7.2 Relationship between Z Transforms and Laplace Transforms

Transcription

1 Chapter 7 Z Transforms 7.1 Introduction In continuous time, the linear systems we try to analyse and design have output responses y(t) that satisfy differential equations. In general, it is hard to solve a differential equation, but we make our job easier by taking Laplace Transforms, and then examining the Laplace Transform Y (s) of the output solution. We gain qualitative information on the nature of y(t) just by observing the locations of the poles in Y (s). In discrete time, the digital filters we implement all satisfy a basic digital filter equation y k+n + a n 1 y k+n a 1 y k+1 + a 0 b m u k+m + + b 1 u k+1 + b 0 u k. which is a difference equation, which is quite different to the differential equations governing our linear filters in continuous time. However, completely analogous to the continuous time case, we don t solve the time domain difference equation directly. We use a transform to gain qualitative information (and quantitative too if we are prepared to do some arithmetic) about the output solution sequence {y k }. The transform we use in discrete time is called the Z Transform. 7.2 Relationship between Z Transforms and Laplace Transforms In continuous time, a linear system is one whose input-output relationship satisfies a differential equation. For example the output y(t) might be related to the input u(t) according to d 2 y(t) dt +2 dy(t) dt + y(t) 3 du(t) dt Now, in general it is too involved to solve this equation in the time domain, and so we move to the frequency domain. That is, we suppose that the input u(t) and output y(t) are sinusoidal in nature. How much of y(t) is due to a sine wave at frequency ω is measured by the Fourier Transform ŷ of y evaluated at ω. That is, the Fourier transform tells us that y(t) has a component in it of the form ŷ(ω)e jωt. Also, we can easily work out that d dt ejωt jωe jωt, d2 dt 2 ejωt (jω) 2 e jωt, (7.1)

2 386 CHAPTER 7. Z TRANSFORMS and so if ŷ(ω)e jωt is a component of y(t) and y(t) satisfies the differential equation (7.1) then we can substitute the component ŷ(ω)e jωt into (7.1) and obtain (jω) 2 ŷ(ω)e jωt + 2(jω)ŷ(ω)e jωt + ŷ(ω)e jωt 3(jω)û(ω)e jωt. Now the e jωt term cancels on both sides of this equation to give a relationship between û(ω) and ŷ(ω) as ŷ(ω) û(ω) 3 (jω) 2 +2jω +1 and this ratio ŷ(ω)/û(ω) is called the frequency response of the system. Quite often this frequency domain knowledge is sufficient for the design of systems, and we avoid the need to solve (7.1). What is missing from the frequency domain analysis is information on the transient response of the system. Frequency response analysis tells us only about the steady state response to a sine wave input. What happens if we want to know how quickly the system recovers after we shock it, say with a step input? Laplace Transform analysis tells us this. Instead of assuming the input and output are sinusoidal, we assume they are exponential. How much of y(t) is due to an exponential e st is measured by the Laplace Transform Y (s) of y. That is, the Laplace Transform tells us that y(t) has a component in it of the form Y (s)e st and (as with Fourier Analysis) we can substitute this into the differential equation (7.1) to obtain s 2 Y (s)e jst +2sY (s)e jst + Y (s)e st 3s U(s)e st to obtain the relationship between Y (s) and U(s) as Y (s) U(s) 3s s 2 +2s +1 H(s) and this ratio is called the Transfer Function of the system. It shows that if the input contains an exponential component U(s)e st then the output will contain a component H(s)U(s)e st. This gives us more information about the nature of the solution of the differential equation (7.1) without requiring it s explicit solution. It is also a generalisation of Fourier analysis since sine wave signals of the form e jωt are special cases of exponential signals e st if we put s jω. We therefore have the following diagram illustrating out understanding of continuous time systems Time Domain d 2 y(t) dt 2 +2 dy(t) dt + y(t) 3 du(t) dt Fourier Transform Laplace Transform Frequency Domain (jω) 2 ŷ(ω) + 2(jω)ŷ(ω)+ŷ(ω) 3(jω)û(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? So far we ve only looked at the sinusoidal steady state analysis of systems in discrete time. What I mean is in discrete time we ve considered the input-output response of our systems to be described by the digital filter difference equation (4.2) of which a particular example is say y k+2 1.5y k u k+1. (7.2) As for the differential equation case in (7.1) it is possible, but involved to find the time domain solution of the difference equation (7.2). Instead, as for the continuous time case, we start by considering only sinusoidal inputs of the form u k Ue jωk ; sampling period (7.3)

