Negative Exponents and Zeros

[Shehezaada]

Hi everyone!

Quick question with regards to a sample problem in a textbook for Z-Transforms. I'm hung up on a question that was done and I'm not quite sure how they arrived at the answer.

H(z) = 6 - 5z^-1 + z^-2
= (3 - z^-1)(2 - z^-1)
= 6 * (z - 1/3)(z - 1/2)
z^2

First question, in the first step, why are the two terms inverted. When I factorize the expression H(z),I usually get ( z^-1 -3)(z^-1 -2)

Second question,I'm not sure how they got from step 2 to step 3.

Thanks for helping guys!

 

[HallsofIvy]

I can't tell you why they happened to write it that way, but that is clearly the same as what you have. (3 - z^-1)(2 - z^-1)= (-(-3+ z^{-1})(-(-2+ z^{-1})= (-3+ z^{-1})(-2+ z^{-1}) because the two negatives cancel, (-1)(-1)= 1. Then you can write it as (z^{-1}- 3)(z^{-1}- 2) because "addition is commutative": a+b= b+ a. Finally, that is the same as (z^{-1}- 2)(z^{-1}- 3) because "multiplication is commutative: ab= ba.

 

[tkhunny]

\(\displaystyle H(z) = 6 - 5z^{-1} + z^{-2}\)

\(\displaystyle H(z) = (3 - z^{-1})(2 - z^{-1})\)

\(\displaystyle H(z) = 3\left(1 - \dfrac{z^{-1}}{3}\right)2\left(1 - \dfrac{z^{-1}}{2}\right)\)

 

[pka]

H(z) = 6 - 5z^-1 + z^-2
= (3 - z^-1)(2 - z^-1)
= 6 * (z - 1/3)(z - 1/2)
*********z^2

I don't why anyone would do it like that.

\(\displaystyle 6-5z^{-1}+z^{-2}=\dfrac{6z^2-5z+1}{z^2}=\dfrac{(3z-1)(2z-1)}{z^2}\).

 

[Deleted member 4993]

Hi everyone!

Quick question with regards to a sample problem in a textbook for Z-Transforms. I'm hung up on a question that was done and I'm not quite sure how they arrived at the answer.

H(z) = 6 - 5z^-1 + z^-2
= (3 - z^-1)(2 - z^-1)
= 6 * (z - 1/3)(z - 1/2)
z^2

First question, in the first step, why are the two terms inverted. When I factorize the expression H(z),I usually get ( z^-1 -3)(z^-1 -2)

Those are equivalent because (a - b)(c - d) = (-1)(b - a)(-1)(d - c) = (-1)(-1)(b - a)(d - c) = (b - a)(d - c)

Second question,I'm not sure how they got from step 2 to step 3.

\(\displaystyle \displaystyle (3 - z^{-1})(2 - z^{-1}) \ = \ \ (3 - \frac {1}{z})(2 - \frac{1}{z})\)

\(\displaystyle \displaystyle = \ \ \frac {3z - 1}{z} \ * \ \frac{2z -1}{z}\)

This is simple algebra I.

Thanks for helping guys!

.

 

[JeffM]

Hi everyone!

Quick question with regards to a sample problem in a textbook for Z-Transforms. I'm hung up on a question that was done and I'm not quite sure how they arrived at the answer.

H(z) = 6 - 5z^-1 + z^-2
= (3 - z^-1)(2 - z^-1)
= 6 * (z - 1/3)(z - 1/2)
z^2

First question, in the first step, why are the two terms inverted. When I factorize the expression H(z),I usually get ( z^-1 -3)(z^-1 -2)

Second question,I'm not sure how they got from step 2 to step 3.

Thanks for helping guys!

First question

6 - 5w + w^2 = (3 - w)(2 - w) = {(- 1)(w - 3)} * {(-1)(w - 2)} = (-1)(-1)(w - 3)(w - 2) = 1 * (w - 3)(w - 2) = (w - 3)(w - 2).

In short (3 - w)(2 - w) = (w - 3)(w - 2). Two different and correct ways to say the exact same thing.

Second question involves knowing what a negative exponent means and a rather odd but legitimate factoring.

\(\displaystyle \left(3 - z^{-1})(2 - z^{-1}\right) = \left(3 - \dfrac{1}{z}\right) * \left(2 - \dfrac{1}{z}\right) = \dfrac{3z - 1}{z} * \dfrac{2z - 1}{z} = \dfrac{3\left(z - \frac{1}{3}\right) * 2\left(z - \frac{1}{2}\right)}{z^2} = \dfrac{6\left(z - \frac{1}{3}\right)\left(z - \frac{1}{2}\right)}{z^2}.\)

 

[Shehezaada]

Guys,

Thank you so much for these replies. I definitely understand now! I'm going back to school (MAsc, Electrical Engineering) after being in industry for the last 20 months..so I really need to brush up on my math skills. Is there any basic textbook that you guys recommend for someone in my situation?

Thanks,
Shivam

 

[lookagain]

z^-2 - 5z^-1 + 6 = 0
1/z^2 + 5/z + 6 = 0
1 + 5z + 6z^2 = 0
(2z - 1)(3z - 1) = 0
z = 1/2 or z = 1/3


Note: \(\displaystyle \ \ \frac{1}{z^2} + \frac{5}{z} + 6 = 0 * \) \(\displaystyle \ \ \ \) has the added restriction that \(\displaystyle \ z \ne 0, \)

\(\displaystyle while \ \ \ 1 + 5z + 6z^2 = 0 \ ** \ \ \ \ does \ \ not.\)


Hypothetically, if ** had given z = 0 as a solution, then you would have had to discard it, because it does not satisfy *.