Question 9.12: Given the analog signal x(t) = e^−at cos(Ω0t)u(t), determine......

Comparing the sampled continuous-time signal x(nT_s) = (e^{−aTs})^n cos((Ω_0T_s)n)u[n] with y[n] we obtain the following two equations:

(i)\:\alpha=e^{-a T_{s}},\;(i i)\:\omega_{0}=\Omega_{0}T_{s}

with three unknowns (a, Ω_0, and T_s) when α and ω_0 are given, so there is no unique solution. However, we know that according to the Nyquist condition

T_{s}\leq{\frac{\pi}{\Omega_{m a x}}}.

Assuming the maximum frequency is Ω_{max} = NΩ_0 for N ≥ 2 (since the signal x(t) is not band-limited the maximum frequency is not known, to estimate it we could use Parseval’s result as indicated in the previous chapter; instead we are assuming that is a multiple of Ω_0) if we let T_s = π/NΩ_0 after replacing it in the above equations we get

\alpha=e^{-a\pi/N\Omega_{0}},\;\;\omega_{0}=\Omega_{0}\pi/N\Omega_{0}=\pi/N.

If we want α = 0.9 and ω_0 = π/2, we have N = 2 and

a=-{\frac{2\Omega_{0}}{\pi}}\log0.9

for any frequency Ω_0 > 0. For instance, if Ω_0 = 2π then a = −4 log 0.9, and T_s = 0.25. Fig. 9.4 displays the continuous and the discrete-time signals generated using the above parameters. The following script is used. The continuous-time and the discrete-time signals coincide at the sample times.