Calculating Poles and Residues of a Z Transform by Hand

[Evo8]

Homework Statement

Consider the following Z tranform
$X(z)=\frac{z^{-1}}{1-2z^{-1}+2z^{-2}}$

Calculate the poles and resides by hand.

Homework Equations

To find residue at pole p1

$(z-p_{1})X(z)z^{k-1}$ Evaluate at $z=p_{1}$

The Attempt at a Solution

OK so I've taken the X(z) and converted to positive power of z by multiplying by z^2/z^2

I get $X(z)=\frac{z}{z^{2}+2z+2}$
I know there is 1 zero at z=o.

For the poles I've used the quadratic equation on the denominator. I come up with
$$p1=\frac{-2\pm\sqrt{-4}}{2} p2=\frac{-2 \pm\ 2j}{2}$$

simplifys to
$$p1=-1-1j$$
$$p2=-1+1j$$


Ok so I am stuck on finding the residues.

The way I see using the above formula I end up with a 0 in the denominator. I know this incorrect. Am I missing some sort of simplification before i evaluate at z=p?

If I take $z^{2}+2z+2$ and substitute $z=-1-1j$ I get 0.

When I use the equation

$(z-p_{1})X(z)z^{k-1}$ Evaluate at $z=p_{1}$

and substitute z with p1 that first term in the parenthesis = o. (p1-p1)=0.

Should something cancel out before i evaluate z=p1so I don't get the 0 in the numerator or denominator? If so I don't see it. I've played with the numbers and moved them around for a couple of days now and I juts don't see it. It doesn't look complicated to me but it never comes out not 0.

any help is appreciated! Thanks!

 

[I like Serena]

Hey Evo8!

The denominator of X(z) can and should be factorized into (z-p1)(z-p2).
Then by multiplying X(z) with (z-p1) that factor cancels after which you can calculate the residu.

 

[Evo8]

Hi I like Serena!

I guess I don't see how it factors out to that then. This is what I would have assumed should be done. I was just under the impression that the denominator does not factor out that way.

When I take (z-p1)(z-p2) and expand this is what I get
$$z^{2}=p_{2}z-p_{1}+p_{1}p_{2}$$

My expanded set of terms is
$$z^{2}+2z+2$$ or $$1-2z^{-1}+2z^{-2}$$

I will google basic factoring and see if I can't figure out what I am missing here. P.s. If you thought my algebra was already sub par, factoring was my LEAST favorite part of algebra ha!

 

[I like Serena]

You solved for the denominator being zero and found p1 and p2.
This means that the denominator factorizes into (z-p1)(z-p2).
The typical verification is by expanding it again and check you get your original denominator.

I'm afraid you did not expand it correctly.
The proper expansion of (z-p₁)(z-p₂) is $z^2 -p_1 z -p_2 z + p_1 p_2$.
When you work that out, you should get $z^{2}+2z+2$.
(Do you?)

Anyway, if you multiply (z-p₁) with $X(z) = {z \over (z-p_1)(z-p_2)}$, you should get $z \over (z-p_2)$, which is what you need to evaluate.

 

[Evo8]

I must have typed the expansion incorrectly. When I look at what I wrote down I have what you put. I don't even know how I got that = sign up there. Anyway thanks for pointing that out.

This makes a little more sense now. So I can literally plug my two poles into the (z-p1)(z-p2)? Now where getting somewhere. I knew it had to be something small that I was forgetting.

Thanks for the help! Hopefully I will be able to continue without issue!

 

[I like Serena]

Oh, the = sign is on the same key as the +, and it's next to the -.
It's a pretty standard typo, it even looks a little bit like a minus sign.
I didn't really think you put the = sign there intentionally. ;)

 

[Evo8]

Ok so after evaluating at expression I am left with this for the residue at p1.
$$
\frac{(-1-1j)(-1-1j)^{k-1}}{(-1-1j)-(-1+1j)}
$$

What is the k? In my book all it says is that there is a dependence on k which is clear here. The example given also uses variable and constants for everything so the result is in terms of k.

Thanks for any hints/help

 

[I like Serena]

Looks good!

I think you have the inverse z-tranform there, that is, $x[k]=\mathcal{Z}^{-1}\{X(z)\}$.

 

[Evo8]

sorry for the delayed response! Thanks so much for your help on this one I like Serena!

Much appreciated as always

 

[I like Serena]

I've looked it up.

The "residue" of a simple pole p1 is (z-p1)X(z) evaluated at z=p1.

The inverse z-transform is the sum of both (z-p1)X(z)z^{-k} evaluated at z=p1 and (z-p2)X(z)z^{-k} evaluated at z=p2, yielding x[k].

 

[Evo8]

That looks right to me. I sometimes end up with repeated poles so my p1 and p2 are the same. So for the first term i get a 0 which brings everything down to 0. I always forget to manipulate and cancel out though. This is where I got lost. This example I was able to complete I believe.

Thanks for posting that up!

 

[I like Serena]

Thanks for posting that up!

:)

In this example you only have simple poles.
Simple poles (typically) only have one factor (z-p1) in the denominator of X(z).

If you have for instance (z-p1)^2 in the denominator, which is what you seem to refer to, you need a more advanced formula.

 

[Evo8]

Indeed. That is exactly the form that where I used the mixed poles method and got hung up. :)

 

[I like Serena]

Okay. ;)As you can see in the wikipedia article, a simple pole has residue:

This is what you've been working with.A pole of order n (in your example you would have n=2) has residue:

More specifically, if you have (z-p1)^2 in the denominator, your residue is:
${d \over dz}((z-p_1)^2 X(z))$ evaluated at z=p1.