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CHAPTER 13 The Digital Processor
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In this regard, recall that if v t is a CT signal, then v t T is the same signal but shifted T seconds to the right of v t on the time axis (note 31 in Appendix). Or, if we substitute t T in place of t we have v t T , which is again the exact same form of signal as v t but now shifted T seconds to the left of v t on the time axis. In the same way, if v nT is a DT sequence, then v nT kT denotes the same sequence but shifted k sample periods to the right of v nT on the nT axis; that is, v nT kT denotes v nT delayed by k sample periods (delayed by kT seconds). Likewise, v nT kT denotes the same DT sequence v nT , but now shifted k sample periods to the left on the nT axis from v nT ; that is, v nT kT starts kT seconds before v nT starts. Consider now a few examples. In Fig. 332 p nT denotes a unit pulse for nT 0; thus, for example, p nT 3T denotes the same pulse but now shifted 3 sample periods to the right of nT 0 to nT 3, as shown below.
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Or, consider the unit-step sequence U nT , illustrated in Fig. 333. The notation U nT 2T would, for example, denote the same sequence as in Fig. 333 but shifted or delayed 2 sample periods to the right; thus
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As another example, the expression v nT p nT T p nT 3T denotes an algebraic sum consisting of a positive unit pulse at nT 1 and a negative or negative-going pulse at nT 3; thus
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As another example, the sum v nT U nT p nT T p nT 3T p nT 4T represents the modi ed unit-step sequence shown below,
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CHAPTER 13 The Digital Processor
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In the above example note that v nT 2 for nT 1, but v nT 0 for nT 3 and nT 4, because at these particular sampling instants the positive and negative sample values cancel each other out. Problem 297 Show graphically the sequence represented by the DT equation v nT U nT U nT T 4U nT 2T U nT 3T U nT 4T Problem 298 Show graphically the sequence represented by the DT equation v nT nT nT 3T 3U nT 7T where nT is the linear-rise sequence of Fig. 334. Now that we ve dealt with the graphical representation of v nT kT in the time domain, let s next consider the corresponding e ect in the z domain. To do this, let us begin by referring back to the time-domain expression of eq. (572). Now suppose all of the sample values v nT remain unchanged but are merely shifted kT seconds to the right; to indicate this, the impulse factor in eq. (572) would become  t nT kT  t n k T which would indicate to us that, for the time-delayed case, n should be replaced by n k in the z-domain expression of eq. (573). This is true, and upon substituting n k in place of n, eq. (573) becomes, for the time-delayed case (where del stands for delayed ), F z del z k
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n k n 1 X n 0
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v nT z n
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because, by the laws of exponents, z z n z k , and, because k and z are independent k of n, the factor z can be put outside, to the left, of the summation sign, as shown. But note that the quantity to the right of z k is, by eq. (573), equal to F z ; thus the last equation above shows that F z del z k F z
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which says that if F z is the z-transform of a given DT sequence, then z F z is the ztransform of the same sequence DELAYED BY k SAMPLE PERIODS. In block diagrams of DT networks a delay is represented by a box labeled z k , which represents some kind of a device capable of producing a delay of k sample periods, as illustrated in Fig. 336.
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