Main answer:
(a) The DT signal x[n] is periodic.
Step-by-step explanation:
To determine if the DT signal x[n] is periodic, we need to check if there exists a positive integer N such that x[n] = x[n+N] for all values of n.
Given that x(t) = 4 cos (18πt + π/4), we can find the period of the CT signal x(t) by dividing the angular frequency of the cosine function (18π) by the fundamental frequency (ω₀) of the signal. In this case, the fundamental frequency is the coefficient of t, which is 18π.
The period T of the CT signal is given by T = 2π/ω₀.
Now, let's determine the period of the DT signal x[n]. The period of a DT signal is equal to the reciprocal of the sampling rate, which is given as ω_s = 22π rad/s.
Therefore, the period of the DT signal x[n] is T_s = 2π/ω_s.
To find the period in samples, we need to divide the period T_s by the sampling interval Δt = 1, since the signal is sampled at a rate of 1 sample per unit time.
Therefore, the period in samples is N = T_s/Δt.
Substituting the values, we have N = (2π/ω_s) / 1 = 2π/ω_s.
Hence, the DT signal x[n] is periodic with a period of 2π/ω_s samples.
(b) To find the numbers A, ω₀, and θ₀, we can use the output y(t) = A cos(ω₀t + θ₀).
Given that the CT signal x(t) = 4 cos (18πt + π/4), we can see that y(t) is a reconstructed waveform of the DT signal x[n] through a perfect D/A converter.
Comparing the two equations, we can see that A = 4, ω₀ = 18π, and θ₀ = π/4.
Therefore, the numbers A, ω₀, and θ₀ are 4, 18π, and π/4, respectively.
I hope this explanation helps! Let me know if you have any further questions.