Z Transforms and the Fibonacci SequenceDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/