3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,


 Ashlyn French
 2 years ago
 Views:
Transcription
1 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [ ] < k = As show i Chapter, the FT does ot always exist. If this occurs, it may be icoveiet to aalye the correspodig sigal/system. I some cases, we ca get aroud the problem by acceptig the impulse fuctio i the frequecy domai as a valid fuctio to represet the sigal i the frequecy domai. A more geeral, ad very ofte more coveiet, solutio is to adopt the socalled Ztrasform (ZT). As you will see i a momet, it is possible to obtai the ZT of sigals whose FT do ot exist. 3
2 The ZT ca be viewed as a more geeral versio of the FT. Its relatioship with the FT is aalogous to that betwee the Laplace trasform ad the Fourier trasform of a cotiuous time sigal. 3. Properties of ZTrasform The ZT of a discrete time sigal x [ ] is X() = xk [ ] k = where is a complex umber. Whe we evaluate the ZT o the uit circle, i.e. we replace i the above equatio by k j = e ω, the the FT of the sigal is obtaied (if it exists i the first place). Example : Determie the ZT of the rightsided expoetial fuctio where u [ ] is the uitstep fuctio. Solutio: x [ ] = au [ ], k k k ( ) X() = xk [ ] = a = a k= k= k= k 3
3 If the above summatio is to exist i the fiite magitude sese, the a < or equivaletly > a. I this case k X() = ( a ) = = ; > a a a k = I the above example, the regio where the ZT exists, i.e. > a, is called the regio of covergece (ROC), the poit = a where the ZT goes to ifiity is called a pole, ad the poit = where the ZT goes to ero is called a ero, The ROC ad the poleero plot of Example are show below (assumig a is real ad positive). Uit circle Im{} ROC pole ero Re{} Circle of radius a Example : Determie the ZT of the leftsided sigal x [ ] =au [ ] 33
4 Solutio: m k k k ( ) ( ) X() = xk [ ] = a = a = a k= k= m= m= m The sum exists oly if a < or < a. I this case m X() = ( a ) = = ; < a a a m = The aswer is similar to that i Example, except for the ROC. Uit circle Im{} ROC pole ero Re{} Circle of radius a Through these two examples, it becomes evidiet that it is very importat to specify the ROC of the ZT. Example 3: Determie the ZT of x [ ] = au [ ] + bu [ ] 3
5 Solutio: Usig the result from Example ad the fact that the ZT is a liear operatio, we have X( ) = + a b The first term exists if > a ad the secod term exists if > b. So the ROC of X() is the overlap of the idividual ROCs, i.e. the exterior of a circle of radius R = max { a, b} As for poles ad eros, we first rewrite X() as X() a+ b ( ) ( ) ( ) = + = a b a b ; > R So the poles are located at ad = ( a+ b)/. = a ad = b; the eros are located at = The ROC ad the poleero plot for a = 3/, b = / is show below. Im{} Uit circle ROC pole Circle of radius a 3 ero Re{} 35
6 Example : Determie the ZT of the sided sigal Solutio x [ ] = au [ ] bu[ ] Usig results from both Examples &, ad the fact that ZT is a liear operatio, we have X( ) = + a b The first term exists if > a while the secod term exists if < b. So the ROC of X() is the overlap of the idividual ROCs. Case I: If a < b, the the ROC is the rig a < < b I this case, the poles are located at = a ad = b, ad the eros are located at = ad = ( a+ b)/; see Example 3. Case II: If a > b, the ROC ad hece X() does ot exist. The ROC ad poleero plot for a = /, b = 3/ is 36
