We mentioned this can easily do something you dont

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Unformatted text preview: ment in gas. The velocity doesn’t change right away. The maximum change is about 1 s later, and then the velocity decays back to zero. The Laplace transform of this response is G(s) = 1 (s + 1)2 EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 33 and the transfer function of the closed loop system is F (s) = G(s) 1 + H (s)G(s) In this case we have unity feedback, so H (s) = 1, and the transfer function is F (s) = 1 k (s+1)2 k + (s+1)2 k (s + 1)2 + k √ √ k √ = k (s + 1)2 + ( k )2 = From the diagram, it is clear that k must be positive. EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 34 This corresponds to the impulse response f (t) = √ √ ke−t sin( kt) What we really are concerned with is the step response. We need to add a initial gain, so that the impulse response sums to 1 (and the step response goes to one). If we do this, the impulse and step responses are 1 k=2 Velocity Impulse Response k=1 0.5 No Feedback 0 k=9 !0.5 !1 0 1 2 3 4 5 6 7 8 9 10 Time, ms EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 1.4 k=9 1.2 Velocity 35 k=2 1 0.8 k=1 0.6 No Feedback 0.4 Step Response 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time, ms We can make the response much quicker with a moderate gain. We can do better with a system in the feedback loop, but that is another course! EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 36 What’s Next? Lot of alternatives • Discrete time linear systems and signals (102B, 264, 265) • More Fourier transforms (261, 262, 366) • Image processing (168, 262, 368) • Medical/Bio Imaging (169, 369A, 369B) • Communications (179, 279, 359) • Random processes, statistical signal processing (178, 278, 378) • Control systems, linear dynamic systems (E105, 263) Many more advanced courses EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 37 The End That’s it! Good luck with the final, and 102B! EE102A:Signal Processing and Linear Systems I; Win 08-09 Pauly 38...
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