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Fs = 100; t = 0:1/Fs:1; x = cos(2*pi*90*t); % appears as 10 Hz plot(t,x) Problems: Frequency response, pole-zero plots, filter types (FIR/IIR).

1. Chapter 2 – Discrete-Time Signals & Systems Typical Problems: Convolution, stability/causality, difference equations.

( f_\textin = 90,\textHz, f_s = 100,\textHz ) ( k=1 ): ( |90 - 100| = 10,\textHz ) → aliased at 10 Hz.

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Digital Signal Processing 4th Proakis SolutionVeuillez remplir les champs obligatoires.Veuillez cocher la case de la confidentialité.Veuillez remplir les champs obligatoires et accepter la case de confidentialité.

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