MATLAB: Help, pls how can i convert from syms to double

inverter

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)

Best Answer

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
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