3 7.2. RELATIONSHIP BETWEEN Z TRANSFORMS AND LAPLACE TRANSFORMS 387 which we assume generates a sinusoidal output at the same frequency Ye jωk (7.4) We then notice that y k+n Ye jω(k+n) e jωn Ye jωk and similarly for u k so that substitution of (7.3) and (7.4) into the filtering difference equation (7.2) gives e jω2 Ye jωk 1.5e jω Ye jωk +0.56Ye jωk 3e jω e jωk and after canceling the common e jωk on both sides we find that the ratio of the strength U of the input sine wave at frequency ω to the strength of the output sine wave due to the input is Y U 3e jω e j2ω 1.5e jω (7.5) This information on how sine wave inputs travel through the discrete time system (7.2) is often sufficient enough for us. However, as for the continuous time case, because it only considers sinusoids in steady state it doesn t tell us anything about the transient response of the system. To find out about this we follow the lead of Laplace Transform analysis and assume that the input has an exponential component u k Ue sk of strength U which produces an exponential output Ye sk of strength Y. Substituting into the difference equation (7.2) then leads to If we denote e s then we arrive at e 2s Ye sk +1.6e s Ye sk +0.64Ye 5k 3e s Ue sk Y U H() (7.6) This transfer function H() tells how exponentials decaying like k e sk map through the system defined by the digital filter difference equation (7.2). You might notice that with s jω we have k e jωk cos ωk +j sin ωk which means the general expression (7.6) will give the sinusoidal steady state frequency response (7.5) for the special choice e jω. The question we haven t answered yet is how in discrete time we judge what the strength U is of the component of the discrete sequence {u k } due to the sinusoidal sequence {e jωk }. The answer is of course provided by Fourier Transforms. More specifically, {u k } is a sequence of numbers not a physical quantity, so it doesn t have a spectrum. However, we are assuming that the numbers {u k } are derived from a physical quantity. This physical quantity is the continuous time signal u(t) which when ideally impulse sampled at period gives the perfectly sampled signal u p (t) k u k δ(t k ) ; u k u(k ) and this perfectly sampled signal u p (t) has a Fourier Transform û p (ω) k u k e jωk

4 388 CHAPTER 7. Z TRANSFORMS which is often called the Discrete Fourier Transform (DFT) of the sequence {u k }. However, you can see it is really more accurate to say it is the continuous time Fourier Transform of the continuous time impulse signal u p (t). There is an advantage with maintaining this physical link in our thinking since we found in section 6.1 on page 6.1 that û p (ω) is the spectrum û(ω) of the underlying continuous time signal u(t) repeated every ω s 2π radians per second: k u k e jωk û p (ω) 1 k û(ω kω s ). Therefore the strength U of the component e jωk in {u k } is U û p (ω) which, if the Nyquist Sampling condition is satisfied, is also U û p (ω) û(ω) the component in the continuous time signal u(t) at frequency ω rad/s. The same sort of arguments apply to the strength Y of e jωk in the sequence {y k }. Again, {y k } has no spectrum since it is just a sequence of numbers, but the impulse train y p (t) k y k δ(t k ) does have a spectrum given by the Fourier Transform ŷ p (ω) of y p (t): ŷ p (ω) k y k e jωk and this is often called the Discrete Fourier Transform of the sequence {y k] }. However, there is really no need to introduce this artificial new quantity called the DFT, since it is really the standard continuous time Fourier Transform of an impulse signal. At any rate, the strength Y of the component e jωk in {y k } is Y ŷ p (ω), the DFT of {y k }. In discrete time we therefore get the following commutative diagram Time Domain y k+2 1.5y k u k+1 Frequency Domain Discrete Fourier X form e j2ω ŷ p (ω) 1.5e jω ŷ p (ω) ŷ p (ω) 3e jω û p (ω) e jω Z Transform 2 Y () 1.5Y () U() This diagram is of exactly the same form as we had in continuous time. The major differences are 1. Sampling has been assumed in order for the discrete time sequences to exist 2. In continuous time we used the symbol e st for a general exponential and made the substitution s jω to get the special case of ( e jωt cos ωt + j sin ωt ) sinusoidal analysis. In discrete time we introduce a new symbol as e s e st t and this means that to get sinusoidal analysis we have to substitute e jω so that k e jω k cos ωk + j sin ωk. 3. In continuous time we just substitute the relationship d p dt p est (s) p e st

