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Even/odd DT signals and periodicity

Short answer

Even: x[-n]=x[n]; odd: x[-n]=-x[n]. A DT signal is periodic with integer period N if x[n+N]=x[n]. For e^(jω0 n), periodicity requires ω0/(2π) rational; if ω0/(2π)=m/N in lowest terms, the fundamental period is N.

Define even and odd discrete time signal. Derive the periodicity condition for discrete time signal.

Even and odd DT signals

  • Even: x[−n]=x[n]x[-n] = x[n] for all nn (symmetric about n=0n = 0). Examples: cos⁡(πn/3)\cos(\pi n/3), {2,1,5‾,1,2}\{2, 1, \underline{5}, 1, 2\}.
  • Odd: x[−n]=−x[n]x[-n] = -x[n] for all nn (antisymmetric, x[0]=0x[0] = 0). Examples: sin⁡(πn/3)\sin(\pi n/3), {−1,0‾,1}\{-1, \underline{0}, 1\}.

Any sequence has xe[n]=12(x[n]+x[−n])x_e[n] = \frac{1}{2}(x[n]+x[-n]) and xo[n]=12(x[n]−x[−n])x_o[n] = \frac{1}{2}(x[n]-x[-n]).

Periodicity condition

A discrete-time signal is periodic with period NN (a positive integer) if x[n+N]=x[n]x[n+N] = x[n] for all nn.

For x[n]=ejω0nx[n] = e^{j\omega_0 n}:

x[n+N]=ejω0(n+N)=ejω0n ejω0N\begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned}

For x[n+N]=x[n]x[n+N] = x[n] we need

ejω0N=1  ⇒  ω0N=2πm,m an integere^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer} ω02π=mN=rational number\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}

So ejω0ne^{j\omega_0 n} is periodic only if ω0/2π\omega_0/2\pi is a rational number. The fundamental period is the smallest positive integer N=2πm/ω0N = 2\pi m/\omega_0, taking mm as the smallest integer that makes NN an integer (with mm and NN having no common factor). The same condition holds for cos⁡(ω0n+θ)\cos(\omega_0 n+\theta) and sin⁡(ω0n+θ)\sin(\omega_0 n+\theta), because they are sums of such exponentials.

This differs from continuous time, where ejω0te^{j\omega_0 t} is periodic for every ω0≠0\omega_0 \neq 0 (period 2π/∣ω0∣2\pi/|\omega_0|). In discrete time nn takes only integer values, so NN must be an integer.

✓ Answer checked by Claude Opus 5.5 (high effort)

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