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Periodicity of sin 6πt, cos 8t and their sum

Short answer

x(t)=sin6πt: T₁=2π/(6π)=1/3 s (periodic). y(t)=cos8t: T₂=2π/8=π/4 s (periodic). Their sum is not periodic because T₁/T₂ = (1/3)/(π/4) = 4/(3π) is irrational, so no common period exists.

x(t) and y(t) are continuous time signals defined as: x(t) = sin 6πt and y(t) = cos 8t. Determine whether: (i) x(t) is periodic, (ii) y(t) is periodic, (iii) x(t) + y(t) is periodic. If periodic, find the fundamental time period of x(t), y(t) and x(t) + y(t).

A CT sinusoid with angular frequency ω\omega is always periodic with T=2π/ωT = 2\pi/\omega. A sum is periodic only if T1/T2T_1/T_2 is rational.

(i) x(t)=sin⁡6πtx(t) = \sin 6\pi t, ω1=6π\omega_1 = 6\pi:

T1=2π6π=13 s⇒periodicT_1 = \frac{2\pi}{6\pi} = \frac{1}{3}\ \text{s} \quad\Rightarrow\quad \text{periodic}

(ii) y(t)=cos⁡8ty(t) = \cos 8t, ω2=8\omega_2 = 8:

T2=2π8=π4 s≈0.785 s⇒periodicT_2 = \frac{2\pi}{8} = \frac{\pi}{4}\ \text{s} \approx 0.785\ \text{s} \quad\Rightarrow\quad \text{periodic}

(iii) x(t)+y(t)x(t) + y(t):

T1T2=1/3π/4=43π(irrational)\frac{T_1}{T_2} = \frac{1/3}{\pi/4} = \frac{4}{3\pi} \quad\text{(irrational)}

No integers m,km, k give mT1=kT2mT_1 = kT_2, so there is no common period.

Answer: x(t)x(t) periodic, T1=1/3T_1 = 1/3 s; y(t)y(t) periodic, T2=π/4T_2 = \pi/4 s; x(t)+y(t)x(t) + y(t) is not periodic (aperiodic).

✓ Answer checked by Claude Opus 5.5 (high effort)

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