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 ztransform 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] ¢ ==} zmF[z] = L f[k  mJu[k  mJzk
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 ztransform 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 ighthand 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 ztransform o f H[z]. f [k  m]zk k =m T his i mportant result relates t he impulse response h[k], which is a timedomain
specification of a system, to H[z], which is a frequencydomain 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]zk ¢ ==} + zm 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 ZTransform
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]zk = 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] zm L f[klzk + zmF[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 ztransform of ku[k  6J directly by using the rightshift 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 ztransform of t he bracketed term by using the rightshift property
[Eq. (11.15b)]. Because u[k]
z~1 = a nd f [k + mJu[kJ = m l z m F[zJ  zm L f [kJzk 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 Jzk 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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