Question

Consider the function f(t)=2Re{eʲπᵗ+2e⁻ʲ²πᵗ}−2. Determine its period, T, its fundamental frequency, ωo​, and plot it over at least two periods.

Answer 1

The period, T, of the function, is 0.5 seconds, and the fundamental frequency, ωo, is 4π radians per second. The function oscillates around the value -2, with a cycle repeating every 0.5 seconds.

Step-by-step explanation:

Finding the Period and Fundamental Frequency:

To determine the period, T, and the fundamental frequency, ωo, of the function f(t)=2Re{eʸπᵗ+2e˃ʸ²πᵗ}−2, we look at the exponents of e in the function. The general form of a complex exponential corresponding to a sinusoid is eʸᵢᵠᵖ1, where ω is the angular frequency. Since the angular frequency (ω) relates to the fundamental frequency (ωo) by ω=2πa, we extract the highest frequency present in the function for the fundamental frequency.

The terms within the function have frequencies of π and 2²π respectively. The higher frequency dictates the period, T, since it corresponds to the fastest repeating part of the signal. So, we use the frequency of the second term, 2²π, to find the period: T = 1/f = 1/2 = 0.5 seconds. Consequently, the fundamental frequency is ωo = 2π/T = 4π radians per second.

Over two periods, the function would complete two full cycles. It starts at -2, oscillates upwards, reaching a maximum, oscillates down, returning to -2, and then repeats this behavior. A visual plot would show the oscillatory behavior around the value -2, with peaks and troughs corresponding to the real part of the complex exponentials.