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Z Transforms and the Fibonacci Sequence


Date: 04/20/98 at 16:08:11
From: karen
Subject: LaPlace + z transforms

My boss is trying to teach me about LaPlace and Z transformations.  
The problem is, he isn't teaching according to how the textbooks teach 
it, so I'm having trouble trying to follow up on my own reading. 

He starts out by saying d/dt = s, so integral dt = 1/s. He also 
mentions zero order hold, and shows me an example:

  F(z) = (symbol that means find the Z transform) [{(1-e^-sh)/s}f(s)]

where (1-e^-sh)/s is the zero order data hold. By the way, I haven't 
found anything on zero order holds or first order holds. 

So:

  (symbol that means find the Z transform of) = Z
  f(s) = 1/s (simple step function, right?)
  z = e^sh

Therefore:

  z^-1 = e^-sh
       = Z[{(1-e^-sh)/s}1/s]
       = (1-z^-1) Z[1/s^2]
       = {(z-1)/z} Z{t}         [in the tables, inverse of 1/s^2 = t]
       = [(z-1)/z] [hz/(z-1)^2]
       = h/(z-1) 
       = y(z)/x(z)
 hx(z) = (z-1)y(z)
 hx(subscript k) = y(subscript k+1) - y(subscript k)
 y(subscript k+1) = y(subscript k) + hx(subscript k)

Also, is that a difference equation? Is a difference equation just 
like a differential equation for discrete functions?

I hope this stuff makes sense. I'd also like to know if you have any 
good sites for zero/first/second order holds.

Thank you.


Date: 04/20/98 at 19:44:47
From: Doctor Anthony
Subject: Re: LaPlace + z transforms

Here is a typical example of using a z-transform to find the 
expression for the nth term of the Fibonacci series. The method is 
reasonably clear if you follow it through.

The difference equation for the Fibonacci series is:

    u(n+2) = u(n+1) + u(n)   with   u(0) = 0,  u(1) = 1

So:

    u(n+2) - u(n+1) - u(n) = 0  

Taking the transforms:

    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

Putting u(0) = 0 and u(1) = 1 this becomes:

    u(z)[z^2 - z - 1] =  z

                 z
    u(z) =  -------------
             z^2 - z - 1

The denominator factorizes to:

    [z-1/2 -sqrt(5)/2)][z-1/2 +sqrt(5)/2]

The trick here 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

and so: 

                            z                     z
    u(z) = 1/sqrt(5)[----------------  - ------------------ ]
                     z-1/2-sqrt(5)/2)      z-1/2+sqrt(5)/2

Then from table of inverse transforms:

       z
    -------  =  a^n    
     z - a

where a is constant. Then:

                              z                       z
    u(z) = 1/sqrt(5) [-------------------- - --------------------]
                       z - (1/2+sqrt(5)/2)    z - (1/2-sqrt(5)/2)

Now from the table of inverse transforms:

    u(n) = 1/sqrt(5)[(1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n]

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
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