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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 ( ) of the system that created y ( t ) when driven by u ( t ). Your deliverables are: 1. A plot of the data-based RAO H ( ω ) versus frequency. 2. The specific transfer function that best describes the data, e.g., G ( ) = 6 . 5 / ( +2 . 2). Recall that H ( ω ) = | G ( ) | 2 . 3. A plot of the step response of G ( ), 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 Wiener-Khinchine relation so as to recover an empirical H ( ω ) = | G ( ) | 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 time-domain 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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