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# SIGNALS AND SYSTEMS EE 421G

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This 107 page Class Notes was uploaded by Adaline Pollich on Friday October 23, 2015. The Class Notes belongs to EE 421G at University of Kentucky taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/228319/ee-421g-university-of-kentucky in Electrical Engineering at University of Kentucky.

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21 Signals and Systems Signals convey information Systems respond to or process information Engineers desire mathematical models for signals and systems in order to solve design problems efficiently and thoroughly Signal encoded information data a dynamic or change in some quantity that has meaning In most cases this is modeled as a function of time or space Noise irrelevant data variability in a quantity that has no meaning or significance In most cases this is modeled as a random variable System a mapping between a set of inputs to a set of outputs an entity that processes signals received at its input and produces another set of signals at the output a process that responds to actions or events at its input by generating actions or events at its output Differential equations or convolution integrals can approximately model many physical systems Kevin D Donahue University of Kentucky 22 Classi cations of Systems 1 Linear vs Nonlinear For all linear systems superposition holds for inputoutput relationships Denote a general system in the following manner ym System H operates on inputx to produce output y The system H is linear if and only if iff for any inputoutput pair y1Hx1 and y2 Hx2 the following statement is also true aly1 a2y2 Ha1x1a2x2 where a1 and 12 are constants Kevin D Donahue University of Kentucky 23 Determine whether or not each system described below is linear Assume inputs and outputs are functions of time denoted byx and y and constants are denoted k O dzx dx 0 y x dlz dt 0 ykx10 o yk1xk2x2 o xd2xx y dzz 2 Constant arameter timeinvariant vs timevar in s stems Kevin D Donahue University of Kentucky 24 A system whose model parameters change with time is considered time varying Note that the output and input will be varying with time for both constant parameter and timevarying systems Systems where outputs differ only by a time shift when the same inputs are applied at corresponding time shifts is a constant parameter system A system is a constant parameter system iff for any input output pair yI Hxf the following statement is also true yl 239 Hxl 7 for all 1 Determine whether the systems below are time varying or time invariant o yl kxl 10 0 yl cos2721xl Kevin D Donahue University of Kentucky 25 3 Instantaneous memor less vs d namic with memor s stems For an instantaneous system the present output value depends only on the present input value In a dynamic system the present output value depends on the present and past input values Dynamic systems usually contain some type of energy storage elements The response of a dynamic system results from two components the initial condition and the input The state of the system refers to the information needed along with the present input to determine the present output Zeroinput response system output due only to system state or initial condition Zerostate response system output due only to the input of the system In general Determine which systems are instantaneous and which are dynamic o y klx kzxz o y x Kevin D Donahue University of Kentucky 26 4 Causal vs Noncausal Systems where the output depends only on the present andor past values of the input are referred to as causal Note that for a causal system the output cannot depend on future input values Systems where the output depends on future input value is referred to as a noncausal system 5 LinpedParaineter vs DistributedParaILeter Svstems In most real systems the interactions between the signal energy and the system elements happen continuously over space ie resistance over a wire In modeling these systems the interaction can be considered to occur at one point in space This is referred to as a lumpedparameter model This is a reasonable model when the dimensions of the elements are small with respect to the energy wavelength When this is not done ie transmission lines the model is referred to as a distributed parameter system Kevin D Donahue University of Kentucky 27 6 ContinuousTime vs DiscreteTime Svstem The input and output for a discretetime system is de ned only at discrete points in time yn Hxn for n e 2 1 0 1 2 If the inputs and outputs are defined over a continuum of time values then the system is a continuoustime system yt Hxt for t E 0oo 7 Analog vs Digital Systems A system whose input and output values take on only a set of discrete values is referred to as a digital system If the values of the input and output can take on a continuum of values then the system is an analog system F or the following digital signal processing system classify the subsystems at each stage Based on the nature of the input and output some subsystems may not t under a continuoustime vs discretetime or analog vs digital system classi cation scheme but a hybrid combination Kevin D Donohue University of Kentucky Elements of a Digital Signal Processing System Analog Signal Discretetime Digital Signal 96611 AW 28 X Signal Quantizer Coder gt W nT an H 10 01 goo Processed Processed Digital Analog Signal Signal Computing Interpolator Hardware and smoothing ya I n Kevin D Donahue University of Kentucky 29 Differential Equation Models for Current and Voltage Systems Capacitors Inductors Resistors il 0 vt L vl Ril v iirdr W E j il V0 W W il jvz39dz39 NH Kevin D Donahue University of Kentucky 210 Example find the inputoutput equation relating input iS to output v0 gt lL M L R1 VOTC V0 i Z R R R Answer S v 2 L 2 i 1 2 v CL 0 L RIC 0 CL R1 0 0r R2 1 I 7 2 IL 0 vc L vc 5 s C RIC IL V0 2 01v r Kevin D D5 0hue University of Kentucky 211 Differential Equation Models for Position and Force Systems Translational Systems Consider Position output and force input in one direction denoted by yt and ft respectively M21580quot f0 M f Linear Spring Stiffness K fl Kyt Linear Dashpot damping coefficient B f I Byl Rotational Systems Consider an angular position output and torque input denoted by 9t and T t respectively Rotational Mass J T l J9 Torsional Spring K TZ K6 Torsional Dashpot B Tl B Electromechanical Systems For a DC motor consider the an angular position output and current input denoted by 6t and it respectively Motor Constant KT T I KT il Kevin D Donahue University of Kentucky 212 Example consider a torsional spring with stiffness K2 ntmrad fastened to the rotor of an armature controlled DC motor with motor constant K 5 ntmA and the rotational mass is J 5 ntmrads2 The friction coefficient for the spinning rotor is B 05 ntm2radsec Find the equation that relates the rotor position to the armature current Assume the polarity of the motor is such that a positive current moves the angular position is a positive direction Describe the motion of the rotor for a step input going from 0 to 2 A Ans K J B Ke 6z 5 eXp Z 205cos2l 00125 sin2t In Matlab gtgt t 02pi20100 gtgt sig 5 exp t205cos2t 0125sin2t gtgt plottsig gtgt title39rotor response39 0 