# How to use Z-transforms to derive formula for Fibonacci seq?

You start with the fibonacci seq:
ysub1 = 1
ysub2 = 1
ysub3 = ysub1+ysub2
then you reformulate this in terms of its z-transforms (these are the discrete versions of Laplace transforms).
You then work out the z-transform representation of the nth term of the sequence.
Then you perform an inverse z-tranform...
show more
You start with the fibonacci seq:

ysub1 = 1

ysub2 = 1

ysub3 = ysub1+ysub2

then you reformulate this in terms of its z-transforms (these are the discrete versions of Laplace transforms).

You then work out the z-transform representation of the nth term of the sequence.

Then you perform an inverse z-tranform to get the generalized ysubn = <some formula with sqrt(5) in it>

I saw this done once. Really really cool. I just cannot seem to get started on it again. Any pointers out there from some 3rd year electrical engineer freshly through a course in digital signal processing?

Thanks for any pointers.

ysub1 = 1

ysub2 = 1

ysub3 = ysub1+ysub2

then you reformulate this in terms of its z-transforms (these are the discrete versions of Laplace transforms).

You then work out the z-transform representation of the nth term of the sequence.

Then you perform an inverse z-tranform to get the generalized ysubn = <some formula with sqrt(5) in it>

I saw this done once. Really really cool. I just cannot seem to get started on it again. Any pointers out there from some 3rd year electrical engineer freshly through a course in digital signal processing?

Thanks for any pointers.

Follow

1 answer
1

Are you sure you want to delete this answer?