5 7.3. CALCULATION OF Z-TRANSFORMS 389 into the differential equation in order to relate 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 we substitute the relationship q p k k+p in order to relate 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. Now, with reference to this last point, the functions Y (s) and U(s) of s are well defined as Laplace transforms. For example, if we have a signal u(t) e st then we know that the function U(s) giving the strength of components in u(t) at e st is the Laplace Transform of u(t): U(s) L{u(t)} 0 e αt e st dt 1 s α But we don t know, in discrete time, how to calculate the equivalent strength function U() that tells us how much of {u k } is due to the exponential { k }. 7.3 Calculation of Z-Transforms The function U() is called the ZTransform of {u k }. Now, remember, that the Fourier Transform ŷ(ω) of y(t) is ŷ(ω) y(t)e jωt dt and that is we want to know the spectral content of a discrete sequence of numbers {y k } we pretend that the {y k } are the weights on impulses. This gives the new signal y p (t) and we take the Fourier Transform ŷ p (ω) of this signal ŷ p (ω) y k e jωk k as a measure of the strength of { e jωk } in {y k }. We follow the same method of argument to get the ZTransform of {y k }. The strength of e st in y(t) is Y (s) given by Y (s) L{y(t)} y(t)e st dt If we want to know the strength of e st the sequence {y k } we first have to form an impulse sampling signal out of the {y k } as y p (t) y k δ(t k ) k

6 390 CHAPTER 7. Z TRANSFORMS and then take the strength as Y p (s) L{y p (t)} k k y p (t)e st dt ( ) y k δ(t k ) e st dt k y k δ(t k )e st dt y k e sk With the notation e s we call the Laplace transform of y p (t) by a new name - the Z Transform of {y k }. Y () Z{y k } y k k. k This is just like the way we renamed the Fourier transform ŷ p (ω) of y p (t) as the Discrete Fourier Transform. We rename the Laplace Transform Y p (s) of y p (t) as the Z Transform Y (). Indeed, if we substitute e jω we get Y ( e jω ) k y k e jωk ŷ p (ω) so the Discrete Fourier Transform (DFT) of a sequence is a special case of the Z Transform of the sequence. Notice that most of the time in text books you will see Y () defined with k starting from 0: y k k since these books implicitly assume all the signals are causal and so 0for k<0. Let s see what the Ztransforms of some common signals are. Throughout all these calculations we use the result that if we have the sum of a geometric series S N N 1 ar k then there is a formula for S N S N a(1 rn ) 1 r Notice now that if r < 1 then r N 0 as N. This allows us to conclude that S k a 1 r Using this formula for the infinite sum of a geometric series we can evaluate the following examples

7 7.3. CALCULATION OF Z-TRANSFORMS 391 Example 7.1 { λ k ; k 0 y k k k λ k k (λ 1 ) k 1 1 λ/ λ ; λ < 1 ; > λ Example { cos ωk ; k 0 (cos ωk ) k ( e jωk + e jωk ) k 1 ( e jω 1) k 1 ( + e jω 1) k { } e jω e jω 1 { 2 1 (e jω + e jω } ) (e jω + e jω )+ 2 ; > 1 That is ( cos ω ) 2 2 cos ω +1 ; > 1

8 392 CHAPTER 7. Z TRANSFORMS Example 7.3 j 2 { sin ωk ; k 0 (sin ωk ) k ( e jωk e jωk ) k j ( e jω 1) k j ( e jω 1) k 2 2 j { } e jω e jω j { 1 (e jω e jω } ) (e jω + e jω )+ 2 ; > 1 That is sin ω 2 2 cos ω +1 ; > 1 Example 7.4 In example 7.1 we found (λ 1 ) k λ ; >λ Now, I can differentiate both sides with respect to λ and obtain But d dλ (λ 1 ) k d dλ λ d dλ (λ 1 ) k ( λ) 2 d dλ λk k kλ k 1 k 1 λ Z{kλk } Therefore if then { kλ k ; k 0 λ ( λ) 2 ; > λ.

