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qr code vb.net library n 1 X n 0 in VS .NET
n 1 X n 0 Scan Code 128 In .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET applications. Drawing Code 128 Code Set B In .NET Using Barcode drawer for VS .NET Control to generate, create Code 128 image in VS .NET applications. X 1 z n z 1 n 1 z z n z 1 1 z 1 n 0
Code 128 Code Set C Recognizer In VS .NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications. Generating Bar Code In .NET Framework Using Barcode printer for VS .NET Control to generate, create barcode image in VS .NET applications. (after multiplying numerator and denominator of the rst fraction by z). But note, by eq. (574), that z n ! 0 as n becomes in nitely great. Thus the nal fraction above has the limiting value of z= z 1 , and hence the ztransform of the unitstep sequence is z 577 F z z 1 Problem 295 Verify, to your satisfaction, that the suggested operation on eq. (576) does lead to the nal result of eq. (577). Another useful DT sequence* is the linear rise of Fig. 334, where T is the constant time between samples, as usual. The ztransform can be found as follows. Note that Fig. 334 is the sampled form of the CT function v t t, so that here v nT nT as you can see from the gure. Thus, upon substituting nT in place of v nT in eq. (573) we have that (remembering that T is constant) F z T Bar Code Scanner In VS .NET Using Barcode scanner for VS .NET Control to read, scan read, scan image in VS .NET applications. Code 128B Drawer In Visual C# Using Barcode generation for .NET Control to generate, create Code 128C image in .NET applications. n 1 X n 0
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Make Code39 In ObjectiveC Using Barcode maker for iPhone Control to generate, create Code 3/9 image in iPhone applications. Making Data Matrix 2d Barcode In ObjectiveC Using Barcode printer for iPhone Control to generate, create Data Matrix ECC200 image in iPhone applications. Let us suppose that, after some experimentation, we decide to try multiplying both sides of eq. (578) by z. Doing this, and noting that zz 1 z0 1, eq. (578) becomes zF z T 1 2z 1 3z 2 4z 3 5z 4 Now take the algebraic sum of the above equation and eq. (578), thus getting 1 z F z T 1 z 1 z 2 z 3 z 4 which, upon multiplying both sides by 1, becomes z 1 F z T 1 z 1 z 2 z 3 z 4 But note that, by eq. (576), the quantity inside the parentheses on the righthand side is equal to the ztransform of the unitstep sequence, which, by eq. (577) equals z= z 1 . Thus, making this substitution into the last equation and then solving for F z , you can verify that the ztransform of the linearrise sequence of Fig. 334 is equal to F z Tz z 1 2 579 Next consider the important continuoustime relationship v t bt , this being called the negative exponential function, where b is a constant. From Fig. 18A (note 13 in Appendix), it follows that in discrete time the negative exponential function would appear in sampled form as in Fig. 335. Fig. 335
Thus the CT function v t bt becomes the DT sequence v nT bnT , and upon substituting this value into eq. (573) we have that F z n 1 X n 0
bnT z n
n 1 X n 0
bT z n
580 and thus
CHAPTER 13 The Digital Processor
h i F z 1 bT z 1 bT z 2 bT z n
for n ! 1 Now note that, since bT is constant, the quantity inside the brackets has the same form as eq. (576) which follows Fig. 333, except now we have bT z in place of z; hence, all we need do is replace z with bT z in eq. (577), and we have that the ztransform of the DT negative exponential sequence is F z bT z z z bT z 1 bT z k z 581 after multiplying numerator and denominator of the rst fraction by bT , then letting k bT . All the foregoing results, plus several more, are summarized in Table 1. Table 1. Number 1. 2. 3. 4. 5. 6. 7. 8. 9. Some zTransforms, n 0, 1, 2, 3, . . . , where k bT , b and T are constants DT signal v nT unit pulse p nT unit step U nT exponential bnT linear rise nT product of (3) and (4) nT bnT sine sin !nT cosine cos !nT damped sine bnT sin !nT damped cosine bnT cos !nT ztransform of v nT F z * 1 z z 1 z z k Tz z 1 2 kTz z k 2 z2 z sin !T 2z cos !T 1 z z cos !T z z2 2z cos !T 1 k sin !T z z2 2k cos !T z k2 z z k cos !T z2 2k cos !T z k2 * We ll use Z to indicate that the ztransform of a DT sequence is to be taken; thus, Zv nT F z , read as the ztransform of a DT sequence v nT is equal to F of Z. Problem 296 This is an interesting and instructive example of the almost magical powers of Euler s formulas (eqs. (153) and (154) in Chap. 6). Corresponding to the CT sinusoidal function v t cos !t, we have the DT sequence v nT cos !nT. By making use of Euler s formulas, and also eq. (580), see if you can derive item (7) in the table. To continue on, in our work it will sometimes be necessary to deal with timedelayed DT signals.

