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# Signals and Systems II ECE 634

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This 13 page Class Notes was uploaded by Miss Felicita Stiedemann on Thursday October 29, 2015. The Class Notes belongs to ECE 634 at University of New Hampshire taught by Daniel Brogan in Fall. Since its upload, it has received 17 views. For similar materials see /class/231685/ece-634-university-of-new-hampshire in Computer Engineering at University of New Hampshire.

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Date Created: 10/29/15

ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 51 The z Transform Daniel S Brogan The z transform is used to convert difference equations into algebraic equations Review What is a difference equation Suppose that you want to estimate the average sea level by taking several measurements at one point on the shore Your measurements will be corrupted by tides and waves From probability theory we can assume that averaging over a larger number of measurements taken at regular intervals will generate a more accurate estimate of the average sea level Let s call our measurements xn and our estimate yn where n is the sample index Using this approach the estimate will be Clearly out estimate can be a function of any number of previous inputs Suppose that instead we want to use a moving average for smoothing Perhaps we want to use only the previous five measurements in our estimate The equation would be yNxN 4xN 3x1v 2xN 1xN This is a lowpass filter Suppose that we assume that data that is closer in time to the present is more important than older data We could alter the coefficient weights to e g yN xN 4xN 3xN 2xN lxN This filter would have a different smoothing response than the previous filter would All of these filters are known as finite impulse response FIR filters because the effect of an input impulse on the estimate will eventually disappear For FIR lters the estimate typically depends only on the current and previous inputs IIR filters on the other hand also depend on previous estimates Thus an input impulse may have a time response that never goes to zero One example of an IIR filter is yN xN 4xN 3xN 2xN lxNyN lyN 2 Note that these filters are causal because they do not depend on future values Noncausal filters can be used in postprocessing data but not with realtime live data The unilateral ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 ztransform that we will study here is only applicable to causal systems Information about bilateral ztransforms can be found in the course text The de nition of the unilateral ztransform The z transform of xn is de ned as X Z 29601quot quot0 where z is a complex variable Likewise the de nition of the inverse ztransform is xnj Xzzquot391dz This counterclockwise integration must have a radius large enough to encompass all of the system poles The region of convergence As with the Laplace transform the z transform pair only applies within a limited region of convergence This should be clear from the above transform de nitions since it would be impossible to take the inverse transform if X z summed to in nity 0 Example 51 7 the region of convergence ROC Find the ztransform and the corresponding ROC for the signal xn aquotu n The z transform de nition states that Xz2aquotunz quot ifz jn 1 J2 3j Rather than working to sum all of these components we can recall that for a geometric series see next example lxx2x m for xltl L l x Thus Daniel S Brogan ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 z X f lt2 Z a or lal lt lzl Thus if z is greater in magnitude than a is each successive term of the sum will be smaller in magnitude than the previous one was 0 Example 52 7 deriving z transform pairs Find the ztransform of a 5n b u n 0 cos nu n d 5n5n 15n 25n 35n 4 From the de nition Xznxnz39quot xo x2x3 z z2 23 0 0 0 a Xaz 1Z ZZ 3 Xaz l forallz l l l b Xbz11z 1z 2z 3 from the previous example X17 2 1 for 1lt lzl Z pn 7 pn c cos nun un e0 e0 l em e m l em e p 1 e1 e j3 Xz 2 3 2 z 2 z 2 z 2 Daniel S Brogan 3 ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 1 em e e 1 ee ee ee Xz 1 2 3 1 2 3 2 z z z 2 z z z From the geometric sum equation for lzl gt eiml1 X 1 z I z 1 z z e39ml z z em 4 2 z em I z e39m 2 z em z e39m lz e39m z em ej eej 1 222 em e39m ZZ 2 J zz cos XEZ 2z emz e39m zz zeme39 1 ZZ chos 1 for Zgt1 1 1 1 1 2423zzz1 d X0z1Z ZZ 3Z 4Z 4 for z 0 The answer can also be found using the geometric sum since Xd z1liiiim iiiim Z Z Z Z 25 26 z7 z8 Whi i Z wz Example 7 deriving more z transform pairs Find the ztransform of a nun b aquot391un 1 From the de nition Daniel S Brogan ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 4 5 i z z z 25 26 z7 XEz1l mz l1 z 2 Z Z Z Z Z Z Z Z 1 1 1 1 1 1 1 1 1 z 31 zquot 222 23 z4 222 23 z4 1 X Z1 1z 2391 2392 2393 2394 1 1 1 1 1 2 3 2 1 2 z z 1 1 2 NZ T z 1 1 12 Z Ty b Xbz a 1u 1 aouz0 6119 6129 6130 Xbzlziz z 11 3 Using the geometric series sum Xbltzgtzlazia Exercise 53b 7 Finding the inverse z transform 2222 llz12 F1nd the 1nverse z transform of X z z lz 23 In order to use our experience with partial fraction expansion Daniel S Brogan ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 XZ222 llz12 A L B C L D J z z lz 23 Z llZ 2lz 22lz 23 Using the methods we learned previously Xz 3 3 1 2 z 2 1 2 2 2 22 2 23 Xz 3322 Z222 2Z22 Using Lathi Table 51 p498 numbers 2 and 10 2R1 222 xn 332quot quot 2quot 2m2quotun xn 332quot 2quot m2quotun Wm 3un 4 xn 3un 3un 4 Exercise E53 7 Inverse z transform