk70 S k S q FIR EX IIR EX Finite Impulse Response ltgt Mn has nite duration Any MA system is also an FIR system7 and vice versa In nite Impulse Response ltgt Mn not nite duration Mn a un a for n 2 0 and la lt 1 is stable and HR EECS 216 LECTURE NOTES TRANSFER ALSO KNOWN AS SYSTEM FUNCTIONS N 3 Y s B s DEF Hs ht Also have Hs DES XES AES as de ned below Poles Roots of Ds 0 Differential Eqn 2770 an 20 bmg X Zeros Roots of Ns 0 Impulse response 6t gt gt ht Transfer functions are associated with zero initial conditions ZSR There are 6 ways to describe LTl systems 1 ht 2 Hs 3 Hjw 4 Differential equation 5 Any input output pair 6 Poles 85 zeros Using transfer functions7 can go from any of these to any other Input t gt gtoutput yt ltgt Hs ltgt ht EggSil Lg EE YsXs HMO N 77 M m E a d y E b B ltgt Hs ltgt POLE plot 7 n dt 7 m dtm v V ZERO 0 WU BltsgtAltsgt 0H 1 DIFFERENTIAL EQUATION s ZEROS 7 EX 1 ha l6 2t6 4mm Hs Jrs POLESrf li ll39 1 3 238 gt Ys326s8Xs2s6 gt j ig68y26x Forced ZSR to te 3tut is yt0 due to zero at 3 eats 3015 Forced step response to tut is H0ut ut Use this below EX 2 j 5j 6y 5j 0 4m Read off Hs g Zeros Ns 82 58 4 0 gt s 17 4 Use llatlab7s roots Poles Ds 82 58 6 0 gt s 27 3 Use llatlab7s roots Hs 0 for sany zero see above while Hs gt 00 for sany pole EX 3 t e gtu f gt gt yt e gtu f etut Compute ht Soln Xss2 Ysi m Hs ht 3etut First term of ytforced response Second termnatural response Hsg gt sYs7Ys 3Xs gt 7313393 Read ht i s 7 EX 4 F1nd the d1fferent1al equatlon 1mplement1ng system H 87W 7 S i s 7 Soln erte Hsimim7 3 2 Ys837822873Xss7 gt 7273y7 EX 5 A system has zero at 17 pole at 737 and a forced step response ut Goal Compute response to tetut Forced step response gt HOil Soln Hs3s 1 X8 1 gtY8 3 Hyt3573tut g g 039 crossmultiply7 and take 4 33 39 371 33 Differential equation Hs3 gt 3y3 f73 EECS 216 LECTURE NOTES INVERSE LAPLACE TRANSFORM MULTIPLE POLES Problem Compute 1NsDs when Ds has a multiple root pole Solution Let W ratio of two polynomials Then N3 A1 A A B B Partlal D3 37PIWHAFW4HAF igNwhere 37122 3 A i 372 372 372 i P 2 P 2 Z Fractlon B2 W P23P2 W l 3 P1K3 P23 PN p3 3173 P37P1KP37P2P37P4m i 372 372 372 K i P 2 P 2 Z And AK sipll32msIN8 p1 lsp1 7 191271911236719571936 And Am digi e unk Am wage p1gtKJspn Kil Ns And A1 S P1KH Huh Before you pass out7 note that computing the residues An uses dig Ds NsDs2 the basic calculus formula N s and Ds are polynomials7 so computing the derivatives is easy Can compute the derivatives do and A K recursively in n Watch the order Compute AK7AK717AK72 A1 in that order Multiply by s p1K before taking derivatives Makes sense Cancel factor 8 p1K in the denominator before doing anything else Expansion B 3 3171 H ND ka Example Compute L 1 Egg iEii L 12I22735 S and S page 515 2327253733 7 A1 A2 B Solutlon S12375 7 73H iltsD2 375 partial fraction expansion where 2 2725 733 2 2725 733 ReSIdues B WIS5 37 A2 lsil 1 s2 3 375 43725 7 2327253733 And A1 g wnpl WEB 5 2327253733 7 5 1 3 So 33733279375 7 31 31 375 partial fraction expansion And 1 567t te t 36a for t gt 0 0 for t lt 0 Note There are several typos in Soliman and Srinath on page 5151 Multiply out your