How should I apply partial fractions here?

[YehiaMedhat]

For the following fraction $$\frac{z}{z^2-3z+2}$$ is it fine to use the partial fraction method like this? $$\frac{A}{z-2} + \frac{B}{z-1}$$Or should I be taking the $z$ in the numerator as a common factor and then assume the solution to be? $$z\left(\frac{1}{z^2-3z+2}\right) = z\left(\frac{A}{z-2} + \frac{B}{z-1}\right)$$Which way is fine? or are they just both fine, because when finding the inverse $\mathcal{z}$ transform for that expression it made a huge difference in the final answer.
They all look the same for me, if I miss something about the fundamentals of partial fractions, please explain it to me, and thanks in advance

 

[Dr.Peterson]

The first method will do what partial fractions are meant to do: give you a sum of proper fractions.

The second will give a sum of improper fractions. But it expresses F(z)/z as a sum of proper fractions, which may be what you really want to do.

The rest depends on what you do with your result. You'll need to show that.

 

[Steven G]

If I had what you wrote in your 2nd method, the first thing I would do is cancel out the z's. Do you think that your initial expression doesn't depend on z?

 

[YehiaMedhat]

The first method will do what partial fractions are meant to do: give you a sum of proper fractions.

The second will give a sum of improper fractions. But it expresses F(z)/z as a sum of proper fractions, which may be what you really want to do.

The rest depends on what you do with your result. You'll need to show that.

Well If I consider the first way, I consume that the whole expression is just equal to other separate fractions as follows:
$$\frac{z}{(z-1)(z-2)} = \frac{A}{z-2} + \frac{B}{z-1}$$$$\therefore z = A(z-1) + B(z-2)$$From that I get two equations with which I can obtain the A, B values:
$$z = Az + Bz,\ z = z(A+B) \rightarrow \ A+B = 1$$And similarly:
$$-A -2B = 0 \rightarrow \ A=-2B$$Solving the two equations together: I get $A\ =\ 2, B\ =\ -1$, so the expression is turned into:
$$\frac{2}{z-2} - \frac{1}{z-1}$$Applying the inverse $\mathcal{Z}$ transform:
$$2u(t-2) - u(t-1)$$
Now if I consider the other way, doing the equations I get the following expression:
$$z\left( \frac{1}{z-2} - \frac{1}{z-1} \right) = \frac{z}{z-2} - \frac{z}{z-1}$$Similarly applying inverse $\mathcal{z}$ transform:
$$2^t - 1$$
This looks to me amazing and weird, or may they're the same expression but written in different symbols and I can't see this subtle relationship.

 

[Dr.Peterson]

Now please show how you get from $\displaystyle\frac{2}{z-2} - \frac{1}{z-1}$ to $2u(t-2) - u(t-1)$, and from $\displaystyle\frac{z}{z-2} - \frac{z}{z-1}$ to $2^t - 1$.

What method are you using? Most of this looks wrong to me (though I don't know a lot about the Z transform, and had to look it up). The second method is closer.

 

[YehiaMedhat]

Now please show how you get from $\displaystyle\frac{2}{z-2} - \frac{1}{z-1}$ to $2u(t-2) - u(t-1)$, and from $\displaystyle\frac{z}{z-2} - \frac{z}{z-1}$ to $2^t - 1$.

What method are you using? Most of this looks wrong to me (though I don't know a lot about the Z transform, and had to look it up). The second method is closer.

It basically comes from the definition:
$$\mathcal{Z}\{f(z)\} = \sum_{t = 0}^{\infty} f(z) z^{-t}$$For the case of $f(z) = 1$, then it's gonna some sort of geometric series which is later dealt like this:
$$\frac{1}{1-\frac{1}{z}} = \frac{z}{z-1}$$For the first one which is $2\ u(t-2) - u(t-1)$, the unit step functions shift the terms in which they exist, the $\mathcal{Z}$ transform of the unit step function which looks like $u(t - n)$ is $\frac{z}{z-1} z^{-n}$.
So I got the first inverse $\mathcal{Z}$ transform like:
$$\frac{2z}{z-2} z^{-1} = 2^{t+1}u(t-1)$$Well, to be honest it would feel even more miserable if I continue, because I have made (once again) really silly typos
I hate this, but it seemed to me fine when I posted it, I feel like I was just blind at that moment!!!!

SORRY GUYS, sorry, really sorry, i don't think it's worth it check my posts again

 

[Dr.Peterson]

I can't be sure you've checked everything and found the correct answer yet; but I hope you see an important lesson:

When you want help, show all your work, not just the parts you know you aren't sure of; by writing it all out for someone else to check, you are likely to catch (at least some of) your own mistakes. Checking your own work as if you were the person on the other end checking (or grading!) your work is, actually, an essential part of mathematics.

I'll let those who are familiar with the topic tell you whether you are still wrong, if they wish.

 

[YehiaMedhat]

I can't be sure you've checked everything and found the correct answer yet; but I hope you see an important lesson:

When you want help, show all your work, not just the parts you know you aren't sure of; by writing it all out for someone else to check, you are likely to catch (at least some of) your own mistakes. Checking your own work as if you were the person on the other end checking (or grading!) your work is, actually, an essential part of mathematics.

I'll let those who are familiar with the topic tell you whether you are still wrong, if they wish.

Yes, exactly, got it
Thank you all guys for helping me so far.