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Stability and Instability. Poles and Zeros
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It is possible for a recursive DT processor to become unstable under certain conditions. The desired condition of stability and the undesired condition of instability can be de ned in general terms as follows. Let a momentary signal, such as the unit pulse of Fig. 332, be applied to the input of a recursive DT processor. If the OUTPUT of the processor dies out and becomes zero as time increases, the processor is STABLE; if, however, the output does not become zero as time increases, the processor is UNSTABLE. Let us take, as an example to illustrate the basic possibilities, the simple recursive processor shown in Fig. 348.
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Fig. 348
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For brevity here we ll denote the unit pulse by 1, as shown in the gure. Note that the above is the same as Fig. 344, except that now the input is given to be the unit pulse, x nT p nT 1, and the multiplier constant is denoted by a. It is the VALUE OF
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CHAPTER 13 The Digital Processor
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THE MULTIPLIER a that determines whether the processor of Fig. 348 is stable or unstable. The explanation is as follows. First remember that, in this case, x nT 0 for all time EXCEPT at t 0, when x nT 1. Thus the output at the instant t 0 is also 1. Then, following this, after T seconds has elapsed, a time-delayed signal, (1) a a, arrives at the input to the adder; thus, at t T, the output is a. Then, after another T seconds has elapsed, a time-delayed signal, now equal to a a a2 , arrives at the input to the adder, so that, at t 2T, the output is a2 . Then, after another T seconds has elapsed, a time-delayed signal, now equal to a3 , is fed back to the input to the adder, so that at t 3T the output is a3 . Continuing on in this way we see that, at any integral multiple of time T, t nT, the output is equal to an ; that is, in Fig. 348, y nT an . Thus the nature of the output sequence in Fig. 348 depends upon the value of the multiplier constant a. Let us discuss this in more detail, as follows. First, note that our de nition of stability or instability, as given above, could also be stated in the following equivalent way. Let a single unit pulse p nT be applied to the input of a (recursive) processor, and let y nT denote the value of the output at any time nT seconds later. Now let L denote the value that y nT would approach if n were allowed to become in nitely great (denoted by writing n ! 1, or loosely, for convenience, simply as n 1 ). We can then say that, in general, a processor is stable if L 0, but is unstable if L is not equal to zero. Let us apply this principle to Fig. 348, where we ve already found that y nT an 591
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Thus, applying the above rule to the particular processor of Fig. 348, we have that
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lim y nT lim an L
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592
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It s apparent that, in this case, the value of L will depend upon the value of the constant multiplier a. As a matter of fact, after some thought we realize that we must consider three separate possibilities for the value of a, these being the cases for a GREATER than 1, a EQUAL to 1, and a LESS than 1. Let us consider each of the three cases as follows. (a > 1): Here the values of the output samples, an , theoretically become in nitely great for n 1. Thus in this case L is certainly not equal to zero, so that Fig. 348 is unstable for a > 1. This is illustrated in Fig. 349. Case II. (a 1): Since 1n 1 we have that L 1, and thus Fig. 348 is unstable for a 1, as illustrated in Fig. 350. Case III. (a < 1): The integral power of a number less than 1 is less than the given number; for instance, if a 1=3, then a2 1=9; a3 1=27, and so on. Thus for this case L becomes equal to zero L 0 as n becomes in nitely great; hence Fig. 348 is stable for a < 1, as illustrated in Fig. 351. In regard to Fig. 349, the output of an actual processor could not, of course, become in nitely great ; instead, in such a case the output would cease increasing and stall when the maximum holding capacity of the digital circuits was exceeded. Now let s return to Fig. 348 and this time apply the z-transform. Let us begin by writing down the nT-domain equation for the gure, which, since it s given that x nT p nT , is y nT p nT ay nT T Case I.
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