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Unformatted text preview: Y (s) = (b3S3 + b2S2 + b ls + b o)X(s) = b3S3X(S) + b 2S2X(s) + b lSX(S) + b oX(s) Signals X (s), s X(s), s 2X(s), a nd s 3X(s) a re available a t v arious points in Fig.
6.20c so t hat Y (s) c an b e g enerated by using feedforward connections t o t he o utput
s ummer, a s d epicted i n Fig. 6.21. Therefore, t he r ealization in Fig. 6.21 h as t he
d esired t ransfer function H (s) i n Eq. (6.71).
t it may seem odd that we first assume~. the ru:istence of x' ar:d generated X, X, x b~ its succ:s~ive
integration, and then in turn generated x from x, x, and x. ThiS. procedure P?ses a dilemma sl~dar
to "Which came first, the chicken or egg?" The problem here IS satisfactOrily resolved by wnt~ng
the expression for x' at the output of the summer in Fig. 6.20b and verifymg that thiS expreSSlOn
is indeed the same as Eq. (6. 75c).
:j:For example, the initial condition x(O) can be incorporated into our r~alization by adding a
constant signal of value x(O) at point Nl. This follows from the fact that fo x(r) dr = x (t)  x(O),
and x (t) = x(O) + x(r) dr. Similarly, the initial condition X(O) should be added at point N2,
and so on. f; 2. D raw t he n feedback connections from t he o utput o f each of t he n i ntegrators
t o t he i nput s ummer. T he n feedback coefficients are ao, a l, a2, . .. , a nl,
respectively, a nd t he feedback connections have negative signs (for s ubtraction)
a t t he i nput s ummer.
3. Draw t he n + 1 feedforward connections t o t he o utput s ummer from t he o utputs o f all t he n i ntegrators a nd t he i nput s ummer. T he n + 1 feedforward
coefficients are bo, b l, b2 , • .. , bn , respectively, a nd t he feedforward connections
have positive signs (for addition) a t t he o utput s ummer. Observe t hat t he connections ak a nd bk s tart from t he s ame p oint. Thus, t he c onnections ao a nd bo
s tart a t t he s ame point, a nd so do connections a l and bl , a nd so on.
Note t hat an is assumed t o b e u nity a nd d oes n ot a ppear e xplicitly anywhere
in t he r ealization. I f an f. 1, t hen H (s) s hould be normalized by dividing b oth i ts
n umerator a nd i ts d enominator by an.
• E xample 6 .18
Find the canonical realization of the following transfer functions.
( a)  5 s+2 ( b) s +5 s +7 ( c) _ s_ ( d) 4s + 28 S2 + 6s + 5
All four of these transfer functions are special cases of H(s) in Eq. (6.70). s +7 ( a) In this case, the transfer function is of the first order (n = 1); therefore, we need
only one integrator for its realization. The feedback and feedforward coefficients are 418 6 C ontinuousTime S ystem A nalysis U sing t he L aplace T ransform 6 .6 F (s) S ystem R ealization 419 y es) ~J
F ig. 6 .23 R ealization o f ao =2 a nd bo =5 7. '!2 b1 = (a) ° ( b) H (s) 1
s (b) F ig. 6 .25 R ealization o f T he r ealization is depicted in Fig. 6.23. Because n = 1 , t here is a single feedback connection
from t he o utput o f t he i ntegrator t o t he i nput s ummer w ith coefficient ao = 2. For n = 1 ,
t here a re n + 1 = 2 feedforward connections i...
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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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