ansvver7 or plug in several values of 87 to check work Matlab gtgt RPK residue2 25 33 1 3 9 5 yields answers R 3 5 1 P5 1 1 K EECS 216 LECTURE NOTES COMPLEX EXPONENTIAL FOURIER SERIES Given Series t is continuous time periodic function Period T gt t t T 00 4 7r T 2 7 7139 3015 Zk700 mke T7 3 fi 2 te 32 ktTdt Discrete in frequencyltgtPeriodic in time Histo rical Note NH Then Where None None MSC Dirichlet Suf cient Conditions for Convergence Bracewell p205 At a meeting of the Paris Academy in 1807 Fourier claimed any periodic function could be expanded in sinusoids Lagrange stood up and said he was wrong Led to Riemann integral Over each period any interval of length T t has a nite number of discontinuities7 maxima7 and minima t is absolutely integrable xtdt lt 00 ESLLN k5j27rktTl 0 for all t at discontinuities of t at ti convergence is to 360517 t 1 l t gtno Fourier series not absolutely integrable t sinlt gtno Fourier series 00 maxima and minima Finite energy 111 0116 period gtMean square convergence MSC weaker T 2 T 2 7r fig72 t2dt lt 00 LIM 257N mkegQ ktTIth 0 Naoo 7T2 PROPERTIES OF FOURIER SERIES H Can also use t a0 2201 ak cos27rktT bk sin27rktT a0 0 f tdt integrate over 1 period use everywhere below 305 cos27rktTdt xk 1ak lbk ak ka 7k 2Rek 2 bk ask 716 2Imk sin27rktTdt Note signs ND Parseval Power f t2dt EEOC ack2 a3 221ai 19 OJ Evenodd 005 2201 bk sin27rktT 505 2200 ak cos27rktT 4 Orthogonality fOT e jg mtTejg mTdt T6m If m7 n y 0 fOT COS27TmTt cos27TquotTtdt fOT sin27TmTt sin27TquotTtdt 6071 t 1 sin7r739 739 0 2 Note Duty cycleT i ifglttggakiT t 0 and fjany eventdt 2 fo my eventdt any any Also even eveneven oddoddeven evenoddodd functions Here Above t is an odd function re ect it about both not each axes Then Instead of computing four integrals7 compute only one integral Try an fig22 t cos277rntdt ifjw0ddevendt 0 this bn fig22 t sin277rntdt ifjw0dd0dddt f0Wevendt again bn 2 fig sinntdt cos7rnthe above result 47 Computation of Complex Exponential Fourier Series Still 3 fT2xte j2ntdt 1 fEW e jmdt f gwa mdt 7T2 E 4 1 ijntO 1 ijntw 4 1 ijwn 4 1 4 1 eas1er i 8jne 47r 8jne O i 4jn1 e 7 2n 7 2jn 1fn odd 1 i L jt L j3t L j5t L j7t th Plug t 2je 6je 10e 14e 18e L jt L j3t j5t j7t j9t 2je 6je 10je 14je 18je gt t sint sin3t sin5t sin7t sin9t in EEOS 216 LECTURE NOTES SAMPLING THEOREM FOR PERIODIC SIGNALS NOTE See DFT Discrete Fourier Transform for more details GIVEN A periodic continuous time signal t such that l t is periodic and real t t T for all t 2 t is bandlimited No frequencies above F HZ 3 t is sampled Given samples t 71A GOAL We can reconstruct t from its samples t 71A IF A lt l2F ltgt Sampling rate gt 2F DERIVATION WITHOUT USING THE FOURIER TRANSFORM l t periodic with period T gt t has the Fourier series expansion 3015 X0X1ej tX2ejtXNejNtXfe jtXfVe jNt Where Xk f0Tte j27rdt and lt F lt Say F Note We will not need to use the formula for X kl No integrals herel Hence t is speci ed by 2N1 complex numbers XLN Xo XN Sample t at t nA so there are 2N 1 samples per period T 2N1AT This and N FT gtA Then setting If nA 7 n 0 2N in the Fourier series gives 2N 1 linear equations in 2N I unknowns Xk