in finding the coefficient of Fourier Series, $a_0, a_n, b_n$. We integrate the periodic function $f(t)$ over the period $T$. That is

$$\frac1T\int^T_0 f(t)\ dt$$ $$\frac2T\int^T_0 f(t)\cos(n\omega t)\ dt$$ $$\frac2T\int^T_0 f(t)\sin(n\omega t)\ dt$$

However, for some function, $a_n$ and $b_n$ can be calculate by just integrate over half of period and multiply the result by 2 that means

$$\frac4T\int^{\frac T2}_0 f(t)\cos(n\omega t)\ dt$$ $$\frac4T\int^{\frac T2}_0 f(t)\sin(n\omega t)\ dt$$

However, I don't know if this approach can be used in all situation or just some special cases. If it can be used only in some special cases, then what is the requirement to use it.

Moreover, for some periodic function having the same positive and negative part for example the sinusoid. Does $a_0$ of this function always return the result of zero?


This can be done when $f$ has some symmetry around $T/2$, the midpoint of the interval $[0,T]$.

If $f(t)=f(T-t)$, then $$\begin{align} \int^T_0 f(t)\cos(n\,\omega\,t)\,dt&=\int^{T/2}_0 f(t)\cos(n\,\omega\,t)\,dt+\int^T_{T/2} f(t)\cos(n\,\omega\,t)\,dt\\ &=\int^{T/2}_0 f(t)\cos(n\,\omega\,t)\,dt+\int^{T/2}_{0} f(T-t)\cos(n\,\omega\,(T-t))\,dt\\ &=2\int^{T/2}_0 f(t)\cos(n\,\omega\,t)\,dt. \end{align}$$ A similar argument shows that $b_n=0$.

I leave to you the case when $f(T-t)=-f(t)$.


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.