Chapter 2 z-transform. X(z) = Z[x(t)] = Z[x(kT)] = Z[x(k)] x(kt)z k = x(kt)z k = x(k)z k. X(z)z k 1 dz 2πj c = 1

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1 One-sided -transform Two-sided -transform Chapter 2 -TRANSFORM X() Z[x(t)] Z[x(kT)] Z[x(k)] x(kt) k x(k) k X() k0 k x(kt) k k0 k x(k) k Note that X() x(0) + x(t ) + x(2t ) x(kt) k + Inverse -transform Z [X()] x(kt) x(k) X() k d 2πj c Where c is a circle with its center at the origin of the plane such that all poles of X() k are inside it Z Transform of elementary functions: Unit step function { (t) 0 t x(t) 0 t<0 X() Z[(t)] k k k0 k0 Region of convergence > Geometric series a + ar + ar 2 + ar 3 + a r r < Exponential Function x(t) { e at 0 t 0 t<0 x(kt) e akt, k 0,, 2, X() Z[e at ] x(kt) k e akt k k0 k0 e at e at

2 See table of -transforms on page 29 and 30 (new edition), or page 49 and 50 (old edition). The -transform X() and its inverse x(k) have a one-to-one correspondence, however, the -transform X() and its inverse -transform x(t) do not have a unique correspondence. Properties and theorems of the -transform Multiplication by a constant: Z[ax(t)] ax() Linearity: Z[αf(k) +βg(k)] αf () +βg() Multiplication by a k : Z[a k x(k)] X(a ) Real translation theorem (shifting theorem): If x(t) 0for t<0 and Z[x(t nt )] n X() Z[x(t + nt )] n [X() n k0 x(kt) k ] Initial value theorem: x(0) lim X() Final value theorem: lim x(k) lim [( k )X()] Real convolution Theorem: let x (t) 0for t<0 x 2 (t) 0for t<0 then k X () X 2 () Z[ x (ht ) x 2 (kt ht )] h0 2

3 INVERSE TRANSFORM Different Methods. Direct division method (Power Series Method) 2. Computational method 3. Partial-fraction-expansion method 4. Inversion integral method Direct division method Express X() inpowersof Example Find Z of X() Solution: x(0) ; x() 2; x(2) 3; x(3) 4 Example 2 Solution: Find Z of X() 0 +5 ( )( 0.2) X() X() x(0) 0 x() 0 x(2) 7 x(3) 8.4 x(4) 8.68 Computational method X() 0 +5 ( )( 0.2) Solution: 0 +5 Let X() U() where U() now, U() u(0) + u() + u(2) u(k) k + 3

4 for U() u(0) u(k) 0, for k, 2, 3 Converting to difference equation now, x(k +2).2 x(k +)+0.2 x(k) 0u(k +)+5u(k) let k 2 ( ) x(0).2 x( ) x( 2) 0 u( ) + 5 u( 2) now, x( ) x( 2) 0 and u( ) u( 2) 0 x(0) 0 Similarly, we find x() 0 We may continue the process to find x(k), k 2, 3, using ( ) Partial Fraction Expansion To find the Z X(), we may expand X() or X() into partial fractions. X() is expanded since each of the expanded terms is generally available in -transform tables. Alternatively, X() may be expanded and use of the shifting theorem may be made. Example let Y () X() a X() a now, thus, Z {Y ()} y(k) a k X() Y () Z {X()} x(k) y(k ) a k { a k k, 2, 3, x(k) 0 k 0 4

5 General procedure for partial fraction expansion: let Given X(), find X() X() If M>N, no adjustment need be made to X(), If N>M, we divide through X() a 0 + a + + a N N b 0 + b + + b M M () c N M N M + c N M N M + + c + c 0 + d 0 + d + + d M M } b 0 + b + + b M {{ M } ψ() Factoring ψ() where we have one repeated pole of order k, callit r,and the rest unique, k+, k+2,, m ψ() A k ( r ) k + A k ( r ) k + + A r + M jk+ A j j (2) Where A j (k j)! [ dk j d k j ( j ) k ψ()] j, j, 2,,k A j ( j ) ψ() j, j k +,k+2,,m Substituting (3) into (2) and multiplying by and taking inverse transform gives us: Z [X()] x(n) Z [ c N M N M+ + c N M N M + + c 2 + c 0 ] k + Z A j M + Z A j ( r ) j r j jk+ x(n) N nm C N M δ(n +(N M +)) + [A N r + A 2 n n r + M jk+ A j n j ] u(n) + + A k n(n ) (n (k 2)) r n k+ (k )! Where the following has been used { } Z n (n ) (n (k 2)) an k+ u(n) ( a) k (k )! where u(n) is the unit step function. 5