Fulton 5 nA 257N Xke7 n 0 2N Sum over different period gt nA 22 Xke7n 0 2N 2N l point DFT 6 We can solve this linear system for the Xk and insert these Xk into the Fourier series in 1 above to get t for ALL values of t BANDLIMITED SIGNAL INTERPOLATION FORMULA 7 Or7 we can note that the solution to this linear system is Xk 3 mnAe7k N N 2N l point DFT 8 Inserting this into the Fourier series and using t fA gives mot Lv 23120 gamma 23120 mmsa m N 7r sin N 7139 T where 8t Zk7N e32 T text p145 9 NOTE This holds for any value of T7 eg7 Tl century 10 Shannon proved this for aperiodic signals think of this as T gt Note As T gt 007 8t gt w sinctA pt in the lecture notes since 2Nle i and sin7r as gt 0 ltgt T gt EX t has periodTIO sec bandlimitF100 HZ Then A since Fourier series of t has 2001 terms gt2001 samples per periodT EECS 216 LECTURE NOTES DIGITAL SIGNAL PROCESSING COMPLETE SYSTEM 05 EX Goal W How Why gt 171 How Why Note 7 How Why Note 21 How hlnl Why 20 How OR OR Continuous time analog signal Audio from microphone signal Digitally lter this signal Lowpass ltered version of Analog lter use EECS 215 ideas Remove frequencies gt FI SY ensures there will be no aliasing Discrete time sampled signal SAMPLING lnl 3305 Ali A INTERVAL We can now use EECS 216 ideas Can recover 5615 from exactly7 due to the anti alias lter Quantized version of Round to nearest of 23 levels Bbits used to represent numbers To send store bits7 not numbers Can7t recover from a n7 but the error is usually negligible Filtered version of input yln hlnl 521m mwin 2391 impulse response of digital lter Lowpass lter gtremove some noise Notch lter gtremove 60 HZ hum Interpolated ynanalog output Use zero order hold constant interpolation Linear interpolation between samples Exact formula Sampling handout ANTIALIAS LOWPASS w V SAMPLING AD CONVER INTERPOLATOR D A CONVERT EECS 216 LECTURE NOTES ALIASING IN A COMPLETE DSP SYSTEM Given Where SAMPLE at I AT5HZ IDEAL SINC A IINTERPOLATOR 05 t cos27rt 2cos87rt l HZ7 4 HZ GOAL Compute Ideal sinc Interpolator 5615 Znpt nTs where pt sinctTs sinct sin7rt7Ttdecaying sinusoid as t gt oo Nyquist Interval Sample Sampling rate5 HZlt2maX frequency of t24Hz gt aliasing Sampling rate5 HZ gt T3Sampling intervall 5 HZSEC011Cl t nTs gt cos047rn 2cosl67rn Alias Ideal Note 2cosl67rn 2cos047rn gt 3cos047rn Note tripled n Ti 5t gt 5615 n 5t 3cos27rt l HZ7 but tripled Original 4 HZ gtaliased l HZ folded across folding freqg25 HZ Now Alias Change t to t cos27rt cos87rt l HZ7 4 HZ cos047rn cos047rn 0 l HZ eliminated Aliasing adds false signals7 interferes with actual signal Now Given Now Get Note Insert ideal antialias lter Lowpass passlt25 HZ7 rejectgt25 HZ ANTli SAMPLE mt AL1As IDEAL SINC A AT5HZ I IINTERPOLATOR 05 Antialias lter eliminates original 2 cos87rt 4 HZ component cos047rn and 5615 cos27rt 1 Hz Aliased false 1 HZ eliminated Original 1 HZ unaffected7 at least Alias since EX Use Acos7r w0n 6 A cosw0 7Tn 6 Acos7r w0n 6 cost is an even function7 and also periodic with period 27F 3cosl77rn g 3cos037rn sinl87rn sin027rn Use to reduce all discrete time signals resulting from sampling For nonsinusoidal signals