HW12_solution

# HW12_solution - ME 132 Solutions # 12 1 Observers (a) Since...

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Unformatted text preview: ME 132 Solutions # 12 1 Observers (a) Since the noise is unknown and zero-mean, the best model we can use is y = bracketleftbig 1 s 2 bracketrightbig u . A state-space realization of this model is x = bracketleftBigg 0 1 0 0 bracketrightBigg x + bracketleftBigg 1 bracketrightBigg u y = bracketleftBig 1 0 bracketrightBig x The observable canonical form of this realization is A o = bracketleftBigg 0 0 1 0 bracketrightBigg , C o = bracketleftBig 0 1 bracketrightBig From the two representations above, the change of basis matrix is T = bracketleftBigg 0 1 1 0 bracketrightBigg If the desired characteristic polynomial for A + LC is s 2 + 1 s + , the observer gain L is K = bracketleftBigg 1 bracketrightBigg To place the observer poles at 2 , 2, the desired characteristic polynomial is s 2 +4 s +4, which means the observer gain is K = bracketleftBigg 4 4 bracketrightBigg (b) The block diagram for the simulations is given in Figure 1. The outputs of the simula- tions are given in Figure 2. Based on the plant model, the second state is the measured output of the actual plant. However, we know that this is not true since the actual state is being corrupted by noise. We will see that as the observer becomes more aggressive, the estimate of x 1 will follow the noise realizations more closely (see Part (d) below). This, of course, is undesirable,the noise realizations more closely (see Part (d) below)....
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## This note was uploaded on 10/20/2010 for the course ME 132 taught by Professor Tomizuka during the Spring '08 term at University of California, Berkeley.

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HW12_solution - ME 132 Solutions # 12 1 Observers (a) Since...

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