ECE 308 -11 Z Transform Rational Z-Transform The inverse of the z-transform Z. Aliyazicioglu Electrical and Computer Engineering Department Cal Poly Pomona Rational Z-Transform Poles and Zeros The poles of a z-transform are the values of z for which if X(z)=∞ The zeros of a z-transform are the values of z for which if X(z)=0 X(z) is in rational function form M −k −−1 M ∑bzk Nz() bbz01+++... bzMk= 0 Xz()==−−1 N =M Dz( ) a01+++ az ... aN z −k ∑azk k=0 Nz() b ( z − zzz)( −−)...( zz) Xz()==0 z−+MN 12 M D()z a012()()() zpzp−− ... zp −N M z − z ∏()k M finite zeros at z = zz12, ,..., zM Nz() NM− k=1 Xz()== Gz N Dz() N finite poles at zpp= , ,..., p ∏()z − pk 12 N k=1 And |N-M| zeros if N>M or poles if M>N at the origin z=0 ECE 307-11 2 1 Rational Z-Transform Poles and Zeros We can represent X(z) graphically by a pole-zero plot in complex plane. Shows the location of poles by (x) Shows the location of zeros by (o). Definition of ROC of a z-transform should not contain any poles. Example: Determine the pole-zero plot for the signal n x()naun= () Im(z) 1 z The z-transform is Xz()== 1− az−1 z− a Re(z) o ax One zero at z1=0 0 One pole at p1=a . p1=a is not included in the ROC ECE 307-11 3 Rational Z-Transform Example.2: Determine the pole-zero plot for the signal ann 07≤ ≤ xn()= a > 0 0elsewhere 8 ,. … 7 −1 88 The z-transform is n 1− (az ) za− Xz() az−1 ==∑() −17 = n=0 1− az z() z− a (zz− )( zz−−)...( zz) Xz()= 12 7 z7 Im(z) j2/π kM The zeros j2/8π j4/8π zaek = zae= zae= o 1 2 o o zae= j14π /8 a Re(z) the poles at p=0 7 o x o 0 o o ROC: the entire z-plane except z=0 o ECE 307-11 4 2 Rational Z-Transform Pole Location and Time-Domain behavior for Causal Signals n For the signal x()naun= () a > 0 x(n) 0<a<1 The z-transform is 1 z Xz()==−1 1− az z− a n One zero at z1=0 . One pole at p1=a. x(n) The signal is decaying 0<a<1 a>1 The signal is fixed if a=1 n The signal is growing if a>1 The signal alternates if a is negative Im(z) Causal signals with poles outside the 0 a Re(z) unit circle become unbounded. o x ECE 307-11 5 Rational Z-Transform Pole Location and Time-Domain behavior for Causal Signals Impulse invariance mapping is z = e s T Im{s} 1 1 Im{z} fmax = ⇒ fs > 1 2π π Re{s} Re{z} -1 1 1 -1 s = -1 ± j ⇒ z = 0.198 ± j 0.31 (T = 1 s) s = 1 ± j ⇒ z = 1.469 ± j 2.287 (T = 1 s) s = j 2 π f Laplace Domain Z Domain Left-hand plane Inside unit circle Imaginary axis Unit circle Right-hand plane Outside unit circle ECE 307-11 6 3 Rational Z-Transform The system Function of a Linear Time –Invariant System The input sequence x(n) The output sequence y(n) The relationship in the z-domain Yz()= HzXz () () Yz() H(z) is obtained Hz()= is called system function X ()z ∞ H ()zhnz= ∑ ()−n We can take the inverse of z-transform to find h(n). n=−∞ H(z) can be obtained from a linear constant-coefficient difference equation. NN yn()=−∑∑ aynkkk () − + bxnk () − kk==10 ECE 307-11 7 Rational Z-Transform H(z) can be obtained from a linear constant-coefficient difference equation. NN yn()=−∑∑ aynkkk () − + bxnk () − kk==10 The z-transform ,,, NN −kk− Yz()=−∑∑ aYzzkk () + bXzz () kk==10 NN −kk− Yz()1+=∑∑ azkk bXzz () kk==10 N bz−k Yz() ∑ k Hz()==k=0 N Xz() −k 1+ ∑azk k=1 ECE 307-11 8 4 Rational Z-Transform 1 Example: yn()=−+ yn ( 1)2() xn 2 The z-transform from the difference equation 1 Yz()=+ Yzz ()−1 2 Xz () 2 The system function Yz() 2 Hz()== 1 Xz() 1− z−1 2 The system has one pole at and one zero at The inverse z-transform n 1 hn()= 2 un () 2 ECE 307-11 9 The inverse of the z-transform The inverse z-transform is formally given by 1 x()nXzzdz= ()n−1 2π j v∫C There are three methods for evaluation of inverse z- transform 1. Direct evaluation by Contour integration (using Cauchy residue theorem) 2. Expansion into a series of term in the