Apply to Fourier series harmonics MSD How Note IF MSD733 fOTt t2dtlIean Square Error Use Parseval7s theorem to add average power in each harmonic Average power of Acosw0n 6 is 1422 1 we 27TR1g IOB1 L gt periodic 2 we 74 077T EECS 216 LECTURE NOTES TRIGONOMETRIC FOURIER SERIES OF PERIODIC SIGNALS THEOREM Let t be a bounded periodic signal with period T Then t can be expanded as a weighted sum of sinusoids with angular frequencies that are integer multiples of we 2 t a0 a1 cosw0t a2 cos2w0t a3 cos3w0t b1 sinw0t b2 sin2w0t b3 sin3w0t This is the trigonometric Fourier series expansion of Or7 we can use only cosines with phase shifts t a0 01 cosw0t Q51 Cg cos2w0t Q52 03 cos3w0t Q53 PROOF Take a high level math course to see this done properly NOTE A Fourier series is a mathematical version of a prism COMPUTATION OF FOURIER SERIES COEFFICIENTS THEOREM Coef cients an7 bn7 on and on can be computed using T T an ttOO t cosnw0tdt and bn f00 t s1nnw0tdt For n 0 we have do tZOT tdtaverage value of t For cosines with phase shifts en Mag 5 Z571 taI171 quot7 n y 0 using an coswtbn sinwt Magi 19 coswt tan 1 a gt 0 Derive using phasors CLnerne jW2 an jbn wag bgle jtaniwziz EXAMPLE OF FOURIER SERIES DECOMPOSITION t i 7T4 for 2k7r lt t lt 2k l7r Square wave withperiod T 7T4 for 2k l7rlttlt 2k7r T27T gtw0 l 2 E 19 f0 t sinntdt Wong sinntdt SH g sinntdt i7 ifn is odd an fir t cosnw0tdt 0 by inspection Set to 7T i 39 I l 0 if n is even at T 81110 3 sln3t 5 Sln5t PARSEVAL S THEORM FOR THIS EXAMPLE Average fOT t2dt i 0 l E2dt L2 This agrees with 7 1 00 00 71 2 Power a Zk1ai bi Zkodd2 1 3L2 5L2 E EEOS 216 LECTURE NOTES PROOF OF THEOREM First we need the following lemma LEMMA The sine and cosine functions are orthogonal functions t0T i t0T i T27 ifz39 j fto coszw0t cosjw0tdt i to s1nzw0t s1njw0tdt i 07 Hz j ftZOT cosiw0t sinjw0tdt 0 even ifz39 j These assume i7j gt 0 PROOF OF LEMMA Adding and subtracting the cosine addition formula cosa l y cosa cosy IF sina siny gives the formulae 2cos cosy cosa y cosa y 2sin siny cosa y cosa y Setting a iwot and y jwot and ftto T dt gives t0T 2ft0T cosiw0tcosjw0tdt to to cosz39 jw0t cosi jw0tdt Since the integral of a sinusoid with nonzero frequency over an integer number i l j of periods is zero7 the rst part of the lemma follows The other two parts follow similarly QED PROOF Multiply the Fourier series by cosnw0t and KOOT dt 00 T t cosnw0tdt a0 cosnw0tdt ft a1 cosnwot coswotdt L a2 cosnw0t cos2w0tdt ft b1 cosnwot sinwotdt ft 192 cosnw0t sin2w0tdt 0 0 0 on 0 from which the an formula follows to T to The formulae for do and bn follow similarly QED COMMENTS 1 to is arbitrary all integrals are over one period 2 we 27 is angular frequency in SEAgol g this is Hertz 3 The sinusoid at frequency we is called the fundamental the sinusoid at frequency nwo is the nth harmonic Harmonics are also called overtones not used much anymore Some say the nth harmonic is at frequency n 1w0 4 The more terms ie7 harmonics we keep in the Fourier series7 the better the approximation the truncated series is to 5 If t has a discontinuity at t 1517 the Fourier series converges to tf lth where 3005 and 3005 