# qr code vb.net library * See note 33 in Appendix. in Visual Studio .NET Generator Code 128C in Visual Studio .NET * See note 33 in Appendix.

* See note 33 in Appendix.
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CHAPTER 13 The Digital Processor
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We continue in this manner until the change in the values of the coe cients becomes as small as we wish, at which point we terminate the series. In the above case, we can continue on for several more terms until we have the good approximation that z 1 0:6z 1 0:36z 2 0:216z 3 0:1296z 4 0:0778z 5 0:0467z 6 z 0:6 which is now in the form of eq. (583), and thus, by direct comparison with (583), we see that the actual sampled values of y t , at t 0; T; 2T; . . . ; 6T, are y 0 1:0 y T 0:6 y 2T 0:36 Now, putting these values into eq. (584), we have that the equation for the sampled function in the time domain is ys t  t 0:6 t T 0:36 t 2T 0:216 t 3T 0:1296 t 4T 0:0778 t 5T 0:0467 t 6T We can now use the above to pictorially show the form of y nT versus nT in the manner previously described. The result is shown in Fig. 337. y 3T 0:216 y 4T 0:1296 y 5T 0:0778 y 6T 0:0467
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Fig. 337. y(nT) versus nT, for
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z . z 0:6
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Given Y z z z 2 1:9z 1
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Find the rst 8 sample values of y t ; that is, y 0 ; y T ; y 2T ; . . . ; y 7T .
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Solution We rst put Y z into the form of eq. (583) by using algebraic long division, as follows, where, to save space, we ve rounded o to two decimal places.
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CHAPTER 13 The Digital Processor
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z 1 1:90z 2 2:61z 3 3:06z 4 3:20z 5 3:02z 6 2:54z 7 z2 1:9z 1 j z z 1:90 z 1 1:90 z 1 1:90 3:61z 1 1:90z 2 2:61z 1 1:90z 2 2:61z 1 4:96z 2 2:61z 3 3:06z 2 2:61z 3 3:06z 2 5:81z 3 3:06z 4 3:20z 3 3:06z 4 3:20z 3 6:08z 4 3:20z 5 3:02z 4 3:20z 5 3:02z 4 5:74z 5 3:02z 6 2:54z 5 3:02z 6 from which we have that the rst 7 terms of Y z in series form are Y z z 1 1:90z 2 2:61z 3 3:06z 4 3:20z 5 3:02z 6 2:54z 7 Now, comparing the above answer with the basic eq. (583), noting that in this case there is no y 0 term, we see that the sample values are y 0 0:00 y T 1:00 y 2T 1:90 y 3T 2:61 y 4T 3:06 y 5T 3:20 y 6T 3:02 y 7T 2:54
Figure 338 is the result of plotting these values versus nT. Fig. 338 thus expresses the z-function z= z2 1:9z 1 in terms of time-domain sampled values of y t . (It so happens that Fig. 338 is a portion of a sine wave in sampled form.)
Fig. 338
CHAPTER 13 The Digital Processor
As you might expect, long division is not the only way to convert a given z-transform into the equivalent time-domain sampled form. Another method, for example, makes use of partial fractions, which is a procedure for writing a given fraction as the sum of a number of fractions in which the individual fractions are each simpler than the original given fraction. The inverse of each such simpler fraction can then be found by direct inspection of a table such as Table 1. The advantage of the partial fraction method is that the answer comes out in exact or closed form, while the long-division procedure comes out in series or open form. Problem 302 Given that Y z 1 z 0:5
nd, by means of long division, the time-domain values of y 0 , y T , y 2T , y 3T , y 4T , and y 5T . Problem 303 Given that Zvs t F z z z2 0:45
nd, by long division, the values of v nT for n 0 through n 9.
The Discrete-Time Processor
The circuitry designed to manipulate DT signals is called a DT (discrete-time) processor, or digital processor if you wish. In dealing with such processors it s common practice to associate the letter symbol x with the input DT sequence and the letter y with the resulting output DT sequence. The symbol (h) will be associated with the processor itself; that is, h will describe the digital circuitry needed to convert a given x input signal into a required y output signal. Thus, if x nT denotes an input sequence and y nT the resulting output sequence, the situation can be represented in block diagram form as shown in Fig. 339, where the box with the h notation contains the required digital circuitry.