variable and 3. Partial-fraction expansion and table lookup. ECE 307-11 10 5 The inverse of the z-transform The Inverse z-transform by Power Series Expansion ∞ −n With given ROC: X ()zcz= ∑ n n=−∞ Example: Using Power Series Expansion to determine the inverse z-transform of 1 >> y=[1]; Xz()= 11.5− zz−12+ 0.5− >> h=[1 –1.5 0.5]; >> n=7; a. ROC: z >1 >> y=[y zeros(1,n-1)]; //y={1,0,0,0,0,0,0} b. ROC: z < 0.5 >> [x,r]=deconv(y,h) x = 1.0000 1.5000 1.7500 1.8750 1.9375 1.9688 r = 0 0 0 0 0 0 1.9844 -0.9844 x(0)== 1,xx (1) 1.5, (2) = 1.75, x (3) = 1.875, x (4) = 1.9375ECE 307-11 11 The inverse of the z-transform a. Since the ROC is the exterior of the circle, we can expect x(n) to be a causal. 371531 1...+++zzzz−−12 − 3 + − 4 + 24 816 31 11−+zz−−12 22 31 1−+zz−−12 22 31 Therefore . zz−−12− 22 3 7 15 31 Xzzzzz( )= 1+++−−12 − 3 + − 4 + ..... 393 zzz−−−123−+ 24 816 244 73 zz−−23− x(n) con be obtained as 44 7217 371531 zzz−−−234−+ xn( )= {1, , , , ,.....} 488 24 816 15 7 zz−−34− ↑ 88 15 45 15 zzz−345−+−− 81616 31 15 zz−45− − 16 16 ECE 307-11 12 6 The inverse of the z-transform b. ROC is the interior of a circle. The signal x(n) is anticausal. The long division in the following way. 26430...zzz234+++ z 5 + 13 zz−−21−+11 22 13−+zz 22 2 32zz− Therefore, 39zz−+23 6 z Xz( )= 2 z23++ 6 z 14 z 4 + 30 z 5 + ..... 76zz23− 72114zzz234−+ x(n) can be obtained as 15zz34− 14 x(n )= {...30,14,6,2,0,0} 15zzz345−+ 45 30 31zz45− 30 ↑ ECE 307-11 13 The inverse of the z-transform The inverse z-transform by Partial Fraction Expansion We usually have X(z) function in a rational function form −−1 M Nz() bbz01+++... bzM Xz()== −−1 N Dz() a01+++ az ... aN z A rational function is called proper if M<N. A rational function is called improper if M≥N. An improper rational function (M≥N) can be written as the sum of polynomial and a proper rational function. Nz() Xz()=+ c cz−−−1() ++ ... c z MN +1 01 Mn− Dz() −12311−− 1 Example: 13++z zz + 1 −1 63 z Xz()= Xz()=+ 1 2 z−1 + 6 51−−12 51 1++zz 1++zz−12− 66 66 ECE 307-11 14 7 The inverse of the z-transform The inverse z-transform by Partial Fraction Expansion Find the poles (denominator roots). We can have distinct real poles, distinct complex poles, multiple-order real poles, and multiple-order complex poles. Distinct Poles: The poles are all different AA12 AN Xz()=+++−11−− ... 1 11−−pz12 pz 1 − pN z We can determine coefficient using different methods. Let’s give an example and use the of the methods ECE 307-11 15 The inverse of the z-transform Example: Determine the partial fraction expansion the following function 1 Xz()= 11.5− zz−12+ 0.5− 1 AA Xz()==+12 (1−− 1zzz−−−111 )(1 0.5 ) 1 − 1 1 − 0.5 z − 1 To find coefficients 11 AzXz=−(1 1−1 ) ( ) = = = 2 1 zz−−11==11(1−− 0.5z−1 ) 1 0.5 11 AzXz=−(1 0.5−1 ) ( ) = = =− 1 2 zz−−11==22(1−−z−1 ) 1 2 We have Assume that signal is causal From the z-transform table 21 Xz()=−−11− n 110.5−−zz x()nun=− 2 (0.5) () ECE 307-11 16 8 The inverse of the z-transform Using MatLab to find inverse z-transform >> syms z n >> iztrans(1/(1-1.5*z^-1+0.5*z^-2)) ans = 2-(1/2)^n >> syms z n >> ztrans(2-(1/2)^n) ans = 2*z/(z-1)-2*z/(2*z-1) ECE 307-11 17 The inverse of the z-transform Multiple – order poles: If X(z) has repeated poles. This time we use a different expansion. Let’s say D(z) contains a repeated poles (1 − pz − 1 ) r . Then Nz() Xz()= −1 r (1− pz ) D1 ( z ) −−+−+121rr AAzAzAzNz12rr− 1 1() =+−−112111 +++... −−−rr + (1)(1)(1)(1)()−−pzpzpzpzDz − − 1 To find A12,,...,AAr 1()dzNzrr−−11 dzNzr−1 () A = A = 11r−1 −1 21−1 z = z = (1)!rdzDz− 1 () p dz D1() z p zNzr−1 () Ar = 1 z−1 = Dz1() p ECE 307-11 18 9 The inverse of the z-transform Example: 1 1 AAAz−1 Xz()= Xz()==++123 (1+−zz−−112 )(1 ) (1+−zzzzz−−−−−1121 )(1 ) (1 + ) (1 − 1 ) (1 − 12 ) Determine the coefficients.