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Unformatted text preview: Sensitivity of Control Systems
An open loop control system operates without feedback and generates the actuating signal in response of an input signal. u C(s) G(s) y If the plant Gs involves variations G, then it induces a relative error of the overall system T s = GsC s: T GC G = = that is the same as the relative error of the original plant. A closed-loop system uses the error signal between the actual output and desired output to generate the actuating signal. Hence measurement of the output is required. T GC G u
- C(s) G(s) y 1 The closed-loop transfer function is C G T s = 1 + ssss : C G If the plant involves small variation, then it induces a relative error of the overall system: C 1 + CGG , C 2GG T 1 + CG2 C G = T G : = 1 + CG2 G1 + CG Hence we have a relative error for the overall system as 1 G T = T 1 + CG G : That is, the relative error of the overall system caused by that of the plant is reduced by a factor of T SG s = 1 + C 1sGs : If the denominator is large for all s on j!-axis, then the system is insensitive to the variation of the plant.
B Sensitivity SA is de ned as ratio of the relative variation of B the e ect to the variation of A the source: B B = B = A B = @ logB : SA A B A @ logA A 2 For open-loop systems with A = T = CG, and B = G,
T SG s = 1: For closed-loop systems with A = T = CG, and B = G, T SG s = 1 + C 1sGs : Feedback System has the ability to improve sensitivity function Consider control system involving unknown disturbance: w u
- C(s) G(s) y Without feedback, the disturbance has direct impact on output as Twy = 1. With feedback, we have T Twy s = 1 + C 1sGs = SG : 3 Feedback System has the ability to attenuate unknown disturbance. Because of the ability to improve sensitivity and to attenuate unknown disturbance, systems are able to track reference signals asymptotically. That is, the tracking error goes to zero asymptotically. Example 1: Operational ampli er. The overall gain is: A G = 1 + Ak ; k = Rf : R1 Hence the sensitivity of G with respect to A is: G A SA = G G = 1 + Ak11 + Ak , kA = 1 +1Ak : A + Ak2 For Ak 1, the sensitivity is 1. Thus the overall gain is insensitive to the variation of A.
Example 2: D.C. Motor. Recall
La va = La dia + Raia + Ke d . dt dt 2 Kiia = J d 2 + b d + TL. dt dt va e , J,b
TL 4 Taking Laplace transform with zero initial condition gives s + W Ia = Vas , KeY s ; Y s = KiIasJ + b s Ra + sLa where Y = L ! = L _ , and W = ,L TL .
1 1 1 Va R a+ sL a Ia Ki sJ+b Y - Ke R a+ sL a A Tvy s = sJ + bsL Ki R + K K = s + 1 s + 1 a+ a i e 1 2 2JLa 1;2 = JRa + bLa JRa + bLa2 , 4JLaKi=A RJ L Ki A = bR + K K ; 1 K aK ; 2 Ra a i e i e a B sL + R Twy s = sJ + bsLa + Ra + K K = s + 1 s + 1 a a i e 1 2 where due to La 0, sL R a B = bR a + RK bR + a K : a + Ki e a Ki e
r The overall transfer functions are 5 A block diagram is as follows:
W Motor Model B A va A ( 1s +1)( 2 s+1) y Open-loop speed control system:
W A B r
Controller va y
D.C. Motor If constant gain C = K is used for control, then Y s = KTvy sRs + Twy sW s:
For w = w1t and r = r1t, we require that that yields K = 1=A. yss = tlim yt = slim0 sY s = CAr + Bw !1 ! 6 Feedback speed control system:
A B W r
- Controller va
D.C. Motor The formula for closed-loop system is KTvy wy Y s = 1 + KTss Rs + 1 +TKTss W s; vy vy KARs + BW s = 1s + 1 2s + 1 + AK 1s + 1 2s + 1 + AK Y s = AKU s + BW s: If w = w1t and r = r1t, then the steady-state is For AK AK yss = slim0 sY s = 1 + AK r + 1 +B w: ! AK
1, yss r. Suppose that 1 = 1=60; 2 = 1=600, A = 10, and B = 50. The refernce speed is 100 rad sec, and the disturbance is w = ,0:1. Compute steady-state motor speed. 7 Solution: For open-loop system, K = 1=A = 0:1, and yss = Ava + Bw = KAr + 50w = 100 , 5 = 95 rad=sec:
For closed-loop system, Since yss ! 100 as K ! 1, we can choose K such that yss is as close as possible to 100 rad sec. Taking K = 9:9 yields AK yss = 1 + AK r + 1 +B w = 1000K , 5 : AK 1 + 10K yss = 99 , 0:05 = 98:95: Conclusion: Feedback has the ability to improve sensitity function, and to attenuate disturbance. The error in output response is less sensitive to the variation of the plant, and impact of unknown disturbance. Another use of feedback is stabilization:
- K 1 s- 1 y We have that the closed-loop poles are roots of s , 1 + K = 0:
By choosing K 1, the closed-loop system is stable.
8 Further discussion on sensitivity: For a closed-loop system The variation of the plant can be a parameter . What is S T = T T ? A simple formula to compute S T is:
T S T = SG S G: CGs T s = 1 + CGs : Example 3: Find S T for the feedback system with C s = 1; Gs = ss 2 : + Solution: First we compute s T SG = 1 +1CG = s2s+ +s + 2 : Then we compute 2 S G = G dG = , ss2+ ss + 2 = s + : d Hence we obtain
T S T = SG S G = , s2 + ss + 2 : 9 Sensitivity in Frequency Domain:
1 jS j!j = 1 + CGj! : We recall inverse Laplace transform: K + K = 2jK j cos! t + 6 K : ,1 L s , j! s + j! o o o Consider sinusoidal input signal
2 6 4 3 7 5 ut = Au cos!ot
applied to a plant Gs. Then the output in s-domain is s Y s = GsU s = Gs s2Aus 2 = Gs s + j!Aus , j! + !o o o K + K + terms generated by Gs: = s , j!o s + j!o The K -value can be computed from sG 1 K = s!j! s , j!oY s = s!j! Au+ j!s = 2 AuGj!o lim lim s o = jK j = AujGj!oj=2; 6 K = 6 Gj!o:
o o 10 Observation: The steady-state response is also sinusoidal. But its amplitude is ampli ed by a factor of jGj!oj, and its phase
is shifted by 6 Gj!o. Example: The gure below shows ut = cos!ot dashed line, and ysss = 2 cos!ot + =4 solid line with !o = 1. Thus Gj!o = 26 45o.
2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 0 2 4 6 8 10 12 14 16 De nition: Gj! is termed frequency response with jGj!j
magnitude response, and 6 Gj!o phase response. 11 Because every signal is a composition of sinusoidal functions, jGj!j and 6 Gj!o characterizes spectral information of the system, and they uniquely determine the behavior of the system. Consider feedback system:
r(t) e(t) D(s) G(s) y(t) (t) Then there hold Some primitive goals are: tracking reference signal in operating frequency range: 0; !r , and noise rejection in frequency range !n; 1. It thus requires ideally 1 j1 + CGj!j = 0; if 0 ! !r ; CGs 1 E s = 1 + CGs Rs; Y s = , 1 + CGs N s: jCGj!j = 0; if ! ! : n j1 + CGj!j 12 ...
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This note was uploaded on 02/03/2012 for the course EE 3530 taught by Professor Chen during the Fall '07 term at LSU.
- Fall '07