gtgtxlabel39seconds39 03 gtgtylabel39radians39 07 0 60 seconds Kevin D Donahue University of Kentucky 121 Transformation of A Single Random Variable Independent fluctuations in a random signal can be modeled with a pdf or cdf If the random signal is passed through a circuit lter amplifier rectifier a new distribution will result for uctuations in the output signal If the operation of the circuit can be expressed as a mathematical representation for a memoryless system the distribution of the output signal can be computed Let gx be a function that maps one RV into another Y gX For the simplest case limit gx to a monotonic function Consider an example of uniformly distributed noise signal between 1 and 1 volt passing through an amplifier with a gain of 10 and saturation at i10 volts Find and sketch the pdf for the output of the amplifier Do the same if the input noise is uniformly distributed between 2 and 2 volts Show first case fyy 2 10uy 10 uy 10 Show second case y uy 10 uy 10 y 10 5y 10 122 To obtain a process for a general transformation between continuous RVs let y0 gx0 where gx0 is a monotonically increasing function then 30 PY S y PgX S gx PX S x FXOC Show that My 1200 dy Likewise show if the case of a monotonically decreasing function then My PY S y PgX 2 gx1 PX S x1 Fix Showfxy 100 dy Examples A zero mean Gaussian noise signal with an average power of 01 watts is passed through a linear amplifier circuit with a gain of 40 dB Find the pdf of the noise at the output What would the pdf be if the amp also added a bias of 5 volts gquot y For either increasing or decreasing fy y fX x gquot y gquot y where a 10 1 y Case 1 eX W m p 202 eXp y m j where 0 10 and m 05 1 Case 2 M m 202 123 If YgX is not monotonic then Xg391Y will not be a function Consider Y X Y9gtlt WQWVW Y a m o a o as u on co co m L m o m 4gt D co 10 B 6 4 2 0 2 4 6 B 10 A 1 2 3 4 5 6 7 8 9 10 Since Xg Y is not a function the derivative d does not ex1st However for the y purpose for mapping one RV to another this means that iX is mapped into the same Y Therefore nonmonotonic functions should be parsed into regions over x where Xg391Y exists as a function Example A zero mean 4 watt white Gaussian signal is passed through a recitfier Find the distribution of the output Showzfyy exp i zj for O S y lt 00 where a 2 124 In general if a function is applied to a single RV YgX the distribution of a transformed RV is computed via fy y fX x where x gfy i 123N is a parsing of the function 1 xgr y into monotonic regions Example An exponentially distributed RV fX x Zexp 2xux is transformed by the function YX42 Find the pdf of Y 2 eXp 8 6Xp2 33X 6 for O S y lt 16 y eXp 8iyj for 16 S y lt oo Show My 51 Periodic Signals A signalfl is periodic iff for some To gt O ffflTo VI The smallest T 0 value that satisfies the above conditions is called the period of ft Consider a signal examined over 0 to 5 seconds shown below What values of T 0 satisfy the condition in the de nition of periodic What is the period of this signal sin2pit sin2pi2t 3 Amplitude Seconds 52 Properties of Periodic Signals The integration over one full period has the same result independent of where the integration limits begin or end lgla Z gla Z Lugla Z The Sinusoid gl sina Z 6 sin27z fl 6 sin277rt 6 Phase 9 Frequency radians per second 0 Frequency Hz f Period T 0 Harmonics Consider a signal consisting of a sum of sinusoids whose frequencies are integer multiples of a fundamental frequency f0 gl C0 C1 sin27z fot 61 C2 sin27z39 2f0Z 62 C3 sin27z39 3f0l 63 where f0 is the fundamental frequency and the sinusoid with frequency kfo is called the k th harmonic in the series and C0 is the zeroth harmonic or DC Examples Find the periods and fundamental frequencies of the signals below 53 a gt 6sin20nt l3cos307t Hint For f0 find the greatest common factor between the frequencies the least common factor will also work Hint For T 0 find the least common multiple between the periods b gt 4 3sin1207l lOsin1507Z 23in2407l c gt 4cos8nl llcos20nl cos10nl d gt 6sin32nl 13cos72nl e gt 6sin20nt l3sin20l Hint Must the sum of periodic sinusoidal signals be periodic Fourier Series 54 A periodic signal fZ of almost any form can be expressed as a series of harmonic sinusoidals Trigforms ft C0 ZflCn cosnaot 6 quot a0 Zia cosna0l 1 sinna0l where C0 a0 a0 C C0809quot b Cquot sin6 n Cquot 1la b 6 quot tan1 b a Exponentialforms f0 exponwm where Dn belongs to the set of complex numbers and D11 Dg n 55 Example Show by plotting the harmonic components of a Fourier series that the periodic extension of rt12rt on the interval 1 1 has Fourier coefficients 1 21 cos7rn b O a a 0 2 7 n 7 gtgtt 40014 Set up time axis to show 4 periods gtgts 5onessizet Initialize summing array with a0 gtgtfor k110 Loop to sum up 10 harmonics gtgts s 21coskpipikquot2coskpit gtgtend gtgtplotts39w39 m up in Him havmumc m1 5 havmumcs upm mm havmumc Up in 5175 havmumc 56 Computing Fourier Series Coef cients trigform Given the Fourier Series representation of a periodic signal fl fl a0 61 cosna0l b sinna0l it can be shown that the coef cients for ft can be computed with 1 2 2 a0 fjfldl an cosnaotdl b smnaotdl Derive the Fourier Series coef cients in trig form for the waveform in the previous example Derive the Fourier Series coef cients in trig form for the periodic extension of pA I on the interval 0 T 0 where 0ltAltT0 a 2 Sinw nA b 2 1 h th t 8 0w a a0 7 quot mz39A 7 quot mz39A 57 Computing Fourier Series Coefficients exponentialform Given the Fourier Series representation of a periodic signal fl f0 exponwor it can be shown that the coefficients for ft can be computed with D Ifrgtexplt jnwrgtdr Note the differences between the Trig and Exponential form 1 Coefficients for the trig for are La for real ft while for the exponential form they are complex 2 The index for summation is from 0 to 00 for the trig form and from 00 to 00 for the exponential form 3 The formulas for the exponential form are more compact and easier to work with than those for the trig form The coefficients for compact trig representation can be determine from the exponential form coefficients C0 D0 and CH Z DJ 6 quot 4D for ngt0 Examples 58 Derive the Fourier Series coefficients in exponential form for the periodic extension of pA I on the interval 0 T 0 where OltAltT0 Plot the magnitudes and phases of the coefficients for T0 10 ms and A 1 ms as a function of frequency f and discuss their meaning nA7r mrA eXp 111 7 Show D 7 for n 0 D0 i quot 117239 A 7 gtgtt0 10e3 Period gtgtdel 1e3 Duration of pulse gtgtn 601 eps 160 Harmonic index eps is close to zero gtgtf0 1t0 Fundamental Frequency gtgtd exp jpidelf0nsinf0npidelpideln Fourier Coef cients gtgtf1gure1 gtgtstemf0nabsd39W39 Stem plot of magnitudes gtgttitle39Magnitude of Fourier Coef cients39 gtgtxlabel39Hz39 gtgtf1gure2 gtgtstemf0nangled39W39 Stem plot of phases gtgttitle39Phase of Fourier Coef cients39 gtgtxlabel39Hz39 59 Magnitude ofFourierCoefficients 100 I I 80 60 40 20 0 6000 4000 2000 Phase of FourierCoefficients 4 I I I I I n n n n n 3 n n n c o n a h o 2 n o n o 0 a o n n n a n o n o n o r n o n o n o 1 o n o n o n n n n n a a n o n a n 0 0 n In a u u 1 v v o u v o v v u 2 0 av 0 u u u v 3 v 4 I I I I I 6000 4000 2000 0 2000 4000 Hz 510 Examples Derive the Fourier Series coef cients in exponential form for the periodic extension