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Adaline Pollich
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This 80 page Class Notes was uploaded by Adaline Pollich on Friday October 23, 2015. The Class Notes belongs to EE 422G at University of Kentucky taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/228327/ee-422g-university-of-kentucky in Electrical Engineering at University of Kentucky.




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Date Created: 10/23/15
EE 422G Notes Chapter 5 Instructor Cheung Chapter 5 The Laplace Transform Dynamic System 51 Introduction xt input yt output 1 System analysis A processor which processes the input signal Linear Dynamic system to produce the output we durum 39l b x I dzquot dz m 0 m anytbo Question Can we determine yt for a given xt Answer Use Fourier transform to convert the ODE into algebraic equation XO Yf Dynamic System Hf YfHfXf ylF 1Yf 2 Two Problems 1 Some common signals do not have a Fourier Transform Example What is the Fourier transform of xt e ur Xf JetuUW Wdt Jleteijmdt does not exist for any f as e w 0 blows up when t a co Even though xr grows unbounded as t gt oo it may still exist as an intermediate step in a larger system Consider the output when xr is fed into a system with impulse response hr 36 2e ur Page 51 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung yt Lama r3e 2e 74 d7 67136 287T 6quot e ut which certainly decays to 0 as t gt oo Initial Condition Problem Say we know the output yt of the above dynamic system is 5 at t0 Nowhere in the Fourier system equations below we could insert this information J YfHfXf yt F 1Yf 3 Solution Laplace Transform Even though xt does not go to zero when t a co but xte enough a 01 may for large Egt We will assume all signals are causal xt 0 for tlt 0 Egt Fourier transform of xte01 T060 e jwdt Txteo39jwtdt Ixte dt where s 039 ja Laplace Transform of xt We will see how Laplace transform takes care of initial conditions later Even though inverse Laplace Transform exists it involves more sophisticated concepts from complex number theory Just like Inverse Fourier Transform we will just use table and Matlab Laplace transform is just as nice as Fourier Transform Ys HsXs initial conditions yt L 1YS Fourier Transform of xtut can obtained by substituting s ja ie setting 039 0 in Xs 1 1 If the Region of Converge of Xs does not include the imaginary axis S a then its Fourier Transform does not exist More later Page 52 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 52 Examples of Evaluating Laplace Transforms using the de nition 1 Step function xtut Lut raw 0 1 w is J0 e d St 731w S to l Assume S S 111111 Hm eiReSte j 1mm l S S l S if Res gt 0 as Res gtHm 0 00 if Res lt0as matte w not sure if Res 0 as lime e mw When Res 0 define 03lms The integral becomes J e mdt r ute dt which is the Fourier transform of ut From chapter 4 we know that Fut i 275w 1a Note that for 03750 the Fourier Transform can be evaluated by substituting sj03 in the expression Us 2 Exponential xt e ut Le ut fe ae dt J eisatdt 0 1 w J e mail S 0t S 0 0 1 1 11th 8 Ha 7 S 0 S 0 1 Resllt if ImSZt hme e e S 0 S 0 1s a if Res gt Rea as e MM H w co if Res lt Re0 as amt H m 1s 00 27r5lmS if Res Rea as lime WWW Page 53 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 3 xt 5 t Recall that the impulse function is represented as a limit of convention functions straddling the origin To incorporate the full delta function we de ne the lower limit of our Laplace integral to be 039 L6t Tame h is 701 e e 87 07 10 2 e o cosat jsinat 1 No constraint on s 07 52B Region of Convergence ROC ROC Pictorial description of the value s where the Laplace Transform of a function exists Anatom of ROC Region where X s W IS infinite Region where Xs is finite but not 39 absolute convergent Region Where Xs IS finite In fact it converges absolutely see below Properties of ROC 1 It is bordered by the RIGHTMOST POLES ofXs A pole is defmed by the complex value s such that the algebraic Xs is infinite 2 If s is in the interior of the ROC then all s with ResRes are inside the ROC 3 It extends to positive infinity Page 54 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung In the interior of ROC not boundary not only does the Laplace integral converge it converges ABSOLUTELY Normal convergent or Xs is nite X s J xte 5 dt lt oo Absolute convergent fl xte dt lt oo We cannot prove that without venturing into a branch of mathematics called the complex variable theory However knowing that Xs converges absolutely inside the ROC we can explain some of the properties For example 2 If s is in the interior of the ROC then all s with ResRes are inside the ROC xte 5 dt xtexp Rest j Imstkit 0 Why b xU exp Rest lexp j Imst dt Note that leXP J39Im5tl VCOSZImStSi112IInSt 1 we have T xte dt xteXP Re3tVt which does not depend on Ims 3 ROC extends to positive infinity Assume p is in the ROC Let q be a complex number with Req gt Re p j lxte q ldt j lxtHexp Reqtpr s xtHexp Reptldt lt oo The second last inequality is due to the fact that l eXP Reqt lSl eXIX R6000 Thus q must also be in the ROC Example X0 z sls3 is s sgtl has two poles at l and 3 as well as a double zero at 4 Its ROC Page 55 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung Final note Fourier Transform if the imaginary axis 3 ja is entirely inside the ROC of X s then the Fourier Transform X f exists If s fa is entirely outside X f does not exist If s fa is the boundary of the ROC the Fourier transform must be evaluated by other means due to the presence of poles e g Try xt ut Page 56 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung Note Using matlab to find Laplace Transform gtgt x1 sym cosomegat x1 cosomegat gtgt X1 laplacexl Laplace Transform X1 ssA2omegaA2 gtgt prettyXl s omega gtgt X2 sym omegas 2omega 2 X2 omegas 20megaA2 gtgt x2 ilaplaceX2 Inverse Laplace x2 sinomegat gtgt dirac0 Delta Function ans inf gtgt dirac045 ans 0 gtgt laplacesym diract ans 1 gtgt heaviside0 Step Function ans NaN gtgt heaviside 13 ans 0 gtgt heavisidel ans 1 gtgt laplacesym heavisidet ans ls Page 57 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 8 Instructor Cheung 84B SteadyState Frequency Response of a Linear DiscreteTime System In this section we study various properties of the discretetime Fourier Transform As we stated earlier it is the same as the continuoustime Fourier Transform of the sampled signal x50 20xnT5t nT X5003 O K We have also claimed that it can be computing by evaluating the Ztransform around the unit circle provided that the unit circle is inside the ROC Definition ofDTFT Jae 7 Xz 20xnre W Note that we use XejmT to indicate the substitution zejmT I 4 I 4 K Let s first show that these two de nitions are equivalent Start with the continuous time Fourier Transform of the sampled signal X5010 20xnr6t nre dt 20xnr e f 6t nrdt 20xnre W Xe Page 532 EE 422G Notes Chapter 8 Instructor Cheung The primarily reason to study Fourier Transform in continuoustime linear system is that the input to a continuoustime linear system is a complex sinusoid of equency 030 the output is also a complex sinusoid of equency 030 with a phase shift and a gain governed by H0030 ContinuousTime linear xt e E System HO yt H Moe l X003 How Yjco I f A Ff 030 w 030 This form of analysis as you know is called the steady state analysis steady state as complex sinusoid is not transient It would be nice if the DTFT can do the same for Discretetime linear system or n DiscreteTime linear n xnT e 7 System HO ynT HeWe T And indeed it is true ynT 2L xnT mThmT Zloejwmrwmmm eWTZO e ijMmT ewnnTHeT Before we go on let s introduce