Lista de exercicio cap.4

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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+a1v1eλ2pv2=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)=23xyt=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= 0TeAtdtB ,temos:

eAT= L-1sI-A-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

L-1sI-A-1= L-1s+2s2+2s+21s2+2s+2-2s2+2s+2ss2+2s+2=L-1s+2s+12+121s+12+12-2s+12+12ss+12+12=L-1s+1s+12+12+1s+12+121s+12+12-21s+12+12s+1s+12+12-1s+12+12= e-TcosT + e-TsinTe-TsinT-2e-TsinTe-TcosT - e-TsinT=e-T(cosT+sinT)e-TsinT-2e-TsinTe-T(cosT-sinT)
E

0TeAtdtB= A-1eAT-IB= 01-2-2-1*e-T(cosT+sinT)e-TsinT-2e-TsinTe-T(cosT-sinT)-1001*11=-1-1210*e-T(cosT+sinT)-1e-TsinT-2e-TsinTe-T(cosT-sinT)-1*11=-1-1210*e-T(cosT+sinT)-1+e-TsinT-2e-TsinT+e-T(cosT-sinT)-1=32-(e-T(cosT+sinT))-(e-T(cosT-sinT))2e-TsinT+e-T(cosT-sinT)-1

Para T=1x1[k+1]x2[k+1]=e-T(cosT+sinT)e-TsinT-2e-TsinTe-T(cosT-sinT)x1(k)x2(k)+32-(e-T(cosT+sinT))-(e-T(cosT-sinT))2e-TsinT+e-T(cosT-sinT)-1u(k)
x1[k+1]x2[k+1]=0,50830,3096-0,6191-0,1108x1(k)x2(k)+1,0471-0,8012u(k)

y[k+1]=23x(k)

Para T=π
x1[k+1]x2[k+1]=e-T(cosT+sinT)e-TsinT-2e-TsinTe-T(cosT-sinT)x1(k)x2(k)+32-(e-T(cosT+sinT))-(e-T(cosT-sinT))2e-TsinT+e-T(cosT-sinT)-1u(k)
x1[k+1]x2[k+1]=-0,043200-0,0432x1(k)x2(k)+1,5648-1,0432u(k)

y[k+1]=23x(k)

4.4 - Encontrar a formacompanheira e forma modal equivalente para a equação

x=-2001010-2-2x+101u y=1-10x

-2-λ001-λ10-2-2-λ=0
-2-λ*-λ*-2-λ--2*1*-2-λ=0
(-2-λ)2*-λ-2λ+4=0
λ2+4λ+4*-λ-2λ-4=0
-λ3-4λ2-4λ-2λ-4=0
-λ3-4λ2-6λ-4=0
λ1=-2; λ2=-1+i; λ3=-1-i

Calculo do autovetor λ1
-2-λ001-λ10-2-2-λ=-2+2001-210-2-2+2=0001-210-20
2b=0→b=0
a+2b+c=0→a=-c;c=-1
v1=10-1

Calculo do autovetor λ2-2-λ001-λ10-2-2-λ=-2-(-1+i)001-(-1+i)10-2-2-(-1+i)=(-1-i)0011-i)10-2(-1-i)
(-1-i)a=0→a=0
(-1-i)b+c=0→c=-1+ib
-2b+-1-i*c=0→ -2b+-1-i*-1+ib=0→ 2b=-1-i→b=-12-12i
v2=0-12-12i1
Calculo do autovetor λ3
É complemento do autovetor λ2
v3=0-12+12i1
Q=1000-12-12i-12+12i-111
P-1= 1000-1212-110
Matriz equivalente
A=PAP-1=100101-1-2-1*-2001010-2-2*1000-1212-110=-2000-110-1-1
B=PB=12-2 C=CP-1=1-1212 D=D=0...