Skip to main content

cos(πn/3) with frequency increased by 2π

Short answer

Increasing the frequency by 2π gives cos((π/3+2π)n)=cos(πn/3) because 2πn is an integer multiple of 2π. The signal is unchanged; DT sinusoids whose frequencies differ by 2π are identical.

The signal x[n] = cos(πn/3) is a periodic signal. What happens if the fundamental frequency of this signal is increased by 2π?

x[n]=cos⁡(πn3)x[n] = \cos\left(\frac{\pi n}{3}\right) has ω0=π/3\omega_0 = \pi/3 and period N=6N = 6. Increase the frequency by 2π2\pi:

y[n]=cos⁡[(π3+2π)n]=cos⁡(πn3+2πn)=cos⁡(πn3)=x[n]\begin{aligned} y[n] &= \cos\left[\left(\frac{\pi}{3} + 2\pi\right)n\right] = \cos\left(\frac{\pi n}{3} + 2\pi n\right) \\ &= \cos\left(\frac{\pi n}{3}\right) = x[n] \end{aligned}

because 2πn2\pi n is an integer multiple of 2π2\pi for every integer nn.

Nothing changes: the new signal is identical to the original, with the same samples and the same period N=6N = 6. DT sinusoids whose frequencies differ by 2π2\pi (or any multiple of 2π2\pi) are the same signal. So the frequency of a DT signal is unique only over an interval of length 2π2\pi (e.g. −π<ω≤π-\pi < \omega \le \pi or 0≤ω<2π0 \le \omega < 2\pi), unlike CT signals, where a higher frequency always oscillates faster. This is the basis of aliasing in sampling.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Signals and Systems chapter 1 questions →