Index in Positon 1 is invalid
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Jan Faber
am 29 Apr. 2022
Beantwortet: Jan Faber
am 29 Apr. 2022
Hello everyone, sadly I have an error that I personally don't really understand, maybe one of you will able to help me, so thanks a lot in advance for that.
So for a task for the university I'm trying to shorten the script a lot by creating a vector through a loop, but I constantly get the error that my Index is invalid.
My aim is to get a 0:14 long vector that contains the following equation for each row depending on n, but sadly it isn't working as planed, i would be really grateful if someone could help me to find my problem here.
for n=0:14
x1=5*sin(2*pi*F1*(kT-n*T));
x2=0.3*cos(2*pi*F2*(kT-n*T));
x=x1+x2;
xv(n,:)=x
end
Error: Index in position 1 is invalid. Array indices must be positive integers or logical values.
Sincerely Jan C. Faber
0 Kommentare
Akzeptierte Antwort
Voss
am 29 Apr. 2022
Indexing in MATLAB starts at 1, so when n == 0, referring to xv(n,:) gives you that error.
One way to fix it is to add 1 to n when you use it as an index, so that n = 0 through n = 14 correspond to rows 1 through 15 of xv:
for n=0:14
x1=5*sin(2*pi*F1*(kT-n*T));
x2=0.3*cos(2*pi*F2*(kT-n*T));
x=x1+x2;
% xv(n,:)=x
xv(n+1,:)=x
end
0 Kommentare
Weitere Antworten (2)
Steven Lord
am 29 Apr. 2022
MATLAB indexing is 1-based not 0-based.
The first element / row / column / etc. of an array in MATLAB is element / row / column / etc. 1 not 0.
0 Kommentare
Siehe auch
Kategorien
Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!