7 Uit circle 3 Im{} ROC pole ero Re{} I summary, what the ZT does is that it describe a DT sigal i terms of a p, a set of eros { }, ad a ROC: set of poles { j } M ( ) i a i i= i ( ) = ; ROC b j ( p ) j = j X I the above equatio, a i ad b j represet respectively the multiplicity of the i th ero ad the j th pole. The FT, if exists, is obtaied by evaluatig the ZT o the uit circle. Specifically, the magitude of the FT depeds o the separatios betwee j the poit e ω ad the poles ad eros accordig to M jω i ( ) = j= X e a jω i e i jω = ; e ROC jω b j e p j 37
8 ROC of the ZT of a fiiteduratio sigal:  The sigal x [ ] is of fiite duratio if x [ ] = for < ad >, where both ad are fiite, with.  I this case, the ZT is  It ca be easily show that ( ) = X x [ ] = ( ) [ ] [ ] max X = x x x = = = where xmax = max x [ ]. So the ZT is fiite for ay fiite but oero value of. If, the the ZT is also fiite at =.  The ROC of a fiite duratio sigal is the etire plae, excludig ifiity ad perhaps. ROC of the ZT of a rightsided sigal:  By a rightsided sigal, we mea x [ ] = for < for some fiite.  I this case, the ZT is X ( ) = x [ ] = 38
9  Let q j = re θ represets a poit o a circle of radius r i the plae. If the ZT exists o this circle, the for ay θ ( ) [ ]( ) o = o X q = x q < j  The ZT o the circle q = re φ, if exists, is q ρ = = q = ( ) = [ ]( ) = [ ] = [ ]( ) X q x q x q x q where ρ = q q Defie ( ) g [ ] = x [ ] q u [ ], f[ ] = ρ u[ ], ad h [ ] = g [ ] f[ ] = gk [ ] f[ k]. k = The it ca be easily show that 39
10 π jω jω h[] = X ( q ) = G( e ) F( e ) dω π π F e ω are the FTs of g [ ] ad f[ ]. It is j where G( e ω j ) ad ( ) jω iterestig to poit out that G( e ) X ( q ) ω= =, ad i geeral jω jω ( ) X ( qe ) G e =. j The importat thig is that G( e ω j ) exists. The FT F( e ω ) will also exist if ρ <, i.e. whe r > r. If this coditio holds, the the ZT o the circle of radius r exists. I summary, the ROC of a rightsided sigal is to the exterior of a circle. The radius of this circle equals the largest pole (i magitude sese). ROC of the ZT of a leftsided sigal:  By a leftsided sigal, we mea x [ ] = for > for some fiite.  The ROC is to the iterior of a circle. The radius of the circle equals the the smallest pole (i the magitude sese). ROC of the ZT of a twosided sigal:  By a sided sigal, we mea a [ ] x that stretches from = to =. 3
11  Sice a sided sigal ca be treated as the sum of a left sided sigal with a ROC of < R ad a rightsided sigal with a ROC of > R +, its ROC is the overlap of the two.  If R > R+, the ROC is Otherwise, the ZT does ot exist. R+ < < R+. Exercise: Show that the ROC must be a cotiuous regio. Exercise: I this exercise, we use the otatio x [ ] Z X( ); ROC: R x to represet the sigal x [ ] ad its ZT X( ) whose ROC is followig R x. Show the (a) ax[ ] + bx[ ] Z ax() + bx( ); ROC: Rx R x where represets itersectio of two ROCs. (b) x X R Z [ ] ( ); ROC: (except o x for the possible addito or delectio of =, or = ) x [ ] X( / ); ROC: R x (c) Z 3
12 (d) (e) (f) Z dx( ) x[ ] ; ROC: R d * * * x [ ] Z X ( ); ROC: R x Z * * x[ ] X (/ ); ROC: / R x x x[ ] x [ ] Z X ( X ) ( ); ROC: R R (g) x x 3. Iverse Fourier Trasform There are a umber of approaches to determie a sigal from its ZT. 3..: By Ispectio A ZT (or a compoet of a ZT) that you will ecouter from time to time is X( ) = ; ROC: > a a This is because may ZTs express the sigal i terms of poles (ad eros). Accordig to a earlier example, the time domai sigal is x [ ] au [ ] =. 3
