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Unformatted text preview: HW/CE# 5 Solution (Where applicable, use the “more exact” tableon page 2 of the ref sheets to answer the
following.) 100 1. Su oseGs =—
pp () .qs+10) is the openloop gain in a unity feedback system. a) What is the phase margin PM and crossover frequency we for the system? 0 ,mp/0 ‘§=05’—9/w=ﬂ3 %=J%0
= $7na4/é b) Sketch the Nyquist plot for the system for a) > 0 in the vicinity of the 1 point. Ail c) Suppose we consider the effects of delay in the feedback path such that
H (s) = e"""' where t,, = 0.041 sec Note that this will not change wc. What is the new phase margin for the system? . ..s[ao%/
CaL¥a1= _ﬁEL, a, )
{(51410) (1) Make a rough sketch for the Nyquist plot for the system with delay added. This need
not be detailed but should show generally what will happen in the vicinity of the 1 point A3 and zero point on the plot. Excel generated plot for 7.5 5 w 5 432 rad/sec ( 1/07} 1?; ‘3’ >
'3' Nyquist PlotG 2 _ 0.2 .0]. ._ e) Estimate Mp, tr, and t5 for the system with delay. Note that although we did not
change, a)" will change in that the phase margin ( and C) changed. 0
My: 33.8 —7 f? 0.3 ——> HP = 34.237. Mn wh
if: {LgVi: alga Sec.
“n K4? *ﬁ .2155: =ité£m = 417,4“, 1) Use MATLAB, Excel, or a program of your own choosing to simulate the response of ‘.
the system (with 0.041 sec delay) to a step input r(t) = u(t). What does your simulation f
yield for Mp, t,, and t5 ? ‘ 671/; We ,D/9f ”{14' Excel Simulation plot: Step Response y(t) Time in Seconds Mp tr ts From 1e) above 37.23% 0.152 sec 1.79 sec Excel Simulation 39.5% 0.136 sec 1.71 sec MATLAB Simulink approach: Block Diagram: Transport
Delay Simulation Results: a} Rmpe «I #191331
QIIMHEIW’S g .4 *2”
:x", ”A . 63*
g. :2
,. ‘..I p: 2. Suppose we have a plant G(s): 21.4 Gm = s(s +1.85) in a unity feedback system. a) Determine for this system: settling time t5, overshoot for a step input Mp, crossover
frequency we, and phase margin PM. 2 , . __ = . .' ' = . n. .
My  N’v‘ .. La)? — (£626, 19;)” Ms" 39 0/“ 02.
ts= 4.43 ._. 5;”:5‘ﬂ‘o g) ”P —— 572.43%
Z‘Jn a
,wb = (9,74% = 44%,...14 ) P7: 22.4
.—_=:_._—. ____’
K(Ts+1) b) Choose a controller: DC (5) = with a selected for a lead compensator. (aTs + 1) Find K, T, and a to realize a new design phase margin of 65.20 and a new crossover
frequency of we = 9.2 rad/sec. 5w L/é[zd&)/r / = K {Z/ﬁP)” > k= 44a”— (ZN/m“ m (a ”cum : MW — (—18» = we», bar
0 IA 0
. / a » _. mg?
)0 M (.5 . Wm /H = (CZ—H.370 = 53.83" See 7/(152’ 94:} C. °¢= /— {“43” .3 (ix/M69 M A) = (J; 3 “7,2 = l...—
Ww “TC/at
.'. '7'— ’ __.L.—. — 0, 3329
22 .ma
ﬂ/JMM M : /¢_ w.» 64.03% __ 54434 “
/Dc (3' i2)/
= /.3/7 . _ , 3 4
g 44:27,» #2:» 2— A50?) x o./od 7‘ _ a . '5’ 32 7
c) What would you estimate for overshoot Mp and settling time tS for the compensated
system?
