Download Topic 8 Notes 8 Applications, stability Jeremy Orloff

Topic 8 Notes
Jeremy Orloff
8 Applications, stability
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Also, there is a more expanded version at http://math.mit.edu/~jorloff/
suppnotes/suppnotes03/s.pdf
Time invariance
Constant coefficient have the property of time invariance. That is, if xp (t) satisfies
P (D)x = f (t) then xp (t − to ) satisfies P (D)x = f (t − t0 ).
Example: x0 + x = e2t has solution xp = e2t /3
⇒ x0 + x = e2(t−4) has solution xp = e2(t−4) /3.
Physically this has to be the case –an exponential decay system doesn’t care what
time it gets started. (Draw input/output graphs)
Stability
Example: Solve the IVP x00 + 2x0 + 3x = cos 2t, x(0) = 2, x0 (0) = 3
Particular solution: complexify x̃00 + 2x̃0 + 3x̃ = ei2t x = Re(x̃).
(Remind yourself why complexification works.)
Using the E.I.T. we find xp =
cos(2t−φ)
√
,
17
where φ = tan−1 (4) in the 2nd quadrant.
Homogeneous solution: char. equation: r2 + 2r + 3 = 0.
√
Roots: r = −1 ± 2 i.
√
√
xh = c1 e−t cos 2 t + c2 e−t sin 2 t
√
√
General solution: x = xp + xh = xp + c1 e−t cos 2 t + c2 e−t sin 2 t
√
Find c1 = 35/17 and c2 = 44 2/17 using the IC (we’ll pretend we did this).
Because of the negative exponent, in the long-term xh (t) → 0. I.e. the initial conditions don’t matter in the long-run.
Definition: Mathematical stability means long-term behavior doesn’t depend (significantly) on initial conditions.
Linear Systems: The system Ly = f is stable if for any IC yh → 0 as t → ∞.
In this case, yh is called the transient.
Linear CC Systems: The system P (D)y = f is stable if all the characteristic roots
have negative real part.
Stability is about the system not the input.
Example: x0 + 2x = f (t) is stable because xh = ce−2t → 0.
Example: A constant coeff. system with roots −2 ± 3i, −3 is stable.
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8 Applications, stability
Example: A constant coeff. system with roots −2, −3, 4 is unstable.
Examples:
1. P (D)y = y 00 + 8y 0 + 7y = f has char. roots -7, -1 ⇒ the system is stable.
2. P (D)y = y 00 − 6y 0 + 25y = f has char. roots 3 ± 4i. The real part is positive ⇒
the system is not stable.
Stability criteria for linear CC systems
1. Stability ⇔ for any IC yh → 0 as t → ∞.
2. Stability ⇔ all char. roots have negative real part.
3. For a first order system P (D)y = y 0 + ky = f : root = −k ⇒ stability ⇔ k > 0.
4. For a second order system P (D)y = y 00 + ay 0 + by = f : stability ⇔ a > 0 and
b > 0 (easy to prove).
5. For a third order system P (D)y = y 000 + ay 00 + by 0 + cy = f : stability ⇔ a, b, c > 0
and ab > c (harder to prove).
Example of an unstable system with positive coefficients
(r + 5)(r − 1 − 100i)(r − 1 + 100i) = r3 + 3r2 + 96r + 505.
6. For higher order systems there is the Routh-Hurwitz criteria, which is described
in the supplementary notes §S.
7. Stability ⇔ all solutions to the homogeneous equation P (D)y = 0 go asymptotically to the homogeneous equilibrium solution y(t) = 0.
Physical stability: An unforced physical system is stable if it always returns to equilibrium.
Example: Damped-spring-mass system: Physical stability matches mathematical
stability. The equilibrium solution is x(t) = 0. The system is modeled by x0 + bx0 +
kx = 0 and since the roots have negative real part x(t) → 0 no matter what the
initial conditions.
Note: 2nd order physical systems, like springs and LRC circuits are always stable.
This is not true of 3rd (and higher) order physical systems.
θ _
d2 θ g
T
Simple pendulum
+ sin θ = 0.
dt2
l
•
(Derivation given in book §2.4 using energy considerations.)
Fgravity
For small θ we use the approximation sin θ ≈ θ (good to two
d2 θ g
decimal places when θ < 15◦ ) to get
+ θ = 0. Thus, the undamped pendudt2
l
lum, with small oscillations, has the same model as the undamped spring!
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