of rt on the interval 0 05 Plot the magnitudes and phases of the coef cients as a function of frequency f and discuss their meaning Show D i for n 0 D l quot 4m 0 4 gtgtt0 5 Period gtgtn 301 eps 130 Harmonic index eps is close to zero gtgtf0 1t0 Fundamental Frequency gtgtd j4pin Fourier Coef cients gtgtd31 14 Must substitute DC term in directly since expression was gtgt simpli ed beyond the point of using L39Hopital39s rule gtgt gure1 gtgtstemf0nabsd39W39 Stem plot of magnitudes gtgttitle39Magnitude of Fourier Coef cients39 gtgtxlabel39Hz39 gtgt gure2 gtgtstemf0nangled39W39 Stem plot of phases gtgttitle39Phase of Fourier Coef cients39 gtgtxlabel39Hz39 Magnitude of Fourier Coefficients I l r l I O TTWO 40 20 40 Hz Phase ofFourier Coefficients 511 31 TimeDomain Analysis of ContinuousTime Systems Linear timeinvariant continuoustime systems LTIC are modeled by linear differential equations dam dzquot quot dz dz a0ybmd fbmild ifbl bof dz dt39 dz where ft and yt are system inputs and outputs respectively and 10 aml b0 bm are constants Inputoutput behavior of any LTIC system is characterized by these constants A more compact notation will be used to represent the above system where Dquot is a derivative operator D aHD a1D a0 yl mem me39 b1D be f I 32 The ZeroInput Response Consider systems excited or driven only by initial conditions ie stored energy D aHD 391 a1D ctoy0 I 0 Solution of first order equation Daoyof 0 Show solution is y0 I ceXp a0l where c is an arbitrary constant If initial condition given at t t0 then y0l yl0exp a0l for t2t0 For a general order assume solution is of the form c expOtl t substitute into zero input response differential equation to show that a system can be characterizes by its characteristic polynomial QM A an711 1all a0 and the solution can be obtained based on the roots of the characteristic eguation QM 0 Let the roots of the characteristic equation of an nth order differential equation be denoted as lllZln 33 Modes of the ZeroInput Response The component of the system s zeroinput response that corresponds to a root of the characteristic equation is referred to a mode or natural freguency a If roots are real and distinct each system mode is an exponential and the total response is described by y0 l c1 exp 1l c2 exp12l cquot expMnl To view examples of how various combination of these modes can behave try running the following script in Matlab t 01 10 De ne Time Axis for function evaluation Loop through 10 different functions for k1 10 c 5rand110 5 Randomly pick 10 coef cients subplot311 plotc Plot them out ylabel39Coef cients39 lam 2rand110 Randomly pick 10 time constants subplot312 plotlam Plot them out ylabel39Time Constants39 subplot313 plott cexplam39t Evaluate and plot function ylabel39Function39 xlabel39seconds39 pause Hold gure until Key is struck end 34 Real and Distinct Roots Example Coef cents Time Constants 1 2 3 4 5 6 7 8 9 10 15 E 1 39 5 E 05 0 I I I I I I I I I 0 1 2 3 4 5 6 7 8 9 10 seconds 35 b If roots are real and repeated each system mode is the product of an exponential and the time variable raised to a power The total response is described by y0l c1 czl cnt 1expllt To view examples of how various combination of these modes can behave try running the following script in Matlab t 0120 Set up Time Axis matrix for function evaluation tm tm tquotk1 Each row is t raised to the next power end Loop through 10 different functions for k1 10 c 5rand110 5 Randomly pick 10 coef cients subplot211 plotc Plot them out ylabel39Coef cients39 lam 2rand11 1 Randomly pick 1 time constant between 1 and 3 and print it out subplot212 plott ctmexplamt Evaluate and plot function ylabel39Function39 xlabel39seconds39 pause Hold gure until Key is struck end Real and Repeated Roots Example time constant 11427 9 Coe cen Func on 8 10 12 14 16 18 20 seconds 36 37 If roots are complex each system mode is an exponentially weighted sinusoid The total response is described by let in a ij l y0l El expal lcmil cosBl t 0 sin81 Z or I Z In To view examples of how various combinations of these modes can behave try running the following script in Matlab t 01 10 De ne Time Axis for function evaluation for k110 Loop through 10 different functions cr 5rand15 5 Randomly pick 5 complex coef cients ci 5rand15 5 c crjci crjci subplot311 plotabsc ylabel39Coef cient Magnitudes39 Plot out magnitudes lamr 2rand15 Randomly pick 10 complex frequencies lami 1 3rand15 lam lamrjlami lamrjlami subplot312 plotabslam Plot out magnitudes ylabel39Time Constant Magnitudes39 subplot313 plott realcexplam39t Evaluate and plot function ylabel39Function39 xlabel39seconds39 pause Hold gure until Key is struck end 2171 y0l quot2ch expmml cm expmml where cm c and l seconds Function 10 Time Constant Magnitudes Coef oent Magnitudes Ix 0 x M 39n 3901 N 39n 01 O I 01 I I I Complex Roots Example 38 39 Example Identify the modes of the system described below D5 2D 10D3 7D2 D 1yl ft Find roots of D5 2D 1OD3 7D2 D 1 0 Use Matlab lam r00ts1 2 10 7 1 1 lam 06224 29286i 06224 29286i 08073 00261 03708i 00261 03708i 310 Solutions to ZeroInput Systems To have a nontrivial solution the system must have some nonzero initial conditions In addition to obtain a unique solution the initial conditions on y0t and all its derivatives must be speci ed up to the order of the differential equation minus 1 Example Consider D5 2D 10D3 7D2 D lyt fl assume that yo0 07 Z 107 Z 17 Z 07 Z 0 for t 2039 Show solution is coefficients 00085 00009i 00085 00009i 16741 00000i 08286 116242i 08286 116242i which implies y0t 00085exp 6224Z cos293t 01 167 exp O8 II 1 156exp003t cos37l 25 for t 20 311 The Unit Impulse Dirac Delta Function The unit impulse function 8t can be defined in the following equation 27 fl0 for altl0ltb Ifl5Z I007 O elsewhere This is sometimes referred to as the sampling property of the impulse function Since this function is defined only in terms of another function it is referred to as a generalized function Other important properties of the unit Impulse function The integral of 8t over any t interval that includes the point where the argument becomes zero is one Mr z0dz 1 Although not a good definition the impulse function can also be described as an d O for l lt O rt ul 5l 00 for Z20 dt lt gt dt lt gt lt gt O for l gt O The impulse function can be consider as a limit of another function altrgtuzJ ur a What is the limit for 5 0 as 8 0 For any 8 what is the area under 63t 312 Examples Evaluate the integral 8 2 00 72 j 51 4 exp 1de 00 Show a WZf1dlf0 foragt 0 a show by using integration by parts Nlm 313 The Unit Impulse Response If the input to a LTIC system is an impulse energy is injected into the system in a moment of time This creates a jump in the stored energy and therefore an instantaneous change in the initial conditions The effect of the impulse is to energize all modes of the system whose behavior can be observed in the impulse response Finding the impulse response The impulse response is similar to the zeroinput response since for t gt 0 5t 0 For a general differential Equation D aHD a1D a0 yl mem bmile 1b1D be f I QDyf PDff The impulse response is ht bn t PDy0 lul where if mltn bn0 and y0t is the zeroinput response where the initial conditions were ME 0 1 y 0 0 y 0 0 y 0 0 Examples Find the impulse response for the given systems D2 