9 7.3. CALCULATION OF Z-TRANSFORMS 393 In fact, we can be a bit more general than this by differentiating p times instead of 1 time to get d p dλ p λ k k 1 λ p k p λ k k (7.7) and also So that we get, on equating (7.7) and (7.8), that if {y k } is defined by { k p λ k ; k 0 then its Z transform is given by Example 7.5 d p dλ p ( λ) p! ( λ) p+1 (7.8) p!λp ( λ) p+1 ; <λ { λ k cos ωk ; k 0 λ k cos ωk k 1 ( λ k e jωk + λ k e jωk ) k 2 1 ( λe jω 1) k 1 ( + λe jω 1) k { } λe jω λe jω 1 ; > 1 { 1 2 λ 1 ( e jω + e jω ) } 2 1 λ (e jω + e jω ) 1 + λ 2 2 So that on simplifying we get ( λ cos ω ) 2 2λ cos ω + λ 2 Example 7.6 { λ k sin ωk ; k 0 (λ k sin ωk k) λ sin ω 2 2λ cos ω + λ 2

10 394 CHAPTER 7. Z TRANSFORMS Example 7.7 and So if Then d dλ λk sin ωk k λ λk sin ωk d ( λ sin ω ) dλ 2 2λ cos ω + λ 2 { kλ k sin ωk ; k 0 sin ω ( 2 λ 2 ) ( 2 2λ cos ω + λ 2 ) 2 λ(2 λ 2 ) sin ω ( 2 2λ cos ω + λ 2 ) 2 Example 7.8 and So if Then d dλ λk cos ωk k λ λk cos ωk ( ) d ( λ cos ω ) dλ 2 2λ cos ω + λ 2 [cos ω (2 + λ 2 ) 2λ] ( 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 We could go on calculating these Ztransforms, but these ones cover the most common cases. They are summarised in table Properties of ZTransforms Now we know how to calculate the functions U() and Y () that gives the strength of the sequence k in {u k } and {y k }. So, for example, if we return to the example of the digital filter y k+2 1.5y k u k+1 we worked out in (7.6) that if {u k } was considered as {U() k } and the output {y k } was considered as {Y () k } then the strength functions Y () and U() were related as U().

11 7.4. PROPERTIES OF ZTRANSFORMS 395 Time Domain Sequence {y k } Z Transform of {y k } { λ k ; k 0 λ ; > λ { cos ωk ; k 0 { sin ωk ; k 0 ( cos ω ) 2 2 cos ω +1 sin ω 2 2 cos ω + 1 ; > 1 ; > 1 { k p λ k ; k 0 { λ k cos ωk ; k 0 { λ k sin ωk ; k 0 { kλ k sin ωk ; k 0 { kλ k cos ωk ; k 0 p!λp ( λ) p+1 ; <λ ( λ cos ω ) 2 2λ cos ω + λ 2 λ sin ω 2 2λ cos ω + λ 2 λ(2 λ 2 ) sin ω ( 2 2λ cos ω + λ 2 ) 2 λ[cos ω (2 + λ 2 ) 2λ] ( 2 2λ cos ω + λ 2 ) 2 Table 7.1: Z transforms of some common discrete time sequences

12 396 CHAPTER 7. Z TRANSFORMS 50 Solution y_k Sample Number k Figure 7.1: Solution {y k } of difference equation. Therefore, if U() is a step function, then from table 7.1 and so U() 1 ( ) ( 1) 70 ( 0.7) 120 ( 0.8) + 50 ( 1) so that Y () is made up of a bunch of strength functions that we know from table 7.1 correspond to the sequences {0.7 k }, {0.8} k and {1 k } respectively so that the output sequence {y k } with strength function Y () is 70(0.7) k 120(0.8) k + 50 ; k 0 which is plotted in figure (7.1). Therefore, using the Ztransforms U() and Y () of the input and output sequences {u k } and {y k } we can work out the response {y k } of a digital filter when driven by {u k }. However, in the calculations we just performed we glossed over a few things. 1. We assumed the Ztransform was linear. That is, we assumed that if and {λ k } λ/( λ) then it was true that (0.7) k 120(0.8) k + 50(1) k

