Fourier Series — Question 2

Question 2 – Solution

Strategy. Centering exposes the pulse width; translating back rotates each sine/cosine coefficient pair.

Step 1: Integrate the centered pulse. The area is $2\pi/3$, so the mean is $1/3$ and $a_0=2/3$. Put $x=c+t$. The sine terms odd in $t$ integrate to zero, leaving $$\boxed{a_n=\frac{2\sin(n\pi/3)}{\pi n}\cos nc,\qquad b_n=\frac{2\sin(n\pi/3)}{\pi n}\sin nc.}$$ These formulas also apply when the pulse crosses the chosen period boundary: integrate instead over a period centered at $c$.

Step 2: Identify genuine missing modes. The amplitude is $$R_n=\sqrt{a_n^2+b_n^2}=\frac{2|\sin(n\pi/3)|}{\pi n}.$$ Hence the entire harmonic vanishes exactly when $3$ divides $n$. For other $n$, $a_n$ alone can vanish because $\cos nc=0$, while $b_n\ne 0$. Discarding such a sine term would remove a real harmonic.

Step 3: Recover the circular center. For $n=1$ the prefactor is positive: $$(a_1,b_1)=\frac{\sqrt 3}{\pi}(\cos c,\sin c),\qquad \boxed{c=\operatorname{atan2}(b_1,a_1)\pmod{2\pi}.}$$ Consistency is exactly $a_1^2+b_1^2=3/\pi^2$ for real measured coefficients. Every pair on that circle corresponds to precisely one center modulo $2\pi$. Using only an ordinary arctangent of $b_1/a_1$ loses quadrant information.

Step 4: Separate location from amplitude. Every $R_n$ is independent of $c$, so amplitude measurements leave every center possible, even if all harmonics are known. At $c=\pi/4$, the pulse edges in $[-\pi,\pi]$ are $-\pi/12$ and $7\pi/12$. The sum is one between them, zero outside, and $1/2$ at each edge. The plotted partial sum uses both coefficient families; the finite oscillations do not alter these limiting edge averages.