10D1yr Dft Ans hZ 10102 eXp 99l 00102 eXp O1Zut D 40yltrgt D 2gtfltrgt Ans hl 5Z 4OeXp 4Ol 2 eXp 4Olut D2 4D 4yl 3fl Ans hZ 3t eXp 2Zut 314 Monday August 27 2007 1102 AM Leda 3 6 Tim SA7315M acct fo i K 6X 95 w 0 1 vac 2 0e 4 o 1 1 t 1 10 1 0 300275990 4 x o 39 XLL 0 ms Delay im umm hc 2 5 m EdiH PSYCJM C W Fvvrkmw39fwjok 0 L t Week2 Page 1 ManclavAugusl 27 2007 1122 AM CD Tum gal scaling acct 39c act t WORM A l Mm 7mm 9 30 at I I j 390 1 X we a 0 WWquot la 1 Q new A141 3 d m h wg vr rth4w Week2 Page 2 Manclav August 27 2007 1128AM TWO StaM45 Lav BMW 360 2 mar xzc3 Cc 5610 W l 145 quot i l abe l 39 t LL f 6y W 39 H 6 ZLtD lift z 39 L C b 411649 W47 A Week2 Page 3 MondayAugust 27 2007 get 2 36106 129 MOdALLDChO39Yk MLth Z H m ssase sf ml 4 W auLdto swat va 39myl LWM oeHa LekHz X069 3 warrior Esra sinw soicl 0350050 00 CAVva 4W In us I mkHuwc ltl l0 kHz 36 meowtn stay x466 A cosmmt mess am 1x 39 W Hvd km L WHm f LE aring 111w w mum cw KW new Week2 Page 4 MondayAugust 272007 l1145 AM ML JL H Ici 3 HMch Ham 1M PM 39inxik All398 mm my cum we DvruLSL 5799 66 1 00 m tso o Mme45 D j 6amp3 out 2 L Fin e Myy 6m 3 61333 lam Week2 Page 5 91 Probability and Statistics Measured signals exhibit deterministic predictable and random unpredictable behavior The deterministic behavior is often governed by a differential equation while the random behavior is modeled in terms of probability distributions and described with statistics Example Additive White Noise Electrons in wires resistors and semiconductor devices are always in motion When no potential is applied across the device electrons move in all directions at different rates Brownian motion The measured current and voltage fluctuations are often referred as noise The noise signal is unpredictable from one time moment to the next however noise signals can sometimes be described by their average power Use Matlab to plot a 60 Hertz sinusoidal voltage deterministic with an average power of 1 Watt measured from a circuit with a noise power of 25 Watts Assume the probabilistic model for the noise is a Normal distribution Show the magnitude spectrum of signal plus noise and histogram of noise Increase noise power to 16 Watts and repeat Explain what you see 92 Matlab Code t 0102314096 time axis sampling rate 4096 Hz f 409602102311024 frequency axis for FFT noise25 sqrt25randn11024 Create noise array with average power 25 watts sig1 sqrt2cos2pi60t Create signal with average power 1 watt figure1 plottnoise25 xlabel Seconds ylabel Volts title Noise Create a histogram of the noise samples with 20 bins n x histnoise2520 figure2 barxn xlabel Volts ylabel Frequency of Occurrence title Histogram of Noise Voltages Create and plot signal plus noise signo sig1 noise25 figure3 plott signo xlabel Seconds ylabel Volts title Signal Plus Noise Plot signal in the frequency domain spec fftsigno409625 Scale by sampling rate and segment duration figure4 plotf1512 absspec1512 xlabel Hertz ylabel Volts title Signal Plus Noise Magnitude Spectrum Increase noise power to 16 watts and plot noise16 sqrt16randn11024 Create noise array with average power 4 watts figure1 plottnoise16 xlabel Seconds ylabel Volts title Noise Create a histogram of the noise samples with 20 bins n x histnoise1620 figure2 barxn xlabel Volts ylabel Frequency of Occurrence title Histogram of Noise Voltages Create and plot signal plus noise signo sig1 noise16 figure3 plott signo xlabel Seconds ylabel Volts title Signal Plus Noise Plot signal in the frequency domain spec fftsigno409625 Scale by sampling rate and segment duration figure4 plotf1512 absspec1512 xlabel Hertz ylabel Volts title Signal Plus Noise Magnitude Spectrum Noise 1 0 5 Seconds Volts o Signal Plus Noise 1 0 5 Seconds Volts Frequency of Occurrence Histogram of Noise Voltages 93 05 0 Volts Signal Plus Noise Magnitude Spectmm 1000 1500 Hertz 2000 2500 Volts Volts Noise 0 15 Seconds Signal Plus Noise 0 15 Seconds Histogram of Noise Voltages 94 Frequency of Occurrence 10 5 0 V olt s Signal Plus Noise Magnitude Spectmm Volts 3 b 500 1000 1500 2000 2500 95 Some Important Statistics Sample mean of N data points xi m xx in Maltab m meanx x is a vector of data points Sample variance of N data points xi s 11 12 in Maltab s2 stdxquot2 x is a vector of data points Sample standard deviation of N data points xi s A my in Maltab s stdx x is a vector of data points Note the average power of a signal with zero mean is close to the sample variance for lar e N the are almost e ual 139 L gt i g a y q Jig N 1 N 96 The histogram of N data points x Datum values are compared to a set of cells or bins which are a subintervals covering the full range of the datum values The number of data that fall within each subinterval are counted and plotted The xaxis corresponds to the cell and the yaxis is the frequency number of datum values within in the cell interval In Matlab nc histxcnumb x is vector of data points cnumb is the number of equal interval cells over which the range of data points is divided n is a vector corresponding to the number of data in each cell and c is a vector containing the midpoints in each cell To plot barcn Pseudorandom Number Generators Complicated and nonlinear deterministic expressions can be used to generate a sequence of numbers that appear random 1 uniformly distributed values over the interval of interest 2 no periodicities or repeated patterns in the sequence 3 small changes in the expression39s parameter seed result in significant changes in the sequences Two random number generators exist in Matlab gtgt randrows columns Generates a matrix of uniformly distributed random numbers between 0 and 1 These numbers can be scaled and translated to any interval Example gtgtur 10rand12000 5 50 uniform numbers between 45 and 55 mm gtgtnx histur10 barxn xlabel39Bin Range39 ylabel39frequency39 gtgtrandnrows columns Generates a matrix of normal distributed numbers with zero mean and standard deviation 1 These numbers can be scaled and translated to any mean and standard deviation Example frequ en Cy w o o gtgtnr 5randn12000 10 Normal numbers with mean 10 and std 05 gtgtnx histnr10 barxn xlabel39Bin Range39 ylabel39frequency39 85 9 95 05 M M5 12 10 1 Bm Range Cominuous me Convolu on 24 Th inpm m kind m Impulsc mpumc M uf a Lumiuuuus imc LTI Mum are given by I LIU Ily39mumagta umnnm me mupw w 1 m 176 12 Conwut me quDul vm in La 411m 1quot n w yu rx1ll lmm 77 41 Functions mum mi mc shnwu m Fig 240 m lt114nd gt0 mm Fig 24m w u limifnr v lt m m m h I my nui uvcdap while mr gt n they Iwcrmp mm 1 chcc or r ltuym A For 7 gt n wz haw vzf1quotquotquotquot17gtcquotquot t a Thxm m mm mm u ompul my m l whau 264 7 By Eq 210 y1zrlxtfmh71v Funmuns hm and m 77 am shown In Fg2 Mb or mm 1gtU AgAm mm Fig Mb we set me or vlt Ur l1md m A y do um nverlm whxln m l gt a quotmy nvc ap from 70 w r A Hum my vlt r m oI or1gt0 clmc m w w m a Thus we can mu m aumm mi 1 I m in ul mm which mme same as Eq 254 u m n w n may Au 7 g 14 Can mm mm ha output m far 5 cnminuuuHimc m svslcm whuw impqu reswm and m Inpul x I By Eu 11 uncuam 7 and 1 sec than my ltH n mm 1 m m m we we In D are my z 39ui no I F um Fig 25u we mamhm I M 7 a mom 1 as a nr1lt0xmd gt0 x mm m gt77 chllap ham to y whuc my gt u may msvlau 0 Hence rm rlt n we han mgtf M Equot39 AVE e M 26641 ym c cquotquotquot r14e 26 1 m u m Combme Eu Mae