So far
•
•
First-order systems (linear)
Second-order systems (linear)
Today
•
•
Higher-order systems (linear)
– when can we approximate as second-order?
Nonlinear systems
– Review of cases we’ve seen
– Linearization
• Example: pendulum
– DC motor nonlinearities
• Example: load connected with gears; saturation, dead zone,
backlash
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
The effect of multiple poles
C(s) =
A
+
s
B (s + ζωn ) + Cωd
2
(s + ζωn ) + ωd2
D
.
s + αr
Images removed due to copyright restrictions.
Please see: Fig. 4.23 and 4.24 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
+
Linear vs Nonlinear (revisited)
Linear spring:
f (x(t)) = Kx(t)
f (t)
Nonlinear spring:
f (x(t)) = K(x)x(t)
f (t)
K(x2 )
K ≡ slope
K(x1 )
x(t)
x1
x(t)
f (t)
f (t)
fc
v(t) ≡ ẋ(t)
Viscous friction:
f (v(t)) = −fv v(t)
x2
v(t) ≡ ẋ(t)
fv
Coulomb friction: f (x(t)) = −fc sgn (x(t))
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
Linear vs Nonlinear (revisited)
e.g., linear spring
Linear system: f (x(t)) = kx(t)
Consider linear combination of inputs
e.g., nonlinear spring
Nonlinear system: f (x(t))
6= kx(t)
p
e.g., f (x(t)) = k x(t).
Consider same linear combination of inputs
a1 x1 (t) + a2 x2 (t)
Then output is same
linear combination of outputs
f (a1 x1 (t) + a2 x2 (t)) = k (a1 x1 (t) + a2 x2 (t)) =
a1 (kx1 (t)) + a2 (kx2 (t)) = a1 f (x1 (t)) + a2 f (x2 (t)) .
2.004 Fall ’07
a1 x1 (t) + a2 x2 (t)
Then output is not the same
linear combination of outputs
p
f (a1 x1 (t) + a2 x2 (t)) = k (a1 x1 (t) + a2 x2 (t)) 6=
p
p
a1 kx1 (t) + a2 kx2 (t) = a1 f (x1 (t)) + a2 f (x2 (t)) .
Lecture 09 – Wednesday, Sept. 26
Linearization
f (x)
δf (x)
Output
~~f (x)
f (x0)
A
Images removed due to copyright restrictions.
Please see Fig. 2.48 and 4.24 in Nise, Norman S. Control Systems Engineering.
δx
4th ed. Hoboken, NJ: John Wiley, 2004.
0
x0
Input
x
x
Figure by MIT OpenCourseWare.
f (x) ≈ f (x0 ) + ma (x − x0 )
where
¯
df ¯¯
ma =
.
¯
dx x=x0
2.004 Fall ’07
Example
Linearize f (x) = 5 cos x near x = π/2.
Answer: We have f (π/2) = 0, ma = −5, so
³
π´
f (x) ≈ −5 x −
(x ≈ π/2)
2
Lecture 09 – Wednesday, Sept. 26
Linearizing systems : the pendulum
2
Jd θ
dt2
L
T
2
θ
T
Mg
L
θ
Mg cos θ
MgL sin θ
2
θ
2
Mg sin θ
Mg
Figure by MIT OpenCourseWare.
The equation of motion is found as
J θ̈ +
M gL
sin θ = T.
2
(We cannot Laplace transform!)
For small angles θ ≈ 0, we have
¯
d sin θ ¯¯
sin θ ≈
× θ = cos θ|θ=0 × θ = 1 × θ = θ.
dθ ¯θ=0
Therefore, the linearized equation of motion is
J θ̈ +
2.004 Fall ’07
M gL
M gL
θ = T ⇒ Js2 Θ(s) +
Θ(s) = T (s).
2
2
Lecture 09 – Wednesday, Sept. 26
Case study: motor with gear load
Some common nonlinearities
XLa
+
•
Saturation
+
ea(t)
ia(t)
Motor circuit
Ra = 8Ω
Armature v (t)
b
circuit
-
Tm(t)
•
Km = 0.5N · m/A
Kb = 0.5V · sec/rad
Kb = 0.5V · sec/rad
Rotor
θm(t)
Figure by MIT OpenCourseWare.
Dead zone
Motor loading
Tm (t) θm (t)
Images removed due to copyright restrictions.
Please see Fig. 2.46 in: Nise, Norman S. Control Systems
Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
Jm
Dm
Figures by MIT OpenCourseWare.
Motor loaded with gears
•
Backlash
N1
Motor
Ja , Da
N2
N
N11 :: N
N22 =
= 25
25 :: 250
250
Ja = 0.02kg · m2
JL = 1kg · m2
Da = 0.01kg · m2 /sec
DL = 1kg · m2 /sec
JL
DL
0.2083
Ωo (s)
=
;
Vs (s)
s + 1.71
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
Θo (s)
0.2083
=
Vs (s)
s (s + 1.71)
Saturation
Images removed due to copyright restrictions.
Please see Fig. 4.29 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
System TF
0.2083
s + 1.71
Input Step function
u(t) ↔
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
1
s
Dead zone
Images removed due to copyright restrictions.
