ELEC2400 Signals & Systems

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1 ELEC2400 Signals & Systems Chapter 7. Z-Transforms Brett Ninnes School of Electrical Engineering and Computer Science The University of Newcastle Slides by Juan I. Yu - July 24, 2003 p.1/40 7. Z-Transforms For continuous time systems, we analysed the response y(t using its Laplace Transform Y (s: The differential equation was transformed in an algebraic equation. The qualitative information on y(t was determined using the poles of Y (s. For discrete time systems, we use the difference equation: y k+n + a n 1 y k+n a 1 y k+1 + a 0 y k b m u k+m b 1 u k+1 + b 0 u k. We do not analyse the output sequence {y k } directly in the time domain, using instead the Z-Transform. Chapter 7. Z-Transforms p.2/40 Relationship between Z-Transforms and Laplace Transforms Consider the continuous time system: d 2 y(t dt + 2 dy(t dt + y(t 3 du(t dt Now in frequency domain, considering sinusoidal components in the input u(t and the output y(t: (jω 2 Y (ωe jωt +2(jωY (ωe jωt +Y (ωe jωt 3(jωU(ωe jωt Now cancelling e jωt on both sides, we get the Frequency Response of the system: Y (ω U(ω 3 (jω 2 + 2jω + 1 which, however, does not consider transient responses of the system. Chapter 7. Z-Transforms p.3/40

2 On the other hand, Laplace Transform considers the exponential components e st of the input u(t and the output y(t. Then: s 2 Y (se jst + 2sY (se jst + Y (se st 3s U(se st and we obtain the Transfer Function of the system: Y (s U(s 3s s 2 + 2s + 1 H(s This is a generalisation of Fourier analysis since sine wave signals e jωt are special cases of exponential signals e st, when s jω. Chapter 7. Z-Transforms p.4/40 Relationship between Z-Transforms and Laplace Transforms Illustrative diagram: d 2 y(t dt 2 Time Domain + 2 dy(t dt + y(t 3 du(t dt Fourier Laplace Frequency Domain (jω 2 Y (ω + 2jωY (ω + Y (ω 3jωU(ω s jω s 2 Y (s + 2sY (s + Y (s 3sU(s This gives us a complete bag of tools to work with in continuous time. What about discrete time? Chapter 7. Z-Transforms p.5/40 Relationship between Z-Transforms and Laplace Transforms Consider, for example, the discrete time system: y k+2 1.5y k u k+1 We assume an input u k Ue jωk ( : sampling period which generates an output at the same frequency Y e jωk. Then: e jω2 Y e jωk 1.5e jω Y e jωk Y e jωk 3e jω e jωk Cancelling e jωk on both sides, we find the discrete time Frequency Response: Y U 3e jω e j2ω 1.5e jω As for the continuous time case, this doesn t tell us anything about the transient response of the system. Chapter 7. Z-Transforms p.6/40

3 Considering an exponential in the input u k Ue sk, which produces an exponential output Y e sk, we have that: e 2s Y e sk + 1.6e s Y e sk Y e 5k 3e s Ue sk If we denote e s then we obtain the Transfer Function: Y U 3 2 H( If s jω then we have k e jωk, and then H( for the special choice e jω gives the sinusoidal steady state frequency response. Chapter 7. Z-Transforms p.7/40 Relationship between Z-Transforms and Laplace Transforms Illustrative diagram: Time Domain y k+2 1.5y k u k+1 Frequency Domain DFT e j2ω Y p(ω 1.5e jω Y p(ω+0.56y p(ω3e jω U p(ω Z e jω 2 Y ( 1.5Y ( U( Similar to the continuous time case, however: Sampling has been assumed in order for the discrete time sequences to exist In continuous time, e st is a general exponential and making s jω we get sinusoidal analysis ( e jωt cos ωt + j sin ωt. In discrete time, e s e st t is introduced, and to get sinusoidal analysis we use k e jω k cos ωk + j sin ωk. Chapter 7. Z-Transforms p.8/40 Relationship between Z-Transforms and Laplace Transforms Illustrative diagram: Time Domain y k+2 1.5y k u k+1 Frequency Domain DFT e j2ω Y p(ω 1.5e jω Y p(ω+0.56y p(ω3e jω U p(ω Z e jω 2 Y ( 1.5Y ( U( d p In continuous time dt p est (s p e st into the differential equation relates the strength Y (s of the component in y(t at e st to the strength U(s of the component in u(t at e st. In discrete time q p k k+p into the difference equation relates the strength Y ( of the component of {y k } at k e sk to the strength U( of the component in {u k } at k. Chapter 7. Z-Transforms p.9/40