and 2 m w can wnl m 25 quotgt1 257 mm is shown m Fig 25er ya 5 Evaluate y1xrleL whcn 1M and M are ahwn in Fig 26 11 by an mahnicnl lmhnique mi N by a g rovhia method n mu Fin M m We mu mum M Ma mm in Immuan um rflluli ru1uIZ ThuL by a 16 hm 37x139h j7xrird1 lvu7u139r 3u1r7u1r7211x gtjluvu1rdr jimmy 72 v dr uhrsmr 77Ir quotTram7277 5m quotmun 7 r 33312 W W lt WWW iltrlt172mgt5 n U wrmws l n therwiso we can zxptew my 5 thmwu 7Uqu 7 3 Ah uuis mtr11721w72 1731417 gt 5uU SJ Jw z my 1 planed m Fig2 xquot hymns u may m 17 b Funclinm Mr W and 1m 77nlmmy Dr Marc wlues or I am skelehnd in m 25 mm Fig 23 w 5 that m and M1777 dn um mum fur lt0 and 1 s mi km ylU In t ltu m I gt5 1 m omanmmla m and M runamp Thus tompuling the am undcr m rectangular pulm 5 Hum immuls we obtain wruch a plmlcd n mg 29 Im I I U I I I I I I 1 I Y Mid quotnun Io Ilto I I 39 7170 2 I o 2 l 4 7 me ux U 1 I I 13911113911 I VHII S JV mm II 1 z NIL LR JU CT LTI System governed by leferen a1 Egua on um Consider a caminuuusnme sysxem whoa inpm x0 and umpul ym m mm by at 11yxr 21m Mme n k 1 osmium a nd yumm m mum condition yonyu and m 4mm 2101 M Exam my in mm af ne mman and mmmm remunscs m ymM mm mlminn which mm d d lquot umr 0 2117 A ume um mgr1r gt0 1m Subnimling Eq 1104mm Eu 2 m we 9mm AMquotquotuI K2quotquot mm mm vw mm A Ka 7 a and K y ly M 09quot rgtn 2m To ohmquot mm w assume m 43 quot Subsmuling ux inm Eu 2 703 um WWHBI39 AruE U rm which we have if 7a m MU 4 3quotquot Culnmxnnyy yltn nu mum gt K yHKrquot nib 1gtD 2106 From Eq 2106 and me auxillaly candi un 340 ya we abmin x gt0 21127 10 I lt0 we have ah u and Eq zmlwmmcs Eq mum Hench may yltu l39mm m mum Lundxliun m in m obuin L g 1 lt a 1 m Comhnung has am and 2075 m mu m alignmad m wmmr yum LLLnrinpul Icmunsu 2m yhU unmlm mange as 2109 when 211 szby 22L Consmcr me symm m th 220 1 5mm nun me syucm i nm linear x y0y g u m Show that the systcm 3 hncar n vU7 x a Recall mm a linear mm m Hus Dummy um um mpm producers zeru mum m mu w MKquot r mm h Wm 1 na see mquot gt1 n Thus mm mm 5 nonlinear u my y u 1M I NJ ixuwn 39AS fullvwa L A 0 and x10 h Iwn myul new mu c mmquot 3 human Tlu a 1n slgnals and m and 51 be m mxrnspondmg onlpuu l39Im n d m HUMer 121m mm Tummazu 2112 MW nw auxiliary mamaquot u 211 founder 19am390111 mm rm Irr m M hy a and aadmg we see um vi 1 myyrrHnmm mums ma Merznuax cquanan mm u z md also mm 11 2111 M my mmmm lt0 Thcwmrc VUY n ma umpm wire pnndms ln dz and thus Ihc gamm u Imsah 222 213 ul rm condnion no u ulm Cnnsidcrmc system in Pmb 220 Show mm the implies that h systcm s timevinvariam Lu mm h be njpunsc m an mun my and mu gu 2m 4 Tm y I 4 mm m le5 4 ml Mn 0 u rm r r gt u c y n man 5 1117 nu yzm mum musfy quothm 11y1lx1l 1115 m and min um Naw mm Eq 211 w m 39 II TI 7 7 7 I 14 mm m u we m mama mm by 21mm haw 7 770 m pquot 11m impulse rzspome hm snuqu satisfy m di creuual mnaliun m m ahu am 212m The homogeneous solulion my w m 2120sam cs m l quot nnhu mu 21213 4 To obtain mm W mum mac Submlming mi inm m 2121 gjm mquot we Hmquot u from when w hm 1 My equot39u 112 m r r m 1m V111m39 u1 1m To nd 01 constant c subsutunns Eu 2 123 mm F4 2120 w obtain 7 uu rl cquot39ul5 II 7 7 dam or face ul u m m 115 39umam Using Eq 1251 and may ma mm mummy vacuum an 5quot d c 39quot l 6l5l 5 mm 1Tnuxuexmpulsg lawns1 5Mquot by mc mn 2120 Con dcr he syslem m th 220 wxm wau Find me Step spouse m nf the system whoquot Mmg m lmpulsn quotsputum h M Find Ah mp rmponsc 1 m h impulse rcspunsc 111 abuincd in Prob 223 4 1 Find me Impulse response Mr from m a 1n th 22 A Kc39WU Swim K LhD Au obtain XII m and hm vma L Thux swing KL hl1 and 39L yquot m Eq 21179 we 0min Lhu smp response uygoiruum 2125 L sing Eqw 112 and 2174 in Flat 2 L1 h step veltpunlte I is given by smf h11hf39xwwmd7 14 which is da mm D Eq 21257 L sxns Em 113 and 2125 In mnnlse response M B aiwn by quotUJ b rlic39 1u1 hlrHE z ry n unwarth ul Using Eqs 12 mu 150 m m 1 1 1 Imam Lir mjrSU mum Icquot39u Thus which ii lhc same as Eq 1 124 CT Fourier Series 54 DunnLine me complex expcnenm Fauncx Enos mpmmmmn for each 0f m following ignals a xlmsmal 1m magma m llcos21 41 III I 1054 sin 71 2 I 5m m r oxmm we get ms W e quot e 39 w Thus m mmplex Fund wcl uicms or L0 mquot n 1 u ukltukh1 by In I Imllar Inshion we haw Thu fundamcnml mauhr flvqucncy mquot of I is 2 mm w ca 2 NW I Z 2 Az imm I 72 H z u in w r w 2ng z Tm m Lumnlcx Fourizr augmciznu M may r An 5 Tu 1 5 Hi 0 w a nymuesun fmm th Lmhc Eundumzmnl permd n ufxis 1V and w mm2 Thus 2 MW 2 mm A A Aym mng Euler39s mum w 1mm 1 rvm341viin6 n qpur fl3mg m I x l a may 417 2 mm 2 2 and an mhcr pk n From mm mm m mndamem period T of w Is 1r and u zwr iz Thus x1unquot l 2 MW 2 w m kw Again nmg Ema lommla 2 gm e and all my 5 55 Cansidu m Dcnadic suuarc wave xm shown Lu Fig Ls I Dcrermme the mmnlex expunenlml Faunur sum of x11 11 Demmmn me uigunummic Fourier series or my A 9 u a I m y mu m kl In 1w Z LM V 1 Using En 51024 We ham I 7 h I m 1 4 mm A T quotquot BMW 1 f munL1 since min m t 1 nm n k 2m n A 2m 4 I 1 w I IV 4 WEEh mm j 3 mm A 3 ch 1 m ms and we chmquot mm A A a 1 xtl7 12 2quot bi rmm Eqs 5 105 514 and 512t m hm A A om N mvw 2RC m quot blmv 39 m zmu39 Suhslimmx quot15 mines in Eq 53 we as sin2m41ul A 2A 39 1 quot39l 39mazmu l 1 51177 AJAquot V3 5 7 0014 44 7 W 2 w Isln 3s 1 5m 58 Cumislct Hm periodwc impulse min 59 shown in Fig 512 and de ned by 2 517 km 5113 51M Fig 5r a7 Denimm me cumplzx expanenli l Fourier series of mm b Dclcnmne he trigonometri Foulin serie of 5r a Lei M 117 MI Zzw m w K E Since an i invnlvcd we Me Eq H mm m delamin the Hunter me icxclm and we mm I J A l u 51 m Iv m MN rm Henruum ge 1 x ml 2 swim E SIN k 1n L h m 391 m V 5mm E 3D3kmnt1 b 3mkwul m Sim Mn iscven hf and by Eq Liam quotA m zivul by 1 r l a ij msmcnskumxr 5116 Tth we 3 5 A I L 7 5117 hi mm up 0 L 512 If 4112 and 21 are periodic signals mm fundamcnml period 1quot and Ihcir complm Fourier mes expressions are mg 3 helm M 2 WM a and am ha amused alt A K t W a 7 U 2 k n W wh re c 13 men by tk 2 Amturm 51291 Now A 1 r n A EMA1L 17 upmm x Tum m is pcrlndlc wun mdamemal puma 1 m 2 mn n 11m Iutnxzrcquot d 1 w VA 1 rms JW 1 rpm mm X quotK 1 am pk W mnwwm y n and he mm m buukuh is mun m a CT Fourier Transform 521 Find rhc Fauncr msrom 0 he signal Fi swan x1e 1 u gt 0 Signal xul um he rewmlen as N r AgtD I m m ya 1 Flg 5 1 W m in mm mmrmm xltmf c ltmtdr439fm39r1 rjmdl 1 rl wyrdlAErwin1rd Heller we lay mu In at 5 m m me39u lrszmm mm at xr a mwn in Fig 54m 512 nd me Fourier transform of me xignax Fig 54901 1 1l u wz From Eq 51381 have uM e H Nva by u dualil nmmny 5 4 we hm 2n n 139 H 11r1quotquot 39 1yre MM xm Km n 1 Fig 519 um 11 and us Fumlcv nansfurm Di ding both smcs by 211 we chum1 5130 a I n The Fauna mfurm 1mm my is shuwn in Fig 549m 519 mm lhu Fuunur namrum uf In Sigmaquot rumiun mm Fig 524 wmclv 1s de ncd n sgllUFr l Oquot 5152 t i l 1 lt D The mum ruminn gn1cm he cxnrnwd sgnm 2um Uamg Eq I 30 we buy 4m 72611 A 47 mm i11 524 5mm lunclion m mu H M 1 Then pnlying m di mmiamn pmpern 555 w mm 2 iuxm yzs12x1m m Ham 2 agume al 5153 andlhcmfolcih 39 39 purcunhg Nun 39 functmn or u Prob 5 41 532 Using me Iimc convoluliun theorem 5 53 nd the inverse Fauna nnsfm39rn nf X031 ma Half From Eq 15451we have n ul u AM 5155 Now gt f 39 quot a W avw My Thus byme um cunvmuunn memem may we have m quot 11Ixz39 ut HumeH quw f drzw39um u Hzmz