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

Therefore we need to combine t he c onjugate poles a

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Unformatted text preview: lace T ransform 6.6 S ystem R ealization 425 (a) + F, ( s) F2 ( s) Y (s) F r(s) ( b) ( b) (a) F ig. 6 .31 O p amp summing and amplifying circuit. ~ (6.79b) RC S T he s ystem a cts as a n i deal i ntegrator w ith a g ain - 1/ R C. F igure 6 .30b a lso shows t he c ompact s ymbol u sed in circuit d iagrams for a n i ntegrator. H (s) = ( _ _ ) 1 100 i ll 100 i ll The Summer C onsider n ow t he c ircuit i n F ig. 6 .31a w ith r i nputs Fl(S), F2(S), . .. , Fr(s). As u sual, t he i nput v oltage Vx(s) ~ 0 b ecause t he g ain o f op a mp ~ 0 0. M oreover, t he c urrent g oing i nto t he o p a mp is very s mall (~ 0) b ecause t he i nput i mpedance ~ 0 0. T herefore, t he t otal c urrent i n t he f eedback resistor R f is Il(S) + I2(S) + ... + Ir(s). M oreover, b ecause Vx(s) = 0, 100 i ll 50 i ll 20 i ll j = 1, 2, . .. , r A lso Y (S) = - Rf [Il(S) == (e) + I2(S) + ... + Ir(s)] [~Fl(S) + ~F2(S) + ... + ~Fr(S)] k lFl(S) + k2F2(S) + ... + krFr(s) F ig. 6 .32 Op amp realization of a second-order transfer function s2!~~!10 (6.80) w here - Rf k;=-- Ri Clearly, t he c ircuit in Fig. 6.31 serves a s ummer a nd a n a mplifier w ith a ny d esired g ain for e ach o f t he i nput s ignals. F igure 6 .31b shows t he c ompact s ymbol u sed in c ircuit d iagrams for a s ummer w ith r i nputs. • E xample 6 .20 Using op a mp circuits, realize the canonical form of the transfer function H (s)- s2 2s+5 + 4s + 10 T he basic canonical realization is shown in Fig. 6.32a. Signals a t various points are also indicated in the realization. Op amp elements (multipliers, integrators, and summers) change the polarity of the o utput signals. To incorporate this fact, we modify t he canonical realization in Fig. 6.32a to t hat depicted in Fig. 6.32b. In Fig. 6.32a, the successive outputs of the summer and the integrators are s 2 X (s), s X(s), and X (s) respectively. Because of polarity reversals in op amp circuits, these outputs are - s2X(s), s X(s), and - X(s) respectively in Fig. 6.32b. This polarity reversal requires corresponding modifications in the signs of feedback and feedforward gains. According to Fig. 6.32a S2 X (s) = F (s) - 4sX(s) - 10X(s) Therefore _ S2X(S) = - F(s) + 4 sX(s) + 1 0X(s) 426 6 C ontinuous- Time S ystem A nalysis Using t he L aplace Transform (a) ( b) F ig. 6.33 Op amp circuits for exercise E6.1l. Because the summer gains are always negative (see Fig. 6.3lb), we rewrite the above equation as - S2 X (S) = - l[F(s)]- 4[-sX(s)]- lOr-Xes)] Figure 6.32b shows the implementation of this equation. The hardware realization appears in Fig. 6.32c. B oth integrators have a unity gain which requires R C = 1. We have used R =100 kO and C = lOJ.lF. The gain of 10 in the outer feedback path is obtained in the summer by choosing the feedback resistor of the summer to be 100 kO and an input resistor of 10 kO. Similarly, the gain of 4 in the inner feedback p ath is obtained by using the corresponding input resistor of 25 kO. T he gains of 2 and 5, required in the feedforward connections, are obtained by using a feedback...
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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.

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