4 The strength of e st in y(t is Y (s, given by: Y (s L{y(t} y(te st dt To know the strength of e st the sequence {y k } we consider again the signal y p (t: Y p (s L{y p (t} ( y k δ(t k e st dt y k δ(t k e st dt y k e sk Replacing e s we define the Z-Transform of {y k }: Y ( Z{y k } y k k Chapter 7. Z-Transforms p.10/40 Calculation of Z-Transforms If we substitute e jω we get Y ( e jω y k e jωk Y p (ω which gives Discrete Fourier Transform (DFT. Most text books define Y (, with k starting from 0: y k k implicitly assuming all causal signals ( 0 for k < 0. Chapter 7. Z-Transforms p.11/40 Calculation of Z-Transforms Recall that the sum of a geometric series: S N N 1 ar k is given by: S N a(1 rn 1 r If r < 1 then r N 0 as N. This yields: S a 1 r This result will be often used to compute Z-Transforms. Chapter 7. Z-Transforms p.12/40

5 Consider the sequence: Then, its Z-transform is: λ k ; k 0 y k k λ k k (λ/ k Which provided λ/ < 1, gives: 1 1 λ/ λ ; > λ Chapter 7. Z-Transforms p.13/40 Calculation of Z-Transforms Consider the sequence Then, its Z-transform is: cos ωk ; k 0 (cos ωk k 1 ( e jωk + e jωk k 2 1 ( e jω 1 k 1 ( + e jω 1 k { } e jω e jω ; > 1 1 { 2 1 (e jω + e jω } (e jω + e jω + 2 Which gives ( cos ω 2 2 cos ω + 1 ; > 1 Chapter 7. Z-Transforms p.14/40 Calculation of Z-Transforms We have found that (λ 1 k λ ; > λ Differentiating both sides with respect to λ > 0: d dλ (λ 1 k d dλ ( λ Therefore: d dλ λk k ( λ 2 kλ k ; k 0 And, in general: k p λ k ; k 0 kλ k 1 k 1 kλ k k λ λ ( λ 2 ; > λ p!λp ; < λ ( λ p+1 Chapter 7. Z-Transforms p.15/40

6 Time Domain Sequence {y k } Z Transform of {y k } λ k ; k 0 λ ; > λ cos ωk ; k 0 ( cos ω 2 2 cos ω + 1 ; > 1 sin ωk ; k 0 sin ω 2 2 cos ω + 1 ; > 1 k p λ k ; k 0 p!λ p ( λ p+1 ; < λ Chapter 7. Z-Transforms p.16/40 Calculation of Z-Transforms Time Domain Sequence {y k } Z Transform of {y k } λ k cos ωk ; k 0 ( λ cos ω 2 2λ cos ω + λ 2 λ k sin ωk ; k 0 λ sin ω 2 2λ cos ω + λ 2 kλ k sin ωk ; k 0 λ(2 λ 2 sin ω ( 2 2λ cos ω + λ 2 2 kλ k cos ωk ; k 0 λ[cos ω (2 + λ 2 2λ] ( 2 2λ cos ω + λ 2 2 Chapter 7. Z-Transforms p.17/40 Using the Z-Transforms U( and Y ( we can now obtain the strength of the sequence k in {u k } and {y k }. Example: Consider the basic filter y k+2 1.5y k u k+1 Then the transfer function is: U( If {u k } is a step function, then U(, and so 1 ( ( 1 70 ( ( ( 1 Chapter 7. Z-Transforms p.18/40

7 Example: (continued 50 Solution y_k 70 ( ( ( 1 Z 1 {Y (} 70(0.7 k 120(0.8 k + 50(1 k ; k Sample Number k However, in the calculations we glossed over a few things: We assumed the Z-Transform was linear. We assumed that the Z-Transform was unique. We assumed that {y k } was causal ( 0 for k < 0. Chapter 7. Z-Transforms p.19/40 Linearity of Z-Transform Suppose {u k } and {x k } are two sequences and α and β two real numbers. Then for the sequence: αu k + βx k The Z-Transform is: y k k α u k k + β αu( + βx( so that the Z-Transform is linear. (αu k + βx k k x k k Chapter 7. Z-Transforms p.20/40 Uniqueness of Z Transform Suppose two sequences {u k } and {x k } with Z-transform U( and X(. If U( X(, then by definition: U( u k k x k k X( U( and X(t are polynomials in 1. The two polynomials are equal if and only if all its co-efficients are equal. Therefore U( if and only if u k for every k. This means each Z-transform is uniquely associated with only one discrete time sequence. Chapter 7. Z-Transforms p.21/40