13 Exercise: Deteremie x [ ] if d X( ) = ; ROC: > a a Aswer: This is simply the sigal x[ ] = au [ ] delayed by d samples (see Part (b) of the last exercise i Sectio 3.). Cosequetly ( d) x [ ] = a u [ d] 3.. Power Series Expasio I its most primitive form, the ZT is a power series i beig x [ ]. coefficiet associated with ( ), with the Cosequetly, the coefficiets i the power series represetatio of the ZT give you the sigal itself. Example: Determie x [ ] whe X( ) = e ; < Solutio: The power series represetatio of X () is m X( ) = e = = ; <!! = m= ( m) 33
14 where! deotes factorial. From this expressio, we ca deduce that x [ ] = u[ ]! ( ) Partial Fractio Expasio We assume that the ZT is give i the form X() = b a M ( c k ) k= ( d i ) i= where ck, k =,,..., M are the eros, di, i =,,..., are the poles, ad a, b are costats. The idea behid the Partial Fractio Expasio techique is to express the above ZT as a weighted sum of terms of the form ( d i ) where m is a iteger. The iverse ZT of the above ca be determied aalytically. Specifically, let m Z f [ ] F ( ) = ( d ) m m m 3
15 where d represets i geeral a pole of X( ). Sice d F F dm ( ) d m( ) = m( ) d, = Fm ( ) dm ( ) d this meas fm[ ] = ( + ) fm [ + ] ; dm ( ) refer to (b) ad (d) i the last exercise i Sectio 3.. Sice f [ ] [ ] du =, so f [ ] = ( + ) du [ + ], f3[ ] = ( + ) du [ + ], f[ ] = ( + 3) du [ + 3] 3, or i geeral fm[ ] = ( + m ) du [ + m] ( m )! 35
16 To describe the partial fractio expasio techique, we cosider three cases, (a) (b) (c) the umber of eros is less tha the umber of poles ad there are o multipleorder poles, the umber of eros is greater tha the umber of poles but still there are o multipleorder poles, ad there are more eros tha poles ad there is oe multipleorder pole. Commets:  A pole is of multipleorder if it is repeated more tha time.  The sceario of more tha oe multipleorder poles ca be cosidered as a straightforward extesio of case (c) above. Case M < ad o multipleorder pole All the poles are differet. I this case, the origial ZT ca be expressed as X() b M ( c k ) k= = = a i= ( d i ) i= d A A A = d d d A i i 36
17 Cosequetly x [ ] = Ai ( di) u [ ] i= The questio that remais to be aswered is: what are the Ai s? If we multiply both sides of X( ) by ( ) d, the result is ( ) ( ) A d A d d X() = A d d ow, if we evalue the above at = d, we simply have ( ) () = d d X = A I geeral, ( i ) d X() = Ai = d i Example: A leftsided sigal has a ZT of X( ) = ( )( ) 37
18 Determie the ROC ad the iverse ZT. Solutio: The ZT ca be expressed as X( ) A = = + ( )( ) ( ) ( ) A where ( ) A = X() = = / ad ( ) A = X() = = / Cosequetly X( ) = + ( ) ( ) Give that x [ ] is a leftsided sigal, the first term must have a regio of covergece of < / ad the secod term must have a regio of covergece of < /. The overall ROC is the overlap of the two, which is simply < /. The iverse ZTs of the first ad secod terms are ( ) / ( ) u[ ] respectively. So / u[ ] ad 38
19 ( ) ( ) {( ) ( ) } u x [ ] = / u[ ] / u[ ] = / / [ ]. Exercise: Repeat the last example for a sided sigal. Case M greater tha or equal to ad o multipleorder pole As i Case, all the poles are differet. I this case, the origial ZT ca be expressed as X M ( c k ) b = = + a k = () Q () ( d i ) ( d i ) i= i= R () where M Q ( ) = B r r= r ad R ( ) are respectively the quotiet ad remaider polyomials (i ) whe / M b a ( c k ) is divided by ( d i ). k = i= 39