Please see: Fig. 4.30 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
System TF
0.2083
s (s + 1.71)
Input Sinusoid
5 sin (2πt) u(t)
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
Backlash
Images removed due to copyright restrictions.
Please see: Fig. 4.31 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
System TF
0.2083
s (s + 1.71)
Input Sinusoid
5 sin (2πt) u(t)
2.004 Fall ’07
Lecture 09 – Wednesday, Sept. 26
Case study solution \1: electro-mechanical model
Motor circuit
XLa
Ra = 8Ω
+
ea(t)
ia(t)
KVL around the DC motor circuit loop
(neglecting the inductance La ) yields
+
Rotor
Ra Ia (s) + Vb (s) = Ea (s).
Armature v (t)
b
circuit
-
Substituting Ia , Vb from the motor equations,
Tm(t)
Ra
θm(t)
Recall DC motor equations
(in the Laplace domain)
Tm (s) =
Km Ia (s);
Vb (s) =
Kb Ωm (s).
Figures by MIT OpenCourseWare.
Tm (t) θm (t)
Jm
Dm
Figures by MIT OpenCourseWare.
˙ (t) ⇔ Ωm (s) = sΘm (s),
Since ωm (t) = θm
we can rewrite the second motor equation as
Vb (s) = sKb Θm (s).
Tm (s)
+ Kb sΘm (s) = Ea (s).
Km
Assume an equivalent load of inertia Jm and
damping Dm , subject to the motor’s torque Tm (s).
Torque balance on this system yields
Tm (t) = Jm θ¨m (t) + Dm θ̇m (t) ⇒
Tm (s) =
Jm s2 Θm (s) + Dm sΘm (s).
To find an equation relating the source Ea (s)
Substituting into the electrical equation from above,
to the motor output angle Θm (s),
¸
·
¢
we must relate the source Ea (s) to the torque Tm (s),
Ra ¡
2
Jm s + Dm s + Kb s Θm (s) = Ea (s)
and the torque Tm (s) to the angle Θm (s).
Km
2.004 Fall ’07
Lecture 09 supplement – Wednesday, Sept. 26
Case study solution \2: load model
N1 : N2 = 25 : 250
Ja = 0.02kg · m2 Recalling the gear loading
JL = 1kg · m2
µ ¶2
Da = 0.01kg · m2 /sec
N1
Jm = Ja +
JL ;
DL = 1kg · m2 /sec
Motor loaded with gears
N2
N1
Motor
Ja , Da
JL
N2
DL
"
s s+
1
Jm
Ra Jm
Ã
Dm +
K m Kb
Ra
!#
We must now relate the equivalent loads Jm , Dm
to the actual load that consists of the motor’s own
armature inertia Ja and compliance Da ,
as well as the load’s inertia JL and compliance DL .
The load is connected to the motor via a gear—pair
of ratio N1 : N2 .
2.004 Fall ’07
Dm = D a +
µ
N1
N2
¶2
DL .
Also note that the load shaft angle is related to the
motor shaft angle via
µ ¶
N1
Θm (s).
Θo (s) =
N2
Figures by MIT OpenCourseWare.
The last equation from the previous page
can be rearranged and rewritten as a transfer function
Kt
Θm (s)
=
Ea (s)
relationships from lecture 2,
After substituting the numerical values, we find
that the transfer function is of the form
Θm (s)
K
0.2083
=
=
,
Ea (s)
s (s + p)
s (s + 1.71)
¢
¡
where the system gain K = 0.2083rad/ V · sec2
and the system pole p = −1.71Hz.
The extra s in the transfer functions’ denominator
indicates that the system includes an integrator.
We can also obtain the TF for Ωo (s) = sΘo (s)
Ωm (s)
K
0.2083
=
=
.
Ea (s)
s+p
s + 1.71
Lecture 09 supplement – Wednesday, Sept. 26
Case study solution \3: torque-speed curve
La
X
Ra = 8Ω
+
ea(t)
ia(t)
Recall the relationship we obtained from KVL
and the motor equations,
+
Rotor
Ra
Tm (s) + Kb sΘm (s) =
Km
Ra
Tm (s) + Kb Ωm (s) =
Km
Armature v (t)
b
circuit
-
Tm(t)
θm(t)
Figures by MIT OpenCourseWare.
Images removed due to copyright restrictions.
Please see: Fig. 2.38 in Nise, Norman S.
Control Systems Engineering. 4th ed.
Hoboken, NJ: John Wiley, 2004.
Ea (s).
Inverse Laplace—transforming,
Ra
Tm (t) + Kb ω(t) = ea (t) ⇒
Km
Tm = −
Kb Km
Km
ωm +
ea .
Ra
Ra
This relationship on the ωm − Tm plane represents
a straight line called torque—speed curve,
with slope −Kb Km /Ra and offset Km /Ra .
ωm = 0 ⇔ Tstall =
Km
ea
Ra
Stall torque;
ea
Kb
No—load speed.
Tm = 0 ⇔ ωno−load =
2.004 Fall ’07
Ea (s) ⇒
Lecture 09 supplement – Wednesday, Sept. 26