8 Time Shifting Property When using: Z{qy k } Z{y k+1 } Y ( we are implicitly assuming that {y k } is causal. However, {y k } might have non-causal components (y k 0 for k < 0 sometimes called initial conditions. Backward shifts in time Z{q n y k } q n y k k y k n k y n 0 + y 1 n ( n y n n + y 1 n (1 n y 0 + y n( n y k k + y k k k1 }{{} init. cond. } {{ } Y ( Chapter 7. Z-Transforms p.22/40 Time Shifting Property Backward shifts in time That is: Z{q n y k } n Y ( + n For n 1, 2, 3,... this becomes: Z{q 1 y k } 1 Y ( + y 1 Z{q 2 y k } 2 Y ( + 1 y 1 + y 2 n y k k k1 Z{q 3 y k } 3 Y ( + 2 y y 2 + y 3.. Z{q n y k } n Y ( + n+1 y y n Chapter 7. Z-Transforms p.23/40 Time Shifting Property An analogous result using Forward shifts in time is: n 1 Z{q n y k } n Y ( n y k k Which for n 1, 2 and 3 is: Z{qy k } Y ( y 0 Z{q 2 y k } 2 Y ( 2 y 0 y 1 Z{q 3 y k } 3 Y ( 3 y 0 2 y 1 y 2 This is not so useful in practice: Note that (for n 3 Y ( will depend on {y 0, y 1, y 2 }, which we are precisely interested to obtain! Chapter 7. Z-Transforms p.24/40

9 Example 1: Consider the case of borrowing money from the bank: we borrow C dollars, we repay in equal installments of P dollars, the interest rate per payment period is 100α% of unpaid principle, y k is the amount owed after k th payment, and u k payment per period. Then, by the time the (k + 1th payment is due: y k+1 y k + αy k u k q (1 + αy k u k. Taking the Z-Transform and using linearity and time shifting properties gives: Y ( y 0 (1 + αy ( U( Chapter 7. Z-Transforms p.25/40 Example 1: (continued y 0 So, we obtain (1 + α U( (1 + α Now y 0 C the amount initially borrowed, and U( corresponds to the constant repayment P. Then: P ; k 0 u k U( P /( 1 Substituting these values gives: C (1 + α P ( 1( (1 + α ( ( P α 1 + C P α (1 + α P ( α + C P (1 + α k ; k 0 α Chapter 7. Z-Transforms p.26/40 Example 1: (continued P ( α + C P (1 + α k ; k 0 α The system is unstable unless α < 0 However banks do not pay you for borrowing money! Of course, α > 0 and we are able to pay off the loan if C P α < Amount owed versus time In the figure: C $10, 000 α 8/12% per month for different payments P. Dollars owing p$60 p$100 p$200 p$ Month Number Chapter 7. Z-Transforms p.27/40

10 Example 2. Suppose a digital filter whose transfer function is: q H(q q 2 1.7q This is equivalent to the input-output relationship: y k+2 1.7y k u k+1 Taking Ztransforms of both sides of this difference equation: ( 2 Y ( 2 y 0 y (Y ( y U( u 0 Which on factoring out Y ( and U( becomes: ( 2 U( + ( 1.7y 0 + y 1 u Here Y ( depends on the unknowns y 0 and y 1! This happens because k 2 was implicitly taken as the division between past and future, and not k 0. Chapter 7. Z-Transforms p.28/40 Example 2. (continued The problem can be avoided rewriting the recursive input-output difference equation as: y k 1.7y k y k 2 u k 1 Equivalently, using the backward shift operator q 1 : H(q q q 2 1.7q q 1 H(q q q 2 Using the ZTransform time shifting properties: Y ( Y ( + y Y ( + 1 y 1 + y 2 1 U( + u 1 Which on factoring out Y ( and U( becomes: U( + ( y y 2 + u U( + ( y y u Chapter 7. Z-Transforms p.29/40 Example 2. (continued If we consider {u k } a unit step at k 0, then: 2 ( 0.9( 0.8( 1 + ( y y 2 ( 0.9( 0.8 Using partial fraction expansion, we obtain: ( ( y y 2 + (40 6.4y y 2 ( 0.8 Then, the step response of the filter is given by: 50 + ( y y 2(0.9 k +(40 6.4y y 2 (0.8 k ; k 0 y 1 y 2 ( 0.9 ; k 1 ; k 2 Chapter 7. Z-Transforms p.30/40

