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11.5. EXAMPLE 227<br />

of solutions (ρ 1 ,ρ 2 ,x 1 ,x 2 ,x 3 ,x 4 ), the polynomials a(s) and b(s) can be determined.<br />

This yields five corresponding transfer functions of order one:<br />

3.37127<br />

G(s) =<br />

−6.48665 + s ,<br />

3.87128<br />

G(s) =<br />

−0.129578 + s ,<br />

396.044<br />

G(s) =<br />

−0.409222 + s , (11.85)<br />

G(s) = 2.97122<br />

4.43913 + s ,<br />

G(s) = −3.1894<br />

0.174884 + s ,<br />

with corresponding H 2 -criterion values 8.24643, 11.1784, 437.85, 8.13223, and 6.16803,<br />

respectively.<br />

The H 2 -criterion values are computed by substituting the values of ρ 1 , ρ 2 , and<br />

x 1 ,...,x 4 in<strong>to</strong> the third order polynomial:<br />

V H = (−0.121789 + 0.00691205i)(1 − (0.666667 + 0.5i)ρ 1 +<br />

(0.194444 + 0.666667i)ρ 2 ) 2 (1+(0.666667 + 0.5i)ρ 1 +<br />

(0.194444 + 0.666667i)ρ 2 ) x 3 1 − (0.121789 + 0.00691205i)<br />

(1 − (0.666667 − 0.5i)ρ 1 +(0.194444 − 0.666667i)ρ 2 ) 2<br />

(1+(0.666667 − 0.5i)ρ 1 +(0.194444 − 0.666667i)ρ 2 ) x 3 2−<br />

1.60977(1 − 0.6ρ 1 +0.36ρ 2 ) 2 (1+0.6ρ 1 +0.36ρ 2 ) x 3 3+<br />

9.74309(1 − 0.333333ρ 1 +0.111111ρ 2 ) 2<br />

(1+0.333333ρ 1 +0.111111ρ 2 ) x 3 4.<br />

(11.86)<br />

It is easy <strong>to</strong> see that only the last two approximations G(s), are feasible because<br />

only those two have a pole in the open left half plane Π − . Since the last approximation<br />

has the smallest real criterion value, this approximation G(s) is the globally optimal<br />

H 2 -approximation of order one. This is the same transfer function as computed before<br />

in Equation (11.77):<br />

G(s) = −3.1894<br />

0.174884 + s . (11.87)<br />

The corresponding values of the solution of the system of equations (11.78) and<br />

the two additional constraints in (11.79) are:<br />

ρ 1 =4.39521<br />

ρ 2 =1.70287<br />

x 1 =2.56458 + 1.52372i<br />

x 2 =2.56458 − 1.52372i<br />

x 3 =2.36141<br />

x 4 =1.54876.<br />

(11.88)

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