Esercizio sul calcolo della risposta forzata

24

X(k+1) =

• 4
• 1

• 1

U(k)=1

k≥0

Evoluzione forzata

A =

• 4
• 1
• 1/4

β =

• 1

X(z) =

• X₁(z)
• X₂(z)

X(z) = H(z)U(z)

zI - A =

• z-4
• 4
• 1/4

(zI - A) a =

• z-1/4
• 1
• z-4

X(z) =

• z^-1/1 (zI - A)^-1 β
• (zI - A)^-1
  • z-4
  • 1
• zU(z)

det(zI-A) =

• (z-1i)(z-4)

U(z)=

• z/z-1

Es. 24

X(k+1) = ^{[1/2 1]}_{[0 1/4]} X(k) + ^[0]_[1] U(k)

U(k) = 1, k ≥ 0

Evoluzione forzata ?

A = ^{[1/2 1]}_{[0 1/4]}

β = ^[0]_[1]

X(z) = ^[X₁(z)]_[X2(z)] = (zI–A)^–1 β U(z) = H(z) U(z)

ovvero X(z) = H(z) U(z)

U(z) = ^z/_z-1

U(k) = 1, k ≥ 0

X(z) = ¹/_{(z–1/2)(z–1/4)}

X[1] = ^z/(z–1/2)

X(k) = H(z) U(z)

X(z) = ^{X₁(z) + X₂(z)}_X2(z)

La matrice H(z):

U(z) = ^z/_z-1

det(zI–A) = (z–1/2)(z–1/4)

→ H(z) = (zI–A)^–1 β

→ ¹/_{(z–1/2)(z–1/4)} ^{1 / (z–1/4)}

X(z) = ¹/_{(z-1/2)(z-1/4)}

(zI–A)² = [A|β]

^H(z)

X(k) = z^{–1k[X₁(z)]}_[X2(z)]

→ x₁(z) = ^z/_{(z–1/2)(z–1/4)(z–4)}

x₂(z) = ^X₂(z)

→ x₁(k) = ?

→ x₂(k) = ?

(Strettamente propria)