by long division Using long division to find the power series in 2391 show that the inverse z transform of zz 05 is 05quot un or 2 quotun 1 052 1 0252 2 Z z 05 05 2 05 05 025Z 1 025Z 1 025Z 1 0125Z 2 0125Z 2 2 3 Xzl Z Z Z Daniel S Brogan ECE634 Signals and Systems I Spring 2009 Lecture 24 March 27 Since Xz xnz39 x0 x2 x3 xn 05quot un Daniel S Brogan ECE634 Signals and Systems 1 Spring 2009 Lecture 3 6 April 24 532 Stability 0 See also Lathi 310 especially the table on p 313 o BIBO and asymptotically stable all poles inside the unit circle 0 BIBO unstable and asymptotically marginally stable no poles outside the unit circle and some simple nonrepeated poles on the unit circle 0 BIBO and asymptotically unstable at least one pole outside the unit circle or repeated poles on the unit circle Daniel S Brogan illl ill H Hquot l n a Figure 321 Characteristic roots location and the corresponding characteristic modes ECE634 Signals and Systems I Spring 2009 Lecture 36 April 24 54 System Realization 0 Using all of the realization models from continuous systems replace ls with 12 That s it o For digital lter implementations rounding errors can move the system poles from the stable region across the unit circle to the unstable region The direct form implementations that we looked at previously are especially susceptible to this Parallel and cascade implementations are generally superior for the implementation of high order lters These forms are quite simple The parallel form places each term of the partial fraction expansion of the overall transfer function in a parallel path and sums the outputs of all the paths For example 422 282 24z28 62 22 22625 2125 21 25 The cascade form factors the overall transfer function into products of lower order terms that are placed in their own blocks For example 422 282 4zz7 Z7 4z 22625 2125 21 25 55 Frequency Response of Discrete Time Systems 0 If the system input to output relationship is 2quot 3 H 2 2quot then letting 2 e9 yields e19quot 3HeQeQquot and cosQn63 H e19 cosQnt9zH eQJ This result is only valid for BIBO and asymptotically stable systems as systems which do not decay cannot reasonably have a frequency response The above equation allows us to find the steadystate response of the system to a sinusoidal input at all frequencies 0 If the digital system has a continuous input cosat sampling this input every T seconds starting at t 0 yields cos ainT Thus we can equate the discrete and continuous frequencies as Q aiT 0 Example 510 Lathi p533 For a system specified by the equation yn 1 08yn xn 1 find the system response to the input alquotl Daniel S Brogan 2 ECE634 Signals and Systems I Spring 2009 Lecture 36 April 24 b cos in 02 6 c a sampled sinusoid cos 1500t with sampling interval T 0001 since we are dealing with the steadystate response initial conditions are all assumed to be zero Thus ynl 08yn xn1cgt zYz 08Yz ZXZ Likewise we could use yn 08yn l xn cgt Yz 082391Yz Xz Rearranging YZ z 1 lelmm 1ogzel Substituting 2 e9 HIE2 1 0186Q l 08cos jsin Q l 08cosS jsin 2 HIEQ1 08cosQ j08sinjQ 89 1 211171 08sinQ H JJl 08cos 2z064sin2QA 1 08 SQ hieQ 1 4sin 2 J Jl l6cosQ064cos2Q064sin2Q Amt 154cos 2 jg 1 1 4sinQ HE J Jl64 l6cosQz tan 5 4cos 2 4sinQ J 5 H 1 4 t 1 6 41 40cosQ an 54cos 2 a 1n610n Daniel S Brogan ECE634 Signals and Systems I Spring 2009 Lecture 36 April 24 4sin0 5 5 4cos0 441 40 5 z tan1 z tan39105 yn1He 1quotm b cos in 02 6 yn1 Hecoslgno2meH 7r 5 4s1n zj yn cos n 02 tan391 7139 6 7239 4l 40cos 5 4cos 6 6 s 4G 7139 71 yn cos n 02 tan 6 J3 4140 5 47 2 2 yn COS n 02 tan l 2 Jl9828cos n l1159 41 20J3 6 5 2x3 6 c The input for this system is cos1500t cos1500nT cos1500n0001 cosl5n cos15n4He15 ynHe15 5 1 4sin15 cos 15n tan m08093cos15n 07021 5 4cos15 yM 41 40cos15 The Periodic Nature of the Frequency Response The continuous cosine wave cosat is a function of two noninteger variables a and t However Q is the only noninteger variable for the discrete cosine wave cos Qn Thus Daniel S Brogan ECE634 Signals and Systems I Spring 2009 Lecture 36 April 24 Daniel S Brogan cosQn cos 9 272m n where m is also an integer variable Since Q is a frequency domain variable this argument extends to the frequency domain as H em H ej927rm In practical terms this means that the frequency response of any discretetime signal repeats every 27239 radianssample Thus it is bandlimited to the range 7239 to 7239 radianssample This means that the discrete frequency F 27239 is limited to the range 12 to 12 From the previous section the apparent frequency is QR 9 272m where 7239 S QR lt 7239 and m is an integer However because cos Qn 6 cosQn t9 the apparent frequency can be represented as IQH in the range 0 to 7239 where the phase is reversed in the cases that QB is negative Aliasing and Sampling Rate Aliasing occurs when a signal at one frequency outside of the fundamental range 7239 S QR lt 7239 appears as signal at another frequency inside the fundamental range In order to avoid this the continuous signal must satisfy 91 th lt 7239 where wk is the highest frequency component in the input Rearranging 7239 w 27239 fh lt l or Tlt 1 lt fh 2T 2f Since the sampling frequency is f lT ECE634 Signals and Systems I Spring 2009 Lecture 36 April 24 fh lt or f gt 2 fh which we know as the Nyquist theorem Example 513 A discretetime ampli er uses a sampling interval T 25 us What is the highest frequency of a signal that can be processed with this ampli er without aliasing mi 1 1 20kHz 2T 2 225ys 2 50m Daniel S Brogan

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