Lecture 3. Digital Signal Processing. Chapter 3. z-transforms. Mikael Swartling Nedelko Grbic Bengt Mandersson. rev. 2016

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1 Lecture 3 Digital Sigal Processig Chapter 3 z-trasforms Mikael Swartlig Nedelko Grbic Begt Madersso rev. 06 Departmet of Electrical ad Iformatio Techology Lud Uiversity

2 z-trasforms We defie the z-trasform of a impulse respose h() as H(z) h()z () We assume a causal impulse respose h(). Causal meas that h() 0 for < 0. The sum is therefore limited to H(z) h()z 0 where z r e jω is a complex umber. Complex umbers are ofte writte as a magitude ad a phase. The trasform H(z) is therefore a complex valued fuctio of a complex valued variable. Example Some examples of z-trasforms directly from the defiitio: Fuctio z-trasform h() H(z) h(0) + h()z + h()z + δ() { } 0... δ( k) z k h( k) z k H(z) h () { } 3 H (z) 3 + z + z h () { } 0 3 H (z) 0 + 3z + z + z 3 z ( 3 + z + z ) Proof for the time delay. y() x( ) Y (z) y()z (3) () x( )z (4) z x( )z ( ) (5) z m x(m)z m (6) z X(z) (7)

3 Example A IIR-system ad its z-trasform. h() u() H(z) z (8) 0 (z ) + z (9) h() a u() H(z) z if z > (ROC) (0) a z () 0 ( ) a z 0 () (a z ) + z (3) a z if z > a (ROC) (4) ROC meas regio of covergece: for which z the sum coverges. For a causal sigal the ROC becomes a regio z R mi. This is the ormal case i this course. Example of z-trasform of o-causal sigal (page 54) Give: ( ) x() for all (5) Fid: The z-trasform X(z) of x(). 3

4 Solutio: X(z) 0 0 x() z (6) ( ( ) z (7) ) z + ( ) z (8) 0 ( ) ( ) z + z (9) 0 ( z ) + z + ( z ) + z (0) z + () z ( ) ( z )( z ) if z < ad z > (ROC) () We are forced to check the ROC for each case idividually. Covolutio becomes multiplicatio Covolutio of the sequeces i time-domai becomes a multiplicatio of the polyomials i z-domai. y() h() x() Y (z) H(z)X(z) (3) Proof: Y (z) y()z (4) h(k)x( k)z (5) k h(k)x( k)z ( k) z k (6) k h(k)z k x( k)z ( k) (7) k H(z)X(z) (8) 4

5 Cascade or Serial couplig of systems Cascadig two systems gives: y () x() h () h () y() The impulse respose of the whole system is the covolutio of the impulse resposes for the idividual systems. h tot () h () h () (9) The system equatio of the whole system is the product of the system equatios for the idividual systems. H tot (z) H (z)h (z) (30) Covolutio i the time-domai is multiplicatio if the z-domai. See earlier proof for covolutio of system ad sigal. Example Calculate iterests usig the z-trasform, cotiued from chapter. determie a formula for the balace o the accout. We had: We ca ow y().05 y( ) + x() (3) x() 00 u() (3) Apply the z-trasform. Y (z).05 z Y (z) + X(z) (33) X(z) 00 Solve for Y (z). Y (z) z (34) X(z) (35).05 z.05 z ( z (36) ) 0.05 z z Apply iverse-trasforms to obtai the time-sequece. The balace at years,, 3, 5, 0 ad 000 is y() 00 (.05 0)u() (38) (37) { } (39) 5