6 Example Find Z {X()} where, X() ( 2 )( 4 ) Solution now, Where Thus X() A A 2 X() A A x(n) Z [ ] + Z [ ] 6 4 δ(n +2)+ 3 4 δ(n +)+ [ 5 2 ( 2 )n 7 6 ( 4 )n ] u(n) 6

7 Inversion integral method Background material: Suppose 0 is an isolated singular point (pole) of F(). Expand F() in a Laurent series about 0 F () a n ( 0 ) n b n + n0 n ( 0 ) n where a n 2πj b n 2πj Γ Γ 2 F () d n 0,, 2, ( 0 ) n+ F () d n, 2, 3, ( 0 ) n+ where Γ and Γ 2 are closed paths around 0 and b F () d() 2πj Γ where Γ is any closed path within and on which F() is analytic except at 0,andb is called the residue of F() at the pole 0. Now F () d Γ F () d + Γ F () d + + Γ 2 F () d Γ m 2πj(b + b b m ) Residue theorem 7

8 Inversion integral X() x(kt) k x(0) + x(t ) + x(2t ) x(kt) k + k0 X() k x(0) k + x(t ) k 2 + x(2t ) k x(kt) + Note, this is the Laurent series expression of X() k around point 0, and x(kt) is the residue x(kt) 2πj c X()k d the inverse integral for the -transform Inverse transform using inversion integral M x(k) x(kt) [residue of X() k at pole i of X() k ] i assuming M poles. The residue K, for simple pole is given by K lim i [( i )X() k ] The residue K, for multiple pole j of order q is given by K (q )! lim j d q d q [( j) q X() k ] 8

9 Example Find Z [X()], where X() 2 ( ) 2 ( e at ) Solution: X() k k+ ( ) 2 ( e at ) Simple pole at e at Double pole at [ 2 k+ ] x(k) residue of i ( ) 2 ( e at ) at pole i K + K 2 where [ K lim ( e at k+ ] ) e a(k+)t e at ( ) 2 ( e at ) ( e at ) 2 K 2 x(kt) (2 )! lim d d [ ( ) 2 k+ ( ) 2 ( e at ) k e e at see steps below at ( e at ) 2 kt T ( e at ) ( e akt ) e at k 0,, 2, ( e at ) 2 ] Steps d v u udv vdu u 2 lim d ( k+ ) d e at (k +) k ( e at ) k+ lim ( e at ) 2 [ (k +) k lim e k+ ] at ( e at ) 2 k + e at ( e at ) 2 k e at + e at ( e at ) 2 ( e at ) 2 k e e at at ( e at ) 2 9

10 Pulse-Transfer Function Difference equation: x(k)+a x(k ) + + a n x(k n) b 0 u(k)+b u(k ) + + b n u(k n) Taking transform X()+a X()+ + a n n X() b 0 U()+b U()+ + b n n U() G() X() U() b 0 + b + + b n n +a + + a n n Pulse Transfer Function Now, Kronecker delta function δ 0 (kt) δ 0 (kt) { for k 0 0 for k 0 Z [δ 0 (kt)] G() is the transform of the response to δ 0 (kt). It is called the pulse transfer function g(k) Z {G()} is called the weighting sequence. 0

11 transform method of solving difference equations Example Solve: x(k +2)+3x(k +)+2x(k) 0; x(0) 0, x() Solution taking the transform 2 X() 2 x(0) x() + 3 X() 3 x(0) + 2X() 0 Substituting initial data X() ( +)( +2) [ ] [ ] Z ( ) k, Z ( 2) k + +2 x(k) ( ) k ( 2) k, k 0,, 2,