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)estdt O =r ioo h (t)estdt+ r OOh(t)estdt = F2(S) + Fl(S) io (6.103) 6 450 ContinuousTime 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)estdt 1 h ( _ t)e st dt 1 h (t)estdt 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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