Z-transform question

[wild crane]

Hi, i think they missed a part, at the red line shouldn't it be: zA(z) -a0 -a1?

[narrow smelt]

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[wild crane]

or wait, if there are just constants they are excluded?

from the "conversion"

that explains why an is just A(z)

[copper axle]

Hm... Let me look up Z-transform. I've heard of it, just haven't used it for recurrence relations. I assume the point is the same as Laplace transform for DE.

[wild crane]

no problem! And yeah, you're correct its like 99% the same thing just different tables(some the same) and rules!

[copper axle]

So, according to this, we have:
𝒵(a(n)) = A(z)
𝒵(a(n + 1)) = zA(z) - a(0)z
𝒵(a(n + 2)) = z^2 A(z) - a(0)z^2 - a(1)z
So, everything seems to be fine on the picture.

[wild crane]

oh yeah i forgot to read that k > 0

and for just a(n) k isnt that

hence its just A(z)

[copper axle]

Well, yeah, we take 𝒵(a(n)) = A(z) just to have a starting point, so to speak.

[wild crane]

Yeah exactly

[copper axle]

And for a(n + k) we just use the properties.

[wild crane]

yeah and we stop at when z^(smt) = z^1?

[copper axle]

Yes.

[wild crane]

Aaaaah!!!!!!!

Tysm LordDarpinger

You're the best

i had this exam before

and funny enough i hadn't graspt this exactly thing

and i failed just cause of that

but this time i know this so now it should be better 😉

[copper axle]

Interesting method! I've usually always just solved recurrence relations using the usual characteristic equation approach, like the linear DEs. Though, in hindsight, it's pretty obvious there should be a counterpart to Laplace transform for recurrence relations.

[copper axle]

You're welcome!

[wild crane]

ahaaaaa

Yeah i agree

+close