20 Sice the degree of the remaider polyomial (i ) is less tha, it meas the secod term i X( ) ca be expressed i the same form as that i Case. Cosequetly, we ca rewrite X( ) as X() = Q () + ( d i ) i= R () A M r B r r= i= i = + i ( d ), where agai ( i ) Ai = d X() = d i Example: Determie the IZT of X( ) = ; < ( )( ) Solutio:  From the ROC, we kow that this is a leftsided sigal.  The deomiator ca be writte as D () = Dividig the umerator C ( ) = by the deomiator, usig log divisio, yields Q () = + 3
21 ad R ( ) = +. 3 This meas + X() = + + ( )( ) 3 A A = + + +, ( ) ( ) where A + + = ( ) = 3 6 ( )( ) = / A + + = ( ) = 3 6 ( )( ) = /  The iverse ZT is thus ( ) ( ) x [ ] = δ[ ] + δ[ ] + u[ ] + u[ ] Case 3 M greater tha or equal to ad multipleorder pole The pole d is repeated s times, i.e. 3
22 d = d = = d. s Each of the remaiig s poles has a order of. This meas the ZT trasform, i product form, is X() = b a k = M ( c k ) s ( p ) ( d i ) i= s+ The ZT, i summatio form, is X() A = + + M s r i m B r r i ( ) d = = i m= ( d ) C m The determiatio of the Ai s ad the ( i ) Br s are the same as before, i.e Ai = d X( ) ; i= s+, s+,...,, = d i ad B r is the rth coefficiets of the quotiet polyomial whe the umerator ( b / a) M ( c ) k= k of X( ) is divided by its deomiator s ( p ) ( d i= s+ i ). The Cm s ca be obtaied by first replacig the by w ad the compute 3
23 ( )!( ) s {( ) ( )} sm d Cm = dw X w sm sm sm d dw ote that ( dw) X ( w ) = ( dw) Bw + A + s r ( dw) Bw ( dw) ( dw) w= d M s s s r i m r r= i= m= ( dw i ) ( dw) M s i r r= i= i = + s + C r= r s r A. ( dw) Commets:  Evaluatig the ( s m) th order derivatives of the first two terms at w= d will always yield ero. s r  Evaluatig the ( s m) Cr dw at w= d will be ero except whe r = m. ote that whe r > m, we will at some poit ed up with differetiatig a costat. O the other had whe r < m, the result after differetiatio will iclude power of the term ( dw ), which whe evaluated at w= d equal to ero. th order derivative of ( ) C m 33
The ztransform. 7.1 Introduction. 7.2 The ztransform Derivation of the ztransform: x[n] = z n LTI system, h[n] z = re j
The Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discretetime LTI systems. 7.
More informationChapter 7: The ztransform. ChihWei Liu
Chapter 7: The Trasform ChihWei Liu Outlie Itroductio The Trasform Properties of the Regio of Covergece Properties of the Trasform Iversio of the Trasform The Trasfer Fuctio Causality ad Stability
More informationCOMM 602: Digital Signal Processing
COMM 60: Digital Sigal Processig Lecture 4 Properties of LTIS Usig ZTrasform Iverse ZTrasform Properties of LTIS Usig ZTrasform Properties of LTIS Usig ZTrasform ve +ve Properties of LTIS Usig ZTrasform
More informationDefinition of ztransform.
 Trasforms Frequecy domai represetatios of discretetime sigals ad LTI discretetime systems are made possible with the use of DTFT. However ot all discretetime sigals e.g. uit step sequece are guarateed
More informationZ  Transform. It offers the techniques for digital filter design and frequency analysis of digital signals.
Z  Trasform The trasform is a very importat tool i describig ad aalyig digital systems. It offers the techiques for digital filter desig ad frequecy aalysis of digital sigals. Defiitio of trasform:
More informationChapter 7 ztransform
Chapter 7 Trasform Itroductio Trasform Uilateral Trasform Properties Uilateral Trasform Iversio of Uilateral Trasform Determiig the Frequecy Respose from Poles ad Zeros Itroductio Role i DiscreteTime
More informationfrom definition we note that for sequences which are zero for n < 0, X[z] involves only negative powers of z.