13 7.4. PROPERTIES OF ZTRANSFORMS 397 and for this to be true it is necessary that Z{y k } Z{70(0.7) k 120(0.8) k + 50(1) k } which means that the Ztransform must be linear 70Z{(0.7) k } 120Z{(0.8) k } + 50Z{(1) k } 2. We assumed that the Ztransform was unique. That is, we assumed that if then only one sequence { {70(0.7) k } can have this Ztransform 3. We assumed that {y k } was causal when in fact it is possible that {y k } may have non-ero values for k<0. Luckily, the first 2 of these assumptions are correct. The last assumption is one we shall have to consider more carefully. Since these assumptions are so important and are necessary whenever we use Ztransform methods to study a digital filter we should state and prove them formally: Linearity of ZTransform Suppose {u k } and {x k } are two sequences and that we make up a sequence {y k } by where α and β are real numbers. Then so that the ZTransform is linear. Uniqueness of Z Transform α αu k + βx k k k k y k k (αu k + βx k ) k u k k + β αu()+βx() k x k k Suppose {u k } and {y k } are 2 sequences with Ztransforms U() and Y (). Suppose that U() Y (). Then by the definition of the Ztransform U() u k k y k k Y () k k Now both sides of this equation are polynomials in 1 whose co-efficients are the samples {u k } and {y k }. Furthermore, 2 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 Ztransform is uniquely associated with only one discrete time sequence.

14 398 CHAPTER 7. Z TRANSFORMS Time Shifting Property So far, we ve been using the idea that Z{qy k } Z{y k+1 } Y () but as we shall see, this implicitly assumes that {y k } is causal. Of course, if {u k } is causal and our digital filter has a causal impulse then the component of {y k } due to {u k } must be causal. However, {y k } might have non-causal components (y k 0for k<0) not due to {u k } since they existed before {u k } was applied to the filter. These non-causal components, sometimes called initial conditions, have an effect on {y k } for k<0thathastobecalculated. In order to perform this calculation it actually turns out to be easier to look at backward shifts in time. Z{q n y k } q n y k k That is y k n k y n 0 + y 1 n 1 + y 2 n n y n n + y 1 n (1 n) + y 2 n (2 n) y }{{} 0 + y y initial conditions n n y k k k1 }{{} initial conditions + y k k. }{{} Y () Z{q n y k } n Y ()+ n For the common examples of n 1, 2, 3 this becomes Z{q 1 y k } 1 Y ()+y 1 n y k k k1 Z{q 2 y k } 2 Y ()+ 1 y 1 + y 2 Z{q 3 3 Y ()+ 2 y y 2 + y 3 There is an equivalent result using forward time shifts which for n 1, 2 and 3 is n 1 Z{q n y k } n Y () n y k k 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

15 7.4. PROPERTIES OF ZTRANSFORMS 399 but this result is not so useful in practice since it implicitly moves the point k 0. If q 3 is the largest time advance in an equation then the Ztransform of the solution Y () will depend on y 0,y 1 and y 2. But these are precisely the things we are trying to calculate since we don t know them! Examples make these points about time shifting, causality and initial conditions clearer. Example 7.9 Let s look at the case of borrowing money from the bank. Suppose we do the following Borrow C dollars. Repay in equal installments of P dollars. Interest rate per payment period is 100α% of unpaid principle. amount owed after k th payment u k payment per period. Then we can write an equation for how much we owe by the time the (k + 1)th payment is due as which using the forward shift operator q becomes y k+1 y k + αy k u k q (1 + α)y k u k. Taking the ZTransform of both sides and using the linearity and time shifting properties of Ztransforms then gives Y () y 0 (1 + α)y () U() so y 0 (1 + α) U() (1 + α). (7.9) Now y 0 C the amount initially borrowed. Furthermore, if I repay a constant amount P (starting from when I borrow the money, but not before) then { P ; k 0 u k and so U() P /( 1). Substituting these values into (7.9) then gives Y () as C (1 + α) C (1 + α) + P α P ( 1)( (1 + α)) { 1 (1 + α) Now, by the uniqueness and linearity properties of Z transforms, there is only one sequence {y k } that can have this Z transform. Reference to table 7.1 and use of the linearity property of Z transforms then tells us that the amount owed must be {1 (1 + α) k} C(1 + α) k + P α P ( α + C P α ) (1 + α) k }

