LTI System
ẋ(t) = A x(t) + B u(t) y(t) = C x(t) + D u(t)
x = state y = output u = input
A = dynamic matrix (mxm) B = input matrix (mxm) C = output matrix (pxm) D = feedthrough matrix (pxm)
Response u(t) = f(t)
x(t) = eAt x0 + ∫0t eA(t-τ) B u(τ) dτ
ZIR: Zero Input Response (FREE) ZSR: Zero State Response (forced)
y(t) = CeAt x0 + ∫0t CeA(t-τ) B u(τ) dτ + D u(t)
eAt = state-transition matrix CeAt = output-transition matrix
Impulse Response u(t) = δDir(t)
x(t) = eAt x0 + H(t)
y(t) = CeAt x0 + W(t)
H(t) = ∫0t eA(t-τ) B δDir(t) dτ = eAt B
W(t) = ∫0t CeA(t-τ) B δsmn(t) dτ + D δsmn(t) = CeAt B + D δ(t)
Change of Coordinates
z = T x ⇒ S → Ṡ : (detT ≠ 0)
ż = Ã z + B̃ u y = C̃ z + D̃ u
Ŵ(t) = W(t) invariant!!
If A is diagonalisable then we can use T = U-1 and Zt(t) = eΛt z0
à = T A T-1 B̃ = T B C̃ = C T-1 D̃ = D
General Response u(t) = f(t)
x(t) = Φ(t) x0 + ∫0t Φ(t-τ) u(τ) dτ
y(t) = Ψ(t) x0 + ∫0t W(t-τ) u(τ) dτ
Φ(t) = eAt Ψ(t) = CeAt
H(t) = CeAt
W(t) = CeAt B + D δ(t)
Linear Algebra
Characteristic Polynomial
PA(λ) = det[A - λI]
Eigenvalues
λi : Are all the solutions of PA(λ) = 0
Eigenvectors
ui : All the vectors such that (A - λiI)ui = 0
so ui belongs to the nullspace of (A - λiI) i.e ker(A - λI)
Algebraic Multiplicity
mA(λi) : Is the multiplicity of the solution of λi in PA(λ) = 0
Geometric Multiplicity
mG(λi) = dim[ker(A - λiI)] = n - rk[A - λiI]
(number of parameters in (A - λiI)ui = 0)
Diagonalizable Matrix
An (m x m) matrix is diagonalizable if and only if mG(λi) = mA(λi) ∀ i
Λ = A [λ1 0 ⋯ 0 0 λ2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ -λm]
U = [u1, ..., um]
eigenvectors matrix
if T = U-1
Λ = TAT-1 = U-1AU
Spectral Form: A = ∑i=1m λi ui uiT
Case of Complex Eigenvalues
If : λi = αi + jωi and ui = ua + jub :
(λi* = αi - jωi ; ui* = ua; - jub;)
- Diagonalisation → [λi 00 λi*]
- Real Blocks (2x2) → [αi ωi -ωi αi]
complex elements
(T-1 = [ui ui*])
real elements
(T-1 = [ua; ub;])
λi ∈ ℝ → Aperiodic natural modes
λi ∈ ℂ → Pseudo periodic natural modes
Foglio delle Trasformate
- Linearità Addizione:
L [α f(t) + β g(t)] = α L [f(t)] + β L [g(t)]
- Linearità Prodotto:
L [f(t) · g(t)] = L [f(t)] · L [g(t)]
- Teorema della Derivazione:
L [d/dt f(t)] = s L [f(t)] - f(0)
- L [δ(t)] = 1
- L [u(t)] = k / s
- L [et] = 1 / s-a
- L [tk / k!] = 1 / sk+1
- L [tk et] / k!] = 1 / (s-a)k+1
- L [sin(ωt)] = ω / (s2 + ω2) = ω/2 j / s-jω + -ω/2 j / s+jω
- L [cos(ωt)] = s / (s2 + ω2) = ω/2 / s-jω + ω/2 / s+jω
- L [sin(ωt+φ)] = s sin φ + ω cos φ / s2 + ω2
- et sin(ωt) = (s-a) / (s-a)2 + ω2
- et cos(ωt) = s-a / (s-a)2 + ω2
Trasformata di funzione con il ritardo:
L [f(t-τ) · g(t-τ)] = L [f(t) · g(t)] · e-sτ
FATTORE BINOMIO (1+jωτ)/(1+jt*ω)
Si fa un'approssimazione intorno alla PULSAVILE DI ROTTURA ω* = 1/|τ| (CUTOFF FREQUENCY)
-1 1 per ωτ → P(s) is STRICTLY PROPER
sτ + aτ-1sτ-1 + ... + a1s + a0
C(s) = dnsn + dn-1sn-1 + ... + d1s + d0 → C(s) is PROPER at maximum
sχ + cχsχ-1 + ... + c1s + c0
Dwr = DcDp + NcNp → poles of Scl are solutions of Dwr(s) = 0 (order max = n+τ)
I want to assign n+τ poles, P1z, P2z,...,Pn+τz → Pcl(s) = (s-P1z)(s-P2z)...(s-Pn+τz)
Pole Assignment: I choose χ = n + τ in order to obtain a solution (at least χ = n+τ, it's possible χ = n-1)
So: C(s) = dn+τsn+τ + ... + d1s + d0⁄sn+τ + cn+τsn+τ-1 + ... + c1s + c0
And (s-P3z)(s-P2z)...(s-Pn+τz) = Dp(s)·(sn+χ + cn+χsn+χ-1 + ... + c1s + c0) + Np(s)·(dnχsn+τ + ... + d1s + d0)
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