We ote that for the past four examples we have expressed the trasform both as a ratio of polyomials i ad as a ratio of polyomials i . The questio is how does oe kow which oe to use? [] X ] from defiitio
More information6.003 Homework #3 Solutions
6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the
More informationSolutions of Chapter 5 Part 1/2
Page 1 of 8 Solutios of Chapter 5 Part 1/2 Problem 5.11 Usig the defiitio, compute the trasform of x[] ( 1) (u[] u[ 8]). Sketch the poles ad eros of X[] i the plae. Solutio: Accordig to the defiitio,
More informationSignal Processing in Mechatronics. Lecture 3, Convolution, Fourier Series and Fourier Transform
Sigal Processig i Mechatroics Summer semester, 1 Lecture 3, Covolutio, Fourier Series ad Fourier rasform Dr. Zhu K.P. AIS, UM 1 1. Covolutio Covolutio Descriptio of LI Systems he mai premise is that the
More informationM2.The ZTransform and its Properties
M2.The ZTrasform ad its Properties Readig Material: Page 94126 of chapter 3 3/22/2011 I. DiscreteTime Sigals ad Systems 1 What did we talk about i MM1? MM1  DiscreteTime Sigal ad System 3/22/2011
More informationGeneralizing the DTFT. The z Transform. Complex Exponential Excitation. The Transfer Function. Systems Described by Difference Equations
Geeraliig the DTFT The Trasform M. J. Roberts  All Rights Reserved. Edited by Dr. Robert Akl 1 The forward DTFT is defied by X e jω = x e jω i which = Ω is discretetime radia frequecy, a real variable.
More informationLecture 3. Digital Signal Processing. Chapter 3. ztransforms. Mikael Swartling Nedelko Grbic Bengt Mandersson. rev. 2016
Lecture 3 Digital Sigal Processig Chapter 3 ztrasforms Mikael Swartlig Nedelko Grbic Begt Madersso rev. 06 Departmet of Electrical ad Iformatio Techology Lud Uiversity ztrasforms We defie the ztrasform
More informationChapter 10: Power Series
Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because
More informationAppendix: The Laplace Transform
Appedix: The Laplace Trasform The Laplace trasform is a powerful method that ca be used to solve differetial equatio, ad other mathematical problems. Its stregth lies i the fact that it allows the trasformatio
More informationFrequency Response of FIR Filters
EEL335: DiscreteTime Sigals ad Systems. Itroductio I this set of otes, we itroduce the idea of the frequecy respose of LTI systems, ad focus specifically o the frequecy respose of FIR filters.. Steadystate
More informationExponential Moving Average Pieter P
Expoetial Movig Average Pieter P Differece equatio The Differece equatio of a expoetial movig average lter is very simple: y[] x[] + (1 )y[ 1] I this equatio, y[] is the curret output, y[ 1] is the previous
More informationChapter 3. ztransform
Chapter 3 Trasform 3.0 Itroductio The Trasform has the same role as that played by the Laplace Trasform i the cotiuoustime theorem. It is a liear operator that is useful for aalyig LTI systems such
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationENGI Series Page 601
ENGI 3425 6 Series Page 601 6. Series Cotets: 6.01 Sequeces; geeral term, limits, covergece 6.02 Series; summatio otatio, covergece, divergece test 6.03 Stadard Series; telescopig series, geometric series,
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationSolutions. Number of Problems: 4. None. Use only the prepared sheets for your solutions. Additional paper is available from the supervisors.
