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by: Adaline Pollich


Adaline Pollich
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This 26 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 34 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 6 Instructor Cheung CarveFILM m OwnPw ri39 4 fraanEV mc lr m 1 Example I A HighPass Filter with zero initial condition 49 W IC 1 0 x I ls Input Output XS YS q Hs xt 109 yt 10 7 105 I 105 1 For any input Xs we can compute YsHsXs We call this output response the ZeroState Response In another words Hs describes the functionality of this highpass lter vairj im M 0 M mud ALBNE This concept applies to more than Just c1rcu1t VJ Fax 15er 2 Example 2 Armature Controlled dc servomotor Example 615 in text Input Ea armature voltage Mpwf Output 9 angular shift 7V Buildin S stem with La lace Transform R L I 39 PMT quot Motor Km SZHCSX Q m rust1 O V V dliim 999 Ea Bank em K50 Load 3 39 J B HQ 9s Km Ego srL Rs B KMKB Page 614 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Key questions to ask 1 How to find and understand Hs 2 How to choose appropriate components so that the system is stable 3 How to compute the frequency response spectrum analyzer 4 How to combine multiple systems together De nition of a Transfer Function Quantitative Description on how the system processes the input to form the output Transfer Function Hs in Laplace transform HS 353 Ys s lersinitislcondiiom A Smgle mput angle output system 8180 system M H Xe z ero initial c onditi on Later we will show how initial conditions can be introduced CVH39MTJ All linear systemchan be described by anODE d yt dltquot1gtyz U i dltmmz dm1 o r P P 39 t b b t er din din l WI 771 1 dim l 060 W W My Taking the Laplace transform and assume zero initial conditions ans Ys an1s 1Ys a0Ys bmszs bm1sm1Xs b0Xs m m l bms bm1s b0 XS or 0 Y S ansnan1s 1a ODE m m l Ys HsXs where Hs bms bmIS 1 quot39 90 516 er Wu ans an1s a0 Timwig slurst be 0 Wquot 1 Remember product in sdomain corresponds to convolution in timedomain Thus the differential equation can also be written in timedomain as a linear ltering t W IXUVZU TdT where htL391Hs is the system impulse response 0 Page 615 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Non zero initial conditions This again n H a 7quot waifquot 2 dt dz d xtb dWon n mil dtmrl a0ytbm b0xt Assume the initial state of the INPUT is zero as we have full control over it x0 x10 xW 0 o In many occasions the system s initial states may not necessarily be zero It is easy in the Laplace transform to incorporate initial conditions Recall from the differentiation theorem we I 39 gt n w Ld y squotYs squot 1yo sHy gto syltHgto yltquot1gto dzquot n th s Y s n 1 order polynomial based on initial conditions Where yi039 denotes the ith order time derivature of y evaluated at t039 Applying this rule to our differential equation and grouping all the terms based on the initial conditions together ansquotYs anilsquot71Ys gels polynomial based on initial conditions bmszsbm4sm Xs b0Xs mm gt it My 7 AmS 39 n4 Let Cs polynomial based on the initial condition What is the degree of Cs b Sm b sm 1b Cs Y S m n HM XS n quot21 ans aHs a0 ans aHs a0 Cs HsXs as aHs a Zero state response Zero input response N Page 616 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Thus there are I WO COlVJIPON39ENTS to the output response Zero State Response ZSR or HsXs It is called ZSR because it is the output response of the system if the initial state condition is zero 2 Zero Input Response ZIR or n CEBI ans an71S a0 It is called ZIR because it is the output response of the system if the input Xs is zero 0 Notice that it has the same denominator as the transfer function implying thame PDEL emailfquot m Separating the output response into two components that are based solely on input and initial conditions allow us to infer unknown initial conditions given a specific input We will see an example in the homewor Do not