Intro Signals&Syst EECS 216
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Date Created: 10/29/15
EEOS 216 SAMPLING THEOREM WHAT IT A bandlimited continuous time signal t can be DOES reconstructed perfectly from the samples nT of t IF the sampling rate exceeds twice the maximum frequency of Claude Shannon7 U M alumnus and 77father of information theory77 Let signal t have spectrum Xw ft 0 for lwl gt B Let t be sampled everyt nT lt gt sampling mte S 2 RADIAN T SECOND Then can reconstruct t 20700 mnT ZfioomnT6t nT gtllt WT IF 5 gt 2BNyquist rate T if w lt B 0 if w gt B So we can implement using gtlt 6t nT gt LPF gt Note that fT low pass lterLPF WHY THIS IS TRUE Note 7K 6t nT Z 26w n5 Fourier series expansion 2 6t nT Z ej27mtT we S Then fej2 tT 2W6w n5 using modulation property of 7 Periodic AND discrete in timelt gtPeriodic AND discrete in frequency fsampledat fZnT6t nT ft Z 6t nT if f t f fz 505 75 iXW 2 25W HS ZXw n5 SIGNAL IN TIME DOMAIN FREQUENCY DOMAIN AAAA AAAE EECS 216 LECTURE NOTES EXAMPLES OF FIR DIGITAL FILTERS FIR Form hn FIRFinite Impulse Response Mn has nite duration and support yn bo n b1n l 92TL 2 bMan OrderM Mn bn for 0 S n S M Mn 0 otherwise Assume Mn is real here Frequency response HEW be 516 b257j2w bMeijMw ejwon awe 11163 m6jn7 implying moon a a Hej 0 coswon argH6j quot l Properties of H 63 Implies Hej periodic in to with period 27T Conjugate symmetry He j Hej Gain is even function Hej He j Phase is odd function argHe jw a7quotgHejw Notch lter Does Why Mn 17 2coswa7 I gt Hej 2 cosw 2cosw0e j Implement using yn 2 cosw0an l n 2 Eliminates component at a single frequency we from input Comb lter Does Why eg sin M i w Eliminate interference at a single frequency7 eg7 60 HZ hum I w 1w hln 171711 gt mew gt e wMg 2 Implement using yn n l n Eliminates components at frequencies ngiL7 i j iil Eliminate interference at harmonics of a single frequency 27T M I Sawtooth wave interference with period M I has these harmonics Ideal LPF Does Why Mn W gtHej l for0lt lwl lt B0 for B lt lwl lt 7T Implement using yn 2700 239 truncate the sum Low pass lter eliminates frequencies above B periodic period 2 Filter out high frequency noise from a low frequency signal smooth APPLICATION TO ELECTROCARDIOGRAM HEART SIGNALS Given Goal tEKG signal periodic you hope at 60 REIENTTSEI Filter out 60 HZ interference from electrical outlet wires in lab Signal Huh t periodl second gtfundamentall HZ gtharmonics273 HZ t co 01 cos27rt 61 02 cos47rt 62 03 cos67rt 63 V h DC FUNDAMENTAL 1ST HARMONIC 2ND HARMONIC Sample Spectrum at 1 kHz gt A 0001 second gt t 0001n gt t 0001n of sampled signal 0001n is periodic in f with period 1 kHz Notch yn 2cos27rn l n 2 eliminates 60 HZ hum EECS 216 CONTINUOUS TIME FOURIER TRANSFORM DEF Xw ft ffooomte j tdt 3015 ffOOOXwej tdw Note 1 Sign change in eijm 2 Integrate over If vs w 3 Factor of Need ff tdt lt oo ltgt t absolutely integrable77 for Xw to exist Huh Splits t into frequency componentsrvpm39sm recombine into Also Regard as Fourier series expansion of t having period T gt 00 Then Aw 2 gt 0 continuous spectrum Units of X to seconds T Also ft t3jw Where L is the 2 sided Laplace transform BASIC PROPERTIES 3015 fjoooXwej tdw gt ht gt yt ff HwXwej tdw Linear faat byt at byt for constants a7 b EX H6lttgt7aeatulttgt17aj1zziz muf n e for w ltlt a 3 Convolution