Unformatted text preview: order. This difficulty can be avoided by
observing that the terms 1 /(8 + 3) and 1 /(8 + 3)2 can be realized with a cascade of two
subsystems, each having a transfer function 1/(8 + 3), as shown in Fig. 6.28. Each of the
three firstorder transfer functions in Fig. 6.28 may now be realized as in Fig. 6.23. •
t::. E xercise E 6.10
Find a canonical, a cascade, and a parallel realization of 3 _(8s + 2) (8+ 5)
+3
1 s+
8  82 + 78 + 10  H( ) _ tIt is possible to realize complex, conjugate poles indirectly by using a cascade of two firstorder
transfer functions. A transfer function with poles  a ± j b can be realized by using a cascade of
two identical firstorder transfer functions, each having a pole at  a. (See Prob. 6.67.) 422 6 C ontinuousTime System Analysis Using t he Laplace Transform 6.6 System Realization 423 Therefore, the op a mp circuit in Fig. 6.29 has t he t ransfer function H (s) = _ Z j(s)
Z (s) F (s) (6.78) By properly chOOSing Z (s) a nd Z j ( s), we c an o btain a variety of transfer functions,
as t he following development shows. ·6
F 1 9.. 2 8 P ar alleI re al'lzat 'Ion 0 f 6.63 78+2)(,+3)"
( .'+378+51 + System Realization Using Operational Amplifiers Y es) I n this section, we discuss practical implementation of the realizations described
in t he previous subsection. Earlier we saw t hat t he basic elements required for t he
synthesis of a n L TIC system (or a given transfer function) are (scalar) multipliers, integrators, a nd summers (or adders). All these elements c an be realized by
operational amplifier (op amp) circuits, as explained below. F~ .Y~s) ~
=&
k= R (a) I rs) +
Yes) +
Yes) (b) F ig. 6.30 (a) Op amp inverting amplifier (b) integrator. F ig. 6 .29 A basic inverting configuration op amp circuit.
Operational Amplifier Circuits T he Scalar Multiplier F igure 6.29 shows a n op amp circuit in t he frequency domain (the transformed
circuit). Because t he i nput i mpedance of t he o p a mp is infinite (very high), all of
t he c urrent I (s) flows in t he feedback p ath, as illustrated in Fig. 6.29. Moreover
Vx(s), t he v oltage a t t he i nput of t he o p amp, is zero (very small) because of t he
infinite (very large) gain of t he op amp. Therefore, for all practical purposes, I f we use a resistor R j in t he feedback a nd a r esistor R a t t he i nput (Fig.
6.30a), t hen Z j(s) = R j, Z (s) = R , a nd Y es) =  I(s)Zj(s)
Moreover, b ecause Vx "" 0, F (s)
I (s) = Z (s)
S ubstitution of t he second equation in t he first yields y es) =  i(~S/ F(s) H (s)=_Rj
R (6.79a) T he s ystem acts as a scalar multiplier (or a n amplifier) with a negative gain !!:.t.. A
positive gain can be obtained by using two such multipliers in cascade o r by ~sing
a single noninverting amplifier, as depicted in Fig. 6.16c. Figure 6.30a also shows
t he c ompact symbol used in circuit diagrams for a scalar multiplier.
T he Integrator I f we use a capacitor G in t he feedback a nd a r esistor R a t t he i nput (Fig.
6.30b), t hen Z j(s) = 1 /Gs, Z(s) = R, a nd 6 424 C ontinuousTime S ystem A nalysis Using t he L ap...
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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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