a common representation of the DTFT based on normalized equency XeimT o o o o o a Dr 4T a W Xequot T Xe 2m r 01 maps me 0TET to re 005 r normalized frequency Page 533 EE 422G Notes Chapter 8 Instructor Cheung Remarks 1 Why called normalized frequency Given the sampling frequency UT and the frequency f 6 r can be more succinctly represented as When using normalized frequency r the DTFT is written as H e m Why ignore the negative frequency For realvalued hnT we can deduce the negative frequency from the positive frequency He 2UhnTeJmT Z hnTe JmT conjugate H e1quot wa j He amp AHequotquot 74mg Thus it is sufficient to show only the positive part only N The followings are a list of DTFT properties and common transform pairs They can be easily deduced from the Z transform tables TABLE 22 FOURIER TRANSFORM THEOREMS Sequence Fourier Transform XEm vlnl We 1 a39n byn uXequot bYe1 2 xn 7 n m an integer e W I39Xu 3 N xn Xel DJ 4 1H MW mew ifxln real I 5 mm jw la 6 39u yn Xe Ye I m 7 Xuyn T Xw qnemumdg 7r 7 ParscvaJ s theorem m 1 8 Z l nl2 E Xequotquot2dw quot7139C 9 Z rrlyn 2i Xz quot Y quot 11m Page 534 EE 422G Notes Chapter 8 Instructor Cheung TABLE 23 FOURIER TRANSFORM PAIRS Sequenc FuurIuTmnsIorm 1 MN z 51quot 7 my film 3 790 H lt 30 Z Miriam A 1 4 nquotu7I IuI I I I rm I N 5 n I VFW Z quotMiAlrrk i4 I n rllln url M lt I PTW I 7 lt I um M Hquot 172m aim L laul lt w 7 m 839 rm X or m lt luller 7 I OgligM lllltdt 1Iz Wm 9 M39 a mIIcmm SlumZ In Wuquot 2 217641 7 mn ancl Iyx II mung1 m V nomm 7 um 2ka 71quot m m 22m I 7x Computing DTFT Just like the continuoustime counterpart it is most common to show the DTFT in terms of its amplitude response HeJ l and phase response AHM However unlike continuoustime where we can draw asymptotic approximation Bode plot by letting mace we can t do that for DTFT as it is periodic Even though there are geometric techniques to draw the frequency responses based on the locations of poles and zeros they are beyond the scope of this course Here we settle with plotting the response using Matlab Page 535 EE 422G Notes Chapter 8 Instructor Cheung Example 1 A simple delay H z 2 1 To get the magnitude and phase response we can use the matlab routine f reqz In freqz the Ztransform is specified by the denominator and numerator coef cients 31 b2z 1 bm 1z quot 611 a2z 1 an lz quot gtgt hw freqzO l 1 Assume Tl gtgt plotwabsh Gives Amplitude Response HZ l l l gtgt Gives Phase Response Unwrap removes phase jumps gtgt plot wunwrap angle h Notice that the amplitude is at gain 1 and the phase is linear A system with response like this is called distortionless as they essentially keep the input intact The negative slope of the phase called the group delay measures the delay of the frequency component In this case izHe T i aif T ie One sample for all frequencies da da Page 536 EE 422G Notes Chapter 8 Instructor Cheung Example 2 General Linear Phase System H z 01022 1 04z 2 022 3 01z 4 gtgt hw freqz 01 02 04 02 01 1 gtgt plot wabs h gtgt plot wunwrap angle h The amplitude and phase responses are l D 9 D E Amplitude 7 Phase D 7 Response 7 Response D E D 5 n 4 D 3 DZ D Notice that the phase is linear with group delay 2T 2 samples for all frequencies It can be shown that the impulse response must be symmetric about the middle sample if it has linear phase Indeed it is true for our lter I 0 Linearphase lter is very important in audio and image applications because 1 In audio the perception of chords requires different frequencies to register at the same time instance A nonlinear phase filter delays those frequencies by different amount making the chord perception dispersed 2 In image color edges require spatial cycles to locate precisely at a particular location A nonlinear phase filter distorts the edges by fattening them Can a filter have zero delay phase Yes but such a filter MUST BE ACAUSAL For example Page 537 EE 422G Notes Chapter 8 Instructor Cheung Due to the symmetry requirement a casual HR filter can never be linear phase See the following example 1052 1 Exam le 3 a s1m le zero Y p p Z 21 063z l00422 2 gtgt hw frqu 1 05 21 063 0042 gtgt plot w abs h gtgt plot wunwrap angle h 1 D a D E Amplitude Phase D 7 Response Response 0 e D 5 D 4 D 3 70 e m 707 D Even though the amplitude response is similar to that of example 2 this HR lter gives a highly nonlinear phase distortion Example 4 A welldesigned HR lowpass filter can provide closed to linearphase performance at least for the passband More in Chapter 9 gtgt ba butter203 gtgt hw freqz ba gtgt plot wabs h gtgt plot wunwrap angle h Amplitude Phase Response 1 Response Page 538 EE 422G Notes Chapter 8 Instructor Cheung Inverse DTFT There are three approaches to recover xnT from XejmT Approach 1 Table lockup or convert back to Ztransform using 2 e or z e quot Exam le H e p 1 056 052 1 Applying inverse Ztransform we get hnT 05quotunT Using the substitution 2 62 we have H z 1 Approach 2 Explicit Inverse DTFT Recall the definition of DTFT Jae 7 Elgar Compare this with the Fourier series representation of a periodic signal yt yr neW k2 They bear a strong resemblance ytis periodic in t with period 21c030 H XemT is periodic in 03 with period 21tT Thus we recognize DTFT is in fact the Fourier series representation in 03 not t of XemT with xnT as the Fourier series coef cients In timedomain we can compute the Fourier coef cients a 0 ijkaibt Yk 2 yte dt Similarly we can compute xnT using the following formula xnT IXemede or with normalized frequency xnT fossXe equotMdr Approach 3 Numerical approximation by first sampling the spectrum with N points and then performing inverse Discrete Fourier Transform We will discuss this method in Chapter 10 Page 539 EE 422G Notes Chapter 8 Instructor Cheung Example Find hnT whose DTFT is H em jan Using the inverse formula we have hnT 55392mef dr Integrating by parts yield 71 0sejquotzquot aliquot r h nT 3927 fem J 12 127m 05 r70 5 This becomes l 1W 11 1 mrm hnT Ee 2e e e lcos7z71 n 7171 cos7m sin7m n 71712 Further simplification can be done n 0 l 2 3 4 5 6 7 8 cosTcnn l ln ln ln ln ln ln ln ln sin1cnTcn2 1 0 0 0 0 0 0 0 0 l 0 39 We can see that M w n and my 2 6n n 1 otherwise 717quot ham 2 cos7m sin771 n 717 0 n 0 1 otherwise Page 540 EE 422G Notes Chapter 8 Instructor Cheung 84A Difference Equation and DiscreteTime Systems XnT DlscreteITiigr e System y HT Definitions of various discretetime properties Properties De nitions Linearity HaX1nT bX21 1T aHX1l 1TbHX21 1T ShiftInvariant HXnTkT ynTkT Casual ynT depends on XkT for kSn BIBO Stable XnT is bounded gt ynT is bounded These are not new concepts you have seen them in continuoustime system In fact one can treat discretetime system as continuoustime system by using the continuoustime surrogates Xst and yst On the other hands a strong motivation to use discretetime system is its ease of implementation using digital logic Thus we want to do all things discrete In this section we want to show The output of all Linear ShiftInvariant LSI discretetime system can be computed by convolution with its IMPULSE RESPONSE Determine if a LSI system is causal and0r BIBO stable based on its impulse response Alternative representation of LSI system difference equation Classification and implementation of impulse responses FIR and IIR i J 4 LSI System Similar to the continuoustime relationship xt Ix15t 239d239 any discrete signal XnT can be written as xnT 2 xkT6nT kT Page 522 EE 422G Notes Chapter 8 Instructor Cheung Feed XnT to a LSI system H ynT HxnT H Zfzm xkT5nT m 2 x U H 501T kT Linearity Define the impulse response hnT H 6nT As H is shift invariant we have hnT kT H6nT kT ynT 2 xkThnT kT