ECEN3513 Signal Analysis Lecture #4 28 August 2006 n Read section 1.5 n Problems: 1.5-2a-c, 1.5-4, & n Quiz Friday (Chapter 1 and/or Correlation) n Lecture 5 assignment n Read 1.6, 1.9, 1.10 n Problems: 1.6-6, 1.9-1, 1.9-3

Autocorrelation n Autocorrelations deal with predictability over time. I.E. given an arbitrary point x(t1), how predictable is x(t1+tau)? time Volts t1 tau

Autocorrelation t1+T t1+T R x (τ) = (1/(T-τ)) x(t)x(t+τ)dt t1 Example R x (1) Take x(t1)*x(t1+1) Take x(t1+ε)*x(t1+1+ ε) x(t1+T-1)*x(t1+T)... Add these all together, then average

Autocorrelation n If the average (R xx (tau)) is positive... u Then x(t) and x(t+tau) tend to be alike Both positive or both negative n If the average (R xx (tau)) is negative u Then x(t) and x(t+tau) tend to be opposites If one is positive the other tends to be negative n If the average (R xx (tau)) is zero u There is no predictability

255 point Noise waveform (Adjacent points are independent) time Volts 0 V dc = 0 v, Normalized Power = 1 watt

R xx (0) R xx (0) n The sequence x(n) x(1) x(2) x(3)... x(255) n multiply it by the unshifted sequence x(n+0) x(1) x(2) x(3)... x(255) n to get the squared sequence x(1) 2 x(2) 2 x(3) 2... x(255) 2 n Then take the time average [x(1) 2 +x(2) 2 +x(3) x(255) 2 ]/255

R xx (1) R xx (1) n The sequence x(n) x(1) x(2) x(3)... x(254) x(255) n multiply it by the shifted sequence x(n+1) x(2) x(3) x(4)... x(255) n to get the sequence x(1)x(2) x(2)x(3) x(3)x(4)... x(254)x(255) n Then take the time average [x(1)x(2) +x(2)x(3) x(254)x(255)]/254

Autocorrelation Estimate of White Noise tau (samples) Rxx 0 0

255 point Noise Waveform (Low Pass Filtered White Noise) Time Volts 23 points 0

Autocorrelation Estimate of Low Pass Filtered White Noise tau samples Rxx 0 23

Correlation Example t x(t) 1 0 3v t y(t+τ) 3-τ 2-τ 3v t x(t) y(t+τ) 1 0 3v So long as τ > 1 area = 0 meaning R XY (τ) = 0

Correlation t x(t) 1 0 3v t y(t+τ) 3 2 3v t x(t) y(t+τ) 1 0 3v τ = 0 area = 0 meaning R XY (0) = 0

Correlation t x(t) 1 0 3v t y(t+τ) 8/3 5/3 3v t x(t) y(t+τ) 1 0 3v τ = 1/3 area = 0 meaning R XY (1/3) = 0

Correlation t x(t) 1 0 3v t y(t+τ) 7/3 4/3 3v t x(t) y(t+τ) 1 0 3v τ = 2/3 area = 0 meaning R XY (2/3) = 0

Correlation t x(t) 1 0 3v t y(t+τ) 6/3 3/3 3v t x(t) y(t+τ) 1 0 τ = 3/3 area = 0 meaning R XY (1) = 0 9v 2

Correlation t x(t) 1 0 3v t y(t+τ) 5/3 2/3 3v t x(t) y(t+τ) 3/3 0 9v 2 τ = 4/3 2/3 area = 3 meaning R XY (4/3) = 3

Correlation t x(t) 1 0 3v t y(t+τ) 4/3 1/3 3v t x(t) y(t+τ) 3/3 0 9v 2 τ = 5/3 1/3 area = 6 meaning R XY (5/3) = 6

Correlation t x(t) 1 0 3v t y(t+τ) 3/3 0/3 3v t x(t) y(t+τ) 3/3 0 9v 2 τ = 6/3 Up to this point, the time bounds of x(t)y(t+τ) existed from t = 2-τ to t = 1 when 1 < τ < 2 (overlap). area = 9 meaning R XY (2) = 9

Correlation t x(t) 1 0 3v t y(t+τ) 2/3 -1/3 3v t x(t) y(t+τ) 2/3 0 9v 2 τ = 7/3 When τ > 2 this is no longer the case. area = 6 meaning R XY (7/3) = 6

Correlation t x(t) 1 0 3v t y(t+τ) 1/3 -2/3 t x(t) y(t+τ) 1/3 0 9v 2 τ = 8/3 3v area = 3 meaning R XY (8/3) = 3

Correlation t x(t) 1 0 3v t y(t+τ) 0/3 -3/3 t x(t) y(t+τ) 0 9v 2 τ = 9/3 3v area = 0 meaning R XY (τ) = 0; τ > 3

Correlation n Asked to solve for an equation for R(τ)? DRAW SOME PICTURES!!!

FIR Filter (a.k.a. MA Filter) x(t) x(t-Δ) Δ delay w1w1 wNwN w2w2 Filter Output y(t) = w 1 x(t) + w 2 x(t-Δ) + w N x(t-(N-1)Δ) Δ delay Δ delay x(t-2Δ) FIR = Finite Impulse Response MA = Moving Average