Final Exam Solution

Page 1

Final Exam Solution, eece 413

1.  + 4θ =Tc +Td θ x1 =θ, x 2 =θ  = −4 x1 +Tc x 1 =θ = x 2, x 2 =θ

a)

 x1   0 x 2 = −4    y = [1

1 x1  0   +  Tc 0 x 2 1

 x1  0 ]  x 2

x = Fx + Gu , y = Hx + Ju

where

1 0 0 F = , G = 1  − 4 0   H = [1 0], J = 0


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