AtenΓ§Γ£o
- π§ Posts ainda estΓ£o sendo revisados
- π Bugs de formataΓ§Γ£o podem ser encontrados
- βοΈ Todos os recursos ainda nΓ£o foram implementados
y[n+1]=(1+i)y[n]+x[n+1]
y[0]=C
x[n]=Οβk=0ββΞ΄[nβΟ(k+1)]
Aplicando transformada Z:
zY[z]βzy[0]=(1+i)Y[z]+zX[z]βzx[0]
zY[z]βzC=(1+i)Y[z]+zX[z]
zY[z]β(1+i)Y[z]=zC+zX[z]
Y[z]βz(1+i)βY[z]=C+X[z]
Y[z](1βz(1+i)β)=C+X[z]
Y[z](zzβ(1+i)β)=C+X[z]
Y[z]=Czβ(1+i)zβ+X[z]zβ(1+i)zβ
Y[z]=Yi0β[z]+Ys0β[z]
Yi0β[z]=Czβ(1+i)zβ
yi0β[n]=C(1+i)n
Exponencial e tals
Ys0β[n]=X[z]zβ(1+i)zβ
Bora ver X[z]
x[n]=Οβk=0ββΞ΄[nβΟ(k+1)]
X[z]=βn=0ββ(Οβk=0ββΞ΄[nβΟ(k+1)])zβn
X[z]=Οβk=0ββ(βn=0ββΞ΄[nβΟ(k+1)])zβn
X[z]=Οβk=0ββzβΟ(k+1)
Ys0β[z]=X[z]zβ(1+i)zβ
Ys0β[z]=Οβk=0ββzβ(1+i)zβzβΟ(k+1)
ys0β[n]=Οβk=0ββ(1+i)nβΟ(k+1) u[nβΟ(k+1)]
tags: Blog