# Find the following fourier series in trigonometric form

1. Nov 23, 2014

### iRaid

1. The problem statement, all variables and given/known data
Find the following fourier series in trigonometric form.

2. Relevant equations
$$y(t)=a_0+\sum\limits_{n=1}^{\infty} a_n cos(n\omega_{0}t)+b_n sin(n\omega_{0}t)$$

3. The attempt at a solution
The graph above is represented by the function:
$$x(t) = \left\{ \begin{array}{ll} -t+1 & \quad 0 < t \leq 1 \\ 0 & \quad 1 < t \leq 2 \end{array} \right.$$

$T_0=2$ and $\omega_{0}=\pi$

To find $a_0$:
$$a_0=\frac{1}{T_{0}}\int_{0}^{T_{0}} x(t)dt = \frac{1}{2}\int_{0}^{1} (-t+1)dt = \frac{1}{4}$$
The integral of 0 can be ignored since it equals 0.

To find $a_n$:
$$a_n=\frac{2}{T_{0}}\int_{0}^{T_0}x(t) cos(n\omega_{0}t)dt=\frac{2}{2}\int_{0}^{1} (-t+1)cos(\pi nt)dt=\int_{0}^{1} -tcos(\pi nt)dt +cos(\pi nt)dt$$

To find $b_n$
$$b_n=\frac{2}{T_{0}}\int_{0}^{T_{0}} x(t)sin(n\omega_{0}t)dt = \frac{2}{2}\int_{0}^{1} (-t+1)sin(\pi nt)dt=\int_{0}^{1} -tsin(\pi nt)+sin(\pi nt)dt$$

I am assuming that the $T_0$ in these equations are for the period of each function, since the function x(t)=0 is from 1 to 2, I only use 1 for $T_0$ for the function x(t)=-t+1.

Now I can find these integrals for $a_n$ and $b_n$ using integration by parts, but I'm wondering if I am going along with this correctly and what I will do once I find it (the summation really confuses me).

Thank you for any help!

Last edited: Nov 23, 2014
2. Nov 23, 2014

### ehild

It looks right, go head. Find an bn, in terms of n. No need to sum, the sum must be the original function:)

3. Nov 23, 2014

### Ray Vickson

Except at x = 0 and x = 2, where the sum will be 1/2 instead of 1 and 0, respectively.

4. Nov 23, 2014

### iRaid

The original function is continuous in negative and positive direction, sorry I should of mentioned that.

5. Nov 23, 2014

### LCKurtz

Wrong assumption. The periodic extension of that function has period $T=2$. Even though the integral for the coefficients is zero for half the period, you still have to use the full period in the trig functions. That is, for example, $sin(\frac {n \pi t}{T}) = sin(\frac {n \pi t}{2})$ and similarly for the cosine.

6. Nov 23, 2014

### ehild

The function must be periodic to have a Fourier series. So it is not continuous at x=2k (for integer k). And Ray is right, the Fourier series returns 1/2 at x=2k instead of f(2k)=0, as the original function.

7. Nov 23, 2014

### vela

Staff Emeritus
The OP was referring to the upper limit of the integral, which can be set to 1 because x(t) is 0 from t=1 to t=2. What he wrote confused me too. The rest of the calculations properly use $T_0=2$.

8. Nov 23, 2014

### LCKurtz

Not sure what you mean by that. If you actually want the periodic extension to represent a continuous function for all $t$ you would have to take the even extension of your function, use $T=4$, and do a half range cosine expansion. Unless you were given instructions in the problem you haven't told us, I doubt that is what is wanted.

9. Nov 23, 2014

### LCKurtz

I don't think so. I don't see $\sin(\frac{n\pi t} 2)$ in those integrands.

10. Nov 23, 2014

### vela

Staff Emeritus
If the period is T=2, then $\omega=2\pi/T = \pi$, so the argument of the trig functions is $n\pi t$.

11. Nov 23, 2014

### LCKurtz

Arghh. You're right. @iRaid: Ignore my posts, except (maybe) post #8.

12. Nov 23, 2014

### iRaid

What I mean is that the function repeats itself forever, it doesn't just go from 0 to 2. What would I end up doing once I figure out the 2 integration by parts? The summation really confuses me and I never understand how examples of these problems get rid of sine or cosine. The examples will say something like "clearly this equals 0 so the sine term can be excluded" but it is not clear to me.

13. Nov 24, 2014

### ehild

You will get an and bn, and then write up the Fourier series as
$y(t)=a_0+\sum\limits_{n=1}^{\infty} a_n cos(n\omega_{0}t)+b_n sin(n\omega_{0}t)$
Perhaps, you find a formula to write up the general terms.
Some of the integrals turn out to be zero if you substitute the boundaries. If you get sin(nπt), for example, it is zero both at t=0 and t=1..Just do the integrals you obtained in the OP and you will see.