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Is there an inverse Z transform for: 1/z1 ? 
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#1
Oct306, 12:13 PM

P: 130

hay guys really struggling to find an inverse Z transform for: 1/(z1)
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 


#2
Feb1010, 01:02 AM

P: 25




#3
Feb1010, 10:29 AM

P: 240

For a casual sequence it will be a Laurent Series: z>1
[tex]\frac{1}{z1}=\frac{1}{z}\frac{1}{11/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[k1]. For an anticasual sequence it will be a simple Taylor Series: z<1 [tex]\frac{1}{z1}=\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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