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Unformatted text preview:  ccs Gucﬁ
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... o 30' l 10.19 The steadystate oﬂ'set error is
KP oﬂ'seterror=w,wu=l— 3+KP =0.2
if K}: = 12. The time constant for this value of K p is
T ___ I ___ 2 = 3 “
c + KP 3 + K p 15
The steadystate response due to the disturbance is
disturbance response = c :10, = 1—; = 0.067 If both inputs are applied, the actual steadystate speed is
w” = command response + disturbance response = (1 — 0.2) + (—0.067) = 0.733 Note that we can make the offset error. the disturbance response, and the time constant
smaller only by making K ,2 larger than 12, but this increases the maximum required torque
and probably the cost of the system. ©2010 McGrawHill. This work is only for nonproﬁt use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful. 10.32 Because C = cos B, the value C = 0.707 corresponds to £3 = 45°, which is a 45° line on
the complex plane. Thus C = 0.707 corresponds to a pair of roots whose real and imaginary
parts have the same magnitude; that is, s = —a :l: ja. The time constant of these roots is
r = l/a. Thus the roots must be 5 = l:tj ifr = 1 and C = 0.707. These roots correspond
to the polynomial equation (3+1—j)(s+1+j)=(s+1)2+1=s’+2s+2=o or
10::2 +203+ 20: 0 Compare this with the system’s characteristic equation obtained from the denominator of the transfer ﬁmction:
1032 + (3 + Kn): + Kp = 0 Thus Kp=20and 3+KD=20, or KD=17. ©2010 McGrewHill. This work is only for nonproﬁt use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful. 10.40 The system's closedloop transfer function can be found from the block diagram. It ‘8 C(s) _ Kp + Km
R(s) ~ 232 + (2+ 1(0):! + KP Thus the characteristic equation is 232 + (2 + K D)s + K p = 0. Because C is speciﬁed to be
less than 1. we can write the following formula for the time constant. 2(2) _ 1 T=2+KD— This gives K D = 2. The damping ratio is given by _2+KD_ 2+2 2 — 2J2Kp — 2J2Kp = JZKP =03 C This gives K p = 2.469.
If the command is a step input with a magnitude m, then R(s) = m/s, and the Final
Value Theorem gives _=m cu=ali_r.r(i,sC(s)=lims( s KplKDS )m
300 282+(2+KD)8+KP Thus the steady state error is zero. so all three speciﬁcations are satisﬁed. ©2010 McGrawHill. This work is only for nonproﬁt use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful. 10.45 a) The closedloop transfer ftmction is C(s) _ 3K}: + 3Kps
R(s)  s2 + 3Kps + 3Kp — 4 The damping ratio is 3K
_ D =
C — 2\/3KP4 0.707
The time constant is 2
= = .1
1' 3KD 0 These two conditions give K p = 68 and K D = 20/3.
b) The closedloop transfer function with the negative rate feedka gain K 1 is 0(8) _ 3Kp
R(s) — 82+3K18+3Kp4 This denominator has the same form as the transfer function in part (a) with K D replaced
by K1. Thus Kp = 68 and K1= 20/3. c) For part (a),
C(s) _ 208 + 204 R(s)  82 + 20.9 + 200
This has numerator dynamics and will overshoot more than the design in part (b), for which C(s) _ 204
R(s) — 32 + 208 + 200 ©2010 McGrawHill. This work is only for nonproﬁt use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful. ...
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 Fall '10
 Meckle

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