23
IDENTIFICATION OF A RESPONSE AMPLITUDE OPERATOR FROM DATA
75
23
Identification of a Response Amplitude Operator
from Data
Load the data file
homework5.dat
from the
course
website
.
The first column is time
t
in
seconds, the second column is a measured input signal
u
(
t
), in Volts, and the third column is
a measured output signal
y
(
t
), also in Volts.
u
(
t
) has no significant frequency content above
2
rad/s
.
Your main task is to estimate the Response Amplitude Operator (RAO)
H
(
ω
) and the
underlying transfer function
G
(
jω
) of the system that created
y
(
t
) when driven by
u
(
t
).
Your deliverables are:
1. A plot of the databased RAO
H
(
ω
) versus frequency.
2. The specific transfer function that best describes the data, e.g.,
G
(
jω
) = 6
.
5
/
(
jω
+2
.
2).
Recall that
H
(
ω
) =

G
(
jω
)

2
.
3. A plot of the step response of
G
(
jω
), making the assumption that the system is causal.
Some hints:
•
The recommended overall procedure is to calculate the autocorrelation function for
each signal, compute the Fourier transforms of these, and then divide according to the
WienerKhinchine relation so as to recover an empirical
H
(
ω
) =

G
(
jω
)

2
.
•
A detailed discussion on use of the FFT in MATLAB is given in the worked problem
entitled
Dynamics Calculations Using the Time and Frequency Domains
. This includes
specification of the frequency vector that goes with an FFT, which is very important
for the current problem.
•
The FFT of an autocorrelation function is supposed to be all real because
R
(
τ
) has
only the cosine phases represented.
Yet look at the signal
fft(cos(t));
and you
will see that it not only has some imaginary parts, but also a lot of content NOT at
frequency one. Some of this appears because numerical Fourier transforms assume that
the timedomain signal is periodic  that the end reconnects with the beginning. This
often gives a discontinuity, which the transform tries to take care of with the extra
stuff you see. Some of this messiness also comes up because the FFT is given only at
discrete frequencies. When your signal actually has other frequencies, the FFT spreads
things out a bit.
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 Spring '05
 GeorgeKocur
 Signal Processing, Wavelength

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