Why is the Fourier coefficent calculated like this?

Hello!

I have a function that is periodic. T = 2. Now I need to find a fourier series of that function, so I need to find coefficent ##a_0 a_k b_k ##

for a0 I have this formula (this is the one we used in class)



Now my x(t) = (-t+1). and when I plug in all of the values I have in the formula I get that ##a_0 = 2 ## but the solution says it should be 1. I checked the solution and what they did is they multiplied the 2/T with 2. I dont understand why they did this? Also they integrated from 0 to 1, I dont understand this either; Shouldnt the Integral be from -1 to 1 ?

For reference here is how the graph of the function looks like.



Thanks in advance!

arhzz said:

Homework Statement: Find the Fourier coefficients
Relevant Equations: Fourier Analysis
View attachment 339696

Now my x(t) = (-t+1).

This is the formula for only the right half of the sawtooth function. IOW, it is correct only for ##t \in [0, 1]##. You also need a formula for the left half, the part in the interval [-1, 0]. Since the formula changes, you'll need separate integrals for each coefficient.

arhzz said:

and when I plug in all of the values I have in the formula I get that ##a_0 = 2 ## but the solution says it should be 1. I checked the solution and what they did is they multiplied the 2/T with 2. I dont understand why they did this? Also they integrated from 0 to 1, I dont understand this either; Shouldnt the Integral be from -1 to 1 ?

For reference here is how the graph of the function looks like.

View attachment 339697

Thanks in advance!

Mark44 said:

This is the formula for only the right half of the sawtooth function. IOW, it is correct only for ##t \in [0, 1]##. You also need a formula for the left half, the part in the interval [-1, 0]. Since the formula changes, you'll need separate integrals for each coefficient.

Oh wait, this describes only the right part. So I need too calculate the coeffient 2 times, one from the intervall from [0,1] and for the intervall [-1 0] again, and than I have the coefficient a0? But is it possible to calculate the coefficient with 1 Integral, because they did it with 1 Integral? Also when I find the 2 different coefficients with two seperate integrals, should I add them to get a0?

arhzz said:

So I need too calculate the coeffient 2 times, one from the intervall from [0,1] and for the intervall [-1 0] again, and than I have the coefficient a0? But is it possible to calculate the coefficient with 1 Integral, because they did it with 1 Integral?

Because of the symmetry, you can get by with one integral, by just multiplying the integral by 2. IOW, this will work:
$$a_0 = \frac 2 2 \int_0^1 1 - t ~dt$$
In addition, this integral can be evaluated by inspection, since it represents twice the area of a triangle whose base is 1 and whose altitude is also 1.

arhzz said:

Also when I find the 2 different coefficients with two seperate integrals, should I add them to get a0?

For the other coefficients in your Fourier series, you might have to use two separate integrals for each coefficient, and add them together. I haven't worked it out, though. It sometimes happens that half of the coefficients are zero, so that might make the work a bit easier.

Mark44 said:

Because of the symmetry, you can get by with one integral, by just multiplying the integral by 2. IOW, this will work:
$$a_0 = \frac 2 2 \int_0^1 1 - t ~dt$$
In addition, this integral can be evaluated by inspection, since it represents twice the area of a triangle whose base is 1 and whose altitude is also 1.
For the other coefficients in your Fourier series, you might have to use two separate integrals for each coefficient, and add them together. I haven't worked it out, though. It sometimes happens that half of the coefficients are zero, so that might make the work a bit easier.

Just to report I have solved the problem and it was just like you said. For the coefficients you need to split it up with the 2 intervals and simply integrate, thankfully a lot cancels out since you plug in 0 for a lot of the function values, it is a bit more integrating but the only difference is the (-t+1) and (t+1) for the different intervals as well as the limits but you pretty much need to integrate once and just adjust the sign, and plug in the appropriate limit.

Thanks for the great help as always!