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Nonlinear Control Sy.. - Free

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10.5. INPUT-OUTPUT LINEARIZATION 279<br />

where<br />

A=<br />

0 1 0 ... 0<br />

0 0 1 0<br />

0 0 0 1<br />

-qo -Q1 .. -9n-1<br />

nxn<br />

, B=<br />

C= [ Po P1 ... Pm 0 ... 11xn' Pin 0.<br />

The transfer function associated with this state space realization can be easily seen to be<br />

H(S) = C(SI - A) 1B =<br />

Pmsm +Pm-lsm-1 + ... +Po<br />

sn+qn-lss-1+...+qo<br />

0<br />

0<br />

0<br />

1<br />

nxl<br />

m 0.<br />

We now calculate the relative degree using definition 10.10. We have<br />

CB = [ Po P1 ... Pm 0 .<br />

y=Ci=CAx+CBu<br />

0<br />

0<br />

IO<br />

1<br />

_ pm, if m=n-1<br />

0, otherwise<br />

Thus, if CB = p.,,,,, we conclude that r = n - m = 1. Assuming that this is not the case, we<br />

have that y = CAx, and we continue to differentiate y:<br />

CAB = [ Po P1 ... Pm 0 ... ]<br />

y(2) = CAi = CA2X + CABu<br />

0<br />

0<br />

J Pm, if m=n-2<br />

l 0, otherwise<br />

If CAB = pm, then we conclude that r = n - m = 2. Assuming that this is not the case,<br />

we continue to differentiate. With every differentiation, the "1" entry in the column matrix<br />

A'B moves up one row. Thus, given the form of the C matrix, we have that<br />

CA'-1B= 1 0 for i=<br />

l pm for i=n - m

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