16 400 CHAPTER 7. Z TRANSFORMS Amount owed versus time p$60 p$ Dollars owing p$ p$ Month Number Figure 7.2: Diagram showing our example problem solution where we calculate how much money we owe on a loan. Notice that this discrete time system is unstable unless α < 0. But negative α corresponds to the bank paying you interest for borrowing money! Of course, the fact that we have instability when we pay the bank interest (α > 0) is not a problem, since if it wasn t, we would never pay off the loan. We just have to make sure that our repayment amount p is big enough that C p α < 0 so that the instability of the system drives us through $0 owed (and not through $+ owed!). A plot of the solution for C $10, 000, α 8/12% per month, and various payment amounts per month is shown in figure 7.2 Let s pick a more difficult 2nd Order example Example 7.10 Suppose I have designed a digital filter whose transfer function H(q) is H(q) q q 2 1.7q so that the input sequence {u k } to the filter is related to the output sequence from the filter {y k } as H(q)u ( k q q 2 1.7q ) u k. (7.10)

17 7.4. PROPERTIES OF ZTRANSFORMS 401 So that the input-output relationship is y k+2 1.7y k u k+1. (7.11) If I take Ztransforms of both sides of this difference equation and use the time shifting properties then I obtain Z{q 2 y k } 2 Y () 2 y 0 y 1 Z{qy k } Y () y 0 Z{qu k } U() u 0 (7.12) 2 Y () 2 y 0 y 1 1.7Y ()() + 1.7y U() u 0 which on factoring out Y () and U() becomes ( ) 2 U()+ ( 1.7)y 0 + y 1 u (7.13) Therefore, we can find what the response {y k } of the filter will be to a step input {u k } by substituting U() 1 into the above equation and then finding what sequence {y k } has the transform Y (). However, as you can see from (7.13), the answer we get will depend on y 0 and y 1. But we don t know them yet, we are trying to calculate them! How did this situation come about? It happens because (7.11) implicitly takes k 2as being the division between past and future, whereas we want it to be at k 0. We can achieve this latter goal by rewriting (7.11) as y k 1.7y k y k 2 u k 1. (7.14) This is completely equivalent to (7.13), since all that really matters is how one sample is related to the samples before it. You can also arrive at (7.14) by dividing both numerator and denominator of H(q) in (7.10) by q 2 to obtain H(q 1 q 1 ) 1 1.7q q 2. Now, if we use the ZTransform time shifting properties on equation (7.14) then we arrive at Z{q 2 y k } 2 Y ()+ 1 y 1 + y 2 Z{q 1 y k } 1 Y ()+y 1 Z{q 1 u k } 1 U()+u 1 Y () Y () 1.7y Y () y y 2 1 U()+u 1 which on factoring out Y () and U() becomes U()+( )y y 2 + u U()+( )y y u

18 402 CHAPTER 7. Z TRANSFORMS This looks better, since Y () will now only depend on the past via y 1,y 2 and u 1. Suppose now, as we suggested before, that {u k } is a unit step at k 0. Then since in this case u ( 0.9)( 0.8)( 1) + ( )y y 2. (7.15) ( 0.9)( 0.8) The only way we know so far to find a {y k } that corresponds to Y () (i.e. to invert the Ztransform Y ()) is to match what we see with Table 5.1. To do this matching we ll have to perform a partial fraction expansion So that A, B, C must satisfy 2 ( 0.9)( 0.8)( 1) A B C 1 A( 0.8)( 1) + B( 0.9)( 1) + C( 0.8)( 0.9) for any value of. Since we get equality for any value of, it must be equal for the special cases 0.8, 0.9, and 1to give So that 0.8 B( )(0.8 1) 0.8 B A( )(0.9 1) 0.9 A 90 1 C(1 0.8)(1 0.9) 1 C 50 2 ( 0.9)( 0.8)( 1) (7.16) Performing similar, but more arithmetically complicated calculations I get [ ( ) 81 ( 0.9( 0.8)( 1) y 1 ( 0.9) + 32 ( 0.8) + 49 ] y 1 (7.17) ( 1) Substituting (7.16) and (7.17) into (7.15) then allows me to rewrite Y () as (I use the fact that I know u 1 0) ( ( y y 2 ) + ( y 1 36y 2 ) Looking at table 7.1 I see that the sequence is uniquely associated with the ZTransform 0.9 ( 1 ) ( ) + ( y y 2 ) 0.8 ). (7.18) { λ k ; k 0 0 ;Otherwise λ

