ch09 - Digital Signal Processing - A Practical Approach...

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1 CHAPTER 9 Additional Problems 9.21 (a) The closed-loop gain of the feedback amplifier is given by (1) We are given the following values for the forward amplification A and feedback factor β : A = 2,500 β = 0.01 Substituting these values in Eq. (1): (b) The sensitivity of the feedback amplifier to changes in A is given by With ( A/A ) = 10% = 0.10, we thus have 9.22 (a) The closed-loop gain of the feedback system is (b) The return difference of the system is Hence the sensitivity of T with respect to changes in G p is (c) We are given: H = 1 and G p = 1.5. Hence for = 1% = 0.01, we require T A 1 β A + ---------------- = T 2500 1 0.01 2500 × + ------------------------------------- 2500 26 ----------- 92.15 == = S T A TT AA --------------- 1 1 β A + 1 26 ----- === T T ------- S T A A A   = 1 26 ----- 0.10 (29 0.0038 0.38% = T G a G p 1 HG a G p + ---------------------------- = F 1 a G p + = S T G p G p G a --------------------- = 1 F --- = 1 a G p + = S T G p F 100 =
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2 The corresponding value of G a is therefore 9.23 The local feedback around the motor has the closed-loop gain G p /(1 + HG p ). The closed- loop gain of the whole system is therefore 9.24 The closed-loop gain of the operational amplifier is We are given Z 1 ( s )= R 1 and Z 2 ( s R 2 . The closed-loop gain or transfer function of the operational amplifier in Fig. P9.24 is therefore 9.25 (a) From Fig. P9.25, we have The transfer function of this operational amplifier is therefore (1) (b) For positive values of frequency ϖ that satisfy the condition G a F 1 HG p ------------ = 99 1.5 ------- 6 6 == T K r G c G p 1 p + (29 1 K r G c G p 1 p + + ------------------------------------------------------------ = K r G c G p 1 p K r G c G p ++ ------------------------------------------------ = V 2 s V 1 s ------------- Z 2 s Z 1 s = V 2 s V 1 s R 2 R 1 ----- = Z 1 s R 1 1 sC 1 -------- + = Z 2 s R 2 = V 2 s V 1 s Z 2 s Z 1 s = R 2 R 1 1 1 + --------------------- = 1 R 2 1 1 R 1 + ------------------------ =
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3 we may approximate Eq. (1) as That is, the operational amplifier acts as a differentiator. 9.26 Throughout this problem, H ( s ) = 1, in which case the open-loop transfer equals G ( s ). In any event, the open-loop transfer function of the feedback control system may be expressed as the rational function P ( s )/( s p Q 1 ( s )), where neither the polynomial P ( s ) nor Q 1 ( s ) has a zero at s = 0. Since 1/ s is the transfer function of an integrator, it follows that p is the number of free integrators in the feedback loop. The order p is referred to as the type of the feedback control system. (a) For the problem at hand, we are given The control system is therefore type 0. The steady-state error for unit step is where That is, Hence, For both ramp and parabolic inputs, the steady-state error is infinitely large. (b) For the control system is type 1. The steady-state error for a step input is zero. For a ramp of unit slope, the steady-state error is where 1 ϖ C 1 ---------- > R 1 V 2 s (29 V 1 s ------------- sC 1 R 2 Gs 15 s 1 + s 3 + --------------------------------- = ss = 1 1 K p + ---------------- K p s 0 lim Hs = K p s 0 lim 15 s 1 + s 3 + = 15 3 ----- 5 == = 1 15 + ----------- 1 6 -- = 5 1 + s 4 + ----------------------------------- = = 1 K v ------
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4 The steady-state error is therefore For a parabolic input the steady-state error is infinitely large. (c) For the control system is type 2. Hence, the steady-state error is zero for both step and ramp inputs. For a unit parabolic input the steady-state error is where The steady-state error is therefore 3/5. For a unit parabolic input the steady-state error is infinitely large.
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ch09 - Digital Signal Processing - A Practical Approach...

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