are the values of t on either side of the discontinuity EECS 216 LECTURE NOTES CONCEPTS BEHIND DISCRETE FOURIER TRANSFORM NOTE See DFT Discrete Fourier Transform for more details Given DFT is a discrete time signal with period N n N for all n XkejQ 39nkN Where Xk 2277 neij27rnkN Huh Fastest oscillating discrete time sinusoid w 7T gt cos7rn 1quot Fourier series of discrete time periodic signal has nite number of terms7 with frequencies 07 7273 N 1e07i7i2i 2w72 Wl If N even7 the component with the highest frequency is w 7T If N odd7 the component with the highest frequency is w m If is real7 then XNk X conjugate symmetry X0 1 N 1mean value of If N is even7 XIV2 1 a2 3 N SIMPLE EXAMPLE WITH N24 Given Goal 59 Then Line Using 247 87 127 1677 87 127 167 247 87 127 16 PeriodN4 Compute DFTFourier series expansion of discrete time periodic NOTE e j2T 1 je j2T 2 1e j2T 3 j XO 124 8 12 16 15 Note this is real X2 24 8 12 16 03 Note this is real X1 g24 8 j 12 1 16j 3 2 X3 Z24 8j 12 1 16 j 3 2339 Xf n 15ej0 3 me way 3 2jejquot3n 7139 37139 spectrum is perlodlc w1th components at 07 i5 l7r7 i7 i27r 213 3 2339 36ej333970 ejm cos7rn ej 4 e T 7 simpli es to 15 72 cosgn 3370 3cos7rn Don7t double at w 07 7T PARSEVAL S THEOREM POWER IS CONSERVED Power Time 2201 2531 Xk2average power of periodic 152 l32jl2 32 I3 2j2 260 since 32j2 13 242 82 122 162 260 They are equal EECS 216 LECTURE NOTES INVERSE ZTRANSFORMS Given Xz W In EECS 216 Assume have M S N E i b0b12bM2M i RATIO OF TWO i RATIONAL z aoza122aNzN1 POLYNOMIALS FUNCTION39 Poles 07p1 pN are roots of aoz ava1 0 assume pn distinct Compute using 77roots77 in Matlab an real gtcomplex conjugate pairs Partial 21 1131 ngN since distinct poles 0 y p1 y 7 pN fraction An z pnXzz evaluated at z pn OR 77residueZ77 in Matlab expansion NOTE an p gt AnH AZ Coest also complex conjugate pairs Causal Xz A0 A1 pr1 AN 27 Term by term7 compute Zilz signal A06n 141 pfuhx ANp un is sum of geometric signals Complex Apquot Apquot 2Apquot cosw0n 6 Where A Aej6p pejw0 conjugate This is much easier than trying to use sines and cosines directly EX 1 Slmple real example Compute inverse Z xform of X 1 Write 22 since 22 3z2 z 1z 2 3 Xz 2Zf1 Zf2 gt 6n 2un EX 2 Slmple complex example Compute inverse Z xform of X 1 y jgt since z2 2z2 z 1jz 1 j 2 A m l A1 f Xe a new jlt1 jgtnuw jlt1 j un EX 3 What if there are multiple poles at the origin 2 0 Use this trick XZ 2322 z4 232z k324 12Z713Z724Z73Zf1 gt 1727 374 gtlt un un 2un 1 3un 2 4un 3 EX 4 Xz 120z 1z 2 z 3z 4z Partial 71 51 2132 203 i 5 Computed as follows fraction A0 Z OXZZz0 1200 10 20 30 40 5 1 expansion A1 z 1XZZZ1 1201 01 21 31 41 5 5 coef A2 z 2Xzzz2 1202 02 12 32 42 5 10 cients A3 2 3XZZz3 1203 03 13 23 43 5 10 compu A4 2 4XZZz4 1204 04 14 24 34 5 5 tation A5 z 5Xzzz5 1205 05 15 2 5 35 4 1 Inverse 6n 5un 102 un 103 un 54 un 15quotun zxform Note this is an unstable signal7 since it blows up as n gt oo EECS 216 LECTURE NOTES EXAMPLES SPECTRA OF ACTUAL SIGNALS Given 36157 0 S t lt T sampled at t nA gt t nA Goal Estimate the spectrum of t from the data set 0 S n S N 1 H Take periodic extension of given 0370 3 t lt T Spectrum of t nA periodic