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

# 647 for a unity feedback system addition of a

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online