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Fig. 336
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In the above, the actual contents of the box could, for example, consist of a series connection of ip- ops, or perhaps an array of charge-coupled devices. It will be a helpful reminder to conclude this section with a summary of certain algebraic operations that apply in the manipulation of z-transform equations. This is done in Table 2, with discussion below.
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CHAPTER 13 The Digital Processor
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Table 2. 1. 2. 3. 4. Some Valid Operations in the z-Domain, where Zv nT F z Zav nT aF z , where a is constant Z v1 nT v2 nT F1 z F2 z Zan v nT F z=a , where a is constant Zv nT kT Zv n k T z k F z , where v n k T 0 for n < k. This is the timedelay theorem.
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First, in the table, item (1) says that if F z is the z-transform of sequence v nT , and if a is constant, then the z-transform of av nT is a times the z-transform of v nT . Next, item (2) says that the z-transform of the sum of two or more sequences equals the sum of the transforms of the individual sequences. This property applies here because we are dealing with linear time-invariant conditions. Next, item (3) can be established as follows. Since, by eq. (573), Zv nT then, Zan v nT
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n 1 X n 0 n 1 X n 0 n 1 X n 0
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v nT z n F z
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v nT z=a n F z=a
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that is, if F z is the z-transform of a sequence v nT , and a is constant, then the ztransform of an v nT is found by replacing z with z=a in the transform of v nT . Problem 299 The notation U t denotes a continuous-time (CT) function called the unit-step function, de ned as equal to 1 for all positive time (including t 0) but equal to zero for all negative time, as illustrated below.
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Given that a CT function v t 3U t 20t is being sampled 100 times per second, what is the z-transform of v nT (Answer: F z z 3z 3:2 = z 1 2 Problem 300 Given v nT U nT U nT T U nT 2T , nd F z . Problem 301 Find F z for the v nT of problem 298.
The Inverse z-Transform
We have rightly said that the solution to a DT problem can be simpli ed if the mathematical work is carried out in the z-domain. After a solution is obtained in the z-domain, the nal step is to inverse transform the answer back into the time domain. In this
CHAPTER 13 The Digital Processor
section we ll consider how an answer in the z-domain can be inverse-transformed into the time domain. We begin our discussion as follows. Going back to eq. (573), we see that a function of z, say Y z , can be expressed in the basic form Y z y 0 y T z 1 y 2T z 2 y 3T z 3 y nT z n 583
where y 0 sampled value of y t at t 0, y T sampled value of y t at t T, y 2T sampled value of y t at t 2T, and so on. Then the corresponding answer to the above equation in the time domain is given by eq. (571); thus (now writing y instead of v ) ys t y 0  t y T  t T y 2T  t 2T y nT  t nT 584
It thus follows that if the answer to a DT problem comes out in the basic form of eq. (583), then there is no di culty in expressing the answer in the time domain, because the sample values of y t , that is, y 0 ; y T ; y 2T , and so on, required in eq. (584), appear directly as the coe cients of the powers of z in eq. (583). The practical di culty, however, is that a solution in the z-domain does not generally come out in the basic form of eq. (583); instead, the solution comes out in the form of the ratio of two polynomials in z, in which case the required sample values of y t cannot be found by simple inspection of the solution (as would be the case if the solution were in the form of eq. (583). It is possible, however, to put Y z directly into the form of eq. (583) by the use of ALGEBRAIC LONG DIVISION,* as the following examples will illustrate. Example 1
Write the function Y z z= z 0:6) in the time domain, that is, in the form of eq. (584).
Solution The rst step is to put the given Y z into the form of eq. (583), which, as mentioned above, can be done by using algebraic long division. For the given Y z the details are as follows. 1 0:6z 1 0:36z 2 0:216z 3 0:1296z 4 z 0:6 j z z 0:6 0:6 0:6 0:36z 1 0:36z 1 0:36z 1 0:216z 2 0:216z 2 0:216z 2 0:1296z 3 0:1296z 3
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