19 7.4. PROPERTIES OF ZTRANSFORMS Response y_k of digital filter Sample Number Figure 7.3: Plot of solution sequence {y k }. and so I can conclude from the rewriting of (7.15) in (7.18) that the step response of my filter is ( y y 2 )(0.9) k +( y y 2 )(0.8) k +( y 1 36y 2 )7 ; k 0 ; k 1 y 1 y 2 ; k 2 If the initial conditions (or non-causal components) for my filter output y 1 and y 2 are ero, then the answer becomes much simpler. { ( 90)(0.9) k + 40(0.8) k + 50 ; k 0 0 ;Otherwise and this response is plotted below: There are 2 important points to come out of this example 1. The d.c. gain of this filter is obviously 50 since a unit magnitude step eventually becomes a d.c. value of 50 at the output. You also know that the frequency response of a digital filter H(q) is obtained by substituting e jω for q. So for H(q) in the example I get a frequency response H(e jω ) e jω e j2ω 1.7e jω

20 404 CHAPTER 7. Z TRANSFORMS So the d.c. gain of the filter is d.c. gain H(e jω ) ω0 H(1) It is arithmetically very complicated to do the calculations to find {y k }. This is mainly due to the partial fraction expansion that we had to do to separate Y () into parts we could recognise from table 5.1. Fortunately, we can in most cases significantly alleviate the arithmetic difficulties mentioned in item 2. We do this by avoiding partial fraction expansion to identify {y k }. Instead we directly calculate {y k } from Y () using a formula for the Inverse ZTransform. 7.5 Inverting ZTransforms Using Contour Integration In the introduction to this chapter we motivated the definition of the Ztransform Y () of {y k } as k y k k (7.19) by first looking at how the Discrete Fourier Transform (Continuous Time Fourier Transform of an impulse sampled sequence) was defined as ŷ p (ω) k y k e jωk (7.20) We can continue using this analogy between ZTransforms and DFT s to derive a formula for the Inverse ZTransform. Specifically, we know that we can invert ŷ p (ω) to recover {y k } by recognising that ŷ p (ω) is periodic, so the righthand side of (7.20) is the Fourier series representation of ŷ p (ω), so the {y k } must be the Fourier co-efficients of ŷ p (ω), so they must be given by 2π 2π 0 ŷ p (ω)e jωk dω (7.21) and this expression is sometimes known as the Inverse Discrete Fourier Transform. Now let s suppose that I make the substitution in (7.20). Then I get ŷ p (ω) e jω k y k k Y () (7.22) If I also make the substitution e jω in (7.21) and also substitute (7.22) in (7.21) I have d dω j ejω j dω d j

21 7.5. INVERTING ZTRANSFORMS USING CONTOUR INTEGRATION 405 and so I get 2π 1 Y () k d j C 1 Y () k d (7.23) 2πj C and this formula is known as the inverse ZTransform. It is a contour integral around C the unit circle. This might look complicated at first, but actually it makes our job of calculating {y k } from Y () much easier since we can use Cauchy s Residue Theorem to calculate contour integrals: 1 f()d 2πj C m n1 Res F () p n Where m is the number of poles of F () inside C, p k is the position of the kth pole, and Res pk F () is the residue of F () at p k. If we apply this to (7.23) we get m Y () Res k (7.24) p n n1 Finally, if the pole in Y () k only occurs once, then there is a very simple formula for the Residue Y () Res k lim ( p n ) Y () k (7.25) p n p n so that (7.24) becomes m lim ( p n ) Y () p n ( m ( p n )Y () n1 n1 k pn ) (p n ) k (7.26) This probably looks quite confusing since it has been presented quite quickly. It becomes clear when we try some examples. Example 7.11 In example 7.10, we arrived at (for ero initial conditions) 2 ( 0.8)( 0.9)( 1) for the Ztransform of the output of a digital filter when fed with a unit step input. Now in example 7.10, we deduced the {y k } corresponding to this Y () by messy partial fraction expansion. Let s try to do it now by using the Contour integral inversion formula (7.23): 1 2πj C k+1 ( 0.8)( 0.9)( 1) d.

