ECEN3513 Signal AnalysisECEN3513 Signal AnalysisLecture #4 28 August 2006Lecture #4 28 August 2006

Read section 1.5Read section 1.5 Problems: 1.5-2a-c, 1.5-4, & 1.5-5Problems: 1.5-2a-c, 1.5-4, & 1.5-5 Quiz Friday (Chapter 1 and/or Correlation)Quiz Friday (Chapter 1 and/or Correlation)

Lecture 5 assignmentLecture 5 assignment Read 1.6, 1.9, 1.10Read 1.6, 1.9, 1.10 Problems: 1.6-6, 1.9-1, 1.9-3Problems: 1.6-6, 1.9-1, 1.9-3

Autocorrelation Autocorrelations deal with predictability over time. I.E. Autocorrelations deal with predictability over time. I.E.

given an arbitrary point given an arbitrary point x(t1)x(t1), how predictable is , how predictable is x(t1+tau)x(t1+tau)??

time

Volts

t1

tau

Autocorrelation t1+Tt1+TRRxx((ττ) = (1/(T-) = (1/(T-ττ))) x(t)x(t+) x(t)x(t+ττ)dt)dt

t1 t1

Example RExample Rxx(1)(1)Take x(t1)*x(t1+1)Take x(t1)*x(t1+1)Take x(t1+Take x(t1+ε)*x(t1+1+ ε)...ε)*x(t1+1+ ε).......x(t1+T-1)*x(t1+T).......x(t1+T-1)*x(t1+T)...Add these all together, then averageAdd these all together, then average

Autocorrelation

If the average (RIf the average (Rxxxx(tau)) is positive...(tau)) is positive... Then x(t) and x(t+tau) tend to be alikeThen x(t) and x(t+tau) tend to be alike

Both positive or both negativeBoth positive or both negative If the average (RIf the average (Rxxxx(tau)) is negative(tau)) is negative

Then x(t) and x(t+tau) tend to be oppositesThen x(t) and x(t+tau) tend to be oppositesIf one is positive the other tends to be negativeIf one is positive the other tends to be negative

If the average (RIf the average (Rxxxx(tau)) is zero(tau)) is zero There is no predictabilityThere is no predictability

255 point Noise waveform(Adjacent points are independent)

time

Volts

0

Vdc = 0 v, Normalized Power = 1 watt

Rxx(0) The sequence x(n)The sequence x(n)

x(1) x(2) x(3) ... x(255)x(1) x(2) x(3) ... x(255) multiply it by the unshifted sequence x(n+0)multiply it by the unshifted sequence x(n+0)

x(1) x(2) x(3) ... x(255)x(1) x(2) x(3) ... x(255) to get the squared sequenceto get the squared sequence

x(1)x(1)22 x(2) x(2)22 x(3) x(3)22 ... x(255) ... x(255)22

Then take the time averageThen take the time average[x(1)[x(1)22 +x(2) +x(2)22 +x(3) +x(3)22 ... +x(255) ... +x(255)22]/255]/255

Rxx(1) The sequence x(n)The sequence x(n)

x(1) x(2) x(3) ... x(254) x(255)x(1) x(2) x(3) ... x(254) x(255) multiply it by the shifted sequence x(n+1)multiply it by the shifted sequence x(n+1)

x(2) x(3) x(4) ... x(255)x(2) x(3) x(4) ... x(255) to get the sequenceto get the sequence

x(1)x(2) x(2)x(3) x(3)x(4) ... x(254)x(255)x(1)x(2) x(2)x(3) x(3)x(4) ... x(254)x(255) Then take the time averageThen take the time average

[x(1)x(2) +x(2)x(3) +... +x(254)x(255)]/254[x(1)x(2) +x(2)x(3) +... +x(254)x(255)]/254

Autocorrelation Estimate of White Noise

tau (samples)

Rxx

00

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

CorrelationExample

t

x(t)

10

3v

t

y(t+τ)

3-τ2-τ

3v

t

x(t) y(t+τ)

10

3vSo long as τ > 1

area = 0 meaning RXY(τ) = 0

Correlation

t

x(t)

10

3v

t

y(t+τ)

32

3v

t

x(t) y(t+τ)

10

3vτ = 0

area = 0 meaning RXY(0) = 0

Correlation

t

x(t)

10

3v

t

y(t+τ)

8/35/3

3v

t

x(t) y(t+τ)

10

3vτ = 1/3

area = 0 meaning RXY(1/3) = 0

Correlation

t

x(t)

10

3v

t

y(t+τ)

7/34/3

3v

t

x(t) y(t+τ)

10

3vτ = 2/3

area = 0 meaning RXY(2/3) = 0

Correlation

t

x(t)

10

3v

t

y(t+τ)

6/33/3

3v

t

x(t) y(t+τ)

10

τ = 3/3

area = 0 meaning RXY(1) = 0

9v2

Correlation

t

x(t)

10

3v

t

y(t+τ)

5/32/3

3v

t

x(t) y(t+τ)

3/30

9v2

τ = 4/3

2/3 area = 3 meaning RXY(4/3) = 3

Correlation

t

x(t)

10

3v

t

y(t+τ)

4/31/3

3v

t

x(t) y(t+τ)

3/30

9v2

τ = 5/3

1/3 area = 6 meaning RXY(5/3) = 6

Correlation

t

x(t)

10

3v

t

y(t+τ)

3/30/3

3v

t

x(t) y(t+τ)

3/30

9v2

τ = 6/3Up to this point, the time bounds of x(t)y(t+τ) existed from t = 2-τ to t = 1when 1 < τ < 2 (overlap).

area = 9 meaning RXY(2) = 9

Correlation

t

x(t)

10

3v

t

y(t+τ)

2/3-1/3

3v

t

x(t) y(t+τ)

2/30

9v2 τ = 7/3When τ > 2 this is no longer the case.

area = 6 meaning RXY(7/3) = 6

Correlation

t

x(t)

10

3v

t

y(t+τ)

1/3-2/3

t

x(t) y(t+τ)

1/30

9v2

τ = 8/3

3v

area = 3 meaning RXY(8/3) = 3

Correlation

t

x(t)

10

3v

t

y(t+τ)

0/3-3/3

t

x(t) y(t+τ)

0

9v2

τ = 9/3

3v

area = 0 meaning RXY(τ) = 0; τ > 3

Correlation Asked to solve for an equation for R(Asked to solve for an equation for R(ττ)?)?

DRAWDRAWSOMESOME

PICTURES!!!PICTURES!!!

FIR Filter (a.k.a. MA Filter)x(t) x(t-Δ)

Δ

delay

w1 wNw2

Filter Output y(t) = w1x(t) + w2x(t-Δ) + wNx(t-(N-1)Δ)

Δ

delay

Δ

delay

x(t-2Δ)

FIR = Finite Impulse ResponseMA = Moving Average