ffjo uy t udu J tyt EX e tute 2tut quot1jw f 1jw117 e tie gtut 4 Time scaling fat for any constant a Try cosw0t Also Time reversal f t X w and ft X w 5 Time delay Jat D Xwe j D Note same signs Also 1 Magnitude same 2 Linear in to phase shift linear phase77 6 Modulation ftejat Xw a Note different signs EX ft cosw0t Xw we Xw we Shifts spectrum This one equation will form the basis of our study of communications 7 Differentiation f ij Assumes no initial condition Why ff Xwdw ff ijwej tdwquot1ijw Note Ift includes 0000001 sin1012t then includes 1000000 cos1012t Use Note ht6tiae atut has frequency response Hw computed above 8 Time multiplication ftat flag fte atut jwia 9 DC frequency0 X0f tdtaverage X0i7r ff Xwdw 10 Duality If fat Xwothen fXt 27Tx w NH CONJUGATE SYMMETRY AND EVEN AND ODD FUNCTIONS 1 t real gt X w Xw Called conjugate symmetry77 of Xw a R6Xw ReX w fjooo t coswtdt even function b ImXw ImX w ff tsinwtdt odd function c Xw X w Fourier magnitude is an even function of w d ArgXw A7quotgX w Argument phase is odd function EECS 216 s PLANE CIRCUIT ANALYSIS BASIC CONCEPT BEHIND s PLANE CIRCUIT ANALYSIS Given Also Goal A circuit with several inductors and capacitors and different sources Non zero initial conditions capacitors charged and inductors juiced Compute any circuit voltage or current as a function of time t for t gt 0 Soln But But Solve constant coef cient differential equation with initial conditions Lots of algebra7 even using Laplace transform have to obtain diff eqn Is there easier way7 like phasors but for any source and initial conds Idea Take Laplace transform of entire circuit Using Zl f sXs ac07 Device Name Resistor Inductor Capacitor Its Formula 11 Ri 11t L 0 After Xform V RI V LsI I CsV 110 Impedance Z R 8L Device Model LiL0 H N Sources have their values replaced with their Laplace transforms Note that a constant source A volts or amps is replaced with g Dependent sources now depend on some circuit variable Vs or I Capacitor I CsV 110 gt V I Q note the sign Can also use the Norton equivalent current source 0110 Note sign difference Voltage sources for L and C initial conditions Now straightforward circuit analysis Like phasors except 87 not jw After compute desired Vs 11t or Is it7 compute 4 Note Vs or Is will be a rational function7 so use partial fractions Given Goal With EXAMPLE RC CIRCUIT DRIVEN BY STEP FUNCTION Series RC circuit driven by step voltage source A 1t W Compute capacitor voltage A R C V Initial capacitor voltageB Soln Then Then Also KVL gt g 812 13 E 0 gt 18 HQS i FEW s R1sO 31R0 Vs g Is g W Compute partial fractions 118 g gt 11t A B Ae tRO for t gt 0 SS TRANSIENT Can rewrite this as 11t AI citRm BaitRC for t gt 0 W W ZSR ZIR EEOS 216 COMPLETE SYSTEM RESPONSE SUMMARY OF VARIOUS SYSTEM RESPONSES The complete response of a system can be decomposed in 2 different ways ZERO INPUT ZERO STATE i TRANSIENT STEADY STATE RESPONSE RESPONSE T RESPONSE RESPONSE W W initial conds from input decays to 0 doesn t decay to 0 The zero state response ZSR can be further decomposed as follows ZERO STATE 7 FORCED NATURAL h M137 IMPULSE RESPONSE RESPONSE RESPONSE W ere RESPONSE W W W from input like input like ht If all poles are in left half plane7 then STEE iTE and NTETAUNSIEEEE Given System described by the linear constant coef cient differential