Problems with declaring the variable right

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Jan Faber
Jan Faber on 15 May 2022
Commented: Voss on 15 May 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

Accepted Answer

Voss
Voss on 15 May 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 Comments
Voss
Voss on 15 May 2022
Sounds good. Let me know if you have any questions about it.
-_

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More Answers (1)

Torsten
Torsten on 15 May 2022
Edited: Torsten on 15 May 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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