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Unformatted text preview: Dynamic Response
1 1st order plant: H s = s + Its impulse response is given by If 0: unstable. For
h(t)
1 ;
t 0. ht = L,1 H s = e, 1t: 0, we have e1
1 t Example 1: H s = 2s + 1 bs = . s2 + 3s + 2 as where bs = 2s + 1 = 2s + 1=2 = z = ,1=2, as = s2 + 3s + 2 = s + 1s + 2 = p1 = ,1, p2 = ,2. Using Partial fraction, 3 1 + : H s = , s+1 s+2 Thus ht = ,e,t + 3e,2t for t 0. 1 For small t, e,t 1 e,2t that gives
ht ,1 + 3 = 2: But 3e,2t dominates response smaller time constant and larger coe cient. As t becomes large, 3e,2t decays to zero a lot faster than ,e,t. Hence ,e,t dominates the response for large t. The impulse response plot below veri es the above analysis.
h(t)
2 1 t
1 2 3 4 2nd order plant with complex roots: prototype
Note that
2 !n H s = 2 2 s + 2
!ns + !n 2 s2 + 2
!ns + !n = s + where !d = !n 1 , 2, = !n and 0 p , j!ds + + j!d 1. 2 Roots location: jd The two poles lie on LHP located n at , j!d. The angle is given by = sin,1 . The smaller the angle,  the closer the poles to the j! axis. The value of measures the distance of the two poles to the j!axis. Using  jd partial fraction, we have that 1 0:5!n 0 1 1 @ , s + + j! A : H s = p 2 j 1 , s + , j!d d The impulse response is thus ! ht = p n 2 e, t sin!dt1t: 1, h(t)
1 et 2 3 t  et
1 3 Summary for Impulse Response:
Single real pole on OLHP:
1 h(t) e1
1 t Single pole at the origin: By taking ! 0 gives 1 impulse response for . s h(t)
1 t Two complex poles on OLHP: h(t)
1 et 2 3 t  et
1 Two imaginary poles: By taking ! 0 gives oscillation impulse response. h(t)
1 2 3 t 1 4 System having pole on j!axis is unstable. Consider H s = 1 as an example. If we excite the system s with unit step, then 1 Y s = H sU s = 2 = yt = t1t s which ends to 1 as t ! 1. Hence even though the impulse response is bounded, the output response to bounded input may not be bounded. Systems having poles on RHP are also unstable. Consider examples 2 !n 1 ; b 2 : a H s = 2 s, s , 2
!ns + !n The impulse responses are plotted below for which both are unstable, since 0; 0, and !n 0. (a) (b) 5 Timedomain Speci cations
A typical step response is shown below:
1+Mp
1 0.9 tp 0.04 1+
0.1 1 +
ts tr t Rise time: tr = the time for the system to reach the vicinity of its new set point = time from 10 to 90. Settling time: ts = the time for the step response to reach within 4 of its steadystate. Overshot: Mp = the maximum amount the system overshots, often in = Mp 100. The peak time: tp = the time for the step response to reach the peak value. 6 Relations to Pole Location
Consider again prototype system: 2 !n H s = 2 : 2 s + 2
!ns + !n Its step response is
0 yt = 1 , e, t @cos !dt + where = !n, and !d = !n 1 , 2. p !d 1 sin !dtA Computation of Mp and tp: The peak is reached at yjt=tp = 0. Since _ 0 2 1 dy , t B A = e @ + !dC sin !dt; dt !d y jt=tp = 0 sin !dtp = 0. We thus obtain _
!dtp = k; k = 0; 1; : The rst peak is at k = 1 which yields !dtp = tp = = p : !d !n 1 , 2 At the above tp value, we have see Figure 3.17, page 128
Mp = ytp , 1 = e, =!d = e,
= p , 1 2 ; 0 1: 7 Computation of rise time: If = 0:5, then 1:8 tr : !n Computation of settling time: By
!d the settling time satis es approximately e,
ts 1 , yt = e, t @cos !dt + 0 1 sin !dtA ; 0:04; = !n: Taking natural logrithm both sides gives !nts = log25 = 3:22 ts = 3:22 = 3:22 : !
