Stability

# Example impulse response for 0 1 like in lab 1

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Unformatted text preview: lity of the system is determined by the signs on ζ ,τ • “Critically damped” If ζ 2 < 1 , these roots are complex, and they will be complex conjugates of each other. • The stability of the system is determined by the signs on ζ ,τ Re( s) = • • −ζ τ Imaginary part leads to oscillations “Underdamped” If τ > 0,ζ > 0 , then the second ­order linear system is stable. Example: Impulse response for 0 <| ζ |≤ 1 : Like in Lab 1, dump in the water quickly. K U ( s) τ 2 s 2 + 2ζτ s + 1 U ( s) = 1 K X ( s) = 2 2 τ s + 2ζτ s + 1 X ( s) = Using Line 19 of Table 3.1, x (t ) = ⎛ t⎞ e−ζ t /τ sin ⎜ 1 − ζ 2 ⎟ τ⎠ ⎝ τ 1− ζ K 2 Point out features of exponential growth or decay of a sinusoid. Now consider the state space version: d 2 x 2ζ dx 1 K + + 2 x = 2 u(t ), x (0) = 0, x '(0) = 0 2 τ dt τ dt τ First we must rewrite this second ­order ODE as a system of first ­order ODEs. Need to define new variables z1 , z2 . Goal, get rid of old x, use new z1 , z2 . z1 ≡ x , z2 ≡ dx dt dz1 = z2 dt dz2 2ζ 1 K = − z2 − 2 z1 + 2 u dt...
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