Why does the function real return expressions containing i ?
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For the following code, CEre is derived using the real function, but it contains i in the expression.
clc;clear all;
syms s;
syms T1 T2 w tau1 tau2 real;
e1=(1-T1*s)/(1+T1*s);
e2=(1-T2*s)/(1+T2*s);
A=[0 1 0 0;
0 0 1 0;
0 0 0 1;
-29.17 -56 -36.7 -10.1];
B1=[-1.55 1 0 0;
-1 -0.3 0 0;
0 0 0.5 0;
-0.7 0 -0.34 -2.6];
B2=[0 0 0 0;
1 1.5 4 0;
0 0 0 0;
-0.33 0 0 -1.1];
B3=[0 0 0 0;
0 0 0 0;
0 0 0 0;
-0.08 -0.7 0 -1];
B4=zeros(4,4); B4(4,3)=-3;
n=size(A,1);
tau3=0.169; tau4=0.26;
cez=numden(simplify(det(w*1i*eye(n)-A-B1*e1-B2*e2-B3*(cos(tau3*w)-1i*sin(tau3*w))-B4*(cos(tau4*w)-1i*sin(tau4*w)))));
% ce=collect(cez,s);
cew=subs(cez,s,w*1i)
CEre=real(cew);
CEim=imag(cew);
After commenting off the function collect, the function real gives a result without the unity of complex i.
0 Kommentare
Antworten (2)
Walter Roberson
am 28 Mär. 2013
I have not tried the above code, but consider
syms w
real(w*1i)
The result may contain "i" because it is possible that w is complex itself.
In this particular case, the result could also be expressed as imag(w), but there are other cases that are more difficult to resolve, such as
real((w + 1i)^2)
1 Kommentar
Ahmed A. Selman
am 28 Mär. 2013
Bearbeitet: Ahmed A. Selman
am 29 Mär. 2013
Check the math, e.g.:
...
TheEq=w*1i*eye(n)-A-B1*e1-B2*e2- B3*(cos(tau3*w)-1i*sin(tau3*w))-B4*cos(tau4*w)-1i*sin(tau4*w);
TheDet=det(TheEq);
TheSimple=simplify(TheDet);
cez=numden(TheSimple);
...etc
Then see what do
TheSimpleEq=simplify(TheEq);
pretty(TheSimpleEq)
give.
Why do you use (1i) instead of (i) only? If it is not declared before, then (i or j) alone are assumed a unity of complex in Matlab.
3 Kommentare
Ahmed A. Selman
am 29 Mär. 2013
If someone wrote a code, he shouldn't have to worry about accidents :)
Walter Roberson
am 29 Mär. 2013
Be safe, program defensively. And make it easier for other people to read.
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