11 Example 2. (continued In particular, if the initial conditions y 1 and y 2 are ero, then: 50 90(0.9 k + 40(0.8 k ; k Response y_k of digital filter Sample Number The d.c. gain of this filter is 50, as expected: d.c. gain H(e jω 1 ω0 H( The (sometimes complex partial fraction expansion will be avoided using an Inverse Z-Transform formula. Chapter 7. Z-Transforms p.31/40 Inverting Z-Transforms Using Contour Integration If we make the substitution: e jω Y p (ω y k k Y ( And also d j e jω dω j dω dω d j Substituying these expressions in the IDFT: 2π 2π 0 Y p (ωe jωk dω 2π C Y ( k d j where the contour C is the unit circle, and simplifying: 1 Y ( k d 2πj This formula is known as the Inverse Z-Transform. C Chapter 7. Z-Transforms p.32/40 Inverting Z-Transforms Using Contour Integration We recall the Cauchy s Residue Theorem: 1 2πj C f(d m Res F ( p n n1 Where {p n } are the m poles of F ( inside C, and Res pk F ( is the residue of F ( at p k. Applying this theorem to the Inverse Z-Transform: 1 2πj C Y ( k d m Y ( Res p n n1 Where {p n } are the m poles of Y (/ inside the unit circle. If every pole p n only occurs once: m Y ( m Res k p n n1 n1 ( p n Y ( k (p n k pn Chapter 7. Z-Transforms p.33/40

12 Example 1 Consider the Z-Transform: 2 ( 0.8( 0.9( 1 By using the Contour integral inversion formula: 1 2πj 3 n1 C k+1 ( 0.8( 0.9( 1 d Res ( 0.8( 0.9( 1 p k k+1 k+1 k+1 k ( 0.9( ( 0.8( ( 0.8( (0.8 k ( ( (0.9 k ( ( (1 0.8( (0.8 k 90(0.9 k + 50 ; k 0 1 Chapter 7. Z-Transforms p.34/40 Inverting Z-Transforms Using Contour Integration From the previous example, we observe that if the poles in Y (/ only occur once, then y k can be obtained applying a simple set of rules: Cancel each pole one by one When you cancel a pole at p n, substitute p n into what is left. Multiply this number by p k n Add all these results up and you ve found the Inverse Z-Transform. Chapter 7. Z-Transforms p.35/40 Inverting Z-Transforms Using Contour Integration Example 2 Consider the Z-Transform: So: Y ( ( 1.5 ( 0.2( 0.5 This has poles at {0, 0.2, 0.5}. Therefore, applying the procedure: ( k ( (0.2k ( (0.5k + + (0 0.2( ( (0.5( (0 k (0.2 k 6.67(0.5 k Now and 0 k 0 for k > 0 so the final answer is: ; k 0 (21.67(0.2 k 6.67(0.5 k ; k 1 Chapter 7. Z-Transforms p.36/40

13 Example 3: Consider the Z-Transform: So: 2 ( ( 1 Y ( k k+1 ( 0.8e j π 5 ( 0.8e j π 5 ( 1 Using Contour integration the inverse Z-transform is: 1 2πj Y ( k d C Y ( k Y ( k Res + Res 0.8e j π 5 0.8e j π 5 Y ( k + Res 1 (0.8 k+1 e j(k+1 π j(k+1 5 (j1.6 sin π 5 (0.8ej π e π 5 ( j1.6 sin π 5 (0.8e j π e j π 5 2 Chapter 7. Z-Transforms p.37/40 Inverting Z-Transforms Using Contour Integration Example 3: (continued Which, after some manipulations, yields: { 4.923(0.8 k 0.8 sin πk π(k + 1 sin 5 5 } Response y_k of digital filter Sample Number Chapter 7. Z-Transforms p.38/40 Inverting Z-Transforms Using Contour Integration Repeated Poles: Suppose Y ( k / has a m poles at p, then the residue at p is given by: Y ( k ( 1 d m 1 Y (k Res ( pm p (m 1! dm 1 p When m 1 this reduces to the previous expression and set of rules. When m > 1, after cancelling the poles at p we have to differentiate m 1 times before substituting p. Chapter 7. Z-Transforms p.39/40

14 Repeated Poles: Example. Consider the Z-Transform: ( + 0.2( Then: So: 1 2πj C Y ( k d Y ( k k ( + 0.2( k Res 0.2 ( + 0.2( Res k 0.2 ( + 0.2( k ( d k d ( ( 0.2 k + ( + 0.2kk 1 k ( ( 0.2 k 6.25(0.2 k + 2.5k(0.2 k ( 0.2k (0.2k + 2k(0.2k ( 0.2k + (2k 1(0.2k Chapter 7. Z-Transforms p.40/40

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