gt I 5155 m quotu H7 u Mm 543 Find the Finder lransfun39n of a gaussian rmlse sums x m denninon 53 xm fl m 5162 Taking an derivaliv of mm sides arm 5162 mm mpm a m w my NW using I mlzgmuuu by DiflS formula f udv s um Em and IleIX 139 1nd marquot1 w mm 1144 imp144 and smug u gt a 11m w gel mm u T 7 in Solving m above lcpamble LliEerzmiu mumquot m rm w 5mm m41quot mm mm A 5 an quotmum cnnsunz1 u cvamaxe A we pmcem as munws Selling u r m Eu 5 I62 and by n chnngc nlvxriahlz M hzvz A 2 7r 4 5 4 Tu f e 7 Subsximring his value 039 A mu Eq 51637 he gcl Xm 39W 5164 2an we have rulino W 5155 we mar mu Fourier LmnsKunn n a gunman pulse Mgnnl is also I gaussim gum in the Inqucncy domzl Figur Ha shows me relauunsmg 11 Eq 5151 mm W u y V Fig 525 Gum puls and in Fnu arunn lu39m 545 Consider a cominuous hmc LT ystcm described by d A d 2ymxm 5165 Usmg mo Fauna mustoxm nd ma ourmn m to each o m fullwwng mum Sign n X1c u1 m An m Taking me Fourier uamfumis 0 Eq 5155 We have fun imm xm Henm 2w me Eq 1155 Y 1 ms Him and V X H i l quot39 39 J u mtum Theyefmc JIKquot Lquot1 ut b From Eq15154 1 Kim 1r6tm I Thus by Eq 566 1nd using lb nanizlraclion uxpansmn rcchniquc w have V X H s 1 m 7 m m7 w w m 1 I x 7 39J 2 1 quot ma 1 5 7quotJ Ef izw 1 W317 whey we med xhe rm mm flm 1 imam Eq 125 Thus 1 m um 7vquot no licd39wn w OMEN that be when mmkm mude 139 mic in mi we beams nf the Fourin lrnnsfnn39n m um Consider me LTI yncm in hob 545 If nhe inpm m is me pzriqd squar 77 oulpul yl Nut um xUD Is the same xl5hown m Fm 5 P101355 Thus mm A u 39 1n and a mm in Eq 51126 We luv m 1 z 7 mm 5quot Xi an mm mm Prob 545 H 1 Hk Hk wrMma mu 7r 2ka Thus m ELI 574wc chum m 39 ym5mm m V I lll2mn quot quot l mu H K 2m 1 up In D INquot H 67 The hammm arm a yUY u glvun by 3 U LU rm Dh E Dh 1me ion m was ur 5 he ampulud at m km hznnanlc wmpnngm ul U By bah illl arm 0 In n and 4A are relmdhy 1 2w 5 may Thus imm Eq 51m wnh P1 0 we ublnin I0 i n 1 Hr wuh m I we nbxnin n 2w 02 10 m2 11 551 Gunsldar an ideal lnwrpnss llar with frequency response Hm I M lt m 0 M gt a The Innur m this lm 5 Sill III w x 11 Find me cuipm ya fur u ltm mm m umpm yI fur a N15 6 h which case does xhc nulput sutfcr dismnion E L117 me Fq u nmpmb S ZIIJue hm mu m M lt imigta x 4i i M a new when n ltm w ham vim XwH1m Xm TMK smm I in When a gt wm hm Vim XwlIiw a quot51 ThuSi Sm i i imam m p in case in that is when a yiixm md me mm docs um picame any ii39l lhh h A 553 Considu an ideal JDwrpau iiimwim frequzncy response 1 M lt w m 0 iwi gt Air Tm input 0 ihh mm is m pcnudlL bquarc wave shown in Fig 5727 Find in output m1 smug A 170 Mm mo WT w in Eq Lil 7llFmb is we gs in V r0 invl sm3m Mm lm 1547 m xiiwnnsc znguiai risquequotng m gmmcr than 4r rad Tnueiom 20 2 ya 5 7 75mm 7 sin3m 554 Comma an ideal iuwpm mm wuii freuuency rcxpmma I m ltmL m o imi gt in Th input in mi mm is X a 141 ma the value ai m such hat mi iicr mm exactly onehair of Ihe noxmalized may of the input signal xm rmm Eq 15 155 1 x 241m 1m w w mu xumm MA 1h nomnlucd L nCl s of I IS 4 1 E x 39dl 41 KI M quot Uxmg Pancul idcmily 16 I11 RomaineI may of ya u 4 x 5 mv n ymum rmm mm W mum 61 Orthogonality Two vectors are orthogonal iff the dot product between them is zero c fx f x cos6 0 Where f denotes the magnitude or length of f equal to the square root of the sum of squares of its elements and 9 is the angle between the vectors This can also be computed by multiplying the two vectors element by element and summing the products Example Determine whether or not the vectors are orthogonal a x1 10f 1 110 b X121f 113 c X 17011 031 Two continuoustime signals are orthogonal over an interval l1 5 iff the correlation between them is zero on that interval C ffzxzdz 0 Determine whether or not the signals below are orthogonal for l e 0 a fl cosa0 I xl sina0 l 0 b fl cosma0 I xt cosna0 I for mn eIntegers 62 Orthogonality Relationships for Sin and Cos 2 mteger mu1t1ples of 7 a 0 a Icosa0 l sina0 ldl O for I 0 b jcosma0 l cosna0 ldl O for T0 integer multiples of 2 7 n im nm eintegers Tn a 2 c jsmma0 l s1nna0 Zdt O for T0 mteger mu1t1ples of l n 7 m nm e1ntegers TD 0 0 0 T d jcosna0 l cosna0 Zdl 3 for T0 mteger mu1t1p1es of 2 7 n emtegers Tn a 0 T e Ismna0 I smna0 ldl 3 for T0 mteger mu1t1ples of 2 7 n e1ntegers Tn a Use the above relationships to derive the formula for the Fourier Series coefficients such that they minimize the mean square difference between a signal xt and its Fourier Series expansion are 2aquot cosna0l 1quot sinna0l In other words find do an xt a0 261 cosnaot b sinna0l dz and b that minimizezj Tu 63 Existence of Fourier Series and Dirichlet Conditions In order for the Fourier Series coefficients to exist for some periodic function ft the following condition must hold f l dt lt 00 In order for the Fourier Series to converge to a periodic function fl the coefficients must exist and fl must have a finite number of maxima minima and discontinuities in one period Distortion Example Consider a pure sine wave at f106 Hz passing through an ampli er circuit The output is sin27rft clipped when the absolute value exceeds sin7z 3 z 0866 Determine the Fourier Series coefficients plot the waveform using different numbers of components and compute the ratio of harmonic energy not located at f106 This is often used as a measure of distortion for amplifier circuits 1 BIBO smbumy Vcrify 11 131130 slabilily condition 121112211er cammuausimc LT1 systems 39 ITI v 39 h 1 A 1h 1 Mr sh 2111 277 Thcn using Eq 21m we have M H f l1 x1rdw ij 11117141711141 f 1h111x17 14skf 11mm sinct 1A1 7 Sk1 from Eq 277 Tllerefmc if quotIt Impulse rcspuns is absolutely inter grahlc lth is f 1 drKltm men M171 skl k and me sysmm is 13150 smbk Z BIBO Stabxhty w 39 m n I impulse responses of me syslcms m gwcn by mm and 131 respectively and hlu39l39ul mm2e mm a b Dcxermin if the overall ystem is 31130 stable 2 1m ym m U1 Fig H7 1 Lu wUJbElhaoulpmul39Lhelirstwsmm ByEq26 wrxrh Then w haw yl w1hi xl xh 1 hzl F Find the Impulse response ht of he overall sysmm shown m Fig 27170 278 279 y11lh 139hl FYUPMV l39hamfoxu u Impulse response of me averall syswm is gwen by my amp1 nus mm the given mm and 112m we have m wryZn A yyd7fj hm 42quot rwl39wm I1 Using me ahavc MIL w have Imam r2Kwe1TAr2f d z177xltm Zu Thus he syslcm EIBO slahlc mpuurrd 2r39cmrul f v n Zequotquotquot u1 7tlr 11 1250 231 3 Inverse Laplace Transform Find lhe inverse Laplace transform of X3 Harm Rcgt0 Wecznwrin 524513v2932j342413 Then x 55 13 5x 13 m1 awn nuzqmnzusy c u r 2 5 Y23 U4 where 11 1 calt 213X l1 553 54 2413 r7727 Thus 4 72 11 V 13 The ROC uf X13 is R6157 gt 039 Thus rn H a ringsidud signal and from Taich 3l WK chum u jc quot39ou 7 w jJEK39l39JWMUI Inserting me idenmy a mossmsmm imu me above expression afler simple mmpuminns we chum x1 Add 11605317 sun31ul l 7 mm sln31ur Alwrnam Solmicn We can wrilc X15 1 MI 5x 13 351 4 I As before by m 331 we ublain clxxltxu 513 m Thenwehavc 55 SI13 1 7 11Ay13 7 4u Thus 1 XJ7S24X37 l 1 3 213Z Than mm Table 31 we chum w quot1 gtt39l39cnslluU quot39s n3111t l lcossin3lul 4 Make sure you know how to compum Routh test and Jury Lest 5 Zrkansform Consider he sequence 011N1agt0 u ulhcrwise Find Xzl and plol the poles and zeros o X1V By Eli 431md using Eq 190 we gm A H M2 3 Mr 455 n F n E4 455 wt sec mm more is a pole or 1N7 iim order ni o and a pal ni z u sinss xln s niic