xnT gtIlthnT Thus We can actually simplify the convolution expression a bit First assume XkT O for klt0 we have ynT 20 xkThnT kT What if the system is also causal If so ynT can only depend on XkT for kSO This implies that hnT kTO f0rkltn lt2 hnTO fornlt0 For a casual LSI system ynT 220 xkThnT kT Notice that the summation reduces from infinitely many terms to only n1 terms Page 523 EE 422G Notes Chapter 8 Instructor Cheung Let s do one convolution by hand 2211 XkT IIIITTIIIIIIIIIIIIIIk 0 21 hkTI IITIIIIIIIIIIIIIIIIkgt 0 yOT EkxkThOkT o o o I o o o o yT EkxkThTkT I I o o o I I y2T kakTh2TkT I I I I I I I I I I I I I I I I gt I I I I I I I I I I I I I I k39 Y4T kakTh4TkT k ynT n I I I I I I I I I I I I I I I L I I I I I I I I I I I I I l I 39 Note that the duration of the output ynT is longer than that of XnT In general if XnT has N samples and hnT has K samples ynT has NKl samples Page 524 EE 422G Notes Chapter 8 Instructor Cheung Ztransform in action Time convolution is easy for finite duration impulse response What if the impulse response is infinitely long Answer Use Ztransform We want to show for any causal LSI system ynT 20xkThnT kT ltgt Yz HzXz ZynT 2i xkThnT kT II Mr S xkThnT kT Zquot Ml l xkTi hnT 167324 xkTihlTzlk ixkTzk ihlTzl xkTz Hz H ZX Z Again we call Hz the TRANSFER FUNCTION of the discretetime system Example l xnT1nunT hnT unT Find x017 hnT Page 525 EE 422G Notes Chapter 8 Instructor Cheung Solution First compute the Z transform Xz and Hz Xz unT i l 72 1 4 Hz unT 2 Then compute their product YltzgtXltzgtHltzgt 1 1 72 3 l 1 l 1 1 12 l gz Apply partial fraction Hz 2 1 1 1 Z 13 i l ZZ 1l Z l l ZZ 1 l EZ 1 Then yltnTgt Z lmz Hequot 4lt ultnfgt DONE Just like Laplace transform we can also determine whether a system is BIBO stable based on the pole locations of the transfer function Recall Hs is BIBO stable if its ROC region includes the imaginary axis And the imaginary axis in the splane maps to the unit circle in zplane Thus we have Page 526 EE 422G Notes Chapter 8 Instructor Cheung Hz 11 1 l 05jl5z l 05 jl5z R R iX k V BIBO Stable Unstable T es of Im ulse Res onses amp Difference E uation Two main types of impulse responses I Finite Impulse Response FIR hnT has finitely many nonzero values Example hnTIITITIIIIIIIIIIIIIII Hz 122391z392 1z3912 Easy to see that if Hz is FIR there is no denominator so it does not have any finite poles gt the corresponding system is always BIBO stable Express it in terms of the input XnT and output ynT Recall Y Z X Z Yz Xz 22 1Xz z 2Xz Taking the inverse Ztransform we have Hzl22 1z 2 ynT xnT2xnT TxnT 2T 1 Page 527 EE 422G Notes Chapter 8 Instructor Cheung This is similar to the approach of recovering the differential equation from the transfer function in continuoustime In discretetime l is called the Difference Equation In general the difference equation of any FIR system can be expressed as follows ynT akxMT kT Why difference equation Great for implementation the followings are from Section 91 92 Example ynT xnT 2xnT T xnT 2T Using simple delay amplifier and summer we can implement this difference equation as XnT delay This structure is called the Direct Form However this is not the only type of implementation Notice that the corresponding Ztransform of the difference equation is Hz l22 1 272 lz 12 Thus the system can also be implemented as applying the following difference equation twice y nT x nTx nT T Page 528 EE 422G Notes Chapter 8 Instructor Cheung This is called the Cascade Form Since any polynomial can be factored into linear and quadratic factors the Cascade Form can be though of as connecting basic linear and quadratic Direct Forms in series Cascade versus Direct Form 1 Direct form uses fewer components 2 Cascade form has the same basic building blocks 3 Quantization in gain coef cients affects all zeros in Direct Form but only individual components in Cascade form II Infinite Impulse Response IIR hnT has finitely many nonzero values 1 2 1 1 Z 11 Z 1 Unlike FIR IIR have poles gt A IIR may not be BIBO stable unless all its poles are within the unit circle Difference Equation Yz lz 1 lz 1 E 1 z 11 z 1 l zzz lwzz z Yz X2 z 1Yz X2 z 2Yz Xz z 1Xz Example hnT 15 l6 unT ltgtHz Taking the Inverse Ztransform ynT X2 ynT T 2 ynT 2T xnT xnT T ynT xnT xnT T ynT T X2 ynT 2T In general the difference equation for any IIR system can be expressed as N M ynT ZakaIT kTZb1ynT lT 160 11 Compare this with the FIR equation ynT ZakatT kT 160 FIR depends only on the INPUT values XnT IIR depends on both the input XnT and the past OUTPUT values ynTT ynT2T Thus IIR contains feedback Page 529 EE 422G Notes Chapter 8 Instructor Cheung HR has more varieties of implementation options than FIR Just like FIR the different implementations are due to different way of writing the Ztransform lz 1 E lH Xampe Z I az lxl zz l Four different ways to write Hz l H z 1z 1 gt FIR gtFeedback Direct Form I l l Z2 2 1 2 2 1 FIR VnT XnTXnTT Feedback ynT VnT7l 2ynTT l12ynT2T Remember the sign change in the feedback gains as compared with those from the transfer function 2 H z 1 1z 1gt Feedback gtFIR Direct Form II 1 l X2 2 1 2 2 Feedback VnT XnT7 12 FIR ynT VnT q ll2VnT2T XnT Direct Form II share delay elements between the Feedback and the FIR see highlighted terms thus using fewer delay elements than Direct Form I Page 530 EE 422G Notes Chapter 8 Instructor Cheung 3 H z Factorized into series of direct form H gt cascade 1 Z 1 Z First block VnT XnTXnTTl 3 VnTT Second block ynT VnTl4ynTT 16 15 1 3Z 1 1 Z 1 ynT VnT wnT VnT 16XnT l3VnTT wnT l 5XnTl 4wnTT 4 H z Partial Fraction gt Parallel Form XnT ynT Just like the FIR cascade and parallel forms are more resilient against quantization noise Also it is easier to build due to the regular patterns Parallel form also has the advantage of having low delay The drawback of cascade and parallel forms is that it needs factorization of the denominator Page 531 EE 422G Notes Chapter 8 Instructor Cheung 83A Z Transform Importance Ztransform to discretetime ltgt Laplace transform to continuoustime Definition Given signal xnT for n 012 ZxnT Xz m xnTz quot x0 xTz 1x2Tz 2 Motivation Let s say xnT unT RS unT is the discretetime step sequence Recall the continuoustime surrogate x50 217 xnT5t nT ZZOEJM 60 nT Taking its Laplace transform 1 Lxr 207JMM60 nT 2 206 l 1 5W Note that the nal expression is NOT a polynomial in s not very convenient to use in practice We use the following substitution 2 e The nal expression will become Nice 1Tz 1 zT 39 In general Lx5t 20xnTequot Zioxmry quot ZxnT Page 515 EE 422G Notes Chapter 8 Instructor Cheung Ma in zeSTTgt0 We are interested in mapping the ROC in splane to the zplane so that we can investigate the stability and other properties directly from the zplane Let 039 Res and a Ims Z e0jaT eoTej39wT gtz fizz UT The real part 039 Res only affects the modulus of z As the ROC in splane is based only on 039 Res let s see how it maps to zplane 039 ltgtltgt a ze T 0 03920 a zeOTl 039oo a lzlze Tzoo In words Negative in nity in splane a Origin in zplane Imaginary axis in splane a Unit circle in zplane Positive in nity in splane a Positive in nity in zplane Pictorially Splane Zplane Page 516 EE 422G Notes Chapter 8 Instructor Cheung Consequences 1 Left half splane Right half splane Imaginary axis in splane ROC in S plane s Res gt p W Inside the unit circle in zplane Outside of the unit circle in zplane Unit circle in zplane a a gt ROC in Zplane 2 z gt r J Hs is BIBO stable if all poles are on the left half plane gt Hz is BIBO stable if all poles are inside the unit circle more later Fourier Transform in S plane is evaluated along joaaXis gt Fourier