eqn n n71 m mil cliff1th anya K 3 b1 3 bmxtgt and Initial Conditions 3107 07 2 0 for differential eqn Goal Compute the response yt to a causal input t 0 for t lt 0 Using Ugi squotYs sn 1y0 050 we obtain the formula smblsm 1 b 701 0 20 3393in Y8 K snalsquot 1aLXs snalsquot 1an ZERO INPUT RESPONSE ZIR SET INPUT0 Def Characteristic eqn 8 a1 8 171 an 0 denominator of Ys Modes Let be the n roots of s alsn 1 an 0 are modes Note The modes of a system are also called natural frequencies of system ZIR zirt Bleplt Bnepnt for t gt 0 Decays to 0 if Rem lt 0 EX Parallel RLC circuit with R3Q7 L4H7 C1712F7 no circuit source Goal Compute the ZIR u if 1160 2 and ic0 l 23960 caujump KCL f00utdt 1 0 Take Li gt ll 0 0 dt 2v u v i 0 Here R3 27 L4H7 0175 gt all 4 311 0 0 12 s 1 3 7 7 Soln Vs tfgjiljgt lt0 3234333s3 331 ut5e 3t103736 not NOW If i0 27 nd initial 110 IC so that ut Ce gt for t gt 0 Huh By choosing IC7 we can cancel out one of the modes of the circuit Soln s 4u0 24 Cs I to cancel 8 1 factor in denominator gt Equate coeff of s gtV0C4V0724 gtV08 gt ut 8e 3tlt EECS 216 l SIDED Z TRANSFORM DEF WHY Zan Xz 2200 3601quot7 even if is noncausal Compare to 2 sided Z transform Xz 20700 To solve difference equations with nonzero initial conditions NH 4a 4b PROPERTIES OF 1SIDED ZTRANSFORM Xz causal gtl sided Z transform X z2 sided Z transform X Unique inverse only for causal signals no info on n lt 0 Equal causal parts but unequal anticausal parts gtsame X Z ROC Always gt lpmax largest pole of XZ7 so donlt bother D gt 0 zxn 13 z DXz 251 nz Dgt0zumnm mxwa Zi3mmro i Shift right gtmust add in terms previously anticausal ii Shift left gtmust subtract off terms now anticausal iii Compare to gs 23 82X8 80 0 SOLVING DIFFERENCE EQNS WITH INITIAL CONDITIONS Solve Take 2 2 321 Solve 2yn 3yn l yn 2 un un l un 2 Chen p264 2Yz 3z 1Yz y 1z z 2Yz 272YZ 3y 1 y 2 Z 1y 1l 3y 1 y 2 Z 1y 1 2 3Z 1 Z 2 y 1zy 2 z 121 z 2 Uz v i 1z 1 z 2 i23271272 YZ 12 zeroist ate response zeroiinput response Plug in Partial fraction y1 27y2 11 YZ 1z 17272 522 1 744271272 232 1z 2172 1 232 1z 2 21 z 11z 1172 1 72 1 1 41 Z 72 ununit step gt Uz Uz 1 1f 727 7 3 141 Note no constant term 52 1 Z lz gegwmm gc4rmm mm natural response forced EECS 216 LAPLACE TRANSFORM Forward Inverse Need Laplace transform Xs of t is Xs t fooo te stdt t 6173 X 863td8 You ll never use this formulaiever t piecewise continuous of exponential order7 for X s to be de ned EX EX EX WW0 f6 WW Si fo a 0 a cm c0swt1t gem eijwt1t 31 12 7 s I I w sjw i sinwtlt 2ijej t 6 3Wt1t Props Vital 0 37a n71 1 6t l eatt Xs a eatcoswtlt 3 5 sXs smug snls 87171340 and Y8 The solution to the linear constant coef cient differential equation dmx n n71 a1lltTiy HanytKltdtm 191 bmtgt Initial Conditions y07 07 l 0 for differential equation7 is n71 dnili iy i 7 0 E as 0 dtquot 1 1 j0 7 snalsquot 1an 7 smblsm 1bm 7 K snalsquot 1an INVERSE LAPLACE TRANSFORM OF RATIONAL FUNCTION Goal Note of RATIONAL if STRICTLY FUNCTION PROPER ORIGINAL i STRICTLY function7 write FUNCTION 7 c PROPER 71 Compute L snalsn1man STRICTLY PROPER mltn If this is not 1 Compute the n poles which solve s alsn l an 0 2 PARTIAL smblsm1bm A1 FRACTION snalsquot 1an 37121 An S Pn Compute where pkmb1pkm 1bm Mimimin smblsm 1 Ak siplsipnbm 8 pklSPk formula Note Note Denominatorpk p1 pk pk71pk pk1pk pn COMPLEX i i Assumes