sequcnce nnd is zero for n lt n Ihc ROC 5 izigt 7 The N mms oi me numerator polynomial 2 4 2 11239 kni L N 1 456 The mo 4 k u cancels hc pol m za The remaining zeros oi Xz at zl JR Mnmm M57 Th unlncro mm is shown in Fig 44 wixh N B inn i m nr cr m Mann uncd ms m 44 Polcrzem plot with N a 6 Zikansfor m A nite sequsnm xn is dc ncd as Xl i532o4 3 T Find Xz and its ROC me ELM 3 and gl vtn rlll we have 1 Xz twp 3 Juzquot l 1 r211x1zx x1zquot lezz3n3zJ 5z1272ar 735 Fsrz mm nr39 39 hrrm39 39 39 39 mlwill cmwugr an Ihm xm Include llmh mulin mer M y ml quotcgaflvr pnwerlt nr 7 Tm 6mm me mull of Funk 62 wt conclud lllal the ROC of X is ll lt zl lt m 7 z transform you can Ignore the le rslded sequmces In parts 01 and 6 Find the z translnrm and me associaled ROC for each bf the following sequences ll Xlnl5nin b Xnunnn c Xnuquot uul I xnun a Xn n39 u n K From Eq 4157 sn H1 all 1 Applying m limeshifting pmpcny 415 we obtain in u lt lzl no gt o Mquot quotVJH zkm lt0 W b From Eq 416 ulnl H lzlgt l Again by he umslllmng properly 418 we ohmquot urn un n gtz quot 7 1ltlzlltw 472 C From Bus 43 Ilnd 4 10 aquotun v lzlgtul BY Eq 420 we obtain nquotquotunlezzz lulltlzlltm 473 1 From Eq 416 ulnl a lzlgt By um imrvur nl nmputy 423 we obtain W l w 1 4 74 lt quotI quot1 lzil 1 2 c From Eqs 45 and 410 quot H 11111 I zgta Again by m Aimemvcrsal properly 423 we ubtam in 1 1 a l H UV Izlt H 475 8 Convoluuon propaty m z wansfoxm Verify um convnlulion properly 426 mm is xnxxzn lezXzz R39RnR2 By de nilinn 235 ytnlxan1w mi nlklink Thus by de nmun Lu YUP innmekAQIw 7 k i nk1 i xztn kxz39quot Noling Iha lhc cm in paremhcscs signal 11h k hell by me Iimcsh n he last cxpvessiml is Ihc xranscml 039 h shined ing propeny u 18 we have YzkxlkllzquotX1gtlk Ikz X2X1zx win an ROC um cumains Ic Imersecliun or he ROC uf X31 and X z It a mm or on mnsnmn cancels 3 vol of the mm m ROC of Y may be larger Thuvi we conclude mm an39ll xzn X zxlz Rugan Impulse response and Inverse Zrtransform The cumm 39 39 LTI xln is unu a Find the impulse response Mn of he Wei 1 Find 1h Output yrr when me input xn is 9mm 2 a XIVquotIHHX1 gtl quot l yln12 uinie mz izigtE Hencc lhc system function 2 is 7 Yz 22 i 1 511 X up Uung pzmaIIraclion cxpansmn we have 2 z 1 where 39nius H16 4 Taking uie inverse zxrensrmm or Him we nbmin Iin 2 Mn 7 4 uu b xn Huiniami Then yzXzHzx where quotnius Taking the inverse Hiansfomi of Hz we obtain Yquot 639 quot8 ul 10 Dxffamce Equauon Fox each uf the allowing difference equations and nssociamd input and mi wndxlinm dclurmine lhe umpm yn a yn yln 1xnwam xm I y 1 b Sybil 4yln 1 yn 2 xn wnh XIII z a xlrllHXz z Hum Taking U11 unilamral mansfmm of lhz given di uence eqununn we 0min 3Yz ru mznylvll z zYlz 1quoty v1 XJZ Subslkunng y I I y 2 2 and Xz ink 11 above expreuion we gel 2 3 42quot 472 2 2 4 or Thus H 43 I swim chccy 41 The ZeroState Response The system state refers to all information required at a point in time in order that a unique solution for the future output can be compute from the input In the case of LTIC systems modeled by differential equations the state is the complete set of initial conditions on the output term If the input of the system consists of the superposition of M functions f0 goof r and yt is the response to individual input components fl then the resultant response is given by W gala r Example Given a system described by D 10 yl f I find the zerostate response for input described by fl Show that complete zerostate response is 8 ya 21 eXp 10Z u l eXp 10Z 8 10 10 I 8 42 Example Find the impulse response in the previous example Should it equal ha 2 1331 pg 10010 l exp 1OZ 8 u 8 Example Find and plot the zerostate response of the system used in the previous ul ul 8 5ut 01 ul 8 01 let 8 2 01 8 8 example for input fl Ans yl 1 eXp 10Zul 1 eXp 10Z 1uZ 1 51 eXp 10Z 1uZ 1 1 eXp 10Z 2uZ 2 35 1 use unitm file from web page t 0011 s 1exp 10tunitt41exp 10t1unitt15 1exp 10t2unitt2 g plotts39w39 xlabel39Seconds39 1 ylabel39yt39 05 title39Response to ft39 11 02 43 Response of linear System to General Input For a general inputfl the output of a linear system can be determined from its impulse response and a special integration operation call convolution Consider a general approximation for any function ft through a series of pulse functions ftwflf wfkAPlZ AA 22 where 00639 m2 W 33 What will happen to p Ah as A goes to zero What will happen tofA as A goes to 0 01 02 03 04 3n 05 07 zero 7 7 Remember from previous examples The response to a sum of inputs is the superposition of their individual responses Apply this to an input in the form of J20 44 Let hA be the response to p then the response to fAZ for a LTIC system is of the form W kAVIA l kAM Then if A goes to zero hA approaches the impulse response hl fAl approaches t and the summation in the above equation approaches an integral yo Ionaea rgtdr where 1 denotes a continuous axis over which the input function and impulse response are multiplied and integrated Note that any point in time can be computed from the above integral which is referred to as the convolution integral An efficient notation from the above integral is W if7hl TdT fl hU 45 Properties of the Convolution Integral 3 117 0 rgtdrzltrgt WI The Commutative Property f1t f2 1 f2 1 f1t The Distributive Property f1t fz I f3t f1t gt f2l f1t gt f3t The Associative Property 21 r L r 1 2 112 m mm The Shift Property Given f1t gt f2l Cl then f1l T gt f2l CZ T f1lf2f7300 Da f ZVLUZFCUZ Z Convolution with an Impulse fl l fl The Width Property If f1l and f2 l have durations T1 and T 2 then the durations off1lf2l is T1 T2 46 Computing the Convolution Integral Analytical Computations For a special class of functions the closedform solution of the convolution integral can be computed These functions can usually be expressed in terms of exponential functions so that the land the C can be separated into factors Example Convolve expalul with expal cosa l 6 ul Ans whinw 6 sin6ut 7 0140 644 6140 a Useful relationships from Euler39s Identity 0 COM eXpJ lt9 eXp J 6 2 Sin6 eXpj 6 expj 3901 0 05 0 1 J For exponential functions mixed with polynomial terms integration by parts is typically used See convolution table 47 Graphical Computations For simple ramp and steplike functions a series of pictures can be drawn out to help perform and understand the convolution operation Example Convolvep1l with 1920 where these are the pulse functions used in the previous examples 190 Tprdr 1 Sketch an example of p1l 239 and 1922 on a 39caxis for llt0 say I 1 What is the convolution for ZltO equal to 2 Sketch an example of p1l 239 and 1922 on a 39caxis for 03231 say I 75 What is the convolution for 05l 1 equal to 3 Sketch an example of p1l 239 and 1922 on a 39caxis for 13232 say I 15 What is the convolution for 1 Z 2 equal to 4 Sketch an example of p1l 239 and 1922 on a 1axis for 23233 say I 225 What is the convolution for 2 l 3 equal to 5 Sketch an example of p1l 239 and 1922 on a 39caxis for 39 say I 4 What is the convolution for 4 equal to 48 Numerical Convolution The most general way to perform convolution is to sample the functions and perform the operation on a digital computer Example Given the impulse response of a LTIC system hl exp 5l sin20 7r lul and a sinusoidal input with Gaussian random noise sl sin18 7 I nlul assume a standard deviation of the noise