confuse ZSR and ZIR with another type of categorization of output response 1 Transient Response 7 The part of the output response that approaches 0 as t 00 2 Forced or Steady State Response what remains in the output as t9oo Let s illustrate their differences with the following example Example 6 7 v ISt order High Pass Filter n 1 Input vt 50052tut output yf V12 1 Initial capacitor voltage v0 1 RC 1 second Page 6 17 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 6 Instructor Cheung Find total response YS Vo VSS R S S YS v0 C YS 1 1jVs i b w RCS 5 S I R Ys1sjVs5 7232 JAN f 7 1K YSS1KS S 1 Z L Hrs 1me 1 It 1s clear from the rst term 1s ZSR and the second 1s ZIR Note that I didn t subsitute v0 l and VSS 5SS24 in the very beginning ofthe calculation IfI do that I will obtain the output 4s2 1 S lS2 4 and I would have no way of differentiating the two components YCS 1 Find zero input response and zero state response v Zero input response L 1 S I Voe 2211 e ut 1 S 1 S SS Zero state res onse L V S L 39 p S l S S l S2 4 Apply partial fraction expansion we got S 5S 7 l 7727j772ji l 2S72 Sl S24 Sl Sij2 Sj2 Sl S24 L 1 S 6 40052t72si112tut S l S 4 2 Find transient and forced response Combining ZIR and ZSR yt 26 4cos 21 2sin 21ut Transient response Which terrns go to zero as t9oo Steady state response Which terrns do not go to zero as t9oo Answer Transient response 26 ut Steady stte response 4cos2t 725m2tu I They are clearly different from ZSR and ZIR Page 6 18 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 6 Time Integration I Assume Y0 2 IX My X5 LXl 700 Then JigDIM XSy0 S S Proof HOMEWORK Example 1 Find the Laplace Transform of 1t d l t l 0 L510R1tEJw1rdr5ut With 10 0 and o Ew1rdr0 We cannot turn this differentialrintegral equation into a pure 2quotdrorder ODE because the second initial condition is specified in terms of the integral Turn everything into Laplace domain L1 r Rm L11d1 5m gt 515 L1390 1215 i1s i j 0 12 dz 5 C5 C5 w s Plugging in the initial conditions 315 1215 i1s E Cs 5 5 LlsZ syLCl 5 Page 5714 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung Example 2 Find the inverse Laplace Transform of X s ml 5 a 1 Snl We first look at the Laplace transform of X s as the a is just a simple frequency shifting see property 2 Observe that Xs A ltgt X0 11t Xs Z ltgt X0 j u1d1 mu X0 0 Xs 3 at j mm di 110 X0 0 1 l t 7 Xs Am ltgt Xt Jw7n1uldl n 11t X0 0 Putting back the frequency shift we have X0 5 amt 17 Z Initial Value Theorem W L 1XSl 3 X0 lim sXs saw This theorem says the behavior of x at t0 transient state can be evaluated by taking 5 gt oo for the Laplace transform sXs lntuitively it makes sense to be able to recover the initial condition directly from the Laplace transform Proof From the time differentiation property we know sXs X0 L1Xr dt riXU e S dt 0 dz Take the limit 5 gt oo on both side Page 5715 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung lim Hm sXs X0 lim Hm I JM 1 Xt lim Hm e SIdt 0 dt Xt eisldt Let s consider the behavior of e as s gt oo Again it is sufficient to consider the real part as it Will dominate the behavior of the functions 1 t0 lim we d H 0 otherwise 35 5 gm m Note that this is not a delta function Main difference unless ft is 00 at t0 0 t finite at t 0 Ohm r mzla fror6r Case 1 Xt is nite at t0 lim Hm sXs X0 Since Xt is finite at t0 Xt is continuous at t0 so X0 X0 Hence lim w sXs X0 sgt Case 2 X Xo X05r lime sXs X0 X0 X0 X0 Proof complete Example 1 A demonstration Where X0 is obvious Xt 2 6 cos cool ut It is evident x0 e 0 cos 00 0 1 Using Laplace transform m erlttgt1 5 2 2 50 500 Page 5716 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung x0 lim sXs lim 500 2 Hm H lt5or 50U hmsz w stz2asa2w02 d520 sd5 llll 2 2 2 Hmd5 2asa 500 d5 25a 1 d25ad5 2 11 lim7l Hm 25 20 Hm d25 ZOO15 H 2 L39 Hospital Rule 10 510 Example 2 Find the initial value of xt given X S 1 105 slO xO lime sXs limes J does not exist 10 10 By simple inspection we know 1 t0 6t 10e 1 ut