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Is there an inverse Z transform for: 1/z-1 ?

by LM741
Tags: 1 or z1, inverse, transform
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LM741
#1
Oct3-06, 12:13 PM
P: 130
hay guys -really struggling to find an inverse Z transform for: 1/(z-1)

doesn't seem to exist in the table of z transforms - so is this in fact possible to invert?? In case you're wondering - this forms part of a tut question.

thanks

John
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msd213
#2
Feb10-10, 01:02 AM
P: 25
Quote Quote by LM741 View Post
hay guys -really struggling to find an inverse Z transform for: 1/(z-1)

doesn't seem to exist in the table of z transforms - so is this in fact possible to invert?? In case you're wondering - this forms part of a tut question.

thanks

John
I'm wondering about something similar. I plotted some points in the transformation and it seems to be a circle but I can't manipulate it to get it in a form where I could find the radius or center.
elibj123
#3
Feb10-10, 10:29 AM
P: 240
For a casual sequence it will be a Laurent Series: |z|>1


[tex]\frac{1}{z-1}=\frac{1}{z}\frac{1}{1-1/z}=\frac{1}{z}\sum^{\infty}_{k=0}z^{-k}=\sum^{\infty}_{k=1}z^{-k}[/tex]

Then by recalling the definition of the Z Transform:

[tex] a[k \leq 0]=0 [/tex]
[tex] a[k \geq 1]=1[/tex]

Or using the step signal it's a[k]=u[k-1].

For an anti-casual sequence it will be a simple Taylor Series: |z|<1

[tex]\frac{1}{z-1}=-\sum^{\infty}_{k=0}z^{k}[/tex]

So

[tex]a[k \geq 1]=0 [/tex]
[tex]a[k \leq 0]=-1[/tex]

Or a[k]=-u[-k]


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