Quiz November 4th, 23 Sigals & Systems (5575) P. Reist & Prof. R. D Adrea Solutios Exam Duratio: 4 miutes Number of Problems: 4 Permitted aids: Noe. Use oly the prepared sheets for your solutios. Additioal
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a elemet subset of the set {,,,
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
More informationChapter 8. DFT : The Discrete Fourier Transform
Chapter 8 DFT : The Discrete Fourier Trasform Roots of Uity Defiitio: A th root of uity is a complex umber x such that x The th roots of uity are: ω, ω,, ω  where ω e π /. Proof: (ω ) (e π / ) (e π )
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationSolution of EECS 315 Final Examination F09
Solutio of EECS 315 Fial Examiatio F9 1. Fid the umerical value of δ ( t + 4ramp( tdt. δ ( t + 4ramp( tdt. Fid the umerical sigal eergy of x E x = x[ ] = δ 3 = 11 = ( = ramp( ( 4 = ramp( 8 = 8 [ ] = (
More informationThe ZTransform. (tt 0 ) Figure 1: Simplified graph of an impulse function. For an impulse, it can be shown that (1)
The ZTrasform Sampled Data The geeralied fuctio (t) (also kow as the impulse fuctio) is useful i the defiitio ad aalysis of sampleddata sigals. Figure below shows a simplified graph of a impulse. (tt
More informationSection 11.8: Power Series
Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i
More informationMATH4822E FOURIER ANALYSIS AND ITS APPLICATIONS
MATH48E FOURIER ANALYSIS AND ITS APPLICATIONS 7.. Cesàro summability. 7. Summability methods Arithmetic meas. The followig idea is due to the Italia geometer Eresto Cesàro (85996). He shows that eve if
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationECES352 Introduction to Digital Signal Processing Lecture 3A Direct Solution of Difference Equations
ECES352 Itroductio to Digital Sigal Processig Lecture 3A Direct Solutio of Differece Equatios Discrete Time Systems Described by Differece Equatios Uit impulse (sample) respose h() of a DT system allows
More informationChapter 4 : Laplace Transform
4. Itroductio Laplace trasform is a alterative to solve the differetial equatio by the complex frequecy domai ( s = σ + jω), istead of the usual time domai. The DE ca be easily trasformed ito a algebraic
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More informationEE Midterm Test 1  Solutions
EE35  Midterm Test  Solutios Total Poits: 5+ 6 Bous Poits Time: hour. ( poits) Cosider the parallel itercoectio of the two causal systems, System ad System 2, show below. System x[] + y[] System 2 The
More informationThe z Transform. The Discrete LTI System Response to a Complex Exponential
The Trasform The trasform geeralies the Discretetime Forier Trasform for the etire complex plae. For the complex variable is sed the otatio: jω x+ j y r e ; x, y Ω arg r x + y {} The Discrete LTI System
More informationModule 18 Discrete Time Signals and ZTransforms Objective: Introduction : Description: Discrete Time Signal representation
Module 8 Discrete Time Sigals ad ZTrasforms Objective:To uderstad represetig discrete time sigals, apply z trasform for aalyzigdiscrete time sigals ad to uderstad the relatio to Fourier trasform Itroductio
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More information4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3
Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges  that is, lims N n
Review of Power Series, Power Series Solutios A power series i x  a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More informationUnit 4: Polynomial and Rational Functions
48 Uit 4: Polyomial ad Ratioal Fuctios Polyomial Fuctios A polyomial fuctio y px ( ) is a fuctio of the form p( x) ax + a x + a x +... + ax + ax+ a 1 1 1 0 where a, a 1,..., a, a1, a0are real costats ad
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationCALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y SlopeItercept Equatio: y m b slope= m yitercept=
More informationf(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim
Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =
More informationAP Calculus Chapter 9: Infinite Series
AP Calculus Chapter 9: Ifiite Series 9. Sequeces a, a 2, a 3, a 4, a 5,... Sequece: A fuctio whose domai is the set of positive itegers = 2 3 4 a = a a 2 a 3 a 4 terms of the sequece Begi with the patter
More informationTEACHER CERTIFICATION STUDY GUIDE
COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra
More informationThe ztransform can be used to obtain compact transformdomain representations of signals and systems. It
3 4 5 6 7 8 9 10 CHAPTER 3 11 THE ZTRANSFORM 31 INTRODUCTION The ztrasform ca be used to obtai compact trasformdomai represetatios of sigals ad systems It provides ituitio particularly i LTI system
More informationThe ZTransform. Content and Figures are from DiscreteTime Signal Processing, 2e by Oppenheim, Shafer, and Buck, Prentice Hall Inc.