Transform in Zplane is evaluated along the unit circle gt XsejmT must be the same as X Se quot ie integral number of full d U A rotation As ail 2n7r ja2JT the Fourier transform is a periodic function in 03 with period 2 mg We have seen this before X003 CDh 03h J after sampling X5003 was03h 05a5coh 03h 03h cos03h cos 0330311 2035 Page 517 EE 422G Notes Chapter 8 Instructor Cheung Examples of Ztransforms Example 84 ztransform of unit pulse 50 l n 0 A xnT 0 n 05n Xzl0z 1l Example 85 zTransform of unit step sequence unT I n20 xnT 0 nlt0 Xzlz 1z 2 1 withROC z z gt 1 l z 1 Typically we write the Ztransform in terms of 2391 rather than 2 as casual sequences contain only the negative powers of z Poles and Zeros in Ztransform l 1 CZ Xz1 hasapoleatzcandazeroatz0 1 0271 1 Z XZ hasazeroatzc andapoleat zoo Properties Time Domain Z Transform l Linearity a1x1nT 612x2 nT a1X1za2X2z 2 Multiply by a aquotxnT X Sj 3 Time Delay xnT mTunT mT m gt 0 z mXz 4 Multiply by n mCnT 2Xz 5 Initial Value Theorem 960 Hme X Z 6 Final Value Theorem xoo 1imH1 z lXz 7 Time Convolution 20xmTynT mT XzYz Page 518 EE 422G Notes Chapter 8 Instructor Cheung Multiply by 61 a zaquotxnT 270aquotxnTz quot 2 mm X EJ n70 a EX Find the Ztransform of ynT e WumT Rewrite ynT e quotunT Given ZunT Z 1 1 Z T W1 1w1 1 Multiply by n zm zix0 xTz 1 x2Tz 2 x3Tz 3 dz dz z xTz 2 2x2Tz 3 3x3Tzquot 0 x0 1 xTz 1 2 x2Tz 2 3 x3Tz 3 ZnxnT EX Find the Ztransform of ynT nTunT 71 znTunT zi T71 T271 2 dz l z l z T Since ZTunTl 71 1 z Initial Value Theorem Xz x0 x1z 1 Having 2 goes to co kills every term except the first Final Value Theorem Final value xoltgt 1i 1111 z 1Xz if Xz does not have any poles on or outside the unit circle except possibly a simple pole at z 1 A formal proof is beyond the scope but here is an informal proof 1 z 1Xz x0 xT x0z 1 x2T xTz 2 x3T x2Tz 3 1 x0 xT x0 x2T xT x3T x2T xoo Page 519 EE 422G Notes Chapter 8 Instructor Cheung 6 used pairs TABLE 3 1 Short Table of zrrrsnsforms n mmm Cun nunmmme Pnir unc un Smple Values z axlsforrn Number t for t gt 0 n for n 2 0 of nn 39 1 n o A x i fnT e 0 n 5 0 5a 1 2 um step l 3 7 Mr awry Kquot 4 I quotT we quotTe Mr 5 sin b sin DAT cos b ms bnT s rm elmr r e 7sin b739z s e s m e 5m 5711 25mg bhf E 9 2 4 cos b g nr ass brzT a rm unnsforrns 5 duvugh 9 a and chan be replaced by mums K m K2 respeczivclyv is was don in Krausfonn 3 Con vergencc of the 27mng reguth pm lt 1 8313 Inverse Z Transforrn Two Basic Methods Express Xz into DefiIiition Form Xz Xm x1z ex Xz 1 22 z 3 4 This implies XO 1 XT 2 X2T 75 X3T 72x4r 1 and zero otherwise 1 ex Long diVision Xz 7m XHT a unT Page 5720 EE 422G Notes Chapter 8 Instructor Cheung 2 Express X 2 into partialfraction from Expand Xz Zakz k 21 all in terms of z 1 ko k0 1 611 k each term has an inverse transform Important before doing partialfraction expansion make sure the z transform is in proper rational inction of 2 1 Example 89 l XZZ z lz 02 1 2 1 02z A B Solution Xzi 1 2 11 02z1 1 2 1 1 022 1 Heaviside s Expansion Method 1 Bl z 1z1 1 1X A 1 Z Z 1 022 1 1 022 1 A 3390 Azizms 1 02 1 02 08 A1 02z 1 ran 0221 1 02 1X 2 Z Z 1 2 1 1 022 1 1 1702z lo 202 1IZ1 A11 f B gt B 025 Xz 12571L2571 3 xnT 125 02502quot l z 1 022 Page 521 EE 422G Notes Chapter 6 Instructor Cheung Applications of the Laplace Transform Applica on in Circuit Analysis 1 Review of Resistive Network 1 Elements V IR V R Ohm s Law Causedby V Causedby V W W21 IIIZ1 0 w R R2 1 Causedby V Causedby V w W 1212sz R1 R 22 V2 Page 61 PDF Created with deskPDF PDF Writer Trial httpmmdocudeskcom EE 4226 Notes Chapter nstmctor Cheung 3 KVL and KCL 7 Select a node for ground Watch out for signs V Vac 0pm ercuxt Voltage R Equwalmt Resxstance I m Short meuxt Current 9 L R Same asbefore Page 672 PDF PDF W terr Tna EE 422G Notes Chapter 6 Instructor Cheung 5 Nodal Analysis and Mesh Analysis R1 WW1 Nodal Analysis Use KCL R2 294 HV2 ViVm Vi0 V81 f I 1 R1 R2 Ra 391 V2nud522 Il p30r2 20 V32 R2 R3 Solve for V and V2 and then calculate other currents and voltages R3 Mesh analysis use KVL V51 2R111R21112R3Il12 V51 R111R412 VS2 Solve for I and 2 2 Characteristics of Dynamic Networks 1 Inductor lira d VLt LE1LI wt 1 0r iLt Z w vL7d7 2 Capacitor C d j m zoo Evan W or vct wic7d7 Page 63 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung 3 Operation Amplifier NonInverting input L t Via Inverting input V0 O T O39 A general opamp model is described above In practice the input resistance Rm is very large gt 1012 Q and the gain A is very large gt105 Thus we will use the ideal model in the analysis 1 Input current 10 0 due to the large input impedence 2 Input voltage difference vit 0 and output voltage vot is dictated by external circuit due to the large gain Example Based on the ideal opamp model V20 V10 1 Also as the opamp does not have any input current applying KCL at the inverting port we have V2tRa VotV2tRb votv2t 1RbRa Plug in l we have votv1t lRbRa This circuit is called NonInverting Ampli er Page 64 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 4226 Notes Chapter 6 Instructor Cheung 4 Mutual Inductor 7 used in transformer Two separate circuits with coupling Clnrents a I 1 zMN 12 Basic equations dz dz v z 1 M 2 1 Ll t dt vl Z1 L2 15 dzl dzz v z M 2 z 2 dz b 7 Very easy to Write very dif cult to analyze Make sure both 1 and z point either away or toward the polarity marks to make the mutual inductance A positive To link the two circuits together introduce a combined Current term 11z2 d39 d39 d39 d39 v1 z L1 1 7M 1 M 1 Mi dz dz dz dz d 7M 1 M i i L1 dz dz 1 2 d v2 z M 1 M dz2 L2 7 2 7M 7 2 dz Mial i2L2 7M 7 2 dz Equivalent Circuit dt dt PDF Created With deSkPDF PDF Writer 7 Trial httpWWWdOCudeSkCom Page 675 EE 422G Notes Chapter 6 Instructor Cheung Example Apply mesh analysis to the following circuit 0 21160 0 R m L DJ rdr REG l a differential equation Using Laplace Transform VS VLS VcS VRS LsIs i0 i vC 0R1s C s s l LsIs Is RIs Cs De ne Generalized Resistors Impedances ZLS LS 3VLS 1SZLS Zcs 2 ms 1sZCs Both capacitor and inductor behave exactly like a resistor 3 V50 ZLS1SZCS1SR1S 3 3 VSS ZLs ZCs R Everything we know about resistive network can be applied to dynamic network in Laplace domain Generalized Ohms Law superposition KVL and KCL Equivalentcircuit Nodalanalysis and mesh analysis Page 66 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom lrvstructor Cheung EE 4226 Notes Chapter 6 3 Laplace transform models of circuit elements What ifthe initial conditions are not zero 1 Capacitor m Cd aw b IsCsVs7CV0 W 14 insult0 CS S de ne Zcs i 14 Zc5151v039 r rnode1 an impedance generalized resistor and a voltage source in series Vs a1so represent it as an impedance and a para11e1 current Altematively you can source Norton equivalent circni W5 BE VERY CAREFUL ABOUT THE POLARlTY OF VOLTAGE SOURCE AND THE DIRECTION OF CURRENT SOURCE Page 677 PDF Writer 7 Tnal EE 422G Notes Chapter 6 Instructor Cheung 2 Inductor L div 0 T b Vt L a V m m 5353 Lt039 Generalized resistor Impedance Z 9 2 LS and Li0 a negative voltage source Lil039 in series I s 39 K Can we obtain VG I LSIS Li 0 Vb From this circuit models Yes Alternatively you can also represent it as an impedance and a parallel current source Norton equivalent circuit 3 Resistor Vs RIs 4Voltage and Current Sources Don t forget to apply Laplace Transform on them vsu gt VSQS FLMUH 13950 gt T 15SLisf 5 OpAmp same ideal model assumption Page 68 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 4226 Notes Chapter 6 Instructor Cheung 6 Mutual Inductance Transformers L M L z m 39139 I U 6 2 gt M 72 vat e Me U Laplace transforrrl model Obtain it by using inductance model Lz MO39 LL M 7340 I LS Laws