with period HZ Have N TA samples of Assume A lt gt N gt 2FT Where Fthe maximum frequency component of nA and its spectrum both periodic gtcompute DFT of Low 4 Two built in Matlab signals Use 1 second segments T 1 Sampling rate8192 HZ gt A 18192 sec gt N U 8192 Assume A lt The Matlab code for each signal looks like load trainXy18192plot40954096fftshiftabsfftX Akpowld 1 Second of Train Whistle Train Whistle Spectrum 800 few spectral lines 600 400 200 1 0 4 1 i L 1 L 5000 0 5000 5000 0 5000 1 Second of Handel Handel Spectrum 400 300 200 100 0 5000 5000 0 5000 Sample Number Frequency in Hz EECS 216 LECTURE NOTES LINE SPECTRUM OF SIGNALSUM OF SINUSOIDS Given Goal t A0 A1 cosw1t 61 AN costh 0N Sum of N sinusoids at N different frequencies Represent amplitudes and phases as scaled impulses vs frequency Using Rez z with z Xkejwkt7 we have t A0 E l1Xkejwkt lee jwkt where Xk Akejgk H COM 4 Can represent a sinusoid at frequency wk 27Tfk by two complex exponentlals at frequencies iwk Note need negative frequencies watch amplitude factor of two Note component at negative frequency wk is complex conjugate of its counterpart component at positive frequency wk Plot impulseslines in frequencies hence llne spectrum EX t 10 14 cos2t l 8cos4t 3 has 5 complex components 10 at w 0 7efj at w i2 4eij3 at w l4 see plot overleaf BEAT SIGNALS SOUND LIKE THEY ARE BEATING What Where Why Easier Identity cosw1t 61 cosw2t 62 2 coswet 6e coswAt 6A we 5w1 wg and wA w1 wg 0e7 0A are de ned similary Apply cosine addition formula and w1 we wA and wg we wA Note that 2 coswet 6e2 coswAt 0A can be written as product ejmtwa eijwcweanejwmwm eijltwmemy Multiply this out gt4 terms gt 2 cosw1t 61 2 cosw2t 62 try itl Sounds like t sounds like a single tone at frequency we m w17 wg whose amplitude vanes slowly with tlme at frequency wA EX 1 Given Given Tuning a piano We wish to tune a piano key to 440 HZ A tuning fork tuned to A above middle C7 which is 440 HZ Piano mistuned to 441 HZ Want to change this to 440 HZ Sol n Sounds like Sol n Strike tuning fork and piano key simultaneously and listen Hear cos27r440t cos27r44lt 2 cos27r05t cos27r4405t A single tone at 4405 HZ whose amplitude varies at 05 HZ Actually l HZ since amplitude is non negative Periodl second Adjust piano key until amplitude variation ceases Period gt EX 2 AM radio WJR 760 kHz transmits a tone EBS at 5 kHZ t 4cos27r7607 000t cos27r5000t 2cos27r75 3937 000t 2 cos27r7657 000t 6j27r765000t 6j27r755000t 67j27r765000t 67j27r755000t Line spectrum 4 components in two closely spaced pairs EEOS 216 LECTURE NOTES BASIC SYSTEM PROPERTIES What Why lnput t gt SYSTEM gt yt output Design the system to lter input DEF A system is LINEAR if the following two properties hold Scaling lf t gt SYS gt y1f7 then at gt SYS gt ayt any constant a NOT true if a varies with time ie7 at Superposition lf 36115 gt SYS gt y1t and 36215 gt SYS gt yg a1t b f2t gt SYS gt ay1t byg linear combinations 15 any constants a7 b NOT true if a or b vary with time ie7 at7 bt I 2115 3W 2 I 2115 205 yt t 1 try it This is an af ne linearconstant system If any nonlinear function of 30157 not linear Nonlinear of just