Chapter11 - CHAPTER 11 11.1 (a) (b) HARDWARE AND SOFTWARE...

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CHAPTER 11 HARDWARE AND SOFTWARE IMPLEMENTATION OF COMMON COMPENSATORS 11.1 (a) Es Ys () = R RC s D D + 1/ = – Ts D D a + 1 ; a = R / R D ; T D = R D C D (b) = – R RR s R C 12 2 1 ++ /( ) = – KTs cD D + + 1 1 a K c = R + ; a = R 1 + ; T D = R 2 C 11.2 Z 1 ( s ) = R 2 + R 1 /(1 + R 1 Cs ); Z 2 ( s ) = R 1 + R 2 2 1 =– Zs 2 1 1 [] - = t at s s + + 1 1 ; t = R 1 C , a = R 2 + < 1 Refer Section 11.2 (Fig. 11.4) for the Bode plot and filtering properties of the lead compensator. 11.3 Z 1 ( s ) = R 1 + R 2 ; Z 2 ( s ) = R 2 + R 1 /(1 + R 1 Cs ) 2 1 = – 2 1 = – t bt s s + + 1 1 ; t = C + , b = R 2 + > 1 Add an inverting amplifier of gain unity. Refer Section 11.2 (Fig. 11.7) for the Bode plot and filtering properties of the lag compensator. 11.4 Z 1 ( s ) = ( R 1 C 1 s + 1) R 3 /[( R 1 + R 3 ) C 1 s + 1] Z 2 ( s ) = ( R 2 C 2 s + 1) R 4 /[( R 2 + R 4 ) C 2 s + 1] 2 1 = – R R 2 1 6 5 - L N M O Q P = K s s s s c + F H G I K J + F H G I K J + F H G I K J + F H G I K J 1 1 1 1 1 2 1 2 t t at bt t 1 = ( R 1 + R 3 ) C 1 ; t 2 = R 2 C 2 ; a = R 1 13 + ; b = R 24 2 + ;
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110 CONTROL SYSTEMS: PRINCIPLES AND DESIGN K c = RRR RR 246 135 13 24 + + F H G I K J 11.5 Z 1 ( s ) = R 1 /(1 + R 1 Cs ); Z 2 ( s ) = R 2 + 1/ C 2 s Es 2 1 () = – Zs R R 2 1 4 3 - L N M O Q P = K c 1 1 1 ++ L N M O Q P Ts D K c = RRC RC RRC 41 1 22 312 + = 39.42; T I = R 1 C 1 + R 2 C 2 = 3.077 T D = RRCC RC 1212 11 2 2 + = 0.7692 These equations give R 1 = R 2 = 153.85 k W ; R 4 = 197.1 k W 11.6 Z 1 ( s ) = R 1 /(1 + R 1 C 1 s ); Z 2 ( s ) = R 2 /(1 + R 2 C 2 s ) 2 1 = -- L N M O Q P R R 2 1 4 3 = RCs 42 31 22 1 1 + + F H G I K J = 2.51 0 345 1 0185 1 . . s s + + F H I K This equation gives R 1 = 34.5 k W ; R 2 = 18.5 k W ; R 4 = 45.8 k W 11.7 By Trapezoidal rule for integration: etd t kT 0 z = T e e T e T e T e k T e kT ( ) ( ) ( ) ( ) ...
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Chapter11 - CHAPTER 11 11.1 (a) (b) HARDWARE AND SOFTWARE...

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