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: X x[ ] For causal sequece, x = 0, < 0: Where is a complex variable All the values of that make the summatio to exist form a regio of covergece.
Problem: Give the sequece, Example fid the trasform of x. Solutio: We kow, Therefore, X Regio of covergece ROC Whe, 2
Problem: Give the sequece, Example 2 fid the trasform of x. Solutio: Therefore, X a a a Regio of covergece Whe, a a 3
Z-Trasform Table 4
Problem: Example 3 Fid -trasform of the followig sequeces. a. b. Solutio: a. From lie 9 of the Table: b. From lie 4 of the Table: 5
Liearity: Example 4 Problem: Z- Trasform Properties Fid - trasform of a ad b are arbitrary costats. Solutio: Usig - trasform table: Lie 3 Lie 6 Therefore, we get 6
Z- Trasform Properties 2 Shift Theorem: Verificatio: = m Sice x is assumed to be causal: The we achieve, 7
Problem: Fid - trasform of Example 5 Solutio: Usig shift theorem, Usig - trasform table, lie 6: 8
Z- Trasform Properties 3 Covolutio I time domai, Eq. I - trasform domai, Verificatio: Usig - trasform i Eq. 9
Problem: Solutio: Give the sequeces, Example 6 Applyig -trasform o the two sequeces, Fid the -trasform of their covolutio. From the table, lie 2 Therefore we get, 0
Iverse - Trasform: Examples Fid iverse -trasform of We get, Example 7 Usig table, Fid iverse -trasform of Example 8 We get, Usig table,
Iverse - Trasform: Examples Fid iverse -trasform of Sice, Example 9 By coefficiet matchig, Therefore, Fid iverse -trasform of Example 0 2
Iverse -Trasform: Usig Partial Fractio Problem: Solutio: Fid iverse -trasform of First elimiate the egative power of. Example Dividig both sides by : Fidig the costats: Therefore, iverse -trasform is: 3
Iverse -Trasform: Usig Partial Fractio Problem: Example 2 Solutio: Dividig both sides by : We first fid B: Next fid A: 4
Example 2 cotd. Usig polar form Now we have: Therefore, the iverse -trasform is: 5
Iverse -Trasform: Usig Partial Fractio Problem: Solutio: Dividig both sides by : Example 3 m = 2, p = 0.5 6
Example 3 cotd. From Table: Fially we get, 7
Partial Fuctio Expasio Usig MATLAB Problem: Example 4 Solutio: The deomiator polyomial ca be foud usig MATLAB: Therefore, The solutio is: 8
Partial Fuctio Expasio Usig MATLAB Problem: Solutio: Example 5 9
Partial Fuctio Expasio Usig MATLAB Problem: Solutio: Example 6 20
Differece Equatio Usig Z-Trasform The procedure to solve differece equatio usig -trasform:. Apply -trasform to the differece equatio. 2. Substitute the iitial coditios. 3. Solve for the differece equatio i -trasform domai. 4. Fid the solutio i time domai by applyig the iverse -trasform. 2
Problem: Example 7 Solve the differece equatio whe the iitial coditio is Solutio: Takig -trasform o both sides: Substitutig the iitial coditio ad -trasform o right had side usig Table: Arragig Y o left had side: 22
Example 7 cotd. Solvig for A ad B: Therefore, Takig iverse -trasform, we get the solutio: 23
Problem: Example 8 A DSP system is described by the followig differetial equatio with ero iitial coditio: a. Determie the impulse respose y due to the impulse sequece x =. b. Determie system respose y due to the uit step fuctio excitatio, where u = for 0. Solutio: a. Applyig Takig -trasform o both sides: o right side 24
Example 8 cotd. We multiply the umerator ad deomiator by 2 Solvig for A ad B: Hece the impulse respose: Therefore, 25
b. The iput is step uit fuctio: Correspodig -trasform: Example 8 cotd. [Slide 24] Do the middle steps by yourself! 26
27 Example 9 Problem: Solutio: Determie the -trasform ad the ROC of the sigal: 43 32 u x 4 3 3 ad 2 2 2 x x x u x u x Let Therefore Applyig -trasform: 4 3 2 X X X 3 ROC:, 3 2 ROC:, 2 2 X X But 3 ROC:, 3 4 2 3 X Fially, Itersectio
28 Example 20 Problem: Solutio: Determie the -trasform ad the ROC of the sigal: u b u x 0 0 l l b b X ROC : > ROC 2 : < b Case a. b < Case b. b > ROCs do ot overlap, so X does ot exist. ROC of X is < < b
29 A A A A A A A A...... 2 3 2 b b b Example 20 cotd. l l b b b b b X