22 406 CHAPTER 7. Z TRANSFORMS The integrand has m 3poles at p 1 0.8,p and p 3 1within C. Cauchy s Residue Theorem says the integral only depends on the position of these poles as follows: 1 2πj 3 k+1 Res p k ( 0.8)( 0.9)( 1). n1 All the poles only occur once, so we can use the simple formula in (7.25) to calculate each residue and we end up with: 3 ( p n ) k+1 lim p n ( 0.8)( 0.9)( 1) k+1 k+1 ( 0.9)( 1) + k ( 0.8)( 1) ( 0.8)( 0.9) n1 0.8(0.8) k ( )(0.8 1) + 0.9(0.9) k ( )(0.9 1) + 1 (1 0.9)(1 0.9) 40(0.8) k 90(0.9) k Which is exactly the same answer we obtained before in example You can see now from the previous example that if the poles in Y ()/ only occur once for each pole, then (7.26) boils down to a simple set of rules 1. Cancel each pole one by one 2. When you cancel a pole at p n, substitute p n into what is left. 3. Multiply this number by p k n 4. Add all these results up and you ve found the inverse ZTransform. Example 7.12 So Y () ( 1.5) ( 0.2)( 0.5) This has poles at 0, 0.2 and 0.5. Therefore, using my simple rule for singly occuring poles I get (0 1.5)0 k ( )(0.2)k + (0 0.2)(0 0.5) 0.2( ) 15(0) k (0.2) k 6.67(0.5) k Now 0 0 1and0 k 0fork>0somyfinalanswer is { ; k 0 (21.67)(0.2) k 6.67(0.5) k ; k 0 + ( )(0.5)k (0.5)( ) 1

23 7.5. INVERTING ZTRANSFORMS USING CONTOUR INTEGRATION 407 Example 7.13 So Y () k 2 ( )( 1) k+1 ( 0.8e j π 5 )( 0.8e j π 5 )( 1) Using Contour integration the inverse Ztransform of Y () then is 1 Y () k d 2πj C Y () k Y () k Y () k Res + Res + Res 0.8e j π 5 0.8e j π 5 1 { } (0.8) k+1 e j(k+1) π 5 j1.6 sin π 5 (ej π 5 1) e j(k+1) π 5 1 j1.6 sin π + 5 (e j π 5 1) 1 0.8e j π 5 2 { } (0.8)k+1 e j(k+1) π 5 1 j1.6 sin π j2im 5 e j + π 5 1 (1 0.8 cos π 5 )2 + (0.8 sin π 5 { } )2 (0.8)k e j π 5 k e j(k+1) π 5 1 sin π Im 5 2(1 cos π 5 ) cos π 5 (0.8) k { (sin π 5 )(1 cos π 5 ) sin πk } π(k + 1) sin { (4.454)(0.8) k sin πk π(k + 1) sin 5 5 A plot of this is shown on the next page. } The only difficulty with this method of inverting ZTransform using Contour integration occurs when a pole is repeated. When this happens, evaluation of the residue at the repeated pole gets a little trickier. To be specific, suppose Y () k / has a pole at p, and that this pole occurs m times. In other words, suppose Y () k / has a term of the form ( p) m in its denominator. Then the residue at p is given by Y () k Res p 2πj (m 1)! lim p d m 1 Y ()k ( p)m dm 1 In the case when m 1, this reduces to the rule we ve already encountered of canceling the pole at p and substituting p into what is left. However, when m is bigger than one, we have to introduce an extra step. After canceling the poles at p, but before substituting in p, we have to differentiate the pole canceled expression m 1 times. Example 7.14 Suppose I want to find the inverse ZTransform of ( +0.2)( )

24 408 CHAPTER 7. Z TRANSFORMS 4 Response y_k of digital filter Sample Number Figure 7.4: Plot of solution sequence {y k }. Then So Y () k 1 Y () k d 2πj C k k ( +0.2)( 0.2) 2 Res 0.2 ( +0.2)( 0.2) 2 + Res 0.2 ( +0.2)( 0.2) 2 k ( 0.2) 2 + d k 0.2 d ( +0.2) ( 0.2) k + ( +0.2)kk 1 k ( +0.2) ( 0.2) k 6.25(0.2) k +2.5k(0.2) k [( 0.2) k (0.2) k +2k(0.2) k] 6.25 [(2k 1)(0.2) k + (0.2) k] k