Extract numerator and denominator from transfer function without manually adding it to the program?

9 次查看(过去 30 天)
Hello im having a trouble where my denominator roots are very long and needed to be extracted without inserting manually in the coding.
Example this is my output
z exp(5) sin(3)
----------------------------------
2
exp(10) z - cos(3) exp(5) z 2 + 1
(exp(5)*sin(3)*((cos(3)^2*exp(10) - exp(10))^(1/2) + cos(3)*exp(5))*((cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5))*(exp(10)*hypergeom([2, -exp(-10)*((cos(3)^2*exp(10) - exp(10))^(1/2) - exp(10) + cos(3)*exp(5))], -exp(-10)*((cos(3)^2*exp(10) - exp(10))^(1/2) - 2*exp(10) + cos(3)*exp(5)), 1/w) + (cos(3)^2*exp(10) - exp(10))^(1/2)*hypergeom([2, exp(-10)*(exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5))], exp(-10)*(2*exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5)), 1/w) - exp(10)*hypergeom([2, exp(-10)*(exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5))], exp(-10)*(2*exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5)), 1/w) + (cos(3)^2*exp(10) - exp(10))^(1/2)*hypergeom([2, -exp(-10)*((cos(3)^2*exp(10) - exp(10))^(1/2) - exp(10) + cos(3)*exp(5))], -exp(-10)*((cos(3)^2*exp(10) - exp(10))^(1/2) - 2*exp(10) + cos(3)*exp(5)), 1/w) + cos(3)*exp(5)*hypergeom([2, exp(-10)*(exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5))], exp(-10)*(2*exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5)), 1/w) - cos(3)*exp(5)*hypergeom([2, -exp(-10)*((cos(3)^2*exp(10) - exp(10))^(1/2) - exp(10) + cos(3)*exp(5))], -exp(-10)*((cos(3)^2*exp(10) - exp(10))^(1/2) - 2*exp(10) + cos(3)*exp(5)), 1/w)))/(2*w*(cos(3)^2*exp(10) - exp(10))^(1/2)*((cos(3)^2*exp(10) - exp(10))^(1/2) - exp(10) + cos(3)*exp(5))*(exp(10) + (cos(3)^2*exp(10) - exp(10))^(1/2) - cos(3)*exp(5)))
(exp(5) sin(3) (#3 + #4) (#3 - #4) (exp(10) #1 + #3 #2 - exp(10) #2 + #3 #1 + cos(3) exp(5) #2
- cos(3) exp(5) #1))/(w #3 (#3 - exp(10) + #4) (exp(10) + #3 - #4) 2)
where
/ 1 \
#1 == F | 2, -exp(-10) (#3 - exp(10) + #4); -exp(-10) (#3 - 2 exp(10) + #4); - |
2 1\ w /
/ 1 \
#2 == F | 2, exp(-10) (exp(10) + #3 - #4); exp(-10) (2 exp(10) + #3 - #4); - |
2 1\ w /
2
#3 == sqrt(cos(3) exp(10) - exp(10))
#4 == cos(3) exp(5)
So im expecting the output to be,
num=[exp(5)*sin(3)]
den=[exp(10) -cos(3)exp(5) .....]
Thank you.

回答(1 个)

Luca Ferro
Luca Ferro 2023-3-22
编辑:Luca Ferro 2023-3-22
supposing that your original function is defined as a transfer function model you can access its properties using dot notation.
num=tfName.Numerator; %tf name is the name of the transfer function variable
den=tfName.Denominator;
The output will be a cell array that you can then acces using curly bracket notation.
Or you coulf stright use the curly bracket notation by:
num=tfName.Numerator{:}; %tf name is the name of the transfer function variable
den=tfName.Denominator{:};
in this case the output will be a numeric array

类别

Help CenterFile Exchange 中查找有关 Matrices and Arrays 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by