How do I rearrange this transfer function to be in negative powers of z using matlab?

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Zack Freeman-Quine
Zack Freeman-Quine 2018 年 3 月 16 日
編集済み: David Goodmanson 2018 年 3 月 23 日
I have this attached transfer function and I want to rearrange it into the standard form with negative powers of z. I wrote this code and it's not giving me values for the coefficients of z^-n like I'd hoped. Is there any way I can do this?
syms a0 a1 a2 a3 a4 b0 b1 b2 b3 b4 z
eqn=((b4*(z^-4)+b3*(z^-3)+b2*(z^-2)+b1*(z^-1)+b0)/(a4*(z^-4)+a3*(z^-3)+a2*(z^-2)+a1*(z^-1)+a0))==(2*(z^4)-5*(z^3)+13.48*(z^2)-7.78*(z)+9)/(4*(z^4)+7.2*(z^3)+20*(z^2)-0.8*(z)+8);
solve(eqn,[a0 a1 a2 a3 a4 b0 b1 b2 b3 b4])
It's just giving me a0=a0 etc

回答 (2 件)

David Goodmanson
David Goodmanson 2018 年 3 月 18 日
編集済み: David Goodmanson 2018 年 3 月 23 日
Hi Zach, if you divide numerator and denominator by z^-4 you obtain
(2 -5*(z^-1) +13.48*(z^-2) -7.78*(z^-3) +9*(z^-4))/(4 +7.2*(z^-1) +20*(z^-2) -0.8*(z^-3) +8*(z^-4))
so
b0 = 2, b1 = -5, b2 = 13.48 etc.
a0 = 4, a1 = 7.2, a2 = 20 etc.

elham kreem
elham kreem 2018 年 3 月 23 日
try to split solution ; like
z1=solve(eqn,a0)
z2=solve(eqn,a1)
.... and so on

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