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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
Lecture 181
Reading:
• Nise: Sec. 1.5
1 Common Inputs Used in Control System Design and Analysis
Two classes of inputs commonly used to characterize the performance of feedback control
systems are:
(a) The “Singularity” Functions: These functions have a discontinuity at time t = 0:
(i) The Dirac Delta (Impulse) Function: The delta function is used to charac­
terize the response of a system to brief, intense inputs. In the figure below, (a)
shows some unit pulses (pulses with unit area so that if the duration of the pulse
is T , its amplitude is 1/T . The Dirac delta function δ(t) is the limiting case of
such pulses as T → 0. Notice that this implies that the amplitude 1/T → ∞.
1 /T
1 /T
d T (t)
1
1 /T
3
1 /T
d (t)
2
0
4
T
1
T
2
T
3
T
t
4
a ) U n it p u ls e s o f d iffe r e n t e x te n ts
0
t
b ) T h e im p u ls e fu n c tio n
The strict definition of δ(t) is
δ(t) = 0, for t �= 0
δ(t) is undefined at t = 0 (lim δ(t) = ∞)
t→0
� 0+
δ(t)dt = 1
0−
(ii) The Unit-Step Function us (t): The unit-step (or Heaviside) function is used
to characterize a system’s transient response to a sudden change.
1
c D.Rowell 2008
copyright �
18–1
u s(t)
1 .0
0
0
t
tim e
The definition is
us (t) = 0 for t < 0
= 1 for t > 0
(iii) The Unit-Ramp Function r(t): The unit-ramp function is used to charac­
terize a system’s ability to follow a time-varying input, and the transient behavior
around a discontinuity in the slope of an input function.
r(t)
1 .0
0
1 .0
0
tim e
t
The definition is
r(t) = 0 for t < 0
= t for t > 0
The singularity functions a related to each other by differentiation and integration, as
shown below:
d (t)
r(t)
u s(t)
in te g r a tio n
0
tim e
t
d iffe r e n tia tio n
in te g r a tio n
1 .0
0
d iffe r e n tia tio n
0
tim e
t
1 .0
0
0
1 .0
tim e
t
(b) Sinusoidal Inputs: The response of linear systems to sinusoidal inputs of the form
u(t) = A sin(ωt + θ)
18–2
u ( t) = A s in ( w t)
A
0
t
p e r io d
2 p
T =
w
is of fundamental importance to control engineering and system dynamics ant will be
studied extensively throughout the course.
Example 1
In practice a machine, described a second-order transfer function G(s), will be
subjected to inputs that change suddenly. Use the unit-step response to deter­
mine how long it will take the machine’s response to settle to a new steady-state
value after a change.
The system’s block diagram is
U (s )
1
s
2
+ 5 s
+ 6
Y (s )
and the governing differential equation is
d2 y
dy
+ 5 + 6y = u(t).
2
dt
dt
The step response assumes that the system is “at rest” at time t = 0, that is
y(0) = 0 and ẏ(0) = 0, and that the input u(t) = us (t).
The solution is
y(t) = yh (t) + yp (t)
where yh (t) is the homogeneous solution, and yp (t) is the particular integral. The
characteristic equation is
λ2 + 5λ + 6 = (λ + 3)(λ + 2) = 0
and
yh (t) = C1 e−3t + C2 e−2t .
Assume yp (t) = K and substitute into the differential equation
0 + 0 + 6K = 1
18–3
or yp (t) = 1/6. The complete solution is
y(t) = C1 e−3t + C2 e−2t + 1/6.
At time t = 0
y(0) = C1 + C2 + 1/6 = 0
ẏ(0) = −3C1 − 2C2 + 0 = 0
giving C1 = 1/3 and C2 = −1/2, so that the system’s response to a unit-step
input is
1
1
ystep (t) = e−3t − e−2t + 1/6.
3
2
The response is shown below, and indicates that it takes this system 2.5–3 seconds
to respond to the step.
0 .2
s s
= 1 /6
R e s p o n s e
y
0 .1 5
0 .1
0 .0 5
in itia l s lo p e is z e r o
0
0
1
2
3
t (s e c )
Example 2
Find the steady-state response of a first-order system to a sinusoidal input u(t) =
A sin(ωt).
u ( t) = A s in ( w t)
1
t s + 1
The differential equation is
y (t)
dy
+ y = u(t)
dt
and assume the complete solution is y(t) = yh (t) + yp (t) as in the previous
example. The characteristic equation is
τ
τλ + 1 = 0
18–4
from which
yh (t) = Ce−t/τ
and assume
yp (t) = K1 cos(ωt) + K2 sin(ωt)
In steady-state we assume that yh (t) = C1 e−t/τ has decayed to zero, and
yss (t) = yp (t) = K1 cos(ωt) + K2 sin(ωt)
Substitution of yp (t) into the differential equation gives
τ ω (−K1 sin(ωt) + K2 cos(ωt)) + (K1 cos(ωt) + K2 sin(ωt)) = A sin(ωt)
or
(ωτ K2 + K1 ) cos(ωt) + (−ωτ K1 + K2 ) sin(ωt)
and comparing coefficients
ωτ K2 + K1 = 0
−ωτ K1 + K2 = A,
or
ωτ A
,
1 + (ωτ )2
K1 =
K2 =
A
1 + (ωτ )2
so that
A
(sin(ωt) − ωτ cos(ωt))
1 + (ωτ )2
�
�
A
1
ωτ
�
= �
sin(ωt) − �
cos(ωt)
1 + (ωτ )2
1 + (ωτ )2
1 + (ωτ )2
A
= �
(cos(φ) sin(ωt) − sin(φ) cos(ωtφ)) .
1 + (ωτ )2
yss (t) = yp (t) =
using the following triangle
1 +
f
2
t)
(w
w t
s in ( f ) =
c o s (f ) =
w t
1 + (w t )
1
1 + (w t )
1
Then the system sinusoidal response is
A
yss (t) = �
1 + (ωτ )2
where φ = tan−1 (ωτ ). We note the following
18–5
sin(ωt − φ)
2
2
(a) The steady-state sinusoidal response is a sinusoid of the same frequency as
the input.
(b) There is a phase shift (lag) between the input and output φ = tan−1 (ωτ ).
(c) The amplitude of the output is a function of the input frequency ω.
(d)
• At low frequencies (ω → 0), the response is yss (t) ≈ A sin(ωt) and the
amplitude approaches that of the input.
• At very high frequencies (ω → ∞), the response is yss (t) ≈ (A/ωτ ) sin(ωt−
π/2) and the response amplitude becomes very small.
√
• When ω = 1/τ , yss (t) = (A/ 2) sin(ωt − π/4), that is the amplitude is
reduced by a factor of 0.707, and the phase shift is 45◦ .
18–6