which blows up at 7 Final Value Theorem If xt and dxt dt are Laplace transformable then llin Xt ling sXs condition ROC of sXs must contain the imaginary axis This theorem says the behavior of x at too steady state can be evaluated by taking 5 gt 0 for the Laplace transform sXs Proof HOMEWORK Example 1 Find the nal value of xt given X s 1 S limHO 51 1 5 S This makes sense as 1 11t Page 5717 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 1 5 Example 2 Find the nal value of xt given X s Using the Final Value Theorem Xoo limHO S 3 0 5 Taking the inverse Laplace transform will get xt es ut which blows up What 3 wrong here The ROC of sXs does not include the imaginary axis 8 Time Convolution Assume LX1f X1s and LX2 t X2s Their convolution can be defined as yt E IX1TXZt TdT 0 and its Laplace transform is Lyt X15X2 s Note You may be more familiar with this form of convolution ya TX1lt2gtX2ltr zgtdz if x1r0 Vtlt0 3ytfxl xzt d 0 if XZ 10 w 3ytJXl1X2t 1d1 0 Therefore Yl jX1lX2t ld NOW we can proceed with the calculation of the integral LjX1tX2 t xldr TjX11X2t xi d2 e stdt Replace t with 00 I X1 1X2 t die 5tdt U 0 Exchange order of int IXI 1 X2 t 1 e stdt 11 0 Page 5718 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom l u u IX 1 WXZ 5 all u X25Tq A rm X16 9 5 This is a partial list ofthe Laplace Transforms of common functions Remember them There are more transform pairs from Table 573 in the book page 220 You do not umu Uu on should be quot 4 I ofthem as homework problems YAELE 53 Exlended Table at SinglerSided Laplace Tvansfurms Lupine Cnmmems an nansronu Darn anon Dtmcx minatmn wuh and um 66 J l 2 1m quottn 3 Dirccreulluarmn 3 9 5 7 W Dl ercnunuan applied ID pmra quot0 Table 571 i 4 r E l 57 4 ms mulllll Y W xampe 5 t 3 Exan i5 1 sum l I w l p 6 l int d 4 Ex 7 l m ll psi an air P in mg me p 7 zxphmysmwum 01 mil rrshlfmnd palr5 PageSVlg rul 39 rul uuuir 7 Trial EE 422G Notes Chapter 5 Instructor Cheung 53 Some Laplace Transform theorems Properties Time Domain Laplace Transform 1 L a1x1ta2x2t 01X1S02X2S meanty anxnt aan 3 2 Frequency Shifting e xt Fs a 3 Time Delay xt aut a e Xs 4 Time Scaling x0a iX ij a a d 5 Time Differentlation Exa sXs x0 2 X l 0 6 Time lntegratlon L x7d7 L xTdT 7 Initial Value Theorem limHy XU lim w 3X0 x0 8 Final Value Theorem lime xt IimH0 sXs xoo 9 Time Convolution xt yt XSYS Using these simple properties We can compute Laplace Transforms of a large number of functions based the Laplace transforms of 5a 1and 6 that we have computed 1 To apply Laplace transform to real circuit problems next chapter You are expected to 1 To be familiar with all the listed properties and transforms 2 To understand the proofs butl will not ask you to do any proof in tests Page 58 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung l Linearity Assume xt a1x1t azx2 t a1 and a2 are time independent X1SLx1tla X2SLx2tl then Xs Lxt a1X1sa2X2s Proof La1x1ra2xr a1x1ra2xre dr J a1x1te a2xte dt al J x1te dt 612 xte dt 01X1S02X2S 1 Find Lcos wot Key to solution express cos wot as linear combination of 5 t ut and 8quot Note that 87 cos a0t jsin a0t cosa0t jsin wot em cosa0tjsina0t 3 cosa0t 6 671 l sa Le We Res gt 0 LEW We Res gt 0 Since Lequotquot Res gt Re0 3 Lcosa0t LeILekur 1 1 1 7 2 sjcq s jwo Aw 2 moo mo S s26q2 What is its region of convergence Ans Intersection of all ROCs and in this case Res gt 0 a 2 Show that Lsinatut 2 2 s a 3 Is Latxt btyt AsXs BSYS and why Page 59 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 2 Complex Frequency shi sshift Theorem Assume yt xte7m XS Lxl NS Lyl Then Ys Xs a Proof Ys yte dt T xtequot dt Xs a Done 1 Find Llcosa0te J Since Lcosa0t 2 sz we then Lcos wot 87 H az s 002 we a 2 Show Lsm more s or we E 154 F39d L 1X 51 i xampe 1n xt 3 S26S13 Solution By completing square 38 sz6s13 s35 s2 6s94 s3 52x2 s32 22 s3222 Xs x0 e 3 cos 21 Seat sin 21 3 Time Delay Theorem Assume Lxt E Lxtu t X s Then Lxt to u t to e 0 X s t0 gt 0 Proof HOMEWORK Page 510 