How to use Z-transforms to derive formula for Fibonacci seq? - z transform stability
Fibonacci It begins with the following:
ysub1 = 1
ysub2 = 1
ysub3 = ysub1 + ysub2
They then formulated in terms of the Z-transform (the versions of the discrete Laplace transform) are.
Then, calculate the Z-transform representation of the nth term of the sequence.
Ysubn After an inverse z-only the general formula = get to \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, some with sqrt (5) it>
I once saw done. Really great. I do not think you can start again working on it. Any pointers, there are some 3rd Recent years electrical engineer with a course in digital signal processing?
Thank you for the advice.
Tuesday, January 12, 2010
Z Transform Stability How To Use Z-transforms To Derive Formula For Fibonacci Seq?
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1 comments:
Here is an example of using a Z-transformation, is to find the
Expression for the nth term of the Fibonacci series. The method is
pretty clear that is followed through.
The difference equation for the Fibonacci sequence is:
u 2 (n) = U (n +1) + u (with n) u (0) = 0, U (1) = 1
So:
u 2 (s) - u (n +1) - u (n) = 0
Among the transformations
z ^ 2 [U (z) - u (0) - U (1) / z] - z [U (z) - u (0)] - U (z) = 0
U (z) [z 2 - z - 1] - z ^ 2 * u (0) - z * U (1) + z * U (0) = 0
Set U (0) = 0 and U (1) = 1 is:
U (z) [z 2 - z - 1] = z
z
U (z) = -------------
z 2 - z - 1
The denominator factorize as:
[z-1 / 2 (sqrt 5) / 2)] [z-1 / 2 + sqrt (5) / 2]
The trick is to express U (z) / z in partial fractions:
1 1
U (z): z = 1/sqrt (5) [---------------- - ----------------]
z-1 / 2 - sqrt (5) / 2 z-1 / 2 + sqrt (5) / 2
This enables:
ZZ
U (z) = 1/sqrt (5 )[---------------- - ------------------]
z-1/2-sqrt (5) / 2) z-1 / 2 + sqrt (5) / 2
According to the table of the inverse transformation:
z
------- S = A ^
Z - A
where A is constant. Then:
ZZ
U (z) = 1/sqrt (5) [-------------------- - ----------------- ---]
z - (1 / 2 + sqrt (5) / 2) z - (1/2-sqrt (5) / 2)
Well, the table of the inverse transformation:
u (n) = 1/sqrt (5) [(1 / 2 + sqrt (5) / 2) ^ - n (1/2-sqrt (5) / 2) ^ n]
-Doctor Anthony, The Math Forum
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