tie M5 VLLS Mczno 124ch 0 l Page 6 9 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Example Find Complex Norton Equivalent circuit given vC 0 O W fart but If E1 gt ls5 855 5 69 1 L b Solution 1 Compute the Short Circuit Current 155 156 15 ZICS Iii I ISlt 5 Straightforward to see 153 7213 To compute Is apply mesh analysis on the left loop S 3 iIAs gtlt15231si12Is IS S S S S IS gtISS721S7i 53 No need to do inverse Laplace transform as tl39ie equivalent circuit is in tl39ie s domaln Page 6 10 PDF Created with deskPDF PDF Writer Trial httpmedocudeskcom EE 422G Notes Chapter 6 Instructor Cheung 2 Find the equivalent impedance z Norrnally we can just kill all the independent sources and combine the 39 usin 39 39 39 39 rules However as there is a dependent source we need to drive it with a test voltage 15 arts Jquot L 5 LC 3 Items V ES CSD Igt Vesms 5 O Z 772510 Vest 5 S 125 S 215 Mesh analysis on the left 100p 15 gtlt 1 Q 315 70 S 7 3 1 QIS 73j15 s 315 0 co 2 OPEN CIRCUIT Z mes 0 So we got an interesting result 5 661S Page 6 1 1 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 4226 Notes Chapter 6 Instructor Cheung Example Find the transfer function Hs V0sVis of the following circuit Assume all initial conditions are zero This is called the SallenKey circuit which we will see again in filter design If I Via Rewrite everything in Laplace domain we have ICI C15 R1 1amp8 A A A v v v V45 van Vsgt We recognize the opamp configuration as a noninverting ampli er so we have R K1 b R a Page 612 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung To find V0 we need Vb which depends on Va All other nodal voltages are known Thus we need two nodal equations Applying KCL at node a we have Va V Va ll7 Va Kll7 0 R 1 R1 2 C1S l l l l 2 77C V 7KC V 7V 1 R1 R2 E R2 R1 1 lt gt Applying KCL at node b we have VFV C2st 0 2 3 LV iKCs V 0 2 R2 0 R2 1 17 Combining equations 1 and 2 by eliminating Va we get 1 b 7 RleClCzsz ch2 R1C2 R1C11 Kls 1 Since V0 KVb we have VU K 14 R1R2C1C2s2 ch2 R1C2 R1C11 Ksl R where K1 b R a Page 613 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 5 3 Some Laplace Transform theorems 9t 1 wad Aid 63 i0 kw 20 Properties Time Domain Laplace Transform ml has as 1 Linearity aixitazxzt a1X1SazXzs M anxnt aans x9 cm Le whirl 2 Frequency Shifting e xt F 5 1 u TLC 9 kmquot 3 Time Delay x07 aut7 a was all 1 aw 1 k S d 4 Time Scaling XOII X 1L P2 hum d 7 JS at mum 5 Time Differentiation E160 sXs 7 x0 463M X l 10m 5 g m Ron 6 Time Integration L xTdT E xTdT M 3 Is u My srgmltkt 7 v M 5 M LT M wk 7 Initial Value Theorem hIHHU J60 hIHHM sXs 7x0 7 A 4 gag 2 8 Final Value Theorem hIHH J60 hmHu SXU X0 7 Squot 1 lt0 9 Time Convolution X yt XSYS L7 1th i1 m ta 3 g H 0 Using these simple properties ZEde 34 T v Y s a 5 K0 t We can compute Laplace Transforms of a large number of functio svbased the 530w jjt i aplace transforms of 61 utand e m that we have computed 3 SK U JL L M er I To apply Laplace transform to real circuit problems next chapter Xm 0 M 79 321 V EMT 5 You are expected to gt83 S 1 DC 39 Q I To be familiar with all the listed properties and transforms 39 WK 2 To understand the proofs but I will not ask you to do any proof in tests WJLJ ow 31 Mvcwi an Mfrs m 1 0mmquot Page 58 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung xmmryrmmznm T amh r T QNcht X3 arXS alxitgl asXa H A i 39r are xiv Assume xt a1x1t a2x2t a1 and 612 are time independent X1SLX1tl XZSLXztl then Xs Lxt a1X1s LIZXZ s l Linearity Proof Llarxmmazxo IM I1X1ILIZXIe39 dt n ralxlte azxte dt Joqomloum U H 7 pm My 339 g a n MEL W IN 11 1 S 12 2 S 1Find Lcosa0t Key to solution express cos wot as linear combination of 51 ut m and 87 Note that if cos7a t jsinewut 005ant JSjnwnt em cosa tjsina t gtcosantewn eew aw 1 Resgtoe ltltj Jo Since Le 1 Resgt7Re0gt girl guy Rely 50 Le 1 Resgt0 0 W sejwn an m M LcostlLewLe39Wl www253qu o1 Rec or iluiwm i 1 1 2 swoon sejwu A W Ktcs gt0 a ro e atxm bum 5 sjansijan Aw 1ltsijwultsjap S Rog 4153 m Roq w m LAND mum What is its region ofconvergence Ans Intersection of all ROCs and in this case Res gt 0 39 mt t 2 Show that Lsmatut sszz Srwlw t 42 50 e gt 3 Is LatxtbtytAsXsBSYS and Why No am New W W A A 777 V 7 i Page 59 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 2 Complex Frequency shi s shi Theorem Assume x yia X6 k wt 53 X0 Lxt YS Lyt Mm m rLL Pine 4 xaw 7 Then E X4S a wry SLWUCD XPmd xzgt m jfx m e t 4t Proof Ys yte alt Ixte t Xs 1 Done U MM xme gtees LHLS J 54 q l du QOC 1 Ldf 9 1 Find Lcosante39 RD J 9 ac c Vt Since Lcosa t S if L0 8t e j quotw S2 239 fetalML My new 5 his I then Lcos cont 7 7 if 3 L7 SH Rt 2 Show Lsinconte m f sa 0U Example 54 Find x0 L XsLquot s 6s 13 1 L L Solution By completing square 5439 1 S Eff t 0 4 8 X0 g DamMM 6 13 L L 15 ng 7 ifs S 5135Li39 7 1 6 HA X H 52gtlt2 WM 1 kr3f22 x0 ees t 39125 W 3 Time Delay Theorem dw f K Assume Lx QELxtutXs 7 i 3 Then Lxt t0ut t0 JQE gs t0gt0 gum an s tezl Proof39 HOMEWORK owlwt were mLtgt x5 1 3L 9t WNW Mom XL t quot 14 e m t W A quot t A x PageDSIO PDF Createml demWriter Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung Example Square wave beginning at l 0 I 39 AR 1 391 1 Azure xsqltrultz72ultr Amman 2ut Z U2ut2TD I I 31 r Z lmtt n m quot T0 T0 2T 33 L 2 2 0M 1 T 2 y Lxsqtl Zie 2 Zle Tw 2leA2 2le ZTE39SH s s s S s lt A 1 1 727121122l 32114m s s s s S i 1 2 gamma c ed39fld 772 3 3 l 7 ROC s W expG Tns eXp7 U Res lt1 s Res gt 0 as exp7 lt1if gt0t 39 C 4 Scaling gingewmnml Q3212 963l gn UL s Assume LS Lxt then L9Eat wgtlt 0 in W 39mm Restriction agt0 xal a times fast if agt1 or slow ifalt1 as xl Proof xmis oat WW Lxat Ixate39 dt r sadw I t gtu ijxre 4fdr setT at 4tnlit 1 x low we 1 s 3 1 Kat r w X t ave 5mm 2x0 5 a ti 3 PM 15 t w a Page 511 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 5 Time Dijfkrential ion Assume XsLxt nal LI5X9 TVAQ VAN t 039 may My v1 Proof Lawn 171 eis dl I e S39dXO dt 0 dt 0 quot 2 Integration by parts MP T b jugdvt ui t jvtdut a Make the follow1ng substitution 1 De nition L a u s mag Rec xts xtgtvt flackle lt co ed 11 JUL 1 d M a nut umer gill 66 l LIVExO 39e dxa S AW In the ROC ofXs we must 751xfxt 651 I have lirnHwe39stxt 0 otherwise the Laplace integral 5 7 M 7 t quotS a O which is the area under e39Stxt 3Q 5quot xle 5 dl from 0 to Foe Wlll become 00 Sx0 n Ms Notes 0 x07 the value of xt when t approaches 0 from the negative side this is the convention recall the definition of L6t Rabl eml 9 on we um o What is the Region of Convergencvekk arjigafpfafx 1me MS g 5 Lo TM 0 What about higher derivatives 53 4 07 37 gm y leltzgttl sLIxlt gtltrle Wm sZXsmlt0exlt gtlt0 f m f Thus1t is easy to show that AWL 5 1 va gag a frml L d 52 was 7 5403 5 1392x0 0 77 M 2077 xltquot1gt0 I r i L A m where x 0 is the kth derivative of xlgReva uated at 039 Consequence Combined with linearity you can now convert any ordinary differential equation AND initial conditions into Laplace Domain Page 512 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom mm 910 q 5 NOT M our EE 422G Notes Chapter 5 Instructor Cheung wa Exam le Solv the followin differentiale uation W w ug gt er w DE JnL39wm OHM1 EaswflVM lgmw mgmw mmmm md w u Nmww Write the differential equation in Laplace Transform 0W 5quot W h39m m 5 dz Wt kw LI L xt sZXs 7 407 7 1500 sZXs 7 s7 0 Ln ranch stU LL Xlt0 rXs 7 x0 sXs 71 mm 7M Lle 7 ulttl 7 1 s 7 5 ss Ms J n1 By linearity we have 112 a sXs7s6xXs765Xs g 183 43 Cs s5gtx 573 5 Danna X6 3L Xs 5 1 saws s7sz 6s 5 We will learn how to do the inverse transform in the next lesson in the meantime we will use matlab gtgt ilaplacesym sA27s7 57 sA26s5 ans 38exp 5tl24exp t712exp 7t Important the Laplace transform