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(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 z-transform of the unit-step 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 z-transform 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
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The above in nite-series form of the answer can be put into closed form by carrying out the following algebraic manipulations.
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* Notations such as v t ; x t , and so on denote continuous-time analog signals, while v nT ; x nT , and so on denote their sampled ( discrete ) form. For convenience we ll often abbreviate continuous time as CT and discrete time as DT.
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CHAPTER 13 The Digital Processor
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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 right-hand side is equal to the z-transform of the unit-step 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 z-transform of the linear-rise sequence of Fig. 334 is equal to F z Tz z 1 2 579
Next consider the important continuous-time relationship v t  bt , this being called the negative exponential function, where b is a constant. From Fig. 18-A (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
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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 z-transform 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 z-Transforms, 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 z-transform 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 z-transform of a DT sequence is to be taken; thus, Zv nT F z , read as the z-transform 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 time-delayed DT signals.
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