0
P” t 6 5" Z’ ) ”(a = g 2
MP : 4, 4 70
’ I  J 1 f i = ’9 Rf, t ‘4)" 0‘ l r
7”,“!!! {1&5 /€ ' t! 5,??? ) g j "I: I a
5:)" n
50 g = +4“ 2 0/74?8 m 95m.
; ”f d) Check your design done in part b) above by: 1) Doing a MATLAB generated Bode plot for Dc(s)G(s). Use MATLAB’s
“margin” function to check the values of we and phase margin for the compensated
system. How did these values compare to your design target values? Bode Diagram Prob 2 PM = 65.2 deg [at 9.2 radrsec) Magnitude (dB) 135 P hase (deg) 4 8]  10" 100 101 10‘ 10 Frequency [radisec] )
(a Design target values matched. 2) Finding closedloop T (s) for the compensated system then doing a MATLAB generated step response to determine overshoot Mp and settling time? How did these
values compare to the values you found in c) above? Amplitude 1.4 1.2 0.8 0.4 0.2 Original
Uncompensated
System: Design Goals:
From 2c T(s) for
Compensated
System Mp 8.04 ‘36 0.5 Step Response Prob 2
settling time 0.954 D .B 1
Time (sec) Mp 52.7% 4.6% 8.04% 12 1.4 5.0 sec 0.5 sec 0.954 3. Suppose we have a plant G(s): 21.4 ' Gm = s(s+l.85) in a unity feedback system. (Same plant as for problem #2 above.) a) What is the steady state error for a unit step input to the closed loop system for this
system? We / fig/em, r, 3;. arm #4.? 2; a(Ts +1)
(aTs +1) input r(t) = t u(t) to the closed loop system is 5 0.05. In essence, ﬁnd a and T for the lag
compensator. b) Add a lag compensator: Dc (3) = so that the steady state error for a ramp /: ’_ / I ' P :. ’1’;
: a .w..  _— W J
3 #136 / +. 404+.) (1.4 p _, V («72 +1) 5 (he/:5?) [inv f5 7‘ an.” (df+l>(+/.8f7 : Aar‘ e ,0: 390 590 §(.(;3+,)(5f/lgf)+0((7§¢I)KZI.V ﬂat
x 2 A72? ~r sd 0(= A72?
{9W .1 <4 NC “I: 35‘ __ 3&9 W.) 7’: <,{§‘DJ—
r 7 2:0 20 c) Check your design done in part b) above by: 1) Doing a MATLAB generated Bode plot for Dc(s)G(s). Use MATLAB’s
“margin” function to check the values of we and phase margin for the compensated
system. These should be the same or close to the values you determined in problem 2a) above.
Bode Diagram
Gm = Int 9113 (at Int radlsec) , Pm = 21 .4 deg (at 4.45 radlsec) 100 50
6‘
B
l.) E O
4:
3
2" 50 4 00 ~90
a
a:
B a, 135
w
a
1:
1 130 ‘
10'
Frequency (radfsec)
PM we
Problem 2a 22.6° 4.44 rad/s Above Plot 21 .40 4.45 rad/s 2) Finding closedloop T(s) for the compensated system then doing a MATLAB generated step response to determine t5 and Mp. These values should be the same or
close to the values you determined in problem 2a) above. Step Response Prob 3
Mp = 55.3% settling time = 5.09 sec Amplitude U 1 2 3 4 5 6
Time (sec)
Mp ts
Problem 2a 52.7% 5.004 sec Plot above 55.3% 5.09 sec 3) Doing a MATLAB generated ramp response for the closed loop system to be
sure that you met the steady state error speciﬁcation. Ramp Response Prob 3
Ramp Error = 0.05 at 9.21 sec System: T
Time (sec): 9.21
Amplitude: 9.15 Amplitude Time (sec) Steady state ramp error = r(t) — y(t) = t — y(t) as t 9 °° = 9.21 — 9.16 = 0.05 at t = 9.21 sec ...
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 Fall '08
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