ofany ODE is a RATIONAL FUNCTION in s a ratio oftwo polynomials X 5 7 M2 39 m M 1 hi 5 05 DQ8 bra 5 28 tr Srb jwwmkms 4 lt1 les ams quot 4 ans 4 MM Page 513 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 4226 Notes Chapter 6 Instructor Cheung 66 Block Diagrams and Feedback Systems What is a block diagram Composition of modular subsystems Want to represent the whole system as Eas Output so that we can study its stability frequency response time response etc Question Can we just study them separate y Basic Block X s Y s 460 E is gm oiitput X 55GS of the block of the block Assumption Ys is determined by input Xs and block transfer flmction Gs The output can have arbitrary fanout and Gs will have no effect on the input These assumptions might not hold in real applications We ll see an example later Cascade connection WCs G1SX s Xs Ws Y s Y s G2sWs 2YltsGzltsGltsgtXlts Y S X S G2 0G1 S aquot amp YsXs 7 Page 632 EE 422G Notes Chapter 6 Instructor Cheung Summer Xxx t W X2 a YoXltsinlts7X3ltsXilts Xxx 7 X40 Example Loading Problem Consider the two RC networks shown in the le diagram below 19 2F 19 2F mt 1F1y1txzt in yzt g m 1FI In yzt H1ltsgt Hzs 2x 2x 1 Using impedances we can easily compute H1v i1 and H2v v Cascade rule implies that the overall system is thus 1 2x 2x H H H 77 1 X m M vl 2xl 2323xl D On the other hand the right diagram of the actual cascade circuit Again it is easy for us to compute the transfer function please work it out yourself HO 2 2x X1v 2x 5xl Question 1 l and 2 are clearly different Why 2 Answer The ideal block model assumes that the input port of the second block draws no current from the output port of the rst block It s clearly not the case here This is called the Leading Problem Question 2 Can you make the circuit behave more like the ideal case Answer Put a bu er or isolating ampli er between blocks Page 633 EE 422G Notes Chapter 6 Instructor Cheung Feedback system ref lence input E s R s 7H s C s j l output Euor signal feedback gain typically Hs1 Gs system to be controlled Let s nd Tr Cr Rr Closedloop transfer function Equation 1 Cr GrEr Equation 2 Er Rr 7 HrCr C I GVRV H5CJ GVRV GVHVCV 195 gt 1 GrHrCr GrR r GU gt T C R 7 5 5 5 1 GrHr This is a negative feedback system For positive feedback We replace Hs by r Hs and the transfer mction is 75 17 GsHs Why feedback Closedloop system is extremely useful in control because it allows a feedback path for adjustment An openloop system like the one below cannot compensate for any disturbances accumulated at the controller and the output Disturbance 1 Disturbance 2 Input or reference utput or Controlled variable On the other hand by using a feedback loop that captures the output the system can adjust the input or reference to compensate for any disturbance Page 634 EE 4226 Nate ChaptErE nstructur Chaung a b R mom A r 1 r reduces 10 every year To combat such decay you try a system with mm formula we have l A EH 7 x l A3 1A3 In order to have Hs A 10 we needto set 3 m 0099 s 1 1u un99 1 1n3 adjust the lam resistance ofthe inverting ampli er Exercise Page 5735 EE 422G Notes Chapter 6 Instructor Cheung Even though the positive feedback uses three times as much component the overall gain Af makes it very insensitive to changes in A l 1 A3 8 We can see this by plotting the gains for the two systems over ten years 1n A 5e 7 4g Single Amplifier V Feedback Amplifier 1 2 3 4 5 e 7 a 9 10 Years Decay in Amplifier Gain Example Approximation of Inverse Given a system Hs if we want to undo its effect we can put form a cascade system like the following This may not always work because 1 Hs may have zeros on the open right half plane 2 The inverse system relies on VERY PRECISE Cancellation of poles and zeros between Hs and Gs not practical in reallife Instead we can approximate it using a negative feedback system Page 636 EE 4226 Notes Chapter 6 Instructor Cheung The overall system is thus HAS A z 1 if lAHsl gtgt1 s Hs Of course we also need to check that Hus is stable Asymptotic Stability of Composite System Consider the following two examples the left system is unstable even though all the subsystems are stable The right system on the other hand is table even though all subsystems are unstable ms wa is to mbine them into a single transfer function There are techniques Nyquist Stability Criterion that can determine the stability of negative feedback systems 39L 39 39 wi iuuLLuvcl 39 quot Idea combine blocks together to form familiar configurations Page 537 EE 4226 Notes Chapter 6 Instructor Cheung Example Find YsXs Ys 6163 G264 Y 1 G1H1 5 G1 G3 GZGA Ys 1GIH1 GIG3 6264 X s 1 G163 6264ij2 1 GIHI HZGIG3 G2G4 1 GlHl Page 638 EE 422G Notes Chapter 7 Instructor Cheung Use State Variable to represent circuits Let s start with a simple example R2 2 LlH a ut C1F I yt OTo NOTE in this chapter we will use ut not to be confused with the step function to denote the input and use xt to denote the state 1 Define the states Similar to the discretecase we define the states based on the memory storage elements For passive circuit the memory storage elements are the capacitors and inductors dvc 1C C dt As the knowledge capacitor voltage and inductor current allows us to infer the capacitor current and the inductor voltage by taking derivatives we de ne state variables to be the capacitor voltage and inductor current x1 v t x2 Note that there is one state variable for each memory storage element and VL LdL J Derive the State and Output equations express the derivatives of the states and the output in terms of the current state and the input ONLY KCL at node a ix2 dt C KVL 72lx15x2iu dt L L L These two equations can be more compactly written in matrix form NJ 0 C WM 0 Wu pg r 1L RLrx UL The above matrix equation is called the State Equation because it relates the change lStorder derivative of the state to the current state and input We can also relate the output to the state variables as follows Output Equation ytl 0x1 x2 Page 75 EE 422G Notes Cllapter 7 Instrllctor Cheung Using this approach We can Write the state variable representation for any circuit In general the state and output equations are always in the following form39 In our previous example We have 0 1C o A B C1 oDo ilL iRL 1L Be very careful about the dimensions of each matrix Let s do another example J Ew flx c vs 3912 c be m T Step 1 De ne the state variable vC as x1 z39L as x Step 2 Express the derivatives 5c x2 and output y in terrns of x1 x2 and input v57 1 t c c 7 i i xq y R21R2 R211 R2352 0 1kx Output Equatlon 2 1 1 1 v 7x 71 1 1 1 81C8lt39mquotzgt8 5717 Rcx1 8x2Rcvs lt1gt 1 1 1 1 R x2ZvLZltvciygtfxiR2x2Zxi 5x2 lt2gt Combining 1 and 2 561 CR1 x2 1 State Equation Page 7 6 EE 422G Notes Chapter 7 Instructor Cheung To summarize For electrical network select Q and Vc as state variables Step 1 Select each iL and vs as state variables di Step 2 For each iL write a KVL IL 2 7 will be included For each v0 write a KCL VC will be included Step 3 Other KCL and KVL and element relation to eliminate other variables other than states iL vc and sources input gtstate equation Step 4 Output equation State Equation from Transfer Functions State Equation gt Tell how to realize and simulate system realization the systems Ns bmsm bmilsm 1 bls b0 Given H s 1 Assume mltn gt BIBO Ds squot aHsquot alsa0 x Ax Bu F1nd u scalar y scalar y Cx Du Such that C s1 A 1 B D the transfer function expression derived directly from the state equations equals the given transfer function NU 13s 1 V0 Assume the denominator polynomial is an nthdegree polynomial define our state vector as a ndimensional vector follows Given a systemHs we can