# Fourier Series: Triangular waveform, Ao calc going wrong somewhere

• Jan 7th 2010, 09:12 AM
emmv
Fourier Series: Triangular waveform, Ao calc going wrong somewhere
The question asked is calculate Fourier series for waveform
$f(t) = 5 + \frac{5}{\pi}\theta for -\pi < \theta < 0$
$f(t) = 5 - \frac{5}{\pi}\theta for -\pi < \theta < 0$

I calculate Ao to be 5 but as you are given the fourier series with the question I can see that that is incorrect and it should be $\frac{5}{2}$

Here is my calculation as I have done it, could someone please tell me where I am making a mistake? Thanks Ed.

$A_o = \frac{1}{2\pi}
\int_{-\pi}^{\pi} \! f\theta \, d\theta$

$A_o = \frac{1}{2\pi}
\int_{-\pi}^{0} \! 5 + \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta + \frac{1}{2\pi}\int_{0}^{\pi} \! 5 - \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta$

$A_o = \frac{1}{2\pi}\displaystyle \left(\int_{-\pi}^{0} \! 5 d\theta + \frac{5}{\pi} \int_{-\pi}^{0} \! \theta \, d\theta + \int_{0}^{\pi} \! 5 d\theta - \frac{5}{\pi} \int_{0}^{\pi} \! \theta \, d\theta \right)$

$A_o = \frac{1}{2\pi}\displaystyle \left( 5[\theta]^0_{-\pi} + \frac{5}{2\pi} [\theta^2]^0_{-\pi} \, + 5[\theta]^\pi_0 - \frac{5}{2\pi}[\theta^2]^\pi_o \right)$

$A_o = \frac{1}{2\pi}\displaystyle \left( 5\pi + 5\pi \right)$

$The \frac{5}{2\pi}[\theta^2] \ cancelling \ each \ other \ out$

$A_o = \frac{10\pi}{2\pi} = 5$
• Jan 7th 2010, 02:29 PM
Drexel28
Quote:

Originally Posted by emmv
The question asked is calculate Fourier series for waveform
$f(t) = 5 + \frac{5}{\pi}\theta for -\pi < \theta < 0$
$f(t) = 5 - \frac{5}{\pi}\theta for -\pi < \theta < 0$

I calculate Ao to be 5 but as you are given the fourier series with the question I can see that that is incorrect and it should be $\frac{5}{2}$

Here is my calculation as I have done it, could someone please tell me where I am making a mistake? Thanks Ed.

$A_o = \frac{1}{2\pi}
\int_{-\pi}^{\pi} \! f\theta \, d\theta$

$A_o = \frac{1}{2\pi}
\int_{-\pi}^{0} \! 5 + \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta + \frac{1}{2\pi}\int_{0}^{\pi} \! 5 - \displaystyle \left(\frac{5}{\pi}\right)\theta \, d\theta$

$A_o = \frac{1}{2\pi}\displaystyle \left(\int_{-\pi}^{0} \! 5 d\theta + \frac{5}{\pi} \int_{-\pi}^{0} \! \theta \, d\theta + \int_{0}^{\pi} \! 5 d\theta - \frac{5}{\pi} \int_{0}^{\pi} \! \theta \, d\theta \right)$

$A_o = \frac{1}{2\pi}\displaystyle \left( 5[\theta]^0_{-\pi} + \frac{5}{2\pi} [\theta^2]^0_{-\pi} \, + 5[\theta]^\pi_0 - \frac{5}{2\pi}[\theta^2]^\pi_o \right)$

$A_o = \frac{1}{2\pi}\displaystyle \left( 5\pi + 5\pi \right)$

$The \frac{5}{2\pi}[\theta^2] \ cancelling \ each \ other \ out$

$A_o = \frac{10\pi}{2\pi} = 5$

I did not read all of your stuff. But, whenever I have seen Fourier expansions it is $\frac{a_0}{2}$

Fourier series - Wikipedia, the free encyclopedia

Could that be the reason? I hope you don't take this as an insult to your intelligence.