Question

For each of the following signals, determine if it is periodic. If yes, specify the fundamental period; if not, explain why:

a. x_1[n] = sin (0.75 π n);
b. x_2[n] = cos (0.75 π n) + sin(0.5 π n + π) + cos(0.3 π n - 0.5 π);
c. x_3[n] = cos (√2 π n + π);
d. x_4[n] = eʲ⁰.⁷⁵ π ⁿ².

Answer 1

The signal x_1[n] = sin(0.75 π n) has a fundamental period of 8/3, x_2[n] = cos(0.75 π n) + sin(0.5 π n + π) + cos(0.3 π n - 0.5 π) has a fundamental period of 40, while x_3[n] and x_4[n] are not periodic because √2 is irrational and the exponent in x_4[n] is a quadratic function of n.

Step-by-step explanation:

Let's determine if the given signals are periodic and find their fundamental period if they are:

• x_1[n] = sin(0.75 π n); This is a periodic signal. The fundamental period, T, is the smallest positive value of n for which the argument of the sine function, 0.75 π n, increases by 2π. Since 0.75 π n = 0.75 π (n+T), T can be found as T = 2π / (0.75 π) = 8/3. Hence, the fundamental period of x_1[n] is 8/3.
• x_2[n] = cos(0.75 π n) + sin(0.5 π n + π) + cos(0.3 π n - 0.5 π); All components are periodic with periods 8/3, 4, and 20/3 respectively. The signal is periodic if there is a common multiple of these periods, which is 40. Thus, the fundamental period of x_2[n] is 40.
• x_3[n] = cos(√2 π n + π); This signal is not periodic because √2 is not a rational number, and thus there is no smallest positive n that makes √2 π n to increase by 2π.
• x_4[n] = eʸ⁰.⁷⁵ π n²; This signal is also not periodic as the exponent is a quadratic function of n, and there is no fixed period that can be repeated indefinitely.