Solution 1

2.51. Periodic. T0 = k

500 , k ∈ Z

2. Aperiodic. If it is periodic, denote the period as T0. We have, sin (4000π(t+ T0)) =sin(4000πt) and cos (11000(t+ T0)) = cos(10000t). Thus, 4000πT0 = 2πk1and 11000T = 2πk2 hold, where k1, k2 ∈ Z . Finally, we get 4π

11 = k1k2

, which isimpossible.

3. Periodic. N0 = k, k ∈ Z

4. Aperiodic. Similar to (2)

2.91.

Px = limT→∞

1

T

∫ T/2

−T/2A2dt = A2.

0 < Px <∞, x(t) is a power-type signal and its power is A2.

2.

Px = limT→∞

1

T

∫ T/2

−T/2A2 cos2(2πf0t+ θ)dt

= limT→∞

1

T

∫ T/2

−T/2

A2

2{1 + cos[2(2πf0t+ θ)]} dt

=A2

2.

0 < Px <∞, x(t) is a power-type signal and its power is A2

2 .

3.

Pu = limT→∞

1

T

∫ T/2

0

1dt =1

2

0 < Pu <∞, u−1(t) is a power-type signal and its power is 12 .

1

4.

Px = limT→∞

1

T

∫ T/2

0

(Kt−

14

)2dt

= limT→∞

1

T2K2

√T

2

= limT→∞

√2K2 1√

T= 0,

Ex = limT→∞

∫ T/2

0

(Kt−

14

)2dt

= limT→∞

2K2

√T

2= ∞.

Thus, x(t) is neither an energy- nor a power-type signal.

2.131. δ(t)

2. 0

3. 0

4.∑∞n=−∞ Λ(t− 2n)

5. Λ′(t)

6. 13δ(t)

7. 16δ(t)

8. 13 cos( 1

3 )δ(t+ 13 )

9. 120δ(t)

10. 115δ(t)

11. δ′(t)

12. 12

13. 1

14. 0

15. 1

16. 1

2

5.51. P (Y = 1) = P (Y = 1|X = 1)P (X = 1) + P (Y = 1|X = 0)P (X = 0) =

0.8× 0.7 + 0.2× 0.3 = 0.62

2. P (Y=1,X=1)P (Y=1) = 0.8×0.7

0.62 = 0.903

5.101. P (X > 7) = Q( 7−4

3 ) = Q1 = 0.159

2. P (0 < X < 9) = P (X > 0) − P (X > 9) = Q(− 43 ) − Q(− 9−4

3 ) = 1 −Q( 4

3 )−Q( 53 ) = 0.86

5.221. ∫ +∞

−∞

∫ +∞

−∞fX,Y (x, y)dxdy

= K

∫ +∞

0

∫ x

0

e−x−ydxdy

= K

∫ +∞

0

(e−x − e−2x

)=

1

2K

= 1.

Thus K = 2.

2.

fX(x) =∫ +∞−∞ fX,Y (x, y)dy =

{2(e−x − e−2x

), x ≥ 0

0, otherwise

fY (y) =∫ +∞−∞ fX,Y (x, y)dy =

{2e−2y, y ≥ 00, otherwise

3. fX,Y (x, y) 6= fX(x)fY (y), thus X and Y are not independent.

4. fX|Y (x|y) =fX,Y (x,y)fY (y) = ey−x, x ≥ y ≥ 0.

5. E[X|Y = y] =∫ +∞−∞ xey−xdx = y + 1.

6. E(X), E(X2), E(Y ), E(Y 2), E(XY ) are 32 ,

72 ,

12 ,

12 , 1, respectively.

ThusCOV (X,Y ) = E(XY )−E(X)E(Y ) = 14 , σ2

X = E(X2)−E(X)2 = 54 ,

σ2Y = E(Y 2)− E(Y )2 = 1

4 , ρX,Y = COV (X,Y )σ2Xσ

2Y

=√55 .

3

5.281. According to the properties of P.229, any set of linear combinations of (X,Y )

are themselves jointly Gaussian. Therefore, Z and W are jointly Gaussian ran-dom variables.This can also be proved using the definition of joint Gaussian random variables.The following website may be of some help.http://cwww.ee.nctu.edu.tw/˜ftchien/class/rp08f/topic2(08).pdf

2.

