Hello Robert,
Suppose X(Z) = 1
---------------------- RoC |z|>1
1-1.5*z^(-1)+0.5*z^(-2)
since this is a right sided sequence, I simply divide the numerator by
denominator & to get,
X(z) = 1+(3/2)*z^(-1)+(7/4)*z^(-2)+...
and the corresponding x[n] as,
x[n] = 1*S[n]+3/2*S[n-1]+7/4*S[n-2] ... ('S' showing impulse)
Now, if the RoC is |z|< 1, meaning X(z) is the ZT of a left sided
sequence and that X(z) be expanded in positive powers of Z. So, first I
convert X(z) from negative power of z to positive powers, to get,
X(Z) = z^2
---------------
0.5-1.5z+z^2
so, when I divide the numerator by denominator, I get
X(z) = 2*z^2+6*z^3+...
and the corresponding x[n] as,
x[n] = 2*S[n+2]+6*S[n+3]+14*S[n+4] ... (S showing impulse)
So, now I repeat my question, if RoC of was 0.5<|z|<1, which makes it a
two-sided sequence, then I would not be able to find its IZT by Power
Series method (I know Partial Fraction can be used)
The only *possible* way (to use Power Series) would be to use partial
fraction to separate the R.S.S. and the L.S.S. and use Power Series
individually
I hope I've put up a clearer question this time around
Reply by robert bristow-johnson●August 27, 20092009-08-27
On Aug 26, 9:14=A0am, "commengr" <communications_engin...@yahoo.com>
wrote:
> Question regarding Power Series method of finding IZT
> -------------------------------------------------------
> Hello. I want to clear a doubt regarding Power Series method for Inverse
> Z-transform
>
> If I have X(z) and I'm given a RoC that says that X(z) is a two-sided
> sequence and I'm supposed to find the IZT by Power Series method. My view
> is that I would not be able to find it using the said method. Since, powe=
r
> series has a provision of going in either right direction (causal, in
> negative powers of Z) and left direction (non-causal, in positive powers =
of
> z). I have not found any example where we are able to go towards both sid=
es
> (referring chapter on Z-transform in the book of Proakis, Oppenhiem)
>
> In order to do so, my approach would be to first divide X(z) into two
> partial fractions. So that I can separate the right sided sequences and
> left sided sequences and perhaps then find the their IZT by Power Series =
(I
> know it would not be worth it and I might as well use the properties to
> find IZT directly).
>
> But do you agree that this is the *only* way to find IZT by Power Series
> of two-sided sequence?
at least for my understanding, you need to be more specific. can you
start with an example of what you want to iZT?
r b-j
Reply by commengr●August 26, 20092009-08-26
Question regarding Power Series method of finding IZT
-------------------------------------------------------
Hello. I want to clear a doubt regarding Power Series method for Inverse
Z-transform
If I have X(z) and I'm given a RoC that says that X(z) is a two-sided
sequence and I'm supposed to find the IZT by Power Series method. My view
is that I would not be able to find it using the said method. Since, power
series has a provision of going in either right direction (causal, in
negative powers of Z) and left direction (non-causal, in positive powers of
z). I have not found any example where we are able to go towards both sides
(referring chapter on Z-transform in the book of Proakis, Oppenhiem)
In order to do so, my approach would be to first divide X(z) into two
partial fractions. So that I can separate the right sided sequences and
left sided sequences and perhaps then find the their IZT by Power Series (I
know it would not be worth it and I might as well use the properties to
find IZT directly).
But do you agree that this is the *only* way to find IZT by Power Series
of two-sided sequence?
Thanks
PS. no hw