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Homework Set 8 Solutions EECS 455 Nov. 24, 2006

1. Problem 5.38, part 3, p. 263

We are given that X(t) = X cos 2!fot + Y sin 2!fot where X and Y are independent Gaussian randomvariables with mean zero and variance σ2.

1. mX(t) = E[X cos 2!fot + Y sin 2!fot] = E[X] cos 2!fot + E[Y] sin 2!fot = 0 since E[X]=E[Y]=0.

2. RX(t+ττττ ,t) = E[X(t+τ)X(t)] = E[(X cos 2!fo(t+τ) + Y sin 2!fo(t+τ) (X cos 2!fot + Y sin 2!fot)] = E[X2] cos 2!fo(t+τ) cos 2!fot + E[XY] cos 2!fo(t+τ) sin2!fot

+ E[YX] sin2!fo(t+τ) cos 2!fot + E[Y2] sin 2!fo(t+τ) sin 2!fot

= σ2 cos 2!fo(t+τ) cos 2!fot + σ2 sin 2!fo(t+τ) sin 2!fot,

since E[XY] = EX EY = 0, and σ2 = EX2 - (EX)2 = EY2 - (EY)2

= σ2 12 cos 2!fo(2t+τ) + σ2

12 cos 2!fot - σ2

12 cos 2!fo(2t+τ) - σ2

12 cos 2!fot

==== σσσσ2 cos 2!fot

3. We first observe that X(t) is WSS. Therefore we can find its power spectral density as follows:

SX(t) = F{RX(τ)} = F{σ2 cos 2!fot} = σσσσ2

2 (δδδδ(f-fo) + δδδδ(f+fo))

4. If σ2X ≠ σ2

Y, then

mX(t) = E[X cos 2!fot + Y sin 2!fot] = E[X] cos 2!fot + E[Y] sin 2!fot = 0 since E[X]=E[Y]=0.

RX(t+ττττ ,t) = E[X(t+τ)X(t)] = E[(X cos 2!fo(t+τ) + Y sin 2!fo(t+τ) (X cos 2!fot + Y sin 2!fot)]

= E[X2] cos 2!fo(t+τ) cos 2!fot + E[XY] cos 2!fo(t+τ) sin2!fot

+ E[YX] sin2!fo(t+τ) cos 2!fot + E[Y2] sin 2!fo(t+τ) sin 2!fot

= σ2X cos 2!fo(t+τ) cos 2!fot + σ2

Y sin 2!fo(t+τ) sin 2!fot,

since E[XY] = EX EY = 0

= σ2X

12 cos 2!fo(2t+τ) + σ2

X 12 cos 2!fot - σ2

Y 12 cos 2!fo(2t+τ) - σ2

Y 12 cos 2!fot

==== σσσσ2

X-σσσσ2Y

2 cos 2!fo(2t+ττττ) + σσσσ2

X+σσσσ2Y

2 cos 2!fot

Notice that in this case, X(t) is not WSS.

2. Problem 5.49, part 1, p. 265. [0,!]

1. X(t) = A cos(2!fot+Θ) where A is a constant and Θ is uniformly distributed on [0,2!].

We have

mX(t) = E X(t) = E [A cos(2!fot+Θ ] = A E cos(2!fot+Θ)

= A ∫-∞

∞ cos(2!fot + θ) p(θ) dθ = A ∫

0

! cos(2!fot + θ)

1! dθ = A ∫

2!fot

!+2!fot

cos(φ) 1! dφ, φ=θ+2!fot

= A 1! sin(φ)| !+2!fot

2!fot = A 1! (sin(!+2!fot) - sin(2!fot)) = A

1! (-sin(2!fot) - sin(2!fot))

= - A 2! sin(2!fot)

Since this is not a constant, X(t) is not WSS. Therefore, it does not have a power spectral densityin the usual sense, and so this problem does not have an answer.

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For completeness, I will also compute the autocorrelation function.

RX(t,t+τ) = E X(t) X(t+τ) = E [ A cos(2!fot+Θ) A cos(2!fo(t+τ)+Θ) ]

= A2 ∫0

! cos(2!fot + θ) cos(2!fo(t+τ) + θ) p(θ) dθ

= A2 ∫0

!

12 (cos(4!fo(t +τ)+2θ ) +cos(2!foτ ))

1! dθ

= A2

2! ∫0

! cos(4!fo(t +τ)+2θ ) dθ +

A2

2! cos(2!foτ )

= 0 + A2

2! cos(2!foτ ) since the cosine goes through 2! radians as θ goes from 0 to !

= A2

2! cos(2!foτ )

Here we see that the autocorrelation function depends only on τ. Though I realized that we needed toassume Θ is uniform on [0,!] in order to make this happen, unfortunately. I neglected to consider whatassumption is needed to make the mean function constant.

Discussion: When the mean and autocorrelation functions are periodic functions of t, then the randomprocess is called "cyclostationary" and one compute the power spectral density by taking the Fouriertransform of the time average of the autocorrelation fuction over t. However, this was not discussed inclass or in the textbook. So this problem will not be graded.

