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Energy/power signals; periodicity of complex exponential

Short answer

Energy signal has finite energy, zero power; power signal has finite nonzero power, infinite energy. x[n]=e^(j(πn/3+π/4)) is periodic because ω₀/2π = 1/6 is rational. Fundamental period N=6 samples.

Define energy and power type discrete time signal. Check whether signal x[n] = e^(j(πn/3 + π/4)) is periodic or not. If it is periodic, state its periodic time.

Energy and power signals

  • Energy signal: a signal with finite energy, 0 < E < ∞, where E = Σ |x[n]|² (sum over all n). Its average power is zero. Example: x[n] = (½)ⁿ u[n], E = 1/(1 − ¼) = 4/3.
  • Power signal: a signal with finite non-zero average power, 0 < P < ∞, where P = lim(N→∞) 1/(2N+1) Σ(n=−N to N) |x[n]|². Its energy is infinite. Example: unit step, and every periodic signal.

A signal cannot be both; some signals (e.g. x[n] = n) are neither.

Periodicity of x[n] = e^(j(πn/3 + π/4))

A discrete exponential e^(jω₀n) is periodic only if ω₀/2π is a rational number, i.e. x[n + N] = x[n] for some integer N.

Here ω₀ = π/3, so

f₀ = ω₀/2π = (π/3)/(2π) = 1/6, which is rational.

So the signal is periodic. Check: x[n + 6] = e^(j(πn/3 + 2π + π/4)) = x[n].

The fundamental period is N = 1/f₀ = 6 samples (the phase π/4 does not affect the period).

This signal has |x[n]| = 1, so it is a power signal with P = 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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