n We thus have the following observations: 1. Thus the larger the distance of the poles to the j!axis, the faster for the step response to settle. 2. The smaller the value, the larger the overshot. 3. The larger the !n value, the faster the setp response to reach steadystate. 8 Example 1: Translation of Speci cations into Pole Locations: tr 0:6s; Mp 10; ts 3s: Solution: To satisfy the speci cations, we need 1:8 3:22 a !n = 3; b = 1:073: t t
For , we have that
r s Mp = e,
= p , 1 2 = r j logMpj : + j logMpj
2 2 Hence by the speci cation, j log0:1j = 0:6; or c = sin,1 36:86o: r 2 + j log0:1j2 Condition a gives allowable region outside of circle radius 2:8, b gives allowable region left of verticle line s = ,1:073, and c gives allowable region below the ray = 36:86o. The intersection is shown in shaded area which one is correct?. (c) (a) (b) (a) (c) (b) 9 Example 2 Prob. 3.27
We notice that the closedloop transfer function is:
Tcs =
K ss+2 1 + ssK +2 = K : s2 + 2s + K Compared to the prototype: we have
2 2 !n Gos = 2 2 s + 2
!ns + !n !n = K = !n = K; p 2 = 2
!n = 2 = = 1; = 1=!n = 1= K: p To achieve 10 overshot, we need j logMpj r j log0:1j = 0:5901: =r 2 + j logMpj2 2 + j log0:1j2 Hence we obtain = 1= K p 0:5901 = K 1=0:59012 = 2:8718: 10 1 1 1 = , : s+ s s s+ Thus we have step response:
Y s = yt = 1 , e, t1t: 1st Oder Systems: Rise time tr : We note that
yt1 yt2 e, t1 t1 = = = = tr = 1 , e, t1 = 0:1; 1 , e, t2 = 0:9; = 0:9; e, t2 = 0:1; or , log0:9; t2 = , log0:1; = log0:9 , log0:1 2:2 t2 , t1 = = : 1 , yts = e, ts = 0:04: Settling time ts: Thus we obtain a similar formula: 3:22 : ts = 11 E ects of Zero
The presense of the zero changes the step response. Consider s 2 !n 1 + !n H s = 2 : 2 s + 2
!ns + !n We have an addition of zero at s = , !n = , . If 1, then the system is close to prototype. If 1, then the zero has the same distance to the j!axis as the poles. The step responses for = 1 dashed line, = 2 dotted line, = 4 dashdotted line, and = 100 solid line are plotted below with = 0:5:
1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 It is observed that as decreases, the overshot increases but the rise time is reduced.
12 The e ect of zero to overshot is also a function of . See the gure below. But in general the closer the zero to the j!axis or the smaller the value, the larger the overshot.
Mp
2 1 = 0.5 = 0.3 = 0.7 2 4 6 8 10 Replacing s by s!n gives s= + 1 1 1 s H !ns = 2 = 2 + : s + 2
s + 1 s + 2
s + 1 s2 + 2
s + 1 Hence the step response is the sum of the original step response and the impulse response divided by . See Figure 3.22 of page 134. If is negative, then there is negative overshot. See Figure 3.23 of page 134. 13 E ects of Addition of Pole:
If an additional pole is present, we have 2 !n : H s = 2 1 + s= s2 + 2
!ns + !n We have an addition of pole at s = , !n = , . If 1, then the system is close to prototype. If 1, then the additional pole has the same distance to the j!axis as the two original poles. The step responses for = 1 dashed line, = 2 dotted line, = 4 dashdotted line, and = 100 solid line are plotted below with = 0:5:
1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 It is observed that as decreases, the overshot decreases but the rise time increases that is opposite to the previous case. 14 The e ect of additional pole to rise time is also a function of . See the gure below. But in general the closer the pole to the j!axis or the smaller the value, the larger the rise time.
nt 8 4 = 0.7 = 0.5 =1 5 10 Conclusion: The adverse e ect of zero is mostly on overshot. The closer the zero to the j!axis, the larger the overshot. The adverse e ect of the addition of pole is mostly on rise time. The closer the additional pole to the j!axis, the larger the rise time. 15 Approximation by Direct Truncation:
Suppose that
s+p with z and p very large compared to locations of the dominant poles. Then 1 0 1 0 10 z z A B 1 + s=z C A Gos @ A Gos: Gs = @ @ p 1 + s=p p This is called approximation by direct truncation. It applies to truncation of more than one pole or zero.
0 s + z1 A Gos Gs = @ 25 104 : Gs = s + 62 + 144 s + 202 + 100 The poles are:
p1;2 = ,6 12j; p3;4 = ,20 10j: Example 3: Consider system Direct truncation gives 5 102 G2s = : s + 62 + 144 16 while the dashed curve is for the second approximate system G2s.
3.5 Step responses: The solid curve is for the true system Gs, 3 2.5 2 1.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 It is seen that the error is due to larger overshot and smaller rise time for the approximate model. Note that direct truncation gives the same nal value. 17 Third Order Approximant:
Since additional pole makes overshot smaller, and rise time longer, we consider 5 102 G3s = s + 62 + 144 1 + s= where = 6. By trial and error, we nd that with = 2:5 the step response is closest possible as shown below:
3.5 3 2.5 2 1.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 18 ...
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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
 Chen

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