help, pls how can i convert from syms to double
조회 수: 1 (최근 30일)
이전 댓글 표시
clc
clear all
syms s
Kw= 0.05;Ka=0.01;
T=0.1; R=500;
f=50; L=2500*10^-6;
X=2*pi*f*L;
d0_1=-0.019; d0_2=0.019; dw1=0.5; dw2=-0.5; dV1=0; dV2=0;
dPavg1=-10; dPavg2=10; dQavg1=0; dQavg2=0;
M=4*X + X^3/R^2;
t=(0:0.2:2)
V1eq=23;
V2eq=23;
O1eq=0.019;
O2eq=-0.019;
a1=(2*X*V1eq/R + 2*V2eq*sin(O1eq - O2eq) + X*V2eq/R*cos(O1eq - O2eq))/4*X + X^3/R^2;
b1=(2*V1eq*sin(O1eq - O2eq) + (X/R)*V1eq*cos(O1eq - O2eq))/M;
c1=(2*V1eq*V2eq*cos(O1eq - O2eq) - (X/R)*V1eq*V2eq*sin(O1eq - O2eq))/M;
d1=-c1;
a2=(2*V2eq*sin(O2eq - O1eq) + (X/R)*V2eq*cos(O2eq - O1eq))/M;
b2=(2*V2eq*X/R + 2*V1eq*sin(O2eq - O1eq) + V1eq*(X/R)*cos(O2eq -O1eq))/M;
c2=(-2*V1eq*V2eq*cos(O2eq - O1eq) + (X/R)*V1eq*V2eq*sin(O2eq - O1eq))/M;
d2= -c2;
a3=(2*(X^2/R^2 + 2)*V1eq - 2*V2eq*cos(O1eq - O2eq) + X/R*V2eq*sin(O1eq - O2eq))/M;
b3=(-2*V1eq*cos(O1eq - O2eq) + X/R*V1eq*sin(O1eq - O2eq))/M;
c3=(2*V1eq*V2eq*sin(O1eq - O2eq) + X/R*V1eq*V2eq*cos(O1eq - O2eq))/M;
d3= -c3;
a4= (-2*V2eq*cos(O2eq - O1eq) + (X/R)*V2eq*sin(O2eq - O1eq));
b4=(2*(2+X^2/R^2)*V2eq - 2*V1eq*cos(O2eq - O1eq) + X/R * V1eq*sin(O2eq - O1eq))/M;
c4=(-2*V1eq*V2eq*sin(O2eq - O1eq) - X/R*V1eq*V2eq*cos(O2eq - O1eq));
d4=-c4;
X1=[d0_1; d0_2; dw1; dw2; dV1; dV2; dPavg1; dPavg2; dQavg1; dQavg2];
A1=[0 1 0 0 0 0 0 0 0 0;
0 0 0 1 0 0 0 0 0 0;
(-Kw*c1)/T (-Kw*d1)/T -1/T 0 (-Kw*a1)/T (-Kw*b1)/T 0 0 0 0;
(-Kw*c2)/T (-Kw*d2)/T 0 -1/T (-Kw*a2)/T (-Kw*b2)/T 0 0 0 0;
(-Ka*c3)/T (-Ka*d3/T) 0 0 (1+Ka*a3)/T (-Ka*b3/T) 0 0 0 0;
(-Ka*c4)/T (-Ka*d4)/T 0 0 (-Ka*a4)/T -(1+Ka*b4)/T 0 0 0 0;
c1/T d1/T 0 0 a1/T b1/T -1/T 0 0 0;
c2/T d2/T 0 0 a2/T b2/T 0 -1/T 0 0;
c3/T d3/T 0 0 a3/T b3/T 0 0 -1/T 0;
c4/T d4/T 0 0 a4/T b4/T 0 0 0 -1/T]
sX1=A1*X1
plot(t,dPavg1)
댓글 수: 3
Walter Roberson
2016년 3월 16일
Note that your dPavg1 is just the scalar value -10 so you would have trouble seeing your plot.
채택된 답변
Javier Bastante
2016년 3월 16일
편집: Javier Bastante
2016년 3월 28일
It's true that you haven't used s but the answer to your question is as easy as this:
>> syms s
>> number=s+s-3/4 %The result will be 3/4 as symbolic
>> number2=double(number) %Now the result is 0.7500
So, the function is double(var)
Hope it's useful. Regards
댓글 수: 1
Javier Bastante
2016년 3월 28일
I wrote syms x instead of syms s. Sorry about that. It's already been corrected
추가 답변 (0개)
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!