Fourier series - DC component, integration problem

[DrOnline]

Homework Statement

Find the Fourier series representation of:

f(t)={-t , -∏<t<0
f(t)={0 , 0<t<∏

This is a piecewise function.

T=2∏ (the period)

Homework Equations

a_{0}=\frac{2}{T}*\int_0^T f(t),dt

The Attempt at a Solution

I need help only with calculating the DC component. Once I get that working, I can continue on my own. Tried multiple times, I keep getting pi/2.

The correct answer should be pi/4.

This is my graph:

Is this graph correct?

I consider the area of the function from -pi to 0 identical if I change it from:

f(t)=-t -∏<t<0

to

f(t)=t 0<t<∏

Is this permitted? Surely it's the same result. Seems alright to me

a_{0}=\frac{2}{2\pi}*\int_0^∏ -t,dt

a_{0}=\frac{1}{\pi}*(\frac{\pi^2}{2}-0)

a_{0}=\frac{\pi}{2}

And the book says it should be pi/4.

What am I doing wrong? Is my graph correct? Is my integration correct? Is the book's solution wrong?

 

[DrOnline]

I think I figured it out... my a₀ seems correct to be ∏/2,

but when I type out the actual Fourier series, the formula is:

a₀/2, making it ∏/4.

If anybody wants to confirm my thinking, that would be helpful!

 

[LCKurtz]

Textbooks are not consistent on the formula for $a_0$ because some formulas assume the constant term in the series is $a_0$ and some use $\frac {a_0} 2$, which causes the formula for $a_0$ to differ by $2$. As long as you are consistent, you should be OK. But there is no need to go to a different interval than that where the function is given. If the constant term of the series is $a_0$ and the period is $2p$ then$$
a_0 = \frac 1 {2p}\int_{-p}^p f(t)\, dt = \frac 1 {2\pi}\left(\int_{-\pi}^0 -t\, dt
+\int_{-\pi}^0 0\, dt\right) = \frac \pi 4$$

 

[DrOnline]

Thank you for helping me with this!

I think perhaps an error has snuck into your maths though, because you write:

a_{0}=\frac{1}{2p}...=\frac{1}{2\pi}...

But p=2∏, so it should be a_{0}=\frac{1}{4\pi} using your formula.

Can you comment on this?

I understand how there are different formulas though, as you say, when I created the question I hadn't realized that I had to divide it by 2 again, but when I did I wanted somebody to confirm my logic.

I also understand, of course, that I don't have to tinker with the range and function, but that saved me having to deal with negative values etc, I should stick with the actual function though, I have to be at a level where that kind of thing doesn't phase me.

 

[LCKurtz]

. If the constant term of the series is $a_0$ and the period is $2p$ then$$
a_0 = \frac 1 {2p}\int_{-p}^p f(t)\, dt = \frac 1 {2\pi}\left(\int_{-\pi}^0 -t\, dt
+\int_{-\pi}^0 0\, dt\right) = \frac \pi 4$$

Thank you for helping me with this!

I think perhaps an error has snuck into your maths though, because you write:

a_{0}=\frac{1}{2p}...=\frac{1}{2\pi}...

But p=2∏, so it should be a_{0}=\frac{1}{4\pi} using your formula.

Can you comment on this?

No, I said the period was $2p$ so $p=\pi$ in your case. And when you do the integral you will have a $t^2$ in the numerator with a limit of $-\pi$ which will give a $\pi^2$ in the numerator.

 

[DrOnline]

Yes of course, I was too quick in reading there, thanks again!