Question

im trying to plot a graph in matlab for a fourier series

f(t) = 1 - t^2
-1 14 years ago

Answer 1

you state that

term = cos(pi*x*n)*(2)*((-1)^(n-1))/(pi*n)^2;

but x is not defined


So Matlab has no idea what to do with it

Answer 2

you probably mean


term = cos(pi*t*n)*(2)*((-1)^(n-1))/(pi*n)^2;

Answer 3

oh and t is a vector


so t^2 doesn't make any sense

Answer 4

type t.^2 instead to tell Matlab to square each element of this vector separately

Answer 5

i added the '.' and changed the variable to 't' and it now plots a parabola :) the question also says to plot on the same axis when N=6 but when i change the value of N the graph stays the same

Answer 6

that's because the first graph doesn't depend on N

Answer 7

notice how you have sumCos, but it's defined by itself but there's nothing defined for it beforehand

Answer 8

how does sumCos start?

Answer 9

i added some code so now its like this,

N=3;
t=-1:.01:1;
y=1-(t.^2);
plot(t,y)
axis([-1 1 0 1])

hold on

kmax = floor((N-1)/2);
sumCos = 0; sumSin = 0;

for n=1:N
term = cos(pi*t*n)*(2)*((-1)^(n-1))/(pi*n)^2;
sumCos = sumCos + term;
end

f = (2/3) + sumCos;
plot(t,f)

i dont know what the bit i added means though (sorry im not very good at matlab) they just included that in one of the examples so i thought i should try it

Answer 10

ok, that does compile and run correctly (or at least a plot comes up)

Answer 11

this is what comes up

Answer 12

yeh that is what i got also, should N=6 give a closer approximation?

Answer 13

it looks like it's approaching 1-t^2, but I'm not really familiar with fourier series

Answer 14

thats ok, thank you so much for all of your help :)

Answer 15

np