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So I have a problem that says "Consider a triangle wave that goes from a minimum value of O volts to a maximum value of V. It has a period T, and is a maximum value at t = 0". And I want to show that $c_n = \frac {a_n}{2} $.

I calculated $a_n = \frac {2V}{\pi^2n^2} ( 1 - (-1)^n)$, which I assume is accurate.

However, solving for $c_n$ I set up the integral as 2 times one leg of the triangle:

$$c_n=2 \frac 1T \int_0^\frac T2 \frac {2V}{T}t e^{\frac {-2\pi int}{T}}\, dt $$

And, after much simplification, I get:

$$c_n = \frac {\cos({\pi n}) (V\pi in - V) -V}{\pi^2 n^2}$$

This appears overwhelmingly far from $\frac {a_n}{2}$, theres even an imaginary part!. Did I make an obvious mistake in my set-up, or is there a way to simplify my expression for $c_n$?

ZacharyC
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1 Answers1

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You didn't define the notation, but I guess you are trying to write the function as $$ f(t)=\sum_{n\geq 0} a_n \cos (2\pi n t/T)+\sum_{n>0} b_n \sin(2\pi n t/T)=\sum_{-\infty<n\leq \infty} c_n e^{2\pi i n t/T} $$

Then you already figure out $b_n=0$ by parity and found $a_n$ by integrating only over half a period and multiplying by two. However this only works because $f$ and $\cos$ are even functions! As $e^{-2\pi i n t/T}$ is not even you have to integrate over the whole period.

$$ c_n=\frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-2\pi i nt/T}dt=\frac{1}{T}\int_{-T/2}^{0}\frac{2V}{T}(t+T/2)e^{-2\pi i nt/T}dt+\frac{1}{T}\int_{-T/2}^{0}\frac{2V}{T}(T/2-t)e^{-2\pi i nt/T}dt $$

I leave you to fill out the details, but it follows from there. :). By the way $\cos(\pi n)=e^{i\pi n}=(-1)^n$.

David Jaramillo
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  • Thank you! That makes sense, I forgot to consider whether the exponential was even. I redid it and it came out right. – ZacharyC Oct 03 '19 at 19:54
  • Hi can you please help me with is question: https://math.stackexchange.com/q/3380406/394202 – Soumee Oct 04 '19 at 17:52