Signal Processing and Linear Systems-B.P.Lathi copy

647 for a unity feedback system addition of a

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Unformatted text preview: a t t his frequency, t he signal fed back is e qual to the i nput signal, a nd t he signal perpetuates itself for ever. In other words, t he signal s tarts g enerating (oscillating) a t t his frequency, which is precisely t he instability. Note t hat t he s ystem r emains unstable for all values of K > 48. T his is clear from t he r oot locus i n Fig. 6.43, which shows t hat t he two branches cross over t o the R HP for K > 4 8. T he crossing point is s = j2.83. T his discussion shows t hat t he same system, which has negative feedback a t lower frequency may have positive feedback a t higher frequency. For this reason, h it) = f (t)u(t) (6.102a) h (t) = f (t)u(-t) (6.102b) T he b ilateral Laplace transform of f it) is given by F (s) = I: f (t)e-stdt O =r i-oo h (t)e-stdt+ r OOh(t)e-stdt = F2(S) + Fl(S) io (6.103) 6 450 Continuous-Time System Analysis Using the Laplace Transform 6.8 The Bilateral Laplace Transform 451 To summarize, t he b ilateral transform F (s) in Eq. (6.103) can be computed from the unilateral transforms in two steps: 1) Split f (t) i nto its causal and anticausal components, h (t) a nd h (t), respectively. (a) 2) T he signals h (t) a nd h (-t) a re both causal. Take the (unilateral) Laplace transform of h (t) a nd add t o i t t he (unilateral) Laplace transform of h (-t), with s replaced by - so T his procedure gives the (bilateral) Laplace transform of f (t). Since h (t) a nd h (-t) are b oth causal, F1(S) a nd F 2(-S) are b oth u nilateral Laplace transforms. Let 0" e1 a nd 0" e2 b e the abscissas of convergence of F1 (s) a nd F2(-s), respectively. This statement implies t hat F1(S) exists for all s w ith R es> 0" e l, F2 ( - s) exists for all s w ith Re s > 0" e 2, a nd F2 (s) exists for all s w ith Re s < -O"e2· Because F (s) = FI(s) + F2(S), F (s) exists for all s such t hat (b) O"el < Re.s < -O"c2 (6.105) (c) o /- crc, D o t- Fig. 6 .48 Expressing a signal as a sum of causal and anticausaJ components. where F1(S) is t he Laplace transform of the causal component h (t). an.d F2(S) is t he Laplace t ransform of the anticausal component h (t). B ut F2 (s) IS given by o F2(S) = i ooh(t)e-stdt 1 h ( _ t)e st dt 1 h (-t)e-stdt 00 = Therefore Region o f convergence for anticausal component o f f ( / ). • Region (strip) o f convergence for the entire f (l). F ig. 6 .49. T he regions of convergence (or existence) of F1(S), F2(S), a nd F (s) are shown in Fig. 6.49. Because F (s) is finite for all values of s lying in the strip of convergence (O"e1 < R es < -O"e2), poles of F (s) must lie outside this strip. The poles of F (s) arising from the causal component h (t) lie t o t he left of the s trip (region) o f c onvergence, and those arising from its anticausal component h (t) lie to its right (see Fig. 6.49). This fact is of crucial importance in finding the inverse bilateral transform. As an example, consider f (t) = e btu(_t) 00 F 2(-S) = Region o f convergence for causal component o f f ( / ). • (d) (6.104) I t is clear t hat F2( - s) is t he Laplace transform of h ( - t), which is caus~l (Fig. 6.48d), so F2( - 8) c an be found from the unilateral transfo...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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