The ZTrasform Cotet ad Figures are from DiscreteTime Sigal Processig, e by Oppeheim, Shafer, ad Buck, 999 Pretice Hall Ic. The Trasform Couterpart of the Laplace trasform for discretetime sigals Geeraliatio
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More information6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
More informationDigital signal processing: Lecture 5. ztransformation  I. Produced by Qiangfu Zhao (Since 1995), All rights reserved
Digital sigal processig: Lecture 5 trasformatio  I Produced by Qiagfu Zhao Sice 995, All rights reserved DSPLec5/ Review of last lecture Fourier trasform & iverse Fourier trasform: Time domai & Frequecy
More informationFinitelength Discrete Transforms. Chapter 5, Sections
Fiitelegth Discrete Trasforms Chapter 5, Sectios 5.250 5.0 Dr. Iyad djafar Outlie The Discrete Fourier Trasform (DFT) Matrix Represetatio of DFT Fiitelegth Sequeces Circular Covolutio DFT Symmetry Properties
More informationProperties and Tests of Zeros of Polynomial Functions
Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by
More informationINFINITE SEQUENCES AND SERIES
11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS
More informationSeptember 2012 C1 Note. C1 Notes (Edexcel) Copyright  For AS, A2 notes and IGCSE / GCSE worksheets 1
September 0 s (Edecel) Copyright www.pgmaths.co.uk  For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk  For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright
More informationAlgebra II Notes Unit Seven: Powers, Roots, and Radicals
Syllabus Objectives: 7. The studets will use properties of ratioal epoets to simplify ad evaluate epressios. 7.8 The studet will solve equatios cotaiig radicals or ratioal epoets. b a, the b is the radical.
More informationFFTs in Graphics and Vision. The Fast Fourier Transform
FFTs i Graphics ad Visio The Fast Fourier Trasform 1 Outlie The FFT Algorithm Applicatios i 1D MultiDimesioal FFTs More Applicatios Real FFTs 2 Computatioal Complexity To compute the movig dotproduct
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationDigital Signal Processing
Digital Sigal Processig Ztrasform dftwave Trasform BackgroudDefiitio  Fourier trasform j ω j ω e x e extracts the essece of x but is limited i the sese that it ca hadle stable systems oly. jω e coverges
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationSummary: CORRELATION & LINEAR REGRESSION. GC. Students are advised to refer to lecture notes for the GC operations to obtain scatter diagram.
Key Cocepts: 1) Sketchig of scatter diagram The scatter diagram of bivariate (i.e. cotaiig two variables) data ca be easily obtaied usig GC. Studets are advised to refer to lecture otes for the GC operatios
More informationDiscreteTime Systems, LTI Systems, and DiscreteTime Convolution
EEL5: DiscreteTime Sigals ad Systems. Itroductio I this set of otes, we begi our mathematical treatmet of discretetime s. As show i Figure, a discretetime operates or trasforms some iput sequece x [
More informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More information(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)
Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig
More informationQuestion1 Multiple choices (circle the most appropriate one):
Philadelphia Uiversity Studet Name: Faculty of Egieerig Studet Number: Dept. of Computer Egieerig Fial Exam, First Semester: 2014/2015 Course Title: Digital Sigal Aalysis ad Processig Date: 01/02/2015
More informationCHAPTER I: Vector Spaces
CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig
More informationPhysics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing
Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam 4 will cover.., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam will cover..9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you
More informationMath 10A final exam, December 16, 2016
Please put away all books, calculators, cell phoes ad other devices. You may cosult a sigle twosided sheet of otes. Please write carefully ad clearly, USING WORDS (ot just symbols). Remember that the
More informationj=1 dz Res(f, z j ) = 1 d k 1 dz k 1 (z c)k f(z) Res(f, c) = lim z c (k 1)! Res g, c = f(c) g (c)
Problem. Compute the itegrals C r d for Z, where C r = ad r >. Recall that C r has the couterclockwise orietatio. Solutio: We will use the idue Theorem to solve this oe. We could istead use other (perhaps
More informationx a x a Lecture 2 Series (See Chapter 1 in Boas)
Lecture Series (See Chapter i Boas) A basic ad very powerful (if pedestria, recall we are lazy AD smart) way to solve ay differetial (or itegral) equatio is via a series expasio of the correspodig solutio
More informationChapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics:
Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals (which is what most studets
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisherprovided material Overview May mathematical
More information( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =!
