Problems with declaring the variable right

5 vues (au cours des 30 derniers jours)
Jan Faber
Jan Faber le 15 Mai 2022
Commenté : Voss le 15 Mai 2022
Good day everyone, sadly I got the following problem with my current task that I don't know how to define the right variable for the current calculation:
for i= 0:14
xkt1(i+1)= xg*sin(2*pi*F1*(k1.*T-i*T));
end
with:
xg= 7*2*5*1*6*3;
xh= 3/10;
F1=3;
F2=100;
T= 1/500;
xkt1=zeros(0:14);
xkt2=zeros(0:14);
k1 is the variable and I tried syms, @(k1) and k1=[ ];, sadly nothing worked and either it had problems of converting it into double, or simply wasnt able to use it correctly, if someone with more knowledge about matlab could help me, I would be really grateful.
Sincerely
Jan C. Faber

Réponse acceptée

Voss
Voss le 15 Mai 2022
Using symbolic k1 and xkt1:
syms k1 xkt1
xg= 7*2*5*1*6*3;
xh= 3/10;
F1=3;
F2=100;
T= 1/500;
for i= 0:14
xkt1(i+1)= xg*sin(2*pi*F1*(k1.*T-i*T));
end
xkt1
xkt1 = 
% evaluate xkt1(1) at k1 = 0, get a symbolic expression:
subs(xkt1(1),k1,0)
ans = 
0
% evaluate xkt1(1) at k1 = 0, get a numeric value:
double(subs(xkt1(1),k1,0))
ans = 0
% evaluate xkt1(2) at k1 = 0, get a symbolic expression:
subs(xkt1(2),k1,0)
ans = 
% evaluate xkt1(2) at k1 = 0, get a numeric value:
double(subs(xkt1(2),k1,0))
ans = -47.4896
Alternatively, using a cell array of anonymous functions:
xg= 7*2*5*1*6*3;
xh= 3/10;
F1=3;
F2=100;
T= 1/500;
xkt1=cell(1,15);
for i= 0:14
xkt1{i+1}= @(k1)xg*sin(2*pi*F1*(k1.*T-i*T));
end
xkt1
xkt1 = 1×15 cell array
{@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))} {@(k1)xg*sin(2*pi*F1*(k1.*T-i*T))}
% evaluate xkt1{1} at k1 = 0:
xkt1{1}(0)
ans = 0
% evaluate xkt1{2} at k1 = 0:
xkt1{2}(0)
ans = -47.4896
  2 commentaires
Jan Faber
Jan Faber le 15 Mai 2022
Thanks a lot, I'll definitely give those options a try, when it should've worked, then I'll mark the answer as accepted.
Sincerely
Jan C. Faber
Voss
Voss le 15 Mai 2022
Sounds good. Let me know if you have any questions about it.
-_

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Plus de réponses (1)

Torsten
Torsten le 15 Mai 2022
Modifié(e) : Torsten le 15 Mai 2022
xg= 7*2*5*1*6*3;
xh= 3/10;
F1=3;
F2=100;
T= 1/500;
fun = @(k1,i) xg*sin(2*pi*F1*(k1-i)*T);
fun(14,0:14)

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