- #1
peripatein
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Hello,
I would like to find the inverse Z transform of the following:
F(z)=1-1.25z-1+0.25z-2/[1-(5/6)z-1+(1/6)z-2]
using (a) partial fractions, and (b) residue theorem
I have obtained different results and hence would appreciate some insight on the discrepancy and how it may be resolved.
(a) Using partial fractions, I obtained the following:
F(z)=(3/2)[4z2-5z+1/(6z2-5z+1)] = (3/2)[10z/(z-1/3) - 2/(z-1/3) - 6z/(z-1/2)]=
=15z/(z-1/3) - 3/(z-1/3) - 9z/(z-1/2) = 6 + 2/(z-1/3) - (9/2)/(z-1/2)
⇔ f(n) = 6δ(n) + 2(1/3)n-1u(n-1) - (9/2)(1/2)n-1u(n-1) =
6δ(n) + [6(1/3)n - 9(1/2)n]u(n-1)
where u(n) is the step function and δ(n) is the dirac function.
(b) Using residue theorem, I obtained the following:
Residue for z=1/2 is -9, residue for z=1/3 is 6, hence:
f(n) = [-9(1/2)n + 6(1/3)n]u(n)
Where am I erring? I'd appreciate any comments.
Homework Statement
I would like to find the inverse Z transform of the following:
F(z)=1-1.25z-1+0.25z-2/[1-(5/6)z-1+(1/6)z-2]
using (a) partial fractions, and (b) residue theorem
I have obtained different results and hence would appreciate some insight on the discrepancy and how it may be resolved.
The Attempt at a Solution
(a) Using partial fractions, I obtained the following:
F(z)=(3/2)[4z2-5z+1/(6z2-5z+1)] = (3/2)[10z/(z-1/3) - 2/(z-1/3) - 6z/(z-1/2)]=
=15z/(z-1/3) - 3/(z-1/3) - 9z/(z-1/2) = 6 + 2/(z-1/3) - (9/2)/(z-1/2)
⇔ f(n) = 6δ(n) + 2(1/3)n-1u(n-1) - (9/2)(1/2)n-1u(n-1) =
6δ(n) + [6(1/3)n - 9(1/2)n]u(n-1)
where u(n) is the step function and δ(n) is the dirac function.
(b) Using residue theorem, I obtained the following:
Residue for z=1/2 is -9, residue for z=1/3 is 6, hence:
f(n) = [-9(1/2)n + 6(1/3)n]u(n)
Where am I erring? I'd appreciate any comments.
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