poles distinct pi y pj CONJUGATE pkHipk gt AkHiAk OJ sna1 sn 1 an Then L 1M Alepltlt Anepntl f formula Note L 1c 06t Alanna Anepnt1t Note Ae jg 37aijw Lil 2A6 coswt 0lt useful formula Several illustrative numerical examples are given overleaf EECS 216 CONTINUOUS TIME FOURIER TRANSFORM DEF Xw ft ffooomte j tdt 3015 ffOOOXwej tdw Note 1 Sign change in eijm 2 Integrate over If vs w 3 Factor of Need ff tdt lt oo ltgt t absolutely integrable77 for Xw to exist Huh Splits t into frequency componentsrvpm39sm recombine into Also Regard as Fourier series expansion of t having period T gt 00 Then Aw 2 gt 0 continuous spectrum Units of X to seconds T Also ft t3jw Where L is the 2 sided Laplace transform BASIC PROPERTIES 3015 fjoooXwej tdw gt ht gt yt ff HwXwej tdw Linear faat byt at byt for constants a7 b EX H6lttgt7aeatulttgt17aj1zziz muf n e for w ltlt a 3 Convolution ffjo uy t udu J tyt EX e tute 2tut quot1jw f 1jw117 e tie gtut 4 Time scaling fat for any constant a Try cosw0t Also Time reversal f t X w and ft X w 5 Time delay Jat D Xwe j D Note same signs Also 1 Magnitude same 2 Linear in to phase shift linear phase77 6 Modulation ftejat Xw a Note different signs EX ft cosw0t Xw we Xw we Shifts spectrum This one equation will form the basis of our study of communications 7 Differentiation f ij Assumes no initial condition Why ff Xwdw ff ijwej tdwquot1ijw Note Ift includes 0000001 sin1012t then includes 1000000 cos1012t Use Note ht6tiae atut has frequency response Hw computed above 8 Time multiplication ftat flag fte atut jwia 9 DC frequency0 X0f tdtaverage X0i7r ff Xwdw 10 Duality If fat Xwothen fXt 27Tx w NH CONJUGATE SYMMETRY AND EVEN AND ODD FUNCTIONS 1 t real gt X w Xw Called conjugate symmetry77 of Xw a R6Xw ReX w fjooo t coswtdt even function b ImXw ImX w ff tsinwtdt odd function c Xw X w Fourier magnitude is an even function of w d ArgXw A7quotgX w Argument phase is odd function EECS 216 SPECTRA FOR MODULATION METHODS Def So Why 1 2 Modulation of a message signal mt Multiplies mt by a sinusoid Modulated signal t mt cos27rfct where fccarm39er frequency So why do we modulate signals mt for communications purposes Transmission mt canlt be radio broadcast t can higher freqs Freq multiplexing Transmit several messages mitt using one Recall Huh Below fmt coswct Mw w6MwwC where fmt Multiply mt by coswct shifts spectrum M w up and down by we fsqut SQUw ftrit TRIw fb0wlt BOWLw DSBSC Each Note Send 3 signals at once by modulating them at different carrier freqs Maximum freq57 bandwidth10 total signal has bandwidth100 3 modulated signals separated by bandwidth2maximum frequency AM Why DSBSC Why Amplitude Modulation adds in carrier cos20t lmt cos20t Receive AM using envelope detector lmtgt0 easy to nd signal Double SideBand Suppressed Carrier excludes carrier mt cos20t Donlt waste power transmitting carrier still use envelope if mt gt 0 Goal So Soln Send two signals leftt and rightt as one single signal stereot Can recover mono signal lefttrightt directly from baseband stereotlefttrighttlefttirighttcos2W38000t works Spectrum SQUw Spectrum TRIw Spectrum BOWLw 1 1 1 05 05 05 M 0 0 0 10 0 1O 10 0 1O 10 0 10 AM spectrum OF bowtcos20t SSB spectrum upper SB 1 0 0 40 20 0 20 40 40 20 0 20 40 DSBSC spectrum OF bowltcos25ttritcos35tsqutcos45t 0 W 50 40 30 20 1 0 0 1 0 20 30 40 50 EECS 216 