Lista de exercicio cap.4

4.2 - Utilizar dois métodos diferentes para encontrar a resposta o impulso de:

x=01-2-2x+11u y=23x
Método 1:
SI=s00s A=01-2-2

sI-A=s00s- 01-2-2=s-12s+2

sI-A-1= s-12s+2-1= 1ss+2-(-1)(2) s+21-2s = 1s2+2s+2s+21-2s= s+2s2+2s+21s2+2s+2-2s2+2s+2ss2+2s+2

Ys=CsI-A-1BUs+DU(s)

OBS.: | xt=Axt+Bu(t)yt=Cxt+Du(t) | x=01-2-2x+11u | y=23x |

Ys=23*s+2s2+2s+21s2+2s+2-2s2+2s+2ss2+2s+2*11Us+0
Ys=2s+4s2+2s+2+-6s2+2s+2 2s2+2s+2+3ss2+2s+2*11Us
Ys=2s-2s2+2s+2 2+3ss2+2s+2*11Us
Ys=2s-2s2+2s+2 +2+3ss2+2s+2*Us
Ys=5ss2+2s+2*1s

Tabela de Transformada de Laplace | eatsenbt=b(s-a)2+b2 | Ys=5*1(s+1)2+(1)2 eatsenbt=b(s-a)2+b2= 1(s+1)2+(1)2 , onde: b=1; -a=1→a=-1

yt=5e-tsin⁡(t) , p/ t≥0

Método 2: xt=0teAt-τBuτdτ p=t-τ eAtv=eλtv eAp=a0A+a1I eλ1pv1=a0λ1v1+a1v1 eλ2pv2=a0λ2v2+a2v2 eλ1p=a0λ1+a1eλ2p=a0λ2+a2 eλ1p-eλ2p=a0λ1-λ2→ a0=eλ1p-eλ2pλ1-λ2 λ1=-1+i ; λ2=-1-i a0=e(-1+i)p-e(-1-i)p-1+i-(-1-i)=e-p+eip-e-p+e-ip2i= e-p(eip-e-ip)2i
Manipulando a formula de Euler

eij-e-ij2i=sinj eij+e-ij2i=cosj e-j=cosj-isinj a0=e-psinp Calculando a1: eλ1p=a0λ1+a1 a1=e(-1+i)p-e-psinp-1+i=e-p*eip-e-psinp-1+i=e-p*(cosp+isinp)+e-psinp-1+i=e-pcosp+e-pisinp+e-psinp- e-pisinp=e-p*(cosp+sinp)
Equação:
eAp=a0A+a1I eAp=(e-psinp)A+e-pcosp+sinpI eAp=(e-psinp)01-2-2+e-pcosp+sinp1001=e-p01-2-2+c+s00c+s= e-pc+ss-2sc+s xt=0teAt-τBu(τ)dt=0te-t-τ(cos(t-τ)+sin(t-τ))e-t-τsin(t-τ)-2e-t-τsin(t-τ)e-t-τ(cos(t-τ)-sin(t-τ))*11u(τ)dτ=0te-t-τ(cost-τ+2sin(t-τ))e-t-τ(cost-τ-3sin(t-τ))uτdτ y(t)=23x yt=230te-t-τ(cost-τ+2sin(t-τ))e-t-τ(cost-τ-3sin(t-τ))uτdτ=0t23e-t-τ(cost-τ+2sin(t-τ))e-t-τ(cost-τ-3sin(t-τ))uτdτ= 0te-t-τ(2cost-τ+4sint-τ+3cost-τ-9sint-τ)uτdτ= 0te-t-τ(5cost-τ-5sint-τ)uτdτ= L {e-t-τ5cost-τuτ*e-t-τ-5sint-τ)uτ= 5*s+1s+12+1*1s-5*1s+12+1*1s= 5s+5-5s+12+1*1s=Ys=5(s+1)2+1 yt=5e-tsin⁡(t) , p/ t≥0

4.3 - Discretizar a equação de estado do problema 4.2 para T=1 e T=π. x=01-2-2x+11u y=23x xk+1T=GTxkT+HTu(kT) SOLUÇÃO:

Sabendo que: GT=eATHT= 0TeAtBdt=