Inverse Z transform with residue method

[Evo8]

Homework Statement

X(z)=z/(z-.7)^2

Use the residue method to calculate x(k)

Hello,

I have this exercise that I am trying to complete. I am not really familiar with the residue method and am having a hard time finding a good example on line. The section in my book does not describe it very well.

Im looking for a pretty basic breakdown of how to do this. Lots of info on inverse Z transforms just not using the residue method.

Thanks for any help guys,

 

[Valkarie]

Homework Equations The residue method is used to calculate the inverse z transform. The formula for the inverse z transform is: X(k) = 1/2π ∫-∞ ∞x(z)e-j(k+1)z. dzThe Attempt at a SolutionThe residue method is used to calculate the inverse z transform of a function. The inverse z transform can be written as: X(k) = 1/2π ∫-∞ ∞x(z)e-j(k+1)z. dzIn this case, the function x(z) is X(z)=z/(z-.7)^2. We can rewrite this as x(z)=A/(z-a) + B/(z-b). Where A=1, B=-1 and a=.7, b=-.7. We can then use the residue method to calculate the inverse z transform of x(z). The formula for the residue method is: X(k) = 1/2πi ∑Res[x(z),z]e-j(k+1)z. Where Res[x(z),z] is the residue of the function x(z) at z. In this case, we have two residues. The first is at z=.7 and the second is at z=-.7. Therefore, the inverse z transform of x(z) can be calculated as: X(k) = 1/2πi (1/2)(e-j(k+.7)) + (-1/2)(e-j(k-.7)) X(k) = 1/2πi (1/2e-j(k+.7)-1/2e-j(k-.7)) X(k) = 1/2πi (1/2e-j(k+.7)-1/2ej(k+.7)) X(k) = 1/2πi (e-j(k+.7))