- #1
karbarca
- 1
- 0
that's my programe :
n=0;
T=2;
Ts=T/200;
t=-T/2:Ts:T/2;
for n=0:0.1:8;
V= 0;
V=V+8/pi^2*(-1)^n*sin(pi*(2*n+1)*t)/(2*n+1)^2;
pause
drawnow
plot(t,V);
end
When T=2, write a Matlab program to visualize the Fourier series (with animation) and investigate the largest error you may have when the number of terms is k (=the highest digit of your student ID number). For example, if you student number is M0001234572, then k=7. In this case, your sum
should run from n=0 to 8.
i need someone to check for me thanks
n=0;
T=2;
Ts=T/200;
t=-T/2:Ts:T/2;
for n=0:0.1:8;
V= 0;
V=V+8/pi^2*(-1)^n*sin(pi*(2*n+1)*t)/(2*n+1)^2;
pause
drawnow
plot(t,V);
end
When T=2, write a Matlab program to visualize the Fourier series (with animation) and investigate the largest error you may have when the number of terms is k (=the highest digit of your student ID number). For example, if you student number is M0001234572, then k=7. In this case, your sum
should run from n=0 to 8.
i need someone to check for me thanks