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Periodic signals; condition for cos(2πf₀n + θ)

Short answer

A periodic signal repeats after a fixed interval; aperiodic does not. For cos(2πf0 n+θ) to be periodic, we need x[n+N]=x[n] ⇒ 2πf0 N = 2πk (k integer). Hence f0 = k/N must be a rational number. The fundamental period N is the denominator in lowest terms.

What is periodic and aperiodic signal? Find the necessary condition for the signal x[n] = cos(2πf₀n + θ) to be periodic.

  • Periodic signal: repeats itself after a fixed interval. CT: x(t+T)=x(t)x(t+T) = x(t) for all tt; DT: x[n+N]=x[n]x[n+N] = x[n] for all nn, NN a positive integer. The smallest such TT or NN is the fundamental period. Examples: sin⁡t\sin t, cos⁡(πn/4)\cos(\pi n/4).
  • Aperiodic signal: does not repeat for any finite period. Examples: e−tu(t)e^{-t}u(t), δ[n]\delta[n], cos⁡(2n)\cos(2n).

Condition for x[n]=cos⁡(2πf0n+θ)x[n] = \cos(2\pi f_0 n + \theta) to be periodic

For period NN:

x[n+N]=cos⁡(2πf0n+2πf0N+θ)\begin{aligned} x[n+N] &= \cos(2\pi f_0 n + 2\pi f_0 N + \theta) \end{aligned}

This equals x[n]x[n] for all nn only if the extra angle is a multiple of 2π2\pi:

2πf0N=2πk  ⇒  f0=kN,k,N integers2\pi f_0 N = 2\pi k \;\Rightarrow\; f_0 = \frac{k}{N}, \quad k, N \text{ integers}

So x[n]x[n] is periodic only if f0f_0 is a rational number. If f0=k/Nf_0 = k/N in lowest terms, the fundamental period is NN.

Example: f0=3/10f_0 = 3/10 gives N=10N = 10. f0=1/(2π)f_0 = 1/(2\pi) (i.e. cos⁡n\cos n) is irrational, so the signal is aperiodic.

✓ Answer checked by Claude Opus 5.5 (high effort)

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