1 / 16

ECEN3513 Signal Analysis Lecture #23 16 October 2006

ECEN3513 Signal Analysis Lecture #23 16 October 2006. Read 6.1, 6.2; Review 6.3 Problems 4.7-2, 6.2-1, 6.3-5 Test #2 on 27 October Quiz 5 results: Hi = 7.8, Lo = 2.9, Ave = 5.35 Standard Deviation = 1.35 Read 6.4, 6.5 (1st section), 6.7 (1st section) Problems: 6.4-3, 6.4-4, 6.5-1

tommy
Download Presentation

ECEN3513 Signal Analysis Lecture #23 16 October 2006

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ECEN3513 Signal AnalysisLecture #23 16 October 2006 • Read 6.1, 6.2; Review 6.3 • Problems 4.7-2, 6.2-1, 6.3-5 • Test #2 on 27 October • Quiz 5 results: Hi = 7.8, Lo = 2.9, Ave = 5.35Standard Deviation = 1.35 • Read 6.4, 6.5 (1st section), 6.7 (1st section) • Problems: 6.4-3, 6.4-4, 6.5-1 • Test #2 on 27 October • Quiz 6 Results: Hi =8.0, Lo = 2.0, Ave. = 5.00Standard Deviation = 2.04

  2. yp(t): Base Periodic Function

  3. Fourier Transform of yp(t)

  4. y(t): Periodic Time Domain Waveform

  5. Fourier Series of y(t)

  6. Linear Time Invariant? yes • δ(t) → h(t)... an impulse response exists • convolution works • y(t) = x(t) * h(t) • transfer function exists • Y(f) = X(f) H(f) • Energy & Power transfer functions exist • |Y(f)|2 = |X(f)|2 |H(f)|2 J/Hz • SYY(f) = SXX(f) |H(f)|2 W/Hz

  7. Linear Time Invariant? no • δ(t) → h(t)... an impulse response exists • convolution gives you the wrong answer • y(t) ≠ x(t) * h(t) • transfer function does not exist • Y(f) ≠ X(f) H(f) • Energy & Power transfer functions don't exist • |Y(f)|2≠ |X(f)|2 |H(f)|2 J/Hz • SYY(f) ≠ SXX(f) |H(f)|2 W/Hz

  8. Autocorrelation (Power Signal) RXX(τ) = lim 1 x(t)x(t+τ)dt T→∞ T T • When in doubt, evaluate this one first • If = 0, evaluate energy signal autocorrelation • SXX(f) = lim |X(f)|2non-zero at same T→∞ T freqs as X(f)! • RXX(τ) ↔ SXX(f) W/Hz • Autocorrelation & Power Spectrum form a Fourier Transform pair

  9. Autocorrelation (Power Signal) +∞ RXX(τ) = lim 1 x(t)x(t+τ)dt = SXX(f) e -j2πfτ df T→∞ T T • RXX(0) = ?? • Average (normalized) power of x(t) • How can you find RXX(0) from SXX(f)? • Evaluate at τ = 0 • The area under power spectrum = average (normalized) power of x(t) • What's SXX(f) for x(t) = 3cos2π100t? • What's SXX(f) for x(t) = 3sin2π100t? -∞ units are W/Hz

  10. Autocorrelation (Energy Signal) RXX(τ) = lim x(t)x(t+τ)dt T→∞ T • RXX(τ) ↔ |X(f)|2 J/Hz • Autocorrelation & Energy Spectrum form a Fourier Transform pair • GXX(f) = |X(f)|2non-zero at same freqs as X(f)!

  11. Autocorrelation (Energy Signal) +∞ RXX(τ) = lim x(t)x(t+τ)dt = |X(f)|2 e -j2πfτ df T→∞ T • RXX(0) = ?? • Average (normalized) energy of x(t) • How can you find RXX(0) from |X(f)|2? • Evaluate at τ = 0 • The area under energy spectrum = average (normalized) energy of x(t) • What's |X(f)|2for x(t) = rect[(t-0.5)/1]? • What's |X(f)|2for x(t) = rect[t/1]? -∞ units are J/Hz

  12. Autocorrelation (Real World) ^ RXX(τ) = 1 x(t)x(t+τ)dt T T • RXX(τ) ↔ SXX(f) W/Hz • FT of autocorrelation yields estimate of Power Spectrum • SXX(f) = |X(f)|2, where X(f) = x(t) e-jωt dt T observation interval ^ ^ ^ ^ ^ T

  13. Autocorrelation (Real World) ^ RXX(τ) = x(t)x(t+τ)dt T • RXX(τ) ↔ GXX(f) W/Hz • FT of autocorrelation yields estimate of Energy Spectrum • GXX(f) = |X(f)|2, where X(f) = x(t) ejωt dt observation interval ^ ^ ^ ^ ^ T

  14. sinusoid + d.c. δ(f-0.1) δ(f+0.1) δ(f)

  15. Magnitude & Phase Plots

More Related