COMM 60: Digital Sigal Processig Lecture 4 -Properties of LTIS Usig Z-Trasform -Iverse Z-Trasform
Properties of LTIS Usig Z-Trasform
Properties of LTIS Usig Z-Trasform -ve +ve
Properties of LTIS Usig Z-Trasform
Properties of LTIS Usig Z-Trasform
Example: h H h H H
Properties of LTIS Usig Z-Trasform
Check stability for the followig systems: EAMPLE : 4 4 H is ustable as it has oe pole at = 4 4. 4 ROC: causal e i 6 4 4 4 4 0 4 0 4 0 0 respose: Impulse 4 4 4 4 h h h h h h x x y y Y Y Y Y Ustable System Impulse respose icreasig Check from DE: From -domai:
H Check stability for the followig systems: EAMPLE : Y y 0.6 Check from DE: Y 0.6 Y 0.6 Impulse 0 h 0 h h 0.6 Y 0.6y respose: From -domai: 0.6 x is stable as it has oe pole at = 0.6 ROC: x 0.6h 0 0.6h0 0.6h 0.6 0.6 0.60.6 0.36 i. e 0.6 causal Stable System Impulse respose decreasig
Check stability for the followig systems: EAMPLE 3: H y Impulse h 0 0.7 Check from DE: 0.7y 0.7h h respose: h 0 h From -domai: 0.7 x 0.7h0 0.7h is stable as it has oe pole at = 0.7 ROC: 0. 7 causal x 0.5x 0.5 0.7h 0 0.5-0.5 0 0.5 0.7 0.5. 0.7. Stable System Impulse respose decreasig causal 0.84
EAMPLE 4: A system is described by the differece equatio i the form y 0.5y x a Fid H ad check the causality ad the stability. b - Fid He jw ad the the magitude ad phase respose SOLUTION a Take the -trasform of both sides of the differece equatio Y 0.5 Y Y 0.5
EAMPLE 4 Cotd. H Y 0.5 0. 5 ROC: 0.5 i. e 0.5 Imagiary Part o 0.0 0.5 Real Part Therefore, the system is causal ad stable
EAMPLE 4 Cotd. b To fid He jw from H, replace by e jw H jw H e 0.5 cos w j si w e H jw si jw jw cosw 0.5 j w e e 0.5 cos cos w j siw w j siw 0. 5 cos w si w cosw 0.5 si w H e jw H e jw cos w si w cosw 0. 5.5 cosw
EAMPLE 4 Cotd. cos w j si w e H jw si jw jw e cosw H e cosw 0.5 j w 0.5 jsiw
The Iverse Z Trasform Simpler methods to obtai THE INVERSE Z TRANSFORM - POWER SERIES EPANSION METHOD - PARTIAL-FRACTION METHOD
. POWER SERIES EPANSION METHOD Express ito a ifiite power series i -. This method is useful whe it is difficult to obtai a closed form expressio for iverse -trasform, the we fid oly the first several terms of x. EAMPLE : Fid x for =0,,, 3, 4, whe is give by : Solutio : 0 5 0. First, rewrite as a ratio of polyomial i -, as follows: 0. 5 0.
EAMPLE : Cotd. Dividig the umerator by the deomiator, we have. 0 7 0. 0 5 0 8.4 3 3 8.68 4 7 7 0.4 3 3 3.4 4 8.4 8.4 3.4.08 3 4 3 4 3.68 5 8.68 8.68 3.68.46 4 5 4 5 3.736 6
That is: 0 7 8.4 3 8.68 4... Thus by obtaiig the iverse -trasform, we have: x 0 7 8.4 3 8.68 4...
EAMPLE : Obtai the values of the first 4 samples of x if: Solutio : 3 3 4 The trasform is already i the form of power series i -. Sice has a fiite umber of terms. It correspods to a sigal of fiite legth. By ispectio All other xk values are ero x0 x x 3 x3 4
. POWER SERIES EPANSION METHOD Cot. Expasio of irratioal fuctios ito power series ca be obtaied from tables. Example : log a, a solutio Usig the power series expasio for : a x 0 x Thus x 0 a 0 The solutio of this problem is i Page 84; Text book
Example : Power series of expoetial: e 0!! 3 3!... x 3...! 3!
Examples of Power Series Expasio of some Fuctios
. PARTIAL FRACTION-EPANSION METHOD Notes: If N is the order of the umerator ad M is the order of the deomiator, the: -If N< M, divide H/, ad make the Partial fractio. -If N>=M, make log divisio ad the make partial fractio
. PARTIAL FRACTION-EPANSION METHOD Cotd. Cosider is give by : m a a b b b m m m m,...... 0 First: Factorie the deomiator of ad fid the poles of...... 0 m P P P b b b... P a P a P a m m
. PARTIAL FRACTION-EPANSION METHOD Cotd. The coefficiet a i are calculated as : P i i i P a For double poles at = P i i P i C P C where Geerally: p i m i r m r p d d r m C! m is max. power of the repeated poles i the problem
EAMPLE : Fid xk if is give by : Solutio : 0 0. We first expad / ito partial fractios as follows : The we obtai From tables we fid 0.5.5 0. 0..5 0. 0. u, 0. u
EAMPLE Cotd. Hece, x. 5 0. u or x. 5 0. ; 0,,,.. I this example, if, rather tha /, is expaded ito partial fractios, The we obtai, 0.5.5.5 0. 0. 0..5
EAMPLE Cotd. Note that the iverse trasform of Z.5.5Z However, by use of the shiftig theorem we fid Also, Z Hece,.5 0. 0..5u.5Z.5 x.5.50..5 0. Which ca be rewritte, x u 0.,,3,...5 0. is ot available from tables 0. u 0,,,... Which is the same as the result obtaied i the same example x 0 0
EAMPLE : Give the Z trasform Where a ad T are costats at e at e Determie the iverse Z trasform x by use of partial fractio expasio method Solutio : The partial fractio of / is foud to be Thus, e e at at
EAMPLE Cotd. Notig that 0,,,... ; Z 0,,,... e e Z at at The iverse trasform of is foud to be 0, u e x or,..,, e T x at at
EAMPLE 3 : Obtai the iverse Z trasform of 4 Solutio : 3 / 64 / 0.5 3 / 4 6 / 0.5 3 / 4 6 / 0.5 4 ad,,, of values Fid the 4 4 u u u x C C B A C C B A
EAMPLE 4 : Cosider :.5 0.5 Determie the iverse Z trasform x by use of partial fractio expasio method Sice the umerator is of the same degree as the deomiator we may divide the umerator by the deomiator.
EAMPLE 5 : Give = Determie the iverse Z trasform x by use of partial fractio expasio method
EAMPLE 6 : a b c 0 Give =, b b a 4ac Determie the iverse Z trasform x by use of partial fractio expasio method
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