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 4226 Notes Chapter 5 Instructor Cheung Example Square Wave beginning at t 0 1 m2uurzuoinguairn 72u772u72Tn 1 11 my T7 Tu gr 2TH 5 xS 117212 I 2124 7212 Z S 2le 27 q S S S S S 1 quotW1 1 1 1 1 4 7271271227 327 3m S S S S S 1 2 ConvergedifWlt 1 77 v 3 13 S S 1 2A 1 7TA s s s 12 2 ROC 5M expk ibs expkiib Res lt1 5Resgt o as exp7 lt1if gt0 4 Scaling Assume Xv mm then xatX Restriction agt0 x e atimes fast ifagt1 or slow ifalt1 as xt Proof q 122 Tamra o ljx1ei7d1 set1at 1 o 1245 a a Page 511 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 5 Instructor Cheung 5 Time Di rentiation Assume XS Lxt Then LEW sXltsgt xlt0 gt dt Proof 1 De nition Lclxtt ef dt e dxt 2 Integration by parts b b j utdvt utvt j vtdut Make the following substitution xt 3 vt e 3 ut Lixt Temdxa dt 0 In the ROC of Xs we must rm have limee39StXt 0 3 lm g ml otherwise the Laplace integral m which is the area under e39stXt 0 9507 5 x1ei dt from 0 to toltgt will become 00 0 sXs x0 Notes 0 3607 the value of Xt whent approaches 0 from the negative side this is the convention recall the definition of L6t What is the Region of Convergence Same as X 3 What about higher derivatives Lx2tJ SLx1tJ x 10 SZXS sx0 x 0 Thus it is easy to show that n squotXs squot1x07 sn72xl0 sxmiz 0 x0171 Where 96 0 is the kth derivative of xt evaluated 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 EE 422G Notes Chapter 5 Instructor Cheung Example Solve the following differential equation d2 d 7 d xa 6Ext 5xt e ut Wlth x0 l and xtti 0 Write the differential equation in Laplace Transform Lxt s2Xs sx0 x 10 SZXS s 0 Lxt SXs x0 SXs l mm M Lle 7 url 1 37 By linearity we have 1 31 s2 7s7 s 7s2 6s 5 We will learn how to do the inverse transform in the next lesson in the meantime we will use matlab szXs s6sXs 65Xs X S gtgt ilaplace sym39 SA27S7 87 SA26S5 39 ans 38exp 5tl24exp t712exp 7t Important the Laplace transform of any ODE is a RATIONAL FUNCTION in s a ratio of two polynomials Page 513 PDF Created with deskPDF PDF Writer Trial httpwwwdocudeskcom EE 422G Notes Chapter 10 Instructor Cheung Chapter 10 Discrete Fourier Transform amp Fast Fourier Transform An assortment of Fourier analysis methods 1 Fourier Series continuoustime periodic signals Mt lt2gt Mlhm Periodic signalxt 47T 27tT 0 2751quot 4751quot l L szl3 Analys1s Xk Jxte 7 dt w 2 Synthesis xt Z Xkejk 15 2 Fourier Transform general continuoustime signals Wtw 4 X0 Analysis Xjw f xte j dt Synthesis xt f Xjwejquotdw 3 Discretetime Fourier Transform general discretetime signals 4443 wAM xnT 275 T 0 275 T Analysis Xej T ZxnTe jquotmT Synthesis xnT Xej T 6 wa Page lOl EE 422G Notes Chapter 10 Instructor Cheung Signals Fourier Transform Characteristics Continuous in t amp Periodic Fourier Series Discrete in 0 Continuous int Continuoustime Continuous in x Fourier Transform Discrete int Discretetime Fourier Continuous in o amp Periodic Transform Discrete in t amp Periodic Discrete Fourier Discrete in o amp Periodic Transform Now we introduce the fourth one the Discrete Fourier Transform DFT N4 4 T 4 T 4 T A V V 39 39t ltgt TTTT TTTT 0 Periodic signal XnT 392 393939 0 37 T 67 27quot N l j n AnalysisXk ZxnTe N k 01N 1 n0 27zk N l in Synthesis xnT iZXke N n 01N 1 N k0 Why DFT FOR APPROXIMATING DTFT OF A FINITEDURATION SIGNAL In reallife computations ALL SIGNALS ARE FINITE the consequence is that you can apply periodic extension to create a periodic discretetime signal H K H N4 N4 4 T iv Vt gt VTlvflvf t Finiteduration signal XnT Periodic signal x nT What is the relation between the DTFT of the finite duration signal XnT and the DFT of the periodic extended signal X nT Page 102 EE 422G Notes Chapter 10 Instructor Cheung Here is the definition of Discrete Fourier Transform DFT again N71 7 n Analysissz mem J N k 01N 1 110 1 N4 1 Synthesis xnT ZXke N n 01N 1 N k0 Compared with the analysis formula of the DTFT of a niteduration xn N71 XeJ T Z xn new quot0 It is easy to see that Xk equals