Using this method for partial fraction simplification led to another solution, how?!!!

[YehiaMedhat]

While solving my $\mathcal{Z}$ transform examples of the lecture, I got stuck (as usual ) with this one:
Get the inverse $\mathcal{Z}$ transform for the following expression: $\LARGE\frac{z^2+1}{z^2-2z}$
My approach:
$$F(z) = \frac{z}{z-2} + \frac{1}{z(z-2)}$$This expression after simplification just looks really simple, the first term is scaled and the second is shifted and scaled, so, the inverse is:
$$\mathcal{Z}^{-1}\left\{ \frac{z}{z-2} + \frac{1}{z(z-2)} \right\} = 2^t + 2^t u(t-2)$$I wish it's not me who's messed the solution, but looking at my professor's approach, which is:
$$\frac{F(z)}{z} = \frac{z^2+1}{z^2(z-2)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z-2}$$And the values of $A,\ B,\ C$ are $-0.25,\ -0.5,\ 1.25$ respectively.
$$\therefore\ F(z) = -0.25 - \frac{0.5}{z} + \frac{1.25z}{z-2}$$$$\therefore\ \mathcal{Z}^{-1} \{F(z)\} = -0.25\delta(t) - 0.5\delta(t-1) +1.25(2)^t$$
Can anyone tell me what on earth does the division over z does with initial steps of the solution?
I just did it without it, and checked my solution to reform it back to the first form and it was just fine, but also looking at the solution of the TA's, they do the same step as my professor, which is weird for me, I feel like this is something concrete in getting partial fractions, but anyway I can't get the same solution, or is it just the same, but with other expression?!!

 

[mario99]

While solving my $\mathcal{Z}$ transform examples of the lecture, I got stuck (as usual ) with this one:
Get the inverse $\mathcal{Z}$ transform for the following expression: $\LARGE\frac{z^2+1}{z^2-2z}$
My approach:
$$F(z) = \frac{z}{z-2} + \frac{1}{z(z-2)}$$This expression after simplification just looks really simple, the first term is scaled and the second is shifted and scaled, so, the inverse is:
$$\mathcal{Z}^{-1}\left\{ \frac{z}{z-2} + \frac{1}{z(z-2)} \right\} = 2^t + 2^t u(t-2)$$I wish it's not me who's messed the solution, but looking at my professor's approach, which is:
$$\frac{F(z)}{z} = \frac{z^2+1}{z^2(z-2)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z-2}$$And the values of $A,\ B,\ C$ are $-0.25,\ -0.5,\ 1.25$ respectively.
$$\therefore\ F(z) = -0.25 - \frac{0.5}{z} + \frac{1.25z}{z-2}$$$$\therefore\ \mathcal{Z}^{-1} \{F(z)\} = -0.25\delta(t) - 0.5\delta(t-1) +1.25(2)^t$$
Can anyone tell me what on earth does the division over z does with initial steps of the solution?
I just did it without it, and checked my solution to reform it back to the first form and it was just fine, but also looking at the solution of the TA's, they do the same step as my professor, which is weird for me, I feel like this is something concrete in getting partial fractions, but anyway I can't get the same solution, or is it just the same, but with other expression?!!

You cannot perform partial fraction decomposition if the numerator and the denominator have the same degree! The numerator must has a lower degree to preform it, so your professor made a trick to increase the denominator degree by 1.

Why do you think your solution is correct and equivalent?

What is the Z-transform of $2^tu(t-2)$?

 

[YehiaMedhat]

Ok, for that one about the partial fractions degree of numerator and denominator, may I haven't heard about, or just forgot.

I thought my solution was equivalent since I did it in a way I think was correct.

The Z-transform for the $2^t u(t-2)$ is:
$$\frac{\frac{z}{2}}{\frac{z}{2}-1} z^{-2} = \frac{z}{z-2} z^{-2} = \frac{1}{z(z-2)}$$

 

[mario99]

Ok, for that one about the partial fractions degree of numerator and denominator, may I haven't heard about, or just forgot.

I thought my solution was equivalent since I did it in a way I think was correct.

Your solution is almost correct.

The Z-transform for the $2^t u(t-2)$ is:
$$\frac{\frac{z}{2}}{\frac{z}{2}-1} z^{-2} = \frac{z}{z-2} z^{-2} = \frac{1}{z(z-2)}$$

The order of applying the properties is wrong. Shifting comes before scaling.

$\displaystyle \mathcal{Z}\left\{2^{t}u(t-2)\right\} = \frac{4}{z(z-2)}$

So

$\displaystyle \mathcal{Z^{-1}}\left\{\frac{1}{z(z-2)}\right\} = \frac{2^tu(t-2)}{4} = \frac{2^tu(t-2)}{2^2} = 2^{t-2}u(t-2)$

If your solution was:

$2^t + 2^{t-2}u(t-2)$

it would be equivalent to the professor's solution.

 

[Steven G]

1st step is always to preform long division (or any equivalent method for a given problem). Now, you can't do division if the degree of the numerator is less that the degree of the denominator (other can reducing).

 

[YehiaMedhat]

1st step is always to preform long division (or any equivalent method for a given problem). Now, you can't do division if the degree of the numerator is less that the degree of the denominator (other can reducing).

Yes, I have seen in my TA's solution that he did the long division first, I thought there's a way to get around it, but it seemed to make more confusion for me about it
But Mr. @Steven G, what do you think, have you seen in any text book or any reference something that indicates that Z-transform has any kind of order to deal with its properties?