nl to be 05 with zero mean find the output of this system for an example of an input signal Use the 39conv39 function in matlab to do the convolution and the 39randn39 function to generate the random noise t 00012 Sampling interval is 001 which is needed to multiply the convolution operation h exp 5tsin20pit plotth39w39 title39impulse response39 xlabel39seconds39 s sin18pit 05randnsizet Add a series of random numbers to the sinusoid plotts39w39 title39Noisy Input39 xlabel39seconds39 sout 001convsh perform convolution tc 00014 make new axis for convolution which is the sum of the two function sizes minus 1 plottc sout 39W39 title39System output39 xlabel39Seconds39 Ndsyhpm 0 0 o o o o 08 06 04 02 02 04 06 08 1 seconds 49 im pulse resp onse seconds System output Seconds 410 System Stability and Roots of the Characteristic Equation Stability is a property of the system that refers to its ability to return to an equilibrium state after being displaced The equilibrium state for linear systems is zero If the system state keeps moving away from its equilibrium state after being displace then the system is referred to as unstable Since the roots of the characteristic equation belong to the set of complex numbers the stability of the system will be related to the position of the roots in the complex plane A LTIC system is 1 Asymptotically stable iff all characteristic roots are in the LHP Unstable iff at least one root is in the RHP Leftha1f Plane LHP Right half Mane RHP or repeated roots exist on the imaginary axis Marginally stable iff no roots exist in the RE or 5aXis RHP while some unrepeated roots exist on the imaginary axis A1l orjoaaxis N OJ 411 Total Response The total response of the system is the sum of its zeroinput response and its w state response These two quantities can be determined independently of each other For an n th order system with input expressed as the superposition of M functions fl impulse response hl and characteristic equation roots 11 x12 2 the total response is given by W g0 eXPMJ f1f fIfMI hU where denotes convolution and rk is an index that accounts for repeated roots If the root 1 is not repeated then rk 0 Recall the classical method for solving the differential equation involves finding the natural and forced solutions then the initial conditions must be applied to the sum of the two solutions to determine the coefficients For the above way of stating the solution the coefficients are determined before the zerostate and zeroinput solutions are combined Thus 0 If inputs change you only need to reevaluate the new convolution integral zero state response the zeroinput response remains unchanged 0 If initial conditions change you only need to reevaluate the new zeroinput response the zerostate response remains unchanged Introduction to Matlab l Starting Matlab File Edit View Web Windaw HElD D l l ggnml lq i ll a L39san Taalbax yam Cache m quothop mammalianquot m M m Tu W 5mm 21 quotmum Helpquot m a gap mm This is the workspace which lists all the variables you are using You may type the commands after the quotgtgtquot symbol This is the command window you can enter commands and data and the results are displayed here Fzmmin mm 924u4 217 m 72 This is the command history window it displays a log of the commands used t 1 4 sun o It might be different if you use an older version of Matlab mine is 701 0 Make sure you have Signal Processing Toolbox Control System Toolbox and Symbolic Math Toolbox To check type gtgt e p and you should see these toolboxes listed among all the help pages signalsi nal Signal Processing Toolbox controlcontrol Control System Toolbox toolboxsymbolic Symbolic Math Toolbox 2 Numbers constants variables and data types 0 Numbers gtgt 4 ans Constants gtgt p i ans 31416 gtgt 34i ans 3 400001 MATLAB variable names must begin with a letter which may be followed by any combination of letters digits and underscores MATLAB distinguishes between uppercase and lowercase characters so A and a are not the same variable Vectors gtgt V 2 3 4 5 6 7 V 3 4 5 6 7 gtgt v4 ans 5 gtgt W 220527 w Columns 1 through 6 20000 25000 30000 35000 40000 45000 Columns 7 through 11 50000 55000 60000 65000 70000 Matrices the formal name of Matlab is MATriX LABoratory so Matlab is very efficient in handling matrices gtgt A l 2 3 4 5 6 7 8 9 A l 2 3 4 5 6 7 8 9 gtgt Al2 ans 2 Strings gtgt w matlab is simple W matlab is simple Symbolic variables gtgt syms x y z gtgt X x X 3 Operations and functions 0 Simple operations all work as they should be Type help ops for a list of operators and help elfun for a list of elementary mathematical functions Almost all of them can be applied to scalar vectors and matrices gtgt 345451AO44 ans 16206 06743i gtgt sin1 2 3 4 ans 08415 09093 01411 07568 gtgt exp1 2 3 4 5 6 7 8 9 ans 10eOO3 00027 00074 00201 00546 01484 04034 10966 29810 81031 0 Entrywise operations are preceded with a gtgt a 1 2 3 4 5 6 7 8 9 lt comment Sign gtgt b 10 11 12 13 14 15 16 17 18 lt no echo gtgt ab ans 84 90 96 201 216 231 318 342 366 gtgt ab ans 10 22 36 52 7O 90 112 136 162 Function calls work pretty much the same way The most powerful aspect of Matlab is a large collection of functions commonly used in many engineering and scientific communities Here are some of the signal processing commands we are going to use in this course We will discuss their usages later r e s 1 due Partialfraction expansion f our 1 e r i f ouri e r Symbolic forward and inverse Fourier Transform 1 ap 1 ac e 1 lap lac e Symbolic forward and inverse Laplace Transform t f Create transfer function for a system bode Magnitude and Phase response of a system p zmap Pole zero plot of a system s 8 Create statespace model for a system s s 2 t f t f 2 s 8 Conversion between transfer function and state space representation 1 sim Simulate time response of LTI models e ig Eigenvalues of matrix expm Matrix exponential c 2 1 Continuous to discrete time z t rans i z t rans Symbolic forward and Inverse Ztransfrom re s i due z Partialfraction expansion in zdomain f i l t e r Discretetime filtering c onv Discretetime convolution but t e r Butterworth filter design chebyl Chebyshev 1 filter design imp i nva r Impulse invariance method for analogtodigital filter conversion b i l i ne a r Bilinear transformation method for analogto digital filter conversion 1 p 2 bp 1p 2 b s Frequency transformation 1 p 2 hp 1p 2 1p f irl fir2 FIR filter design f re q s f re q 8 Numerical Fourier transform f f t Fast Fourier Transform Symbolic operations most of the basic operations can be applied directly to symbolic variables To actual evaluate an expression of symbolic variables use the subs command gtgt syms a b c gtgt s sina bexp c S sina bexp c gtgt a05b9cl gtgt subs s ans 117289 gtgt a0b0c0 gtgt subs s ans 0 4 Miscallaneous Plotting xO00lzl yexpx XO 0 Ol l plotxexpx r xexp2Xl39g39 xlabel x ylabel y title Simple plot Simple plot 8 l l Saving the workspace gtgt save everythingmat save workspace to everythingmat gtgt load everythingmat load everthingmat Saving what you type and the matlab output to a text le gtgt diary footxt output amp response goto footxt gtgt m gtgt diary off stop logging You can also save all your commands in a le and execute it later on The file should have a m extension

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