implement this with the following block diagram gm wo gm x 0 10 00m am im gm am 00 00 gm 3 E 3 E E E E I E E xmn m m am 00001xd0 xnt 541m xna 5w 7 xna Page 77 EE 422G Notes Chapter 7 Instructor Cheung Note that our definition already provide almost the complete form of a state equation except that we don t have an expression for 5c I That has to come from the system In the first part of our system 1 I V0 0 Us DSVs r39 SquotVs an1squot 1Vs alsVs a0Vs Applying inverse Laplace Transform dquot a 1 d ut dtn vt 61H f vt a1 E vt a0vt 5cquot t a xn t a1x2t a0x1t 5911 U ayHxn t a1x2 t a0x1t Now plug it back to our system XI 0 0 0 0 x1 x2 0 0 l 0 0 x2 0 W x a0 a1 a2 a3 an1 x 1 Let Ns b s quot blsW1 bwls bm The output equation can be written as Y0 NSVS bmsmVs bmilsm 1Vs blsVs b0Vs In time domain t b vtb vt bivtbvt y mdtm mil quot39 ldt 0 dr39 bmxm1tbmrlxmt b1x2tbox1t x1 b0 b1 bm sz x Page 78 EE 422G Notes Chapter 7 Instructor Cheung 4s23s l Exam leH s p 6s37s2s5 First we normalize Hs so that the leading coefficient of the denominator 2 2 l l szS 3 7 2 l 5 s 3 s By inspection we can just write out the state equations polynomial is l H s XI 010x1 0 xi 0 0 l x20u x 7imx 1 x1 y x2 x 3 Where do we go from here After obtaining the state variable representation it will allow us to l Analyze its stability controllability and observability a The eigenvalues of our state matrix A are precisely the poles of the linear system A system is called controllable if regardless of its initial state we can drive its state to any value within finite time by using a suitable input A system is controllable if and only if the following matrix is invertible U BABABZyABr A system is called observable if we can determine the initial state given any output A system is observable if and only if the following matrix is invertible O C CA CA 2 CA quot 1 Page 79 EE 422G Notes Chapter 7 Instructor Cheung 2 Simulate it with a computer by discretization x Ax Bu To Simulate we can follow the procedure below y 2 CK Du 1 At t0 xt 0 or can be any initial state 2 Compute dynamics x0 Axt But 3 Compute output yt Cxt Dut 4 Compute new state by Taylor Series Expansion xtAt xtxtAt 5 I t At 6 Go to step 2 This implementation assumes the input is constant and the state vector is linear within the sample interval tt At While the rst assumption is not too unreasonable we can improve upon the second one by combining step 2 and 4 together to compute XI AI eAA XQ eAA IA 1Bul This involves a matrix exponential which can be computed by using Taylor Series eA I AtiA2z2 iA3I3 2 3 Butl am afraid we are running out of time For further reference on the use of StateVariable Representation please consult the excellent text Linear System Theory and Design by CT Chen Page 710 EE 422G Notes Chapter 8 Instructor Cheung Roadmap for DiscreteTime Signal Processing Continuoustime Signal Xt cos21cft 1 1 1 1 1 1 05 0 2 05 1 1 1 1 1 1 1 0 02 04 06 08 1 12 14 16 Discretetime Si nals Section 81 1 1 1 1 111 1 1 1 1 11 1 1 1 1 1 1 1 051 1 1 1 1 1 1 1 a 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 O 1 1 1 l 0 02 04 06 08 1 12 14 16 18 2 Sampling Period T seconds or sampling frequency lT Hz Denoted as XnT or simply Xn for n0l2 If T is too big we have no idea of the original continuoustime signal 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 5 1 1 1 1 1 1 1 1 1 1 1 1 39 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 O 1 1 1 7D 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 T1 1 1 1 1 1 1 1 1 1 1 1 0 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 11 1 1 1 1 1 1 1 1 1 J 1 d 1 11 1 J 1 01 Page 51 EE 422G Notes Chapter 8 Instructor Cheung So increase the sampling frequency but by how much Surprising result by Nyquist Section 82 A continuoustime signal having no frequency components above fh Hz can be completely recovered by samples taken at a sampling rate greater than or equal to 2f11 Hz This is called the Nyquist rate 0 39l nT Do Reconstruction formula xl Zn0 W1T Zno x TSincl 717T Quantization 1 Section 82 2 Not only time can be discretized signal level can too and must to be stored in computer xnT 103422364353 z 1034 A 32bit computer can store 232 4294967296 levels Is it good enough Similarities between continuoustime and discretetime analysis PROPERTIES CONTINUOUSTHVIE DISCRETETIME System Properties Linearity causality Linearity Causality Stability 8 4 Stability S t mE ti n alfferentlal Equatlon Difference Equation 84 VS 6 qua Eyltrgt3yltrgt xltrgt ynTyn1T xrnT Complex Laplace Transform Ztransform 83 Frequency S 0 nT Transform Xs I0 xle dz Xz Zn0xnTz Page 52 EE 422G Notes Chapter 8 Instructor Cheung Transfer function Hs 2s s24s1 H Z Z 22 1 2 2 42 1 1 Fourier Transform ContmuousTime Fourier Transform CTFT Xjw xte f dt Discretetime Fourier Transform DTFT 84 Xij 20xnTe W Fourier Series Xk l T ContmuousTime Fourier Series for Periodic Signals T 271 2 Jkit f7 xte T dt Discrete Fourier Transform DFT for FiniteDuration Signals 413102104 N71 71quot Xk Zxne N 110 Synthesis Technigue Filter Designs Again Lowpass Highpass Bandpass Bandreject Basic strategy start with analog prototype and then map to discretedomain 1 In nite Impulse Response IIR Filter Design Section 94 impulse nT 1 response goes on forever hnT E n 012 2 Finite Impulse Response FIR Filter Design Section 95 nite impulse response hnT 0 5 10 2 Computational Technigue Fast Fourier Transform Section 104 is a fast algorithm to compute DFT The basic idea is to the recursive nature of the discretetime complex exponentials to reduce the complexity of the algorithm XO XZ X4 x6 XI X3 XS x 7 X 0 Xl X 2 X 3 X 4 XS X 6 X 7 a Result of one decimation of the time samples Page 53 EE 422G Notes Chapter 8 Instructor Cheung Chapter 8 DiscreteTime Signals and Systems Part one AID and WA 82A AnalogtoDigital Conversion Commuous tima Discretetune Dmmwmg Digital signai A A n A mam quD sugmz SigKai SW Sam lin O eration Sampling Model xt 4021 51771739 1 n v smmrmwm Here our sampling mction pt is a periodic signal composed ofa n Y w train of delta mctions quot Mmulmnllnr no Suw nldm P0 Z 50 nT A mn l l I I l 0 T 2T 3T 4T 5T m mmmmmmm Why 16 instead ofjust using the values of W at 0 T 2T 7 1 Almost the same 2 7 0 XHT 7 xnT trnT tnT Page 54 EE 422G Notes Chapter 8 Instructor Cheung 2 The impulse retains energy of the samples behaves like a real signal I xst2dt ZxnT2 j 60 nTdt 20 xnT2 For analysis purpose we use x50 to represent the discretetime signal xnT x50 is the continuoustime surrogate of xnT Spectrum of Xsja and its relationship to X ja Multiplication in Time ltgt Convolution in Frequency xxtxt pa lt3 Xto39w mejltw vgtPltjvgtdv From Table 42 Ch4 P 1w Firm 5t nT ZM aky another impulse train in frequency domain with period a 2 21 our sampling frequency Thus w l m XXJQ LXl w 0 39 Zmw5U mwsdv Zmxjw jm wx Compare this with the impulse train XS 1w becomes a periodic spectrum with period ms Pictorially if X fa looks like KG 0 What does XS 1w looks like Page 55 EE 422G Notes Chapter 8 Instructor Cheung Two cases Case 1 oh S successive copies of X jw do not overlap X5003 ms03h oasoaswh oah 03h 0350311 ms oasoah 2035 It is trivial to see that we can recover X jw by multiplying X S jco with the ideal lowpass filter with bandwidth cos 2 A H003 V 1352 0352 Case 2 oh gt o ZS success1ve copies of X 1w overlap X5003 20s oas 03h 03h 035 2035 No way to reconstruct X jw Nyquist sampling theorem A signal having no frequency components above 03h rads is completely specified by samples that are taken at a uniform sampling rate greater than or equal to 20311 rads This is called the Nyquist Sampling Rate Page 56 EE 422G Notes Chapter 8 Instructor