Show that the Fourier series for the square wave function $$f(t)=\begin{cases}-1 & -\frac{T}{2}\leq t \lt 0, \\ +1 & \ \ \ \ 0 \leq t \lt \frac{T}{2}\end{cases}$$ is $$f(t)=\frac{4}{\pi}\left(\sin\left(\frac{2\pi t}{T}\right)+\frac{\sin(\frac{6\pi t}{T})}{3}+\frac{\sin(\frac{10\pi t}{T})}{5}+\cdots\right)$$

I understand that the general Fourier series expansion of the function $f(t)$ is given by $$f(t)=\frac{a_0}{2}+ \sum_{r=1}^{r=\infty}\left(a_r\cos\left(\frac{2\pi r t}{T}\right)+b_r\sin\left(\frac{2\pi r t}{T}\right)\right)$$ But what happened to the $$\frac{a_0}{2}$$ term at the beginning of

$$f(t)=\frac{a_0}{2}+ \sum_{r=1}^{r=\infty}\left(a_r\cos\left(\frac{2\pi r t}{T}\right)+b_r\sin\left(\frac{2\pi r t}{T}\right)\right)$$ for the general Fourier series expansion?

From advice, I've been told that the constant term can be found by integrating $f(t)$ such that $$\int_{t=-\frac{T}{2}}^{t=\frac{T}{2}}f(t)\mathrm{d}t= \int_{t=-\frac{T}{2}}^{t=\frac{T}{2}} \left(\frac{a_0}{2}+ \sum_{r=1}^{r=\infty} (a_r\cos\frac{2\pi r t}{T}+b_r\sin\frac{2\pi r t}{T})\right)\mathrm{d}t$$ from here could someone please show me the steps involved in showing that $$\frac{a_0}{2}=0$$

Many thanks,



First, your function considered on each of the intervals $[0,T/2[$ and $[-T/2,0[$ separately, is just a constant function. It's the whole that is non-constant. So, when you integrate, since you can separate out your integration over the different integration intervals, on them, you are just integrating a constant function.

So, $f$ didn't disappear, $f$ is just equal to $1$ over the interval $[0,T/2[$.

Second, your function is also odd. The constant term is found by simply integrating the function over an interval symmetric around the origin.

$$a_0=\frac{2}{T}\int_{t=-\frac{T}{2}}^{t=\frac{T}{2}}f(t)\,\mathrm{d}t=\frac{2}{T}\int_{t=-\frac{T}{2}}^{t=0}f(t)\,\mathrm{d}t+\frac{2}{T}\int_{t=0}^{t=\frac{T}{2}}f(t)\,\mathrm{d}t \\ =\frac{2}{T}\int_{t=-\frac{T}{2}}^{t=0}-1 \, \,\mathrm{d}t+\frac{2}{T}\int_{t=0}^{t=\frac{T}{2}} 1 \, \,\mathrm{d}t = 0 \; .$$

Therefore the integral is zero.

EDIT: $$\begin{eqnarray}\int_{t=-\frac{T}{2}}^{t=\frac{T}{2}}f(t)\,\mathrm{d}t & = & \int_{t=-\frac{T}{2}}^{t=\frac{T}{2}} \left(\frac{a_0}{2}+ \sum_{r=1}^{r=\infty} a_r\cos\frac{2\pi r t}{T}+b_r\sin\frac{2\pi r t}{T}\right)\,\mathrm{d}t\\ & = & \int_{t=-\frac{T}{2}}^{t=\frac{T}{2}} \frac{a_0}{2}\,\mathrm{d}t+ \sum_{r=1}^{r=\infty} a_r \int_{t=-\frac{T}{2}}^{t=\frac{T}{2}}\cos\frac{2\pi r t}{T}\,\mathrm{d}t+\sum_{r=1}^{r=\infty}b_r\int_{t=-\frac{T}{2}}^{t=\frac{T}{2}} \sin\frac{2\pi r t}{T}\,\mathrm{d}t \\ & = & \int_{t=-\frac{T}{2}}^{t=\frac{T}{2}} \frac{a_0}{2}\,\mathrm{d}t+ \sum_{r=1}^{r=\infty} a_r \cdot 0+\sum_{r=1}^{r=\infty}b_r\cdot 0 \\ & = & \int_{t=-\frac{T}{2}}^{t=\frac{T}{2}} \frac{a_0}{2}\,\mathrm{d}t \\ & = & \frac{a_0}{2}\cdot T \end{eqnarray}$$

  • $\begingroup$ Thanks for your reply, I'm still a bit confused could you explain in a bit more detail? I don't understand why "Therefore the integral is zero." By "constant term" are you referring to $$\frac{a_0}{2}$$ Why is $f(t)$ equal to 1? I know that $f(t)$ is 1 on that interval but $f(t)$ appears in the integrand. Sorry this is really simple to you, it isn't simple to me. $\endgroup$
    – BLAZE
    Sep 11 '15 at 4:32
  • 2
    $\begingroup$ No problem. I'll further elaborate my answer. $\endgroup$ Sep 11 '15 at 5:44
  • $\begingroup$ Thanks, that makes more sense, only thing I still don't understand is why $\frac{a_0}{2}=\frac{2}{T}\int_{t=-\frac{T}{2}}^{t=\frac{T}{2}}f(t)\mathrm{d}t$ or put in another way; why can the constant term $$\frac{a_0}{2}$$ be found by simply integrating the function over an interval symmetric around the origin? $\endgroup$
    – BLAZE
    Sep 11 '15 at 19:54
  • 2
    $\begingroup$ Look at your Fourier series for $f$. Integrate both sides. Because the $\sin$ and $\cos$ get integrated over a (or several) full period(s), they integrate to zero. But not the constant term. $\endgroup$ Sep 12 '15 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.