.8,.9: Taylor ad Maclauri Series.8. Although we were able to fid power series represetatios for a limited group of fuctios i the previous sectio, it is ot immediately obvious whether ay give fuctio has
More informationU8L1: Sec Equations of Lines in R 2
MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (D) Cosider the lie passig through A (,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie
More informationMath 2784 (or 2794W) University of Connecticut
ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really
More informationPAPER : IITJAM 2010
MATHEMATICSMA (CODE A) Q.Q.5: Oly oe optio is correct for each questio. Each questio carries (+6) marks for correct aswer ad ( ) marks for icorrect aswer.. Which of the followig coditios does NOT esure
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science. BACKGROUND EXAM September 30, 2004.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Departmet of Electrical Egieerig ad Computer Sciece 6.34 Discrete Time Sigal Processig Fall 24 BACKGROUND EXAM September 3, 24. Full Name: Note: This exam is closed
More informationMAXIMALLY FLAT FIR FILTERS
MAXIMALLY FLAT FIR FILTERS This sectio describes a family of maximally flat symmetric FIR filters first itroduced by Herrma [2]. The desig of these filters is particularly simple due to the availability
More informationPLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)
Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages
More informationThe Phi Power Series
The Phi Power Series I did this work i about 0 years while poderig the relatioship betwee the golde mea ad the Madelbrot set. I have fially decided to make it available from my blog at http://semresearch.wordpress.com/.
More informationCurve Sketching Handout #5 Topic Interpretation Rational Functions
Curve Sketchig Hadout #5 Topic Iterpretatio Ratioal Fuctios A ratioal fuctio is a fuctio f that is a quotiet of two polyomials. I other words, p ( ) ( ) f is a ratioal fuctio if p ( ) ad q ( ) are polyomials
More informationSignal Processing. Lecture 02: Discrete Time Signals and Systems. Ahmet Taha Koru, Ph. D. Yildiz Technical University.
Sigal Processig Lecture 02: Discrete Time Sigals ad Systems Ahmet Taha Koru, Ph. D. Yildiz Techical Uiversity 20172018 Fall ATK (YTU) Sigal Processig 20172018 Fall 1 / 51 Discrete Time Sigals Discrete
More informationMath 210A Homework 1
Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the CauchyRiema equatios? Solutio. a) A fuctio f : G C is called
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationUnit 6: Sequences and Series
AMHS Hoors Algebra 2  Uit 6 Uit 6: Sequeces ad Series 26 Sequeces Defiitio: A sequece is a ordered list of umbers ad is formally defied as a fuctio whose domai is the set of positive itegers. It is commo
More informationMAT 271 Project: Partial Fractions for certain rational functions
MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project,
More informationFundamental Concepts: Surfaces and Curves
UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat
More informationChapter 7: Numerical Series
Chapter 7: Numerical Series Chapter 7 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals
More informationEE / EEE SAMPLE STUDY MATERIAL. GATE, IES & PSUs Signal System. Electrical Engineering. Postal Correspondence Course
SigalEE Postal Correspodece Course 1 SAMPLE STUDY MATERIAL Electrical Egieerig EE / EEE Postal Correspodece Course GATE, IES & PSUs Sigal System SigalEE Postal Correspodece Course CONTENTS 1. SIGNAL
More informationANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A.
013 ΜΑΘ Natioal Covetio ANSWERS (1) C A A A B (6) B D D A B (11) C D D A A (16) D B A A C (1) D B C B C (6) D C B C C 1. We have SOLUTIONS 1 3 11 61 iiii 131161 i 013 013, C.. The powers of i cycle betwee
More information2 Geometric interpretation of complex numbers
2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that
More informationChapter 1. Complex Numbers. Dr. Pulak Sahoo
Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia Email : sahoopulak1@gmail.com 1 Module2: Stereographic Projectio 1 Euler
More informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More information