t OK 22 1 3yt 5 6930 2115 sintt 2115 sint 2115 t 2115 03009305 1 A system is TIMEINVARIANT if this property holds lf t gt SYS gt gt7 then t T gt SYS gt yt T any constant time delay T NOT true if T varies with time eg7 Tt NOT HOW yt 3t 2 2115 sint 2115 tt U W 150305 20 052 20 a3215 2115 a315 If t appears anywhere other than in 30157 not time invariant Else OK DEF Note A system is CAUSAL if it has this form for some function yt FT7T S t depends only on present and past inputs Physical systems must be causal But DSP lters need not be causal DEF A system is MEMORYLESS if yt present input only DEF ie LTl system is BIBO STABLE iff Let t gt SYSTEM gt yt If at S M for some constant M7 then yt S N for some N Every bounded input Bl produces a bounded output BO77 HOW Where BIBO stable ltgt ff hTdT S L for some L absolutely integrable lmpulse 6t gt SYS gt htimpulse response of the system EX Prove A discrete time invariant system has these two responses Q7073 gt SYS gt 9717072 and 9707071 gt SYS gt 17271 The system is nonlinear Proof By contradiction Suppose the system is linear But then 9707071 gt SYS gt 1727 1 implies Q707 3 gt SYS gt 37673 since we know it is time invariant Then Q7 07 3 produces two outputs EECS 216 LECTURE NOTES CONTINUOUSTIME AND DISCRETETIME SINUSOIDS DEF A sinusoidal signal has the standard form t Acoswet Where A Z 0amplitude we 2 0frequency S 7Tphase shift Freq fe we 2Wcircular or cyclic frequency in HertZcycles per second Note Sinusoid t Acos27rfet Q5 has cyclic frequency fe Hertz Period T fi 3 gt t t T for all 15 using cosa cosa 219 DC DC constant signals we fe 0 T oo low frequency limit RMSX fOT A2 cos2wet Q5dt fOT A721 cos2wet 2Q5dt using cos2a 1 cos2 and Mcost 0 by inspection DC we 0 gt 3015 AcosQ5 gt RMS AcosQ5 Z 0 note A Z 0 Note Can convert other forms to this standard form Some examples t 3 cos7t01 3cos7t7r01 using cosa cosac7r t 3sin7t 01 3cos7t g 01 using sina cosa 3 t 3cos 7t 01 3cos7t 1 7T 01 using cos a cos NH xt Continuous time sinusoids are periodic with period T i 2 f0 xn Discretetime sinusoids are NOT periodic UNLESS fe is rational we 2w ltgt has period T7 provided is reduced to lowest terms Why Acoswen Q5 Acoswen T Q5 ltgt weT 27Tk ltgt fe EX we 27F gt T 1 we 2027T gt T 100 we 2 gtnonperiodic Freq Acoswen Q5 Acoswe 2k7rn Q5 gt we itself is periodic As we increases from 0 to 7T7 oscillation rate increases As we increases from Tr to 2W7 oscillation rate decreases Huh Frequency range 7T 3 we 3 27F ltgt 7T 3 we 3 0 since we periodic WLOG Restrict discrete time frequencies to range 7T 3 we 3 7T Note we 0 gt coswen 1 we 7T gt coswen 1quot SUM OF MULTIPLE SINUSOIDS If t has period Tee and freq fee yt has period Ty and freq fy And TeeTy is a rational number otherwise t yt not periodic Then t yt has period TeeyLeast Common Multiple of Tee and Te Then t yt has freq feeyGreatest Common Divisor of fee and fy Note Always works for sinusoids rarely fails for general periodic signals How 1 Reduce to lowest terms 2 Teey NTee MTy Why 05 1 Txy 3105 Txy 05 1 NTOU 1 2105 MTy y EX t cos037rt cos7rt gt fe 7 gt fee 6 10 gt eey 60

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