Signal Processing and Linear Systems-B.P.Lathi copy

# 1112a by c omputing ikj i n b oth cases for k 0 1 2

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Unformatted text preview: t is n ot very useful unless we w ant to know only t he first few terms o f t he sequence f[k]. R epeated application of this property yields ¢==}~ [~F[zJ + f [-I]] + f[-2J f [k - 2]u[k] 6.. E xercise E I1.3 Using l ong division t o find t he power series in z -l, s how t hat t he inverse z-transform o f z/(z - 0.5) is (0.5)k u[k] o r ( 2)-k u[k]. 'V Relationship Between 681 11.2 Some P roperties o f z -Transform 1 = z2F[z] a nd h[k] and H[z] ¢ ==} z-mF[z] = L f[k - mJu[k - mJz-k k=O k =-oo F or causal systems, t he limits on t he s um are from k = 0 t o 0 0. T his equation shows t hat t he t ransfer function H[z] is t he z-transform of t he impulse response h[k] of a n LTID system; t hat is H[z] Recall t hat f [k - mJu[k - mJ = 0 for k < m, s o t hat t he limits on t he s ummation o n t he r ight-hand side can be taken from k = m t o 0 0. Therefore 00 Z { I[k - m]u[k - m]} = L (11.14) E xercise E ll.4 Redo Exercise E9.5 b y t aking t he inverse z-transform o f H[z]. f [k - m]z-k k =m T his i mportant result relates t he impulse response h[k], which is a time-domain specification of a system, to H[z], which is a frequency-domain specification of a system. The result is parallel t o t hat for LTIC systems. /:::,. ( 1l.16c) 00 Z { I[k - mJu[k - m]} h[k]z-k ¢ ==} + z-m L f[-kJz k k =l Proof: H[z] = L ( 1l.16b) m f [k - m]u[k] For an LTID system, if h[k] is its unit impulse response, then in Eq. (9.57b) w e defined H[z], t he s ystem transfer function, as h[k] 1 + : ;f[-IJ + J[-2J 00 = L f[rJz-(r+m) r =O 'V 1 = -F[z] zm 1 1.2 S ome properties o f t he Z-Transform To prove Eq. (11.16c), we have T he z -transform properties are useful in t he derivation o f z -transforms o f many functions a nd also in t he solution o f linear difference equations with constant coefficients. Here we consider a few i mportant p roperties of t he z -transform. 00 00 Z { I[k - m]u[k]} = L f[k - m]z-k = L f[r]z-(r+m) r =-m k=O Right Shift (Delay) If f[k]u[k] ¢ ==} F[z] m = t hen f [k - l]u[k - 1] ¢ ==} a nd f [k - m]u[k - m] and f [k - l]u[k] ¢ ==} 1 - F[z] z 1 ¢ ==} 1 z - F[z] z-m L f[-klzk + z-mF[zJ k =l ( IU5a) Left Shift (Advance) - F[z] zm + f [-I] ( IU6a) If ( IU5b) J[kJu[k] ¢ ==} F[zl t hen f [k + IJu[kJ ¢ ==} zF[zJ - zf[OJ ( 1l.17a) 682 11 D iscrete- Time S ystems A nalysis Using t he z - Transform R epeated a pplication o f t his p roperty y ields f [k + 2Ju[kJ = 11.2 683 S ome P roperties o f z -Transform We cannot find the z-transform of ku[k - 6J directly by using the right-shift property [Eq. (11.15b)J. So we rearrange it in terms of (k - 6)u[k - 6J a s follows: z { z (F[zJ - zf[OJ) - f[l]} f[k1 = ku[k] - [(k - 6)u[k - 6] = z2 F[zJ - z2 f[OJ - zf[lJ (11.17b) + 6u[k - 6]] We can now find t he z-transform of t he bracketed term by using the right-shift property [Eq. (11.15b)]. Because u[k] z~1 = a nd f [k + mJu[kJ = m -l z m F[zJ - zm L f [kJz-k u[k - 6J k=O Also, because ku[k] P roof: B y d efinition = 00 Z { I[k + mJu[k]} = L f[k =~ 1 (11.17c) (Z~1)2 k=O z 1 = 1 ~ (z _ 1)2 = z5(z _ 1)2 1 (k - 6)u[k - 6] + m Jz-k 1 z z _ 1...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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