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Determine which of the following signals is periodic. If a signal is periodic, determine its period.

(a) x[n] = eʲ⁽²πⁿ/⁵⁾
(b) x[n] = sin(πn/19)
(c) x[n] = n eᶦπⁿ
(d) x[n] = eʲⁿ

User Xandros
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Final answer:

Out of the four signals, (a) x[n] = eʹ⁺⁴(˙πn/˕) and (b) x[n] = sin(πn/19) are periodic, with periods of 5 and 38 respectively. Signals (c) and (d) are not periodic.

Step-by-step explanation:

To determine which of the following signals is periodic and, if they are, to determine their period, we will examine each one individually:

  • a) x[n] = eʹ⁺⁴(˙πn/˕): This signal is periodic with a period of 5. This is because when the input n is increased by 5, the value inside the exponent increases by $2\pi$, which completes one full cycle of the complex exponential, bringing the signal back to its original value.
  • b) x[n] = sin(πn/19): This is a periodic signal with a period of 38. The sine function repeats every $2\pi$, so the input must change by an amount that causes the argument of the sine to increase by $2\pi$. Since $\frac{\pi}{19}$ is in the argument, we need to multiply it by 38 to achieve an increment of $2\pi$.
  • c) x[n] = n eᴇπ⁴: This signal is not periodic because the n multiplies the complex exponential and causes the amplitude to increase with n, preventing the signal from repeating.
  • d) x[n] = eʹ⁺⁴: This signal is not periodic as it has an ever-increasing exponent with n, meaning the signal does not repeat at regular intervals.

Calculations for amplitude, period, frequency, and velocity of waves

  • To calculate the amplitude of the wave, we look at the coefficient in front of the sine or cosine function in the wave equation.
  • The period of the wave can be shown using the angular frequency $\omega$ from the wave equation. For example, if $\omega = 1.57 s^{-1}$, the period $T$ is the inverse $(2\pi/\omega)$.
  • Frequency of the wave $f$ is the inverse of the period $T$.
  • The velocity of the waves $v$ can be calculated by multiplying the frequency $f$ by the wavelength $\lambda$.
User Zaven Nahapetyan
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