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2.004 Dynamics and Control II
Spring 2008
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Massachusetts Institute of Technology
Department of Mechanical Engineering
2.004 Dynamics and Control II
Spring Term 2008
Solution of Problem Set 2
Assigned: Feb. 15, 2008
Due: Feb. 22, 2008
Problem 1:
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(a)
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Assuming an ideal dc motor:
P owerin = P owerelect = vb im = Kv Ωm im
P owerout = P owermech = Tm Ωm = Km im Ωm
P owerin = P owerout ⇒ Kv Ωm im = Km im Ωm ⇒ Kv = Km
(b)
Using the data sheet:
Km = 30.2
Kv =
mNm
Nm
= 30.2 × 10−3
A
A
1
1 V
1
=
rpm =
317 V
317 rpm
317 ×
V
2π rad
60
s
= 30.1 × 10−3
V
rad
s
Problem 2:
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Using an approach similar to that we used in class (with the same basic mass/friction linear
model):
1
(a)
dv
+ Bv = Fp (t) + Fd (t)
dt
Ignore Fd , assume Fp (t) = Ke φ(t) and employ v = dx
:
dt
m
m
dx
d2 x
+B
= Ke φ(t)
2
dt
dt
Ke
X(s)
=
Φ(s)
ms2 + Bs
(b-c)
(d)
In the steady state,
dx
dt
d2 x
dx
m 2 +B
= Ke Kp (xpark − x)
dt
dt
= 0 hence:
xs.s. = xpark
(e)
2
Position Control: m ddt2x + B dx
= Ke Kp (xpark − x)
dt
dv
Velocity Control: m dt + Bv = Ke Kp (vdesired − v)
In the position control vs.s. = 0, all the left terms of the equation are zero and a zero
error can be held. On the other hand for the velocity control, dv
| = as.s. = 0 but
dt
s.s.
vs.s. 6= 0, the drag force is not zero, and then vs.s. =
6 vdesired .
Problem 3: Nise, Chapter 5, Problem 4.
2
(1)
Split G3 and combine it with G2 and G4 . Also use feedback formula on G6 loop:
G1
_
R(s) +
G2+G3
_
+
G5
G6
C(s)
G6
C(s)
1+G6
+
G4+G3
G7
(2)
Push G2 + G3 to the left past the pickoff point:
R(s) +
_
_
G1
G2+G3
+
G5
+
1+G6
G4+G3
G2+G3
G7
(3)
Using the feedback formula and combining parallel blocks:
R(s) +
_
G2+G3
1+ G1(G2+G3)
G4+G3
G2+G3
+ G5
G6
1+G6
C(s)
G7
3
Figures by MIT OpenCourseWare.
(4)
Multiplying the blocks of the forward path and applying the feedback formula:
T (s) =
G6 G4 +G6 G3 +G6 G5 G3 +G6 G5 G2
1+G6 +G3 G1 +G2 G1 +G7 G6 G4 +G7 G6 G3 +G6 G3 G1 +G6 G2 G1 +G7 G6 G5 G3 +G7 G6 G5 G2
Problem 4:
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Zo
Zs
�
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1
Ro Cs
1
Ro
= Ro ||
=
1 =
Cs
Ro + Cs
Ro Cs + 1
Ro
= R + Ls + Zo = R + Ls +
Ro Cs + 1
Voltage Division:
R
o
Vo
Zo
Ro Cs+1
=
=
Ro
Vs
Zs
R + Ls + Ro Cs+1
Ro
=
2
LRo Cs + (L + RRo C)s + (R + Ro )
G(s) =
Problem 5: Input impedance does not characterize a system uniquely. In other words, sys­
tems with different characteristics, like below examples, can have the same input impedance.
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(a)
Za = R + Ls + 2R || (Ls + 2R)
2R(Ls + 2R)
= R + Ls +
2R + (Ls + 2R)
2RLs + 4R2
= R + Ls +
4R + Ls
0.5RLs + R2
= R + Ls +
R + 0.25Ls
4
�
(b)
Zb = 2R + Ls + R || (0.25Ls)
R(0.25Ls)
= 2R + Ls +
R + 0.25Ls
R(0.25Ls)
= R + Ls + R +
R + 0.25Ls
0.5RLs + R2
= R + Ls +
R + 0.25Ls
= Za
5