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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 301
Reading:
• Nise: 10.1
1
Sinusoidal Frequency Response
1.1
Definitions
Consider a sinusoidal waveform
f (t) = A sin (ωt + φ)
A
A s in ( w t + f )
A s in ( f )
0
-A
t
T =
2 p
w
where
A is the amplitude (in appropriate units)
ω is the angular frequency (rad/s)
φ is the phase (rad)
In addition we can define
T the period T = 2π/ω (s)
f the frequency, (f = 1/T = ω/2π) (Hz)
1
c D.Rowell 2008
copyright �
30–1
The Euler Formulas: We will frequently need the Euler formulas
ejωt = cos (ωt) + j sin (ωt)
e−jωt = cos (ωt) − j sin (ωt)
or conversely
�
1 � jωt
e + e−jωt
2
�
1 � jωt
e − e−jωt
sin (ωt) =
2j
cos (ωt) =
1.2
The Steady-State Sinusoidal Response
vm (t)
F (t)
vm (t)
0
t
In p u t
F (t)
m a s s
m
fr ic tio n B
tr a n s ie n t
re s p o n s e
s te a d y -s ta te
re s p o n s e
o
t
R e s p o n s e
Assume a system, such as shown above, is excited by a sinusoidal input. The total response
will have two components a) a transient component, and a steady-state component
y(t) = yh (t) + yp (t).
We define the steady-state component as the particular solution yp (t). Let the system dif­
ferential equation be
an
dn y
dn−1 y
dy
dm u
dm−1 u
du
+
a
+
.
.
.
+
a
+
a
y
=
b
+
b
+ . . . + b1
+ b0 u
n−1
1
0
m
m−1
n
n−1
m
m−1
dt
dt
dt
dt
dt
dt
with a complex exponential input
u(t) = ejωt .
Assume a particular solution yp (t) to be of the same form as the input, that is
yp (t) = Aejωt
and since
dk yp
= A(jω)k ejωt
dtk
substitution into the differential equation gives:
�
�
an (jω)n + an−1 (jω)n−1 + . . . + a1 (jω) + a0 Aejωt
�
�
= bm (jω)m + bm−1 (jω)m−1 + . . . + (b1 jω) + b0 ejωt
30–2
or
A=
bm (jω)m + bm−1 (jω)m−1 + . . . + b1 (jω) + b0
an (jω)n + an−1 (jω)n−1 + . . . + a1 (jω) + a0
Examination of this equation shows its similarity to the transfer function H(s), in fact
A = H(s)|s=jω = H(jω)
so that the steady-state response yss (t) is
yss (t) = yp (t) = Aejωt = H(jω)ejωt ,
(1)
or in other words, the steady-state response to a complex exponential input is defined by
the transfer function evaluated at s = jω, or along the imaginary axis of the s-plane. Note
that H(jω) is in general complex.
u (t) = e
L in e a r S y s te m
jw t
H (s )
y
s s
( t) = H ( jw ) e
jw t
We now extend this argument to a real sinusoidal input, for example u(t) = cos (ωt) =
(ejωt + e−jωt )/2. The principle of superposition for linear systems allows us to express the
response as the sum of the two responses to the complex exponentials:
yss (t) =
�
1�
H(jω)ejωt + H(−jω)e−jωt
2
We now proceed as follows:
• We show that H(−jω) = H(jω) where H(jω) denotes the complex conjugate (see the
Appendix), so that
�
1�
jωt
−jωt
yss (t) =
H(jω)e + H(jω)e
2
(2)
• We break up H(jω) into its real and imaginary parts,
H(jω) = � {H(jω)} + j� {H(jω)}
H(jω) = � {H(jω)} − j� {H(jω)}
and use the Euler formula to write
ejωt = cos (ωt) + j sin (ωt)
e−jωt = cos (ωt) − j sin (ωt)
• We combine the real and imaginary parts of Eq. (2) to conclude
yss (t) = � {H(jω)} cos(ωt) − � {H(jω)} sin(ωt)
30–3
(3)
• We then use the trig. identity
a cos θ − b sin θ =
√
a2 + b2 cos(θ + φ)
to write Eq. (3) as
yss (t) = |H(jω)| cos (ωt + � H(jω))
(4)
where
�
�
�2 {H(jω)} + �2 {H(jω)}
�
�
� {H(jω)}
H(jω) = arctan
� {H(jω)}
|H(jω)| =
0
Equation (4) states the answer we seek. It shows that
• The steady-state sinusoidal response is a sinusoid of the same angular frequency
as the input,
• The response differs from the input by (i) a change in amplitude as defined by
|H(jω)|, and (ii) an added phase shift � H(jω).