3. Problem 5.50, p. 265. Just the α ≠ β case.

We are given that X(t) is a stationary random process with RX(τ) = e-α |t| where α>0. It is the input to

an LITS with impulse response h(t) = e-βt u(t), where β>0 and β≠α. The output Y(t) is WSS since

the input is WSS. Its power spectral density is

SY(f) = SX(f) |H(f)|2 = F{RX(t)} |F{h(t)}|

2 = F{e

-α |t| } |F{e

-βt u(t)}|2

= F{2α

α2+4!2f2 } | 1

β+j2!f |2 =

2αααα(αααα2+4!2f2)(ββββ2222+4!2f2)

4. White Gaussian noise Nw(t) with power spectral density S Nw

(f) = 5 is passed through a linear time-invariant filter with frequency response

H(f) = 1, 3≤f≤8

0, else

Let M(t) denote the output. Find Pr(M(t)>1).

Note: There is an error in the stated frequency response. The frequency response of any filter with a realimpulse response must have a real part that is even. Thus, the frequency response should have been

H(f) = 1, 3≤|f|≤8

0, else

Since the input to the filter is Gaussian with mean 0, the output is also Gaussian with mean 0. Hence,

Pr(M(t)>1) = ∫1

∞ pM(m) dm = ∫

1

∞

1

√2!σM exp{-

(m-EM)2

σ2M

} dm, let n = m-EM

σM = mσM

= ∫1/σM

∞

1

√2!σM exp{-

n2

2 } σM dn = ∫1/σM

∞

1

√2! exp{-

n2

2 } dn = Q(1

σM)

We now find σM.

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σ2M = EM2 - (EM)2 = EM2 = RM(0) = ∫

-∞

∞ SM(f) df = ∫

-∞

∞ S

NW

(f) |H(f)|2 df

= ∫-8

-3 5 × 1 df + ∫

3

8 5 × 1 df = 50

Finally,

Pr(M(t)>1) = Q(1

σM) = Q(

1

√50) = Q(0.14) ≅≅≅≅ 0.44 from the Q-function table

5. A zero-mean white Gaussian noise with power-spectral density No/2 passes through an ideal low passfilter with bandwidth B.

a. Find the autocorrelation function of the output process Y(t)

RY(τ) = F-1 (SY(f)) = F

-1 (SX(f) |H(f)|

2 ) = F

-1 (

No

2 | Π(f/2B) |

2 ) =

No

2 F

-1 ( Π(f/2B))

= No

2 2B sinc(2Bτ)

b. Assuming τ = 1/(2B) find the joint probability density function of the random variables Y(t) andY(t+τ). Are they independent?

Since the input to the filter is a Gaussian random process the output is a Gaussian random process.Therefore, Y(t) and Y(t+τ) are jointly widesense stationary. To find their joint density, we need onlyfind their means and their covariance. Their means are zero since the input to the filter has zero mean.Their covariance is

cov(Y(t),Y(t+τ)) = E Y(t)Y(t+τ) - EY(t) EY(t+τ) = E Y(t)Y(t+τ) = RY(τ)

= No

2 2B sinc(2Bτ) =

No

2 2B sinc(2B 1

2Β) = No

2 2B sinc(1) = 0

Therefore, for this value of τ, Y(t) and Y(t+τ) are uncorrelated. Since they are also Gaussian. we camsay they are independent. Thus, their joint density is simply the product of their marginal densities.Each is zero mean with variance

σ2Y = RY(0) =

No

2 2B sinc(0) = BNo

The joint density is

pY(t)Y(T+ττττ)(y1,y2) = 1

√2!BNo exp{-

y21

2BNo}

1

√2!BNo exp{-

y22

2BNo} =

12!BNo

exp{- y

21+y

22

2BNo}

6. (a) Find the band of spectral occupancy, bandwidth and bandwidth efficiency of M-ary pulse positionmodulation. Assume M = 2k. Assume the source is the usual binary source with independentequiprobable letters that emits Rb = 1/Tb bits/second. Assume the "pulses" are rectangular.

In M-ary PPM the M signals are rectangular pulses of duration Tp = T/2k = kTb/2k. By the convention

given in class, each signal has BSO [0, 1

2Tp]. Since these are all the same, the modulation has

BSO = [0, 1

2Tp] = [0,

2k-1

kTb] = [0, Rb

2k-1

k ]

The bandwidth is W = Rb 2k-1

k and the bandwidth efficiency is

ηηηη = RbW =

k2k-1

(b) Compare the bandwidth efficiency to that of PAM, as found in class.

For PAM we found η = 2k. For k = 1, PAM has twice the bandwidth efficiency as PPM. As kincreases the bandwidth efficiency of PAM increases linearly, while that of PPM decreases in anessentially exponential fashion. Thus PAM is much more bandwidth efficient when k is not small.