Content License: Creative Commons Attribution Non Commercial 4.0 International (CC-BY-NC-4.0)Credit must be given to the creatorOnly noncommercial uses of the work are permittedDownloadsDownload🖍️ CorrectionExercice 1¶Exercice 2¶Exercice 3¶Pour le signal x(t)=e−2tu(t)x(t) = e^{-2t} u(t)x(t)=e−2tu(t) :Px=0,Ex=14.P_x = 0, \qquad E_x = \frac{1}{4}.Px=0,Ex=41.Pour le signal y[n]=12nu[n]y[n] = \frac{1}{2^n} u[n]y[n]=2n1u[n] :Py=0Ey=43.P_y = 0 \qquad E_y = \frac{4}{3}.Py=0Ey=34.