XejmT eValuated at a Question 1 What does it mean Recall that XejmT is periodic with period equal to the sampling frequency fS or 21cT thus Xk represents the kth sample when sampling XeJ T with N samples per period For example if N6 we have HeimT CD 27T 0 27r T Question 2 Is there any loss in information do we need something similar to the Nyquist theorem in Frequency domain No as long as the number of samples per period is the same as or larger than the duration of the signal Why Xk is also the coef cient for the Discretetime Fourier series of the periodic signal X nT Page 103 EE 422G Notes Chapter 10 Instructor Cheung Interpretation Discretetime Fourier series Recall the continuoustime Fourier Series for a periodic signal Xt with period NT Anl 39 J 1 INT z z 39dz ays1s kz x 6 mo m N kzl Synthesis xt ZXke NT I Now we sample Xt with sampling period T the resulting discretetime sequence XnT is also periodic with period N NT W t Periodic signal Xt Now we consider the continuoustime surrogate Xst of XnT x z 20xnT5t nT This is also a periodic signal with period NT so we compute its Fourier Series coef cient by integrating the product of one single period with eXp j 21ckNT 71k A 1 NT N71 ijin Xk W 0 Zn0xnT5t nTe dt 1 N71 lmquot 2 7 xnTe N NT quot 0 If we define X k NTXk then it coincides with the analysis formula of DFT N71 j n Xk Z xn Te N 110 Page 104 EE 422G Notes Chapter 10 Instructor Cheung Plugging it back to the Fourier Series synthesis formula we have 27116 J 1 w 7 1 w n x t 2 ZXke 11 2 xnT 2 ZXke N N km N km This is almost the same as the synthesis formula for DFT 271k in N71 xnTiZX eJN n01N 1 N k k0 Unlike continuoustime Fourier series which needs infinitely many harmonics the discretetime needs only N of them The reason is that the discretetime eXponentials are themselves periodic in N as well eX 39 2 kNn eX 39n3927m eX 39n P J N P J N J P J N 2766 Let s consider the case when N4 The real part Rea1exp jn cosT n 5 51er1111111111111111Hn 0 kl cosTm2 0 n k2 W1111111111t 0111111111 151 NHHHHHHHH Page 105 EE 422G Notes Chapter 10 Instructor Cheung The imaginary part Imexpjn sin M n k0 sin0nun k1 sinTm2 0 n k2 sin7tn 9 9 9 9 9 9 9 9 9 gt 0 n k3 sin3nn2 n 0 k4 sin21tn un I W Q Q Q WW FQ W rf 0 As you can see exp n has four distinct sequences before it starts repeating 270 27139 7139 2722 273 37139 eX n uneX n eX n eX n eX 7meX n eX n 4 4 2 4 4 2 DFT Theorem states that ANY periodic sequence Xn with period 4 can be written as a linear summation of these four sequences the coef cients are the DFT of Xn xn X0un X1 exp 71 X2 eXp7m X3 eXp 37 n Page 106 EE 422G Notes Chapter 10 Instructor Cheung Example 1 4 m a J4 N71 j n X0 xnTe 4 11114 110 N4 7 39En 7 39E0 7 211 7 39le 7 39Es X1ZxnTeJ4 J4 eJ4 J4 eJ4 l j lj0 110 N4 7 39n 7 39410 7 39E1 7 39412 7 39413 X2 xnTe 4 e 4 e 4 e 4 e 4 1711710 110 N71 7n 761 7611 7612 7773 X3 xnTeJ4 J4 J4 J4 J4 1j l j0 110 Properties of the DFT lLinearity Axn Byn H AXk BXk 2TimeShift xn 7 m a Xke f2 quotN XkWNquotquotquot 3 Frequency Shift xne2 quotquotN H Xk 7 m 4 Parseval s Theorem Nil 2 1Nil 2 Zxn N ZIXUCH n0 160 5 Circular convolution xn yn H XkYk Page 107 EE 422G Notes Chapter 10 Instructor Cheung l X l n ltgt Xk l T l yn n 1 Yk 4 l l n ltgt XkYk But convolVing Xn and yn should result in XnT ynT I You may recall that convolVing two nite sequences of length N will result in a sequence of length 2Nl Thus it is OBVIOUSLY WRONG TO EXPECT THAT MULTIPLYING THEIR NPOINT DFT S WILL GIVE THE RIGHT ANSWER I I s H O W When multiplying the two Npoint DFT an operation called circular convolution occurs which is equivalent to computing a Npoint summation on the periodic extensions of the two original sequences Page 108 EE 422G Notes Chapter 10 Instructor Cheung So what should you do in to order to compute Xnyn in the frequency domain Answer Zeropad the original signal to 2N1 and take a 2N1 DFT xkT yTkT 2N 1pt IDFT XkYk I T n Page 109


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