Thursday, September 21, 2000

Homework: Due Monday, September 25, 2000

(assigned 9/11) #1.1, 1.2, 1.3, 1.12, 1.20, 1.23

(assigned 9/18) #1.21, 1.26

Prof. Yates began class by doing a problem from the homework:

Problem 1.3 (pg. 78)



·  If Y(t) is Ergodic, it’s also strictly stationary and wide sense stationary.

·  When trying to show if something is WSS, SS, ergodic, or not, try hard to show that it’s NOT because it usually isn’t

Part (b) is Y(t) stationary?

Analysis:

1.  Does E[Y(t)] = a constant?

·  Yes, recall E[Y(t)] = 0

However, we still don’t know if it’s stationary, but we haven’t proven that it’s not.

2.  Does RY(t, 0) = var(Y(t)) = E[(Y2(t)] = a constant independent of t? (Here we are checking the second moment)

·  No, if we put in a different t, the variance changes (recall a is a function of t)

Þ Therefore Y(t) is NOT stationary

3.  Does RY(t, t) = RY(t) ? Is it WSS?

Other questions to ask if necessary.

4.  fY(t)(y) = independent of t?

Part (C) is Y(t) ergodic?

To be ergodic, it needs to be stationary, thus it is NOT ergodic.

Question from classmate: in a previous lecture we proved that a certain function, X(t), was stationary.

Recall (9/14/00): X(t) = cos (2pf0t + q) “random phase sinusoid” where q is uniform over [0, 2p]

If we again answer the questions above:

1.  Yes

2.  Yes

3.  Yes

4.  Maybe true

Similar to problem 1.1:

X(t) = Asin2pf ct

·  Just by looking at X(t), we see it is nonstationary

*Prof. Yates said to skip reading 1.10 – 1.13 and that we’ll come back to that later.


Problem 1.20 (pg. 83)

X(t) is stationary, zero mean, Gaussian process with autocorrelation function RX(t)

Y(t) = X2(t)

RX(t) = E[X(t)X(t+ t)]

Part (a) E[Y(t)] = E[X2(t)] t=0 = RX(t)|t=0 = RX(0)

Part (b) E[Y(t)] = RX(0)

Autocov = CY(t+ t) = E[(Y(t) – E[Y])(Y(t+t) – E[Y])]

= E[(Y(t)Y(t+t)] – (E[Y]) 2

= RY(t, t) – (E[Y]) 2

= E[(Y(t)Y(t+t)] - RX2 (0)

= E[X2 (t)X2 (t+t)] - RX2 (0)

·  RX(0) = s2 and [Y(t+t)] = E[X2(t+t)] = E[X2]

X1 = X(t) jointly Gaussian random variable

X2 = X (t+t)

Þ Therefore E[X2 (t)X2 (t+t)] = E[X12X22]



Equation 1.85 (pg. 57)

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