Fourier Series — Question 9

Question 9 – Solution

Strategy. Resolve the products into actual harmonics before inferring frequency content, period or energy.

Step 1: Resolve the two products. The identities $\cos x\cos 3x=(\cos 2x+\cos 4x)/2$ and $\sin x\sin 3x=(\cos 2x-\cos 4x)/2$ give $$\boxed{s_{a,b}(x)=\cos 3x+\frac{a+b}{2}\cos 2x+\frac{a-b}{2}\cos 4x.}$$ The mean and every sine coefficient are zero. The neighboring frequencies, often called sidebands, are $2$ and $4$, not frequency $1$ from the modulation alone.

Step 2: Invert the coefficient measurements. The linear equations give $$\boxed{a=C_2+C_4,\qquad b=C_2-C_4.}$$ Requiring $C_4=0$ and $C_2=d$ fixes $a=b=d$ uniquely. The two modulation terms reinforce the lower frequency and cancel the upper frequency when their amplitudes agree.

Step 3: Establish the least period. For $d=1/2$ the signal is $s=\cos 3x+\tfrac 12\cos 2x$. If $T>0$ is a period, translating this finite expansion and using orthogonality forces $3T$ and $2T$ each to be integer multiples of $2\pi$. Subtracting gives $T$ an integer multiple of $2\pi$. Conversely $2\pi$ works. Thus the fundamental period is $\boxed{2\pi}$. A nonzero first harmonic is sufficient for this period in a $2\pi$ polynomial, but is not necessary: frequencies $2$ and $3$ already enforce it.

Step 4: Measure power and peak amplitude. Orthogonality gives $$\boxed{P=\frac 12\left(1+\frac{(a+b)^2}{4}+\frac{(a-b)^2}{4}\right) =\frac 12+\frac{a^2+b^2}{4}.}$$ At $a=b=1/2$, $P=5/8$. Also $|s|\leq|\cos 3x|+\tfrac 12|\cos 2x|\leq 3/2$, with equality at $x=2k\pi$. For negative equality both cosines would have to be $-1$; the equations $3x=(2j+1)\pi$, $2x=(2k+1)\pi$ are incompatible by parity. Positive equality requires both to be $1$, giving exactly $x=2k\pi$.