Today’s plan

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Today’s plan
•
1st order system response: review
– the role of zeros
•
2nd order system response:
– example: DC motor with inductance
– response classifications:
• overdamped
• underdamped
• undamped
•
Linearization
– from pendulum equation to the harmonic oscillator
2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
1
Review: step response of 1st order systems
steady state
(final value)
Nise Figure 4.3
2.004 Spring ’13
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Lecture 07 – Thursday, Feb. 21
2
Steady-state
•
•
Note that as t➞∞, the exponential decays away, and the step response
tends to 1, i.e. the value of the driving force.
More generally, if we consider a flywheel-like viscously damped system with
equation of motion expressed as 1st order linear time-invariant ODE
J!˙ (t) + b!(t) = f (t)
and step excitation
f (t) = F0 step(t)
[where f(t) and F0 have units of torque,]
the step response is
F0 ⇣
1
!(t) =
b
the angular velocity as t➞∞ tends to
•
•
e
t/⌧
!1 =
⌘
,
t>0
✓
J
⌧=
b
◆
F0
b
This long-term value is known as the system’s steady state
In systems where the output physically represents a velocity, the steady
state is also known as terminal velocity.
2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
3
Steady state in the Laplace domain: the final value thm.
•
It turns out that we can predict the steady state of a system directly
in the Laplace domain by using the following property known, for
obvious reasons, as the final value theorem:
limt!1 g(t) = lims!0 sG(s).
which holds generally if g(t) and G(s) form a Laplace transform pair.
•
In the case of the flywheel, the transfer function is
For the step response we have
It is easy to verify that
F (s) =
lims!0 s⌦(s) =
1
⌦(s)
=
F (s)
Js + b
F0
F0
) ⌦(s) =
s (Js + b)
s
F0
b
in agreement with the result for the steady state of this system
in the previous page
2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
4
A word on zeros
Transfer function
with a zero
H(s) =
1
1
s+z
=s·
+z·
D(s)
D(s)
D(s)
Derivative
operator
Amplification
(“gain”)
Example: compare the step response of the two systems below:
H0 (s) =
2.004 Spring ’13
1
s+p
H(s) =
Lecture 07 – Thursday, Feb. 21
s+z
s+p
5
Step response without zero vs with zero
U (s) =
1
s
F0 (s)
1
H0 (s) =
s+p
) f0 (t) =
F (s)
s+z
H(s) =
s+p
Example:
1.8
1.6
1
1
p
e
✓
pt
1
s
1
s+p
,
Please verify
yourselves this
partial-fraction
expansion!
◆
t > 0.
F (s) = sF0 (s) + zF0 (s)
) f (t) =
d
f0 (t) + zf0 (t) = e
dt
3
without
zero
2.5
1.4
jω
1.2
f0 [a.u.]
p = 0.5;
z = 1.0
2
1
1 1
=
F0 (s) =
s s+p
p
1
σ
0.8
pt
+
z
1
p
e
pt
,
t > 0.
with
zero
jω
2
f [a.u.]
1
U (s) =
s
1.5
σ
1
0.6
-0.5
0.4
-1
-0.5
0.5
0.2
0
−2
2.004 Spring ’13
0
2
4
t [sec]
6
8
10
Lecture 07 – Thursday, Feb. 21
0
−2
0
2
4
t [sec]
6
8
6
10
Zero on the right-hand side?
jω
jω
1
σ
0.9
0.25
0.6
0.2
f [a.u.]
0.4
0.5
−0.2
0.3
−0.4
0.2
−0.6
0.1
−0.8
2
4
t [sec]
Example:
6
2
1.8
1.6
p = 0.5;
1.4
8
−1
−2
10
2
2.5
jω
1.2
f0 [a.u.]
0
3
without
zero
(lhs)
1.0
0
0.4
0
-0.5
0.6
0.7
0
−2
σ
0.8
1
σ
0.8
4
t [sec]
6
8
with
zero
(lhs)
10
jω
2
f [a.u.]
-0.5
0.8
f [a.u.]
1
1.5
σ
1
0.6
-0.5
0.4
-1
-0.5
0.5
0.2
0
−2
2.004 Spring ’13
0
2
4
t [sec]
6
8
10
Lecture 07 – Thursday, Feb. 21
0
−2
0
2
4
t [sec]
6
8
7
10
The general 2nd order system
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
8
The general 2nd order system
Nise Figure 4.10
2.004 Spring ’13
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Lecture 07 – Thursday, Feb. 21
9
The general 2nd order system
Nise Figure 4.11
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
10
The underdamped 2nd order system
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
11
The underdamped 2nd order system
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
12
The underdamped 2nd order system
forced response,
sets steady state
Nise Figure 4.10
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
13
Transients in the underdamped 2nd order system
Nise Figure 4.14
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
14
Transient qualities from pole location in the s-plane
Nise Figure 4.19
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
15
Linearization
•
This technique can be used to approximate a non-linear system
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
16
Linearizing systems: the pendulum
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2.004 Spring ’13
Lecture 07 – Thursday, Feb. 21
17
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2.04A Systems and Controls
Spring 2013
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