solve a function without solving for the variable within the function

Question

Christopher Piland 2020-5-18

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/526232-solve-a-function-without-solving-for-the-variable-within-the-function

I am trying to make a Routh Array solver. It works very well when you put integers into it but if I have a variable (like "K" for gain) I can't seem to get it to not solve for K but rather present "K" as part of its answer. I have the code I am working on below. I would like the answers with "K" int them to come out with K in them.

%% 4 element Routh Array
clc
clear all
%S^4 solver
syms k; %Holde K in the array as itself
var=((0.07*k)+0.015)
K=solve(var)
% load the Routh array 
%a0s +  a1sn + a2sn + a3sn + a4sn
%Example s^4 + 3s^3 + 3s^2 + 2S + 1
a0=1; a1=2.3; a2=1.37; a3=2.65; a4=K; aout=0;
%S4
a0=1 ; a2=3 ; a4=1 ;
S4=[a0 a2 a4];
%S3
a1=3 ; a3=2 ; aout=0;
S3 = [a1 a3 aout];
%S2
b1=((a1*a2)-(a0*a3))/a1; b2=((a1*a4)-(a0*aout))/a1;
S2 = [b1 b2];
%S1
c1=((b1*a3)-(a1*b2))/b1; 
S1 = [c1];
%S0
d1=((c1*b2)-(b1*0))/c1;
S0=[d1];
disp('4 Pole S4 Routh Array output NORMAL')
disp(S4)
disp(S3)
disp(S2)
disp(S1)
disp(S0)
ia0=a4; ia1=a3; ia2=a2; ia3=a1; ia4=a0;
%S4
ia0=1 ; ia2=3 ; ia4=1 ;
iS4=[ia0 ia2 ia4];
%S3
ia1=3 ; ia3=2 ; aout=0;
iS3 = [ia1 ia3 aout];
%S2
b1=((ia1*ia2)-(ia0*ia3))/ia1; b2=((ia1*ia4)-(ia0*aout))/ia1;
iS2 = [b1 b2];
%S1
c1=((b1*ia3)-(ia1*b2))/b1; 
iS1 = [c1];
%S0
d1=((c1*b2)-(b1*0))/c1;
iS0=[d1];
disp('4 Pole S4 Routh Array output INVERTED')
disp(iS4)
disp(iS3)
disp(iS2)
disp(iS1)
disp(iS0)