6 Solutio to secod order differece equatios First order differece equatios were solved i chapter. Usig the z-trasform we ca ow easily solve higher order differece equatios. We solve for 0 ad assume that both y() ad x() are zero for egative (system i rest). Give a secod order differece equatio: y().7y( ) + 0.8y( ) x( ) x( ) (40) We ca z-trasform each term i the equatio ad we get (assumig both x() ad y() are causal). Y (z).7z Y (z) + 0.8z Y (z) z X(z) z X(z) (4) Solve for Y (z). Y (z) z z.7z X(z) H(z)X(z) (4) + 0.8z Usig tables of formulas for z-trasforms we ca also easily determie y() ad h(). I most cases we are oly lookig for the system properties from H(z). Example: Fiboacci sequece (page 0) The Fiboacci sequece is a sequece where a value is the sum of the two previous values. x() { } (43) Ca we fid a closed form equatio for this sequece? y() y( ) + y( ) where y(0) ad y() (44) Solutio usig impulse respose. y() y( ) + y( ) + δ() (45) Apply the z-trasform. Y (z) z Y (z) + z Y (z) + (46) z z (47) Partial fractio expasio gives where Y (z) p p A p z + A p z (48) ( + 5 ) ( 5 ) A A 5 5 (49) (50) 6

7 After iverse z-trasform the closed form equatio for the sequece becomes y() A p + A p for 0 (5) This ca also be solved as a differece equatio with iitial values (later i the course, see page 0). Iverse z-trasform For most of the cases, use tables of formulas. Defiitio (page 8) The defiitio of the iverse z-trasform is y() πj Y (z)z dz (5) C where C is a closed path aroud the origi ad iside the ROC. Solved usig residual calculus. Polyomial divisio (page 83) Y (z).5z + 0.5z (53) + 3 z z z 3 + (54) y() { } (55) Use table of kow trasforms This is the method we will use the most. First order Y (z) 0.9z (56) y() 0.9 u() (57) Secod order with real poles by partial fractio expasio Y (z) 3 z + z z z (58) ( ) y() u() u() (59) 7

8 Secod order with complex poles by table of formulas si ( ) π 4 z Y (z) cos ( ) π 4 z + 4 z (60) ( ) ) π y() si( 4 u() (6) Split the system ito first ad secod order polyomials by partial fractio expasio. For secod order polyomials, check first ad foremost if the poles are real or complex valued. More o this i the ext lecture Solvig differece equatios with iitial values First order differece equatios were solved earlier ad we assumed that the iitial values were zero for < 0. We ca use the z-trasform eve if y() 0 for < 0. We solve for 0 ad assume x() 0 for < 0 but y() 0 (system ot i rest). We defie the oe sided z-trasform as Y + (z) y()z 0 eve if y() 0 for < 0. Usig this defiitio the z-trasform of time shift is differet. Assumig y 0 () y( ) the z + -trasform becomes Y 0 + (z) y 0 ()z 0 (6) (63) y( )z (64) 0 y( ) + y( )z ( ) z (65) z Y + (z) + y( ) (66) We ow have to apply the iitial values y() for < 0 to get the correct aswer. Usig the oe sided z-trasform we ca ow solve differece equatios with iitial values. See the example i the book for a solutio of the Fiboacci-sequece (page 07). Commo: x() X + (z) (67) (68) 8

9 k x( k) z k X+ (z) + x( )z (69) [ x( k) + x( k + )z + + x( )z k+ + z k X + (z) ] (70) Example First order differece equatio with iitial values. Give: y() ay( ) + x() with a give iitial value for y( ). Fid: y() for 0. Solutio: Y + (z) a [z Y + (z) + y( ) ] + X(z) (7) az X(z) + a y( ) (7) az Iverse trasform with the give values of X(z) ad y( ). Direct form II (page 65) We ca illustrate differece equatios graphically. A simple diagram is show below, called Direct form II or caoical form. More about this i chapter 9. Give: System draw o the form: w() x() + + y() b 0 z a w( ) b Fid: The coectio betwee x() ad y(). Solutio: Itroduce the helper variable w() ad the otatios X(z), W (z) ad Y (z). Determie the sigal w(). W (z) a z W (z) + X(z) (73) X(z) (74) a z 9

10 Determie the output sigal y(). Y (z) b z W (z) + b 0 W (z) (75) b 0 + b z X(z) (76) a z H(z)X(z) (77) Therefore is Y (z) H(z)X(z) where H(z) b 0 + b z a z (78) 0