Power Spectral Density (PSD) 6.011, Spring 2018 Lec 18 1 iid - PowerPoint PPT Presentation

Power Spectral Density (PSD) 6.011, Spring 2018 Lec 18 1

iid signal x[n], uniform in [-0.5,+0.5] 2

y[.] obtained by passing x[.] through resonant 2 nd -order filter H(z), poles at ±0.95e^{j π /3} 3

Extracting the portion of x(t) in a specified frequency band x ( t ) H ( j v ) y ( t ) H ( j v ) ¢ ¢ 1 - v 0 v 0 v 4

Questions (warm-up for Quiz 2!) WSS process x [ · ] with C xx [ m ] = ρδ [ m − 1] + δ [ m ] + ρδ [ m + 1] . δ What is the largest magnitude ρ can have? WSS process x ( · ) with mean µ x and PSD S xx ( j ω ). What is its FSD? Zero-mean WSS process x ( · ) with 1 S xx ( j ω ) = 1 + ω 2 and let y ( t ) = Z + x ( t ), where Z has zero mean, variance σ 2 , and is uncorrelated σ with x ( · ). What are µ y and S yy ( j ω )? 5

Periodograms (e.g., a unit-intensity “white” process) M = 1, T = 50 M = 1, T = 50 M = 1, T = 50 M = 1, T = 50 4 4 4 4 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 0 0.5 1 0 0.5 1 0 0.5 1 0 0.5 1 v /(2 p ) v /(2 p ) v /(2 p ) v /(2 p ) M = 4, T = 50 M = 4, T = 200 4 4 CT case: X T ( jω ) ↔ x ( t ) windowed to [ − T, T ] − 3 3 2 2 | X T ( j ω ) | 2 1 1 Periodogram = 2 T 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 v /(2 p ) v /(2 p ) M = 16, T = 50 M = 16, T = 200 4 4 X N ( e j Ω ) ↔ x [ n ] windowed to [ − N, N ] − DT case: 3 3 2 2 | X N ( e j Ω ) | 2 Periodogram = 1 1 6 2 N + 1 0 0 v /(2 p ) v /(2 p )

Periodogram averaging (illustrating the Einstein-Wiener-Khinchin theorem) M = 1, T = 50 M = 1, T = 50 M = 1, T = 50 M = 1, T = 50 4 4 4 4 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 0 0.5 1 0 0.5 1 0 0.5 1 0 0.5 1 v /(2 p ) v /(2 p ) v /(2 p ) v /(2 p ) M = 4, T = 50 M = 4, T = 200 4 4 3 3 2 2 1 1 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 v /(2 p ) v /(2 p ) M 16, T 50 M 16, T 200 4 4 3 3 2 2 7 1 1 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 v /(2 p ) v /(2 p )

Periodogram averaging (illustrating the Einstein-Wiener-Khinchin theorem) M = 1, T = 50 M = 1, T = 50 M = 1, T = 50 M = 1, T = 50 4 4 4 4 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 0 0.5 1 0 0.5 1 0 0.5 1 0 0.5 1 v /(2 p ) v /(2 p ) v /(2 p ) v /(2 p ) M = 4, T = 50 M = 4, T = 200 4 4 3 3 2 2 1 1 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 v /(2 p ) v /(2 p ) M = 16, T = 50 M = 16, T = 200 4 4 3 3 2 2 1 1 8 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 v /(2 p ) v /(2 p )

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