H(jω) is known as the frequency response function. |H(jω)| is the magnitude of the frequency
response function, and � H(jω) is the phase.
A m p litu d e
u (t)
y (t)
a m p litu d e r a tio :
|u (t)|
|H ( jw ) | =
|y (t)|
|u (t)|
|y (t)|
t
T im e
0
D t
p e r io d T
p h a s e s h ift:
30–4
Ð H ( jw ) = - 2 p D t
T
Note that if |H(jω)| > 1 the sinusoidal input is amplified, while if |H(jω)| < 1 the input is
attenuated by the system.
Example 1
The mechanical system
vm (t)
m a s s
m
F (t)
fr ic tio n B
has a transfer function
vm (s)
1
=
F (s)
ms + B
where m = 1 kg, and B = 2 Ns/m. Find the steady-state response if F (t) =
10 sin(5t).
H(s) =
1
s+2
so that the frequency response function is
1
2 − jω
H(jω) = H(s)|s=jω =
= 2
jω + 2
ω +4
H(s) =
Then
1
|H(jω)| = √
,
2
ω +4
With ω = 5 rad/s,
�
� ω�
H(jω) = arctan −
.
2
vss (t) = 10 |H(jω)| sin(5t + � H(jω)
10
= √ sin(5t − arctan 2.5)
29
= 1.857 sin(5t − 1.1903)
Example 2
Plot the variation of |H(jω)| and � H(jω) from ω = 0 to 10 rad/s.
From above
1
|H(jω)| = √
, and
2
ω +4
These functions are plotted below:
�
30–5
� ω�
H(jω) = arctan −
.
2
0 .5
F r e q u e n c y R e s p o n s e M a g n itu d e
0 .4
0 .3
0 .2
0 .1
fre q u e n c y (ra d /s )
F re q u e n c y R e s p o n s e P h a s e (d e g )
0
0
0
2
4
6
8
1 0
-1 0
-2 0
-3 0
-4 0
-5 0
-6 0
-7 0
-8 0
-9 0
Note that
• As the input frequency ω increases, the response magnitude decreases.
• At low frequencies the phase is a small negative number, but as the frequency
increases the phase lag increases and apparently is tending toward −90◦ at
high frequencies.
30–6
Appendix: Evaluation of H(−jω).
We start with
H(jω) =
so that
H(−jω) =
Note that
bm (jω)m + bm−1 (jω)m−1 + . . . + b1 (jω) + b0
an (jω)n + an−1 (jω)n−1 + . . . + a1 (jω) + a0
bm (−jω)m + bm−1 (−jω)m−1 + . . . + b1 (−jω) + b0
an (−jω)n + an−1 (−jω)n−1 + . . . + a1 (−jω) + a0
(jω)k
= (−1)k/2 ω k
j(−1)(k−1)/2 ω k
(−jω)k = (−1)k/2 ω k
−j(−1)(k−1)/2 ω k
k even
k odd
k even
k odd
Thus in both H(jω) and H(−jω)
• The terms with even powers of ±jω in the numerator and denominator of H(jω) and
H(−jω) generate real terms, while
• the terms with odd powers of ±jω generate imaginary terms.
With these substitutions, comparison of H(jω) and H(−jω) shows
•
The real terms (even powers of ±jω) are the same, while
• The imaginary terms (odd powers of ±jω) have opposite signs
leading to the conclusion
H(−jω) = H(jω).
30–7