IMPULSES AND IMPULSE RESPONSE CONTINUOUS TIME IMPULSES Def Math Def Def An impulse is the limiting case of a constant area high and fast pulse An impulse has zero width in nite height7 and nite area under it Mathematicians Impulses are distributions or generalized functions Don7t call them Dirac delta functionsiDirac would sue for defamation The impulse response of a system is simply its response to an impulse Impulse6 t gt gt htimpulse response makes sense The step response of a system is simply its response to a step function steplt gt SYSTEM gt ststep response is easier to compute CONTINUOUS TIME IMPULSE RESPONSE OT Response yt to a pulse t of Width A7 height A How to compute the response to an input that doesn7t exist physically Consider a series RC circuit driven by a high and fast voltage pulse Sourcestepatlt gt ytstl e tROIt familiar Step up7 then down tlt 1t A gt yt st st A Scale previous tilt 1t A gt yt ist st 1 A For t gt A7 this becomes yt ZI e tROH i e O mROH BAR0 giftR0 A 1 getR0 eetaw for t gt 0 BOX using ex II ImmforajARC ltlt I andarea1 A 1 note the units is independent of A and Z7 as long as A ltlt RC PHYSICAL INTERPRETATION OF IMPULSE AND IMPULSE RESPONSE H 01me Take Norton equivalent Impulsive current source Shot of charge Rl AA charges capacitor up to R 1AA R lo q Cu Capacitor voltage decays Impulsive source like an initial condition Currentm How can the capacitor voltage jump Because the current is in nite 1 A m DurationiA 1 and durationA don7t matterionly chargeproduct PROPERTIES OF IMPULSE AND IMPULSE RESPONSE H Impulse responseht step response For above circuit7 ht I e tROU e tRO for t gt 0 For above circuit7 CHIkaN Mt 5 1H8 5 1M 5 1 fab 6t edt area under 6tl if a lt c lt b ff t6t adt a sifting property t6t a a6t a 6at 1 BitR0 31R0 W ff 6t adt 1 1lts0gt i6t both have area M M 6tlt and are unde ned EECS 216 LECTURE NOTES FORWARD OR DIRECT ZTRANSFORM DEF Xz 2200 nz quot compare to t fooo te stdt Huh Or Finite duration signal gtpolynomial X with coe icientsan In nite duration ncoei cients of Laurent power series X Finite length signal Q7 17 47 27 5 Note 0 3 and nite duration5 gt Xz 3 12 1 4z 2 22 3 5z 4 324 Z3 422 22 5z4 Z6n D Z D for any integer delay D Z 0 Z6n 1 Expon ential Zaquotuhd 33oaquotz ZZJGZ U 11 az 1Kz a EX22 YWVA3 TEXaza arwhnj T Sinu soids Zcosw0nun Zejw0 un Ze jw0quotun since linear 22720044110 17271 cosw0 i 7 Z2722 cosw01 39 1722 1 cosw0z 2 1 1 1 1 i 5176 02 1 517673 02 1 i Linear Delay EX Note 7 RATIONAL 3221z4 z i 22 271 i i FUNCTION 42372223274 W Z 7174 ulnl If Xz and D Z 07 then Zan D Z DXz zmw um mr1 zgr if1f Note that un un 2 5n 5n 17 so these are consistent ii 22 Also Note z um ud Convolutionlt gtpolynomial multiplication gtllt X log1 2171 if you recognize this Eigen funcs of LTI Note 2 gt gt z in gt scaled 2 out 2200 hnz quot hngtkz Z hizn i z Z hz39z i Hej HZZejw Note 2 plays the same role that 6 plays in continuous time Hztransfer function Hej lzejw frequency response EX Soln 1 1 2 Note Compute step response to of LTI system with Mn 27 37 1 We need to compute yn 27 37 1 gtllt Do this two ways yn 26n 36n 1 16n 2 gtllt Using un gtllt 5n un7 yn 2un 3un 1 un 2 27 1 try it has duration2 Yz 2 3z 1z 2f 2 271 gt yn 2 1 This system has wiped out the step input This seldom happens
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