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Unformatted text preview: which uses zero-input/zero-state
separation of t he i nput) t o t he electrical engineering community. 8 +1
(8 2 + 58 + 6) Y(8) - (28 + 11) = - 4
8+ so t hat
(8 2 + 58 + 6) Y (8) = (28 + 11) + '-...-'
i nitial c ondition t erms 8+1
8 +4 :r " '--..--'
i nput t erms T herefore Y (8) = 28 + 11
82 + 58 + 6 "-v-"'
z ero-input c omponent 8 +1
(8 + 4)(8 + 58 + 6) + -:-----:-;-;-;;---::--:-:2
, ... # z ero-state c omponent 6 7 5]
3 /2 ]
= [ 8 +2-8+3 + 8 +2+8+3-8+4 Exercise E 6.6
d2 d dt~ + 4 ft + 3y(t) = T aking t he i nverse transform of this equation yields z ero-input r esponse 393 df 2 di + I (t) for t he i nput I (t) = u (t). T he i nitial conditions are y (O-) = 1 a nd y (O-) = 2.
Answer: y (t) = ~(l + g e- t - 7e- 3t )u(t) \ 7 z ero-state r esponse lH 20 lH 20 Comments on Initial Conditions at 0 - and at 0+
T he initial conditions in Example 6.9 are y(O-) = 2 a nd y(O-) = 1. I f we let
t = 0 in t he t otal response in Eq. (6.44), we find y(O) = 2 a nd y(O) = 2, which
is a t o dds with t he given initial conditions. Why? T he reason is t hat t he initial
conditions are given a t t = 0 - (just before t he i nput is applied), when only t he zeroinput response is present. T he z ero-state response is t he result of t he i nput f (t)
applied a t t = o. Hence, this component does not exist a t t = 0 -. Consequently,
t he initial conditions a t t = 0 - are satisfied by t he zero-input response, not b y t he
t otal response. We can readily verify in this example t hat t he z ero-input response
does indeed satisfy the given initial conditions a t t = 0 -. I t is t he t otal response
t hat satisfies t he initial conditions a t t = 0+, which are generally different from the
initial conditions a t 0 -.
T here also exists a L:+ version of t he Laplace transform, which uses t he initial
conditions a t t = 0+ r ather t han a t 0 - (as in o ur present L:_ version). T he L:+
version, which was in vogue till t he early sixties, is identical t o t he L:_ version
except t he l imits of Laplace integral [Eq. (6.18)] are from 0+ t o 0 0. Hence, by
definition, t he o rigin t = 0 is excluded from t he domain. This version, still used in
some m ath b ooks, has some serious difficulties. For instance, t he Laplace transform
of 6(t) is zero b ecause 8(t) = 0 for t 2: 0+. Moreover, this approach is p ractically
useless in t he t heoretical s tudy of linear systems because t he response obtained by
this method c annot be separated into zero-input a nd zero-state components. As
we know, the zero-state component represents t he system response as a n explicit
function of t he i nput, and without knowing this component, it is n ot possible to
assess t he effect of t he i nput on t he s ystem response. The L:+ version can separate
t he response i n t erms of t he n atural a nd t he forced components, which are not
as interesting a s t he zero-input a nd t he z ero-state components. Note t hat we can +
.! 5 (a) F ( b) ( c) o 1-- F ig. 6 .8 Analysis of a networ...
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