Cheung The phenomenon of sampling under the Nyquist rate is called ALIASING In time domain 1 At the times of sampling indicated by vertical lines the two signals shown both A agree with the observations 3 o 5 Either of these signals could 7 produce the indicated observations shown by circled points so the two signals are said to be 0 quotaliasedquot 0 250 5 0 750 1000 TlME In frequency domain Xt Sine wave of f Hz X003 2nf 2139Ef J xst sampling Xt at fslt2f X5003 4 4 4 4 4 4 4 4 E i E E i i E E 39 39 I 4391th l 39 2lnf 2nf 39 39 21tf 2ith 39 39 41kg 39 W If our ear can be model as a low pass lter with bandwidth around f Hz T EARGm r this is what we will hear the original plus unwanted lower frequech components EARj 0Xsj 0 ll ll 2139Ef 2139Ef Page 5 7 EE 422GNotes Chapter 8 Downsample by a factor of two look at the low frequency panern around the If sampling frequency is occurs This processing is called antialiasing Instructor Cheung Page 58 EE 422G Notes Chapter 6 Instructor Cheung More about ZSR and ZIR 1 Finding unknown initial conditions Given the following circuit with unknown initial capacitor voltage V0 lS V 0 s Hs Y Input g Output X S Ys X s Xt 109 yt i 1 103 103 1 Simple mesh analysis yield Xs s lOs 10v0 Y X S lOSl S lOSl Immediately we know that the transfer function Hs is 103 H S 103 1 and the ZSR and ZIR are respectively 103 Y X m0 IOSH S 10v0 Y 3 lOsl So if you use a step function xt utas input you get an exponential decay output yt e m ua Page 619 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung 2 Find out about an unknown system xw Yltsgt How do we find an unknown system and initial conditions Answer Input a test signal and measure the output The use the IO relationship to figure out the system Example The output of the system is Y s 10 51 for an input X s l S S So is the transfer function Hs Y0 SS Xs 103 1 NO we ve forgotten the Initial Conditions It should be clear that ONE pair of IO signals will not be suf cient we need two Why As we have learnt the total output response of a system is YsHsXs DU If we have two IO pairs we have CS Y1SHSX1SDS 1 ms HsXzs 2 Subtracting 2 from 1 eliminates the initial conditions and obtains the transfer function Y1S Y2S X1S X2S The initial conditions can then be computed by substituting Hs back to l or 2 HS Try the above example again with another IO pairs Xs 1 and Y s S Page 620 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Stability of systems We first need to introduce the concepts of BOUNDED and TRANSIENT signals There are two types of wellbehaved signals 1 Bounded xt SM lt00 for all t 2 Transient lime xt 0 Are these signals bounded Are they transient 1 60 2 costut 3 tsintut 4 eT39ut 5 e ut Answers 1 6t Not bounded and transient 2 costut Bounded and not transient 3 tsintut Not bounded and not transient 4 equotut Bounded and transient 5 e ut Not bounded and not transient Can you tell from their Laplace transforms A sufficient condition for a signal to be a transient signal The poles on its Laplace Transform must be on the open left half plane This condition is the same as saying the ROC contains the imaginary aXis gt the ROC contains the origin 2 fr xt dt lt oo gtlim xt 0 law A bounded signal must satisfy the following two conditions 1 Its Laplace transform must be proper NX S 2 2 The poles must either be a on the open left half plane or b on the imaginary aXis AND simple ie multiplicity l ie if Xs degree NXs ltdegreeDX 3 Page 621 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Reason for Condition 1 Consider 60 L 39t 1 is not proper For general Xs if degree NXs 2degree DXs applying long diVision RX DX Thus Xt is not bounded due to the delta functions Xs ansquot an1squot 1 a0 1 L 1Xs andlquott an15 quot t a05t S Reason for Condition 2 If all the poles are on the open left half plane we know the signal decays to zero If there are poles on the imaginary aXis there are two cases Case 1 Poles are simple ie multiplicity 1 such as Lcost ss2l In this case the signal is oscillating but still bounded Case 2 Poles are not simple ie multiplicity gt 1 such as Ltsintls2l2 In this case the signal is not bounded Now we come back to study the stability of a system All systems can be classi ed into stable or unstable There are three different types of stable systems 1 BoundedInput BoundedOutput Stable BIBO System 2 Asymptotic Stable System 3 Marginally Stable System Here is a Venn diagram that describes the classification Page 622 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung BIBO describes systems behavior subjected to general input assuming zero initial conditions Asymptotic marginally and unstable systems refer to the longterm behaVior of a system under any initial conditions but no input JM Marginally Asymptotically Unstable Stable Stable Recall Given a system H s and inputXs the most general form of S C s N 3 out ut is Y X P S DU DU S The rst term is the ZIR and the second term is ZSR BIBO Stability Definition Assume zero initial state a system Hs is BIBO stable if it outputs a bounded output for any bounded input A BIBO system Hs must satisfy the following two conditions 1 If Hs degreeNs S degreeDs S 2 The poles of Hs must be on the open left half plane NU 133 As there is zero initial condition the output Ys is just X s To ensure Ys is bounded we need to check two things Page 623 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung l degreenumerator of Ys lt degreedenominator of Ys degreenumerator of Ys degreeNs degreenumerator of Xs degreedenominator of Ys degreeDs degreedenominator of Xs Since Xs is bounded degreenumerator of Xs lt degreedenominator of Xs thus all we need is degreeNs S degree s J Ys has either open left half plane poles andor simple poles on joaaXis As Xs is bounded its poles must either be on the open left half poles or simple on the imaginary axis To ensure Ys HsXs satisfied the same criteria all the poles of Hs must be on the open left half plane Hs cannot have any pole on the imaginary aXis not even simple one because the input Xs might also have a simple pole at the same location and the resulting Ys will have DOUBLE imaginary poles making it nonbounded Asymptotic Stable Definition Assume zero input a system is asymptotic stable if it gives a transient output for ANY initial state A system is asymptotically stable if all the poles of the Laplace transform are on the open left half plane S This is easy As there is no input the output is Y s CS In general Ys will S D 1 13s have the same poles as unless there are cancellations of poles and zeros Thus 1 has open left half 13s to ensure yt is transient all we need is to ensure that plane poles Also it is easy to see that BIBO stability gt Asymptotic stability Page 624 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Marginally Stable Definition Assume zero input a system is marginally stable if it gives a bounded output for ANY initial state A system is asymptotically stable if all the poles of the Laplace transform are either on the open left half plane or simple on the imaginary aXis S Cs To ensure Y s D is bounded we need to check S l degreeCs lt degreeDs This is always true See page 615 2 Poles of Ys are either on the open left half plane or simple on the imaginary aXis Also it is easy to see that BIBO stability gt Asymptotic stability gt Marginally stability Unstable system Definition Assume zero input a system is unstable if it gives a unbounded output for some initial state An unstable system is a system that is NOT marginally stable Page 625 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom


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