22 Frequency Response for Linear Systems

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ME 132, Spring 2005, UC Berkeley, A. Packard
22
22.1
Frequency Response for Linear Systems
Theory for Stable System: Complex Input Signal
Consider the linear dynamical system
ẋ(t) = Ax(t) + Bu(t)
y(t) = Cx(t) + Du(t)
(99)
We assume that there are n states, m inputs, and q outputs (so A ∈ Rn×n , B ∈ Rn×m ,
C ∈ Rp×n , D ∈ Rp×m ).
If the system is stable, (ie., all of A’s eigenvalues have negative real parts) it is “intuitively”
clear that if u is a sinusoid, then y will approach a steady-state behavior that is sinusoidal,
at the same frequency, but with different amplitude and phase. In this section, we make this
idea precise.
Take ω ≥ 0 as the input frequency, and (although not physically relevant) let ū ∈ C m be a
fixed complex vector. Take the input function u(t) to be
u(t) = ūejωt
for t ≥ 0. Then, the response is
x(t) = eAt x0 + 0t eA(t−τ ) Bu(τ )dτ
R
= eAt x0 + eAt 0t e−Aτ B ūejωτ dτ
R
= eAt x0 + eAt 0t e(jωI−A)τ dτ B ū
R
Now, since A is stable, all of the eigenvalues have negative real parts. This means that all
of the eigenvalues of (jωI − A) have positive real parts. Hence (jωI − A) is invertible, and
we can write the integral as
x(t) = eAt x0 + eAt
Rt
0
e(jωI−A)τ dτ B ū
¯t
= eAt + eAt (jωI − A)−1 e(jωI−A)τ ¯¯ B ū
h
0
i
= eAt x0 + eAt (jωI − A)−1 e(jωI−A)t − I B ū
h
i
= eAt x0 + eAt e(jωI−A)t − I (jωI − A)−1 B ū
h
i
= eAt x0 + eAt ejωt eAt − I (jωI − A)−1 B ū
h
i
= eAt x0 + ejωt − eAt (jωI − A)−1 B ū
h
i
= eAt x0 − (jωI − A)−1 B ū + (jωI − A)−1 B ūejωt
Hence, the output y(t) would satisfy
h
i
y(t) = CeAt x0 − (jωI − A)−1 B ū + C (jωI − A)−1 B ūejωt + Dūejωt
In the limit as t → ∞, the first term decays to 0 exponentially, leaving the steady-state
response
h
i
yss (t) = D + C (jωI − A)−1 B ūejωt
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ME 132, Spring 2005, UC Berkeley, A. Packard
Hence, we have verified our initial claim – if the input is a complex sinusoid, then the steadystate output is a complex sinusoid at the same exact frequency, but amplified by a complex
gain of D + C (jωI − A)−1 B.
The function G(jω)
G(jω) := D + C (jωI − A)−1 B
(100)
is called the frequency response of the linear system in (99). Hence, for stable systems, we
have proven
u(t) := ūejωt ⇒ yss (t) = G(jω)ūejωt
More precisely, this is the frequency response from u to y, so we might also write G yu (jω) to
indicate what is the input (u) and what is the output (y). Gyu can be calculated rather easily
using a computer, simply by evaluating the matrix expression in (100) at a large number of
frequency points ω ∈ R.
22.2
MIMO Systems: Response due to real sinusoidal inputs
In the case where the system has multiple inputs and outputs, it is a bit more complicated
to write out the response due to sinusoidal inputs at a fixed frequency, since the different
inputs may all have different magnitudes and phases. Suppose that there are m inputs, and
q outputs (so B ∈ Rn×m , C ∈ Rp×n , D ∈ Rp×m ). Take a ∈ Rm , b ∈ Rm , and ω ≥ 0.
Consider the input
u(t) = a cos ωt + b sin ωt
Note that this is the real part of a complex input, namely
h
u(t) = Re (a − jb)ejωt
i
Hence, the steady state output must be the real part of a function, specifically,
h
y(t) = Re G(jω)(a − jb)ejωt
i
So, in summary: To determine the steady-state response due to an input
u(t) = a cos ωt + b sin ωt,
let c ∈ Rq and d ∈ Rq be real vectors so that
c − jd = G(jω)(a − jb)
Then, the steady-state output is
yss (t) = c cos ωt + d sin ωt
ME 132, Spring 2005, UC Berkeley, A. Packard
22.3
212
Experimental Determination
Since the frequency response has the interpretation as a representation of the frequencydependent amplitude gain and phase shift between a sinusoidal input and steady-state output, it is easy to obtain experimentally for a given physical system. This can be done by
performing several different forced-response experiments, with the forcing being a sinusoid,
each experiment using a different frequency.