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AtenΓ§Γ£o

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EquaΓ§Γ£o de montante com aportes

y[n+1]=(1+i)y[n]+x[n+1]y[n+1]=(1+i)y[n]+x[n+1]

y[0]=Cy[0]=C

x[n]=Ο„βˆ‘k=0∞δ[nβˆ’Ο(k+1)]x[n] = \tau\sum_{k=0}^\infty\delta[n-\rho(k+1)]

Aplicando transformada Z:

zY[z]βˆ’zy[0]=(1+i)Y[z]+zX[z]βˆ’zx[0]zY[z]-zy[0]=(1+i)Y[z]+zX[z]-zx[0]

zY[z]βˆ’zC=(1+i)Y[z]+zX[z]zY[z]-zC=(1+i)Y[z]+zX[z]

zY[z]βˆ’(1+i)Y[z]=zC+zX[z]zY[z]-(1+i)Y[z]=zC+zX[z]

Y[z]βˆ’(1+i)zY[z]=C+X[z]Y[z]-\frac{(1+i)}{z}Y[z]=C+X[z]

Y[z](1βˆ’(1+i)z)=C+X[z]Y[z](1-\frac{(1+i)}{z})=C+X[z]

Y[z](zβˆ’(1+i)z)=C+X[z]Y[z](\frac{z-(1+i)}{z})=C+X[z]

Y[z]=Czzβˆ’(1+i)+X[z]zzβˆ’(1+i)Y[z]=C\frac{z}{z-(1+i)}+X[z]\frac{z}{z-(1+i)}

Y[z]=Yi0[z]+Ys0[z]Y[z]=Y_{i0}[z]+Y_{s0}[z]

Yi0[z]=Czzβˆ’(1+i)Y_{i0}[z]=C\frac{z}{z-(1+i)}

yi0[n]=C(1+i)ny_{i0}[n]=C(1+i)^{n}

Exponencial e tals

Ys0[n]=X[z]zzβˆ’(1+i)Y_{s0}[n]=X[z]\frac{z}{z-(1+i)}

Bora ver X[z]X[z]

x[n]=Ο„βˆ‘k=0∞δ[nβˆ’Ο(k+1)]x[n] = \tau\sum_{k=0}^\infty\delta[n-\rho(k+1)]

X[z]=βˆ‘n=0∞(Ο„βˆ‘k=0∞δ[nβˆ’Ο(k+1)])zβˆ’nX[z]=\sum_{n=0}^\infty(\tau\sum_{k=0}^\infty\delta[n-\rho(k+1)])z^{-n}

X[z]=Ο„βˆ‘k=0∞(βˆ‘n=0∞δ[nβˆ’Ο(k+1)])zβˆ’nX[z]=\tau\sum_{k=0}^\infty(\sum_{n=0}^\infty\delta[n-\rho(k+1)])z^{-n}

X[z]=Ο„βˆ‘k=0∞zβˆ’Ο(k+1)X[z]=\tau\sum_{k=0}^\infty z^{-\rho(k+1)}

Ys0[z]=X[z]zzβˆ’(1+i)Y_{s0}[z]=X[z]\frac{z}{z-(1+i)}

Ys0[z]=Ο„βˆ‘k=0∞zzβˆ’(1+i)zβˆ’Ο(k+1)Y_{s0}[z]=\tau\sum_{k=0}^\infty \frac{z}{z-(1+i)}z^{-\rho(k+1)}

ys0[n]=Ο„βˆ‘k=0∞(1+i)nβˆ’Ο(k+1) u[nβˆ’Ο(k+1)]y_{s0}[n]=\tau\sum_{k=0}^\infty(1+i)^{n-\rho(k+1)}\space u[n-\rho(k+1)]

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