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Period of 3sin(7πn/9 − 2) + cos(πn/13 + 2)

Short answer

First term has angular frequency 7π/9 giving N1=18; second term ω=π/13 gives N2=26. Fundamental period N=lcm(18,26)=234 samples, fundamental frequency ω0=2π/234=π/117 rad/sample.

Determine the fundamental period and fundamental frequency of the periodic signal x[n] = 3 sin((7π/9)n − 2) + cos((π/13)n + 2)

A sum of DT sinusoids is periodic if each ωi/2π\omega_i/2\pi is rational; N=LCM(N1,N2)N = \text{LCM}(N_1, N_2). Phase shifts do not change the period.

Term 1: 3sin⁡(7π9n−2)3\sin\left(\frac{7\pi}{9}n - 2\right), ω1=7π9\omega_1 = \frac{7\pi}{9}

ω12π=718  ⇒  N1=2πm7π/9=18m7  ⇒  N1=18 (m=7)\frac{\omega_1}{2\pi} = \frac{7}{18} \;\Rightarrow\; N_1 = \frac{2\pi m}{7\pi/9} = \frac{18m}{7} \;\Rightarrow\; N_1 = 18\ (m=7)

Term 2: cos⁡(π13n+2)\cos\left(\frac{\pi}{13}n + 2\right), ω2=π13\omega_2 = \frac{\pi}{13}

ω22π=126  ⇒  N2=26 (m=1)\frac{\omega_2}{2\pi} = \frac{1}{26} \;\Rightarrow\; N_2 = 26\ (m=1)

Fundamental period:

N=LCM(18,26)=LCM(2⋅32, 2⋅13)=2⋅9⋅13=234N = \text{LCM}(18, 26) = \text{LCM}(2\cdot 3^2,\ 2\cdot 13) = 2\cdot 9\cdot 13 = 234

Fundamental frequency:

ω0=2πN=2π234=π117 rad/sample,f0=1234 cycles/sample\omega_0 = \frac{2\pi}{N} = \frac{2\pi}{234} = \frac{\pi}{117}\ \text{rad/sample}, \qquad f_0 = \frac{1}{234}\ \text{cycles/sample}

Answer: N=234N = 234 samples; ω0=π/117\omega_0 = \pi/117 rad/sample (f0=1/234≈0.00427f_0 = 1/234 \approx 0.00427 cycles/sample).

✓ Answer checked by Claude Opus 5.5 (high effort)

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