E[X] = E[Y ] = 0, E[X2] = E[Y 2] = σ2, E[XY ] = ρE[X]E[Y ] = ρσ2.

From above, we have

E[ZW ] = E[−X2 sin θ cos θ +XY cos2 θ −XY sin2 θ + Y 2 sin θ cos θ]

= sin θ cos θ(E[Y 2]− E[X2]

)+ (cos2 θ − sin2 θ)E[XY ]

= ρ(cos2 θ − sin2 θ

)σ2,

E[Z] = E[W ] = 0.

Thus, COV (Z,W ) = E[ZW ] − E[Z]E[W ] = ρ(cos2 θ − sin2 θ

)σ2. Since

Z and W are jointly Gaussian random variables, independence is equivalent touncorrelatedness, i.e., COV (Z,W ) = 0.Finally, we have cos 2θ = 0 which leads to θ = 1

2kπ + π4 , k ∈ Z .

5.40

1. SX(f) = N0

2 and H(f) =

{1, |f | ≤ B0, otherwise

Then we have SY (f) = SX(f)|H(f)|2 =

{N0

2 , |f | ≤ B0, otherwise

According to Wiener-Khinchin Theorem,RY (f) = F−1SY (f) = N0 sin(2πBτ)2πτ

.

2. RY (f) = 0 for τ = 12B . According to Property 1 of P.247, Y (t) is also a

Gaussian process. For jointly Gaussian random variables Y (t) and Y (t+τ), un-correlatedness is equivalent to independence. Hence, the two random variablesare independent.

E[Y (t)] = E[Y (t+τ)] = 0, and δ2Y = RY (0) = N0B, thus the joint probabilitydensity function of Y (t) and Y (t+ τ) can be given by

fY (t),Y t+τ (y(t), y(t+ τ)) =1

2πN0Be−

y2(t)+y2(t+τ)2N0B

4

5.581. The output bandpass process has non-zero power content for frequencies in the

band 49× 106 ≤ |f | ≤ 51× 106. The power content is

P =

∫ −49×106−51×106

10−8(

1 +f

108

)df +

∫ 51×106

49×10610−8

(1− f

108

)df

= 2× 10−2.

2. The output process Nt can be written asN(t) = Nc(t) cos(2π50× 106t)−Ns(t) sin(2π50× 106t),where Nc(t) and Ns(t) are the in-phase and quadrature components, respective-ly, given by

Nc(t) = N(t) cos(2π50× 106t) + N̂(t) sin(2π50× 106t)

Ns(t) = N̂(t) cos(2π50× 106t)−N(t) sin(2π50× 106t)

The power content of the in-phase component is given by

E[|Nc(t)|2] = E[|N(t)|2] cos2(2π50× 106t) + E[|N̂(t)|2] sin2(2π50× 106t)

= E[|N(t)|2] = 2× 10−2

where we have used the fact that E[|N(t)|2] = E[|N̂(t)|2]. Similarly we findthat E[|Nc(t)|2] = 2× 10−2.

3. The power spectral density of Nc(t) and Ns(t) is

SNc(f) = SNs(f) =

{SN (f − 50× 106) + SN (f + 50× 106), |f | ≤ 50× 106

0, otherwise

where SNc(f) =

{10−8, |f | ≤ 106

0, otherwise . The power content of SNc(f) can be

found easily as PNc = PNs =∫ 106

−106 10−8df = 2× 10−2.

4. The power spectral density of the output is given by

SY (f) = SX(f)|H(f)|2 = 10−6(|f | − 49× 106)(10−8 − 10−16|f |) for49× 106 ≤ |f | ≤ 51× 106

Hence, the power content of the ouput is

PY = 10−6∫ −49×106−51×106

(−f − 49× 106)(10−8 + 10−16f))df

+10−6∫ 51×106

49×106(f − 49× 106)(10−8 − 10−16f))df

= 10−6(

2× 104 − 4

3102)

The power spectral density of the in-phase and quadrature components of theoutput process is given by

SYc(f) = SYs(f) = 10−6((

(f + 50× 106)− 49× 106) (

10−8 − 10−16(f + 50× 106)))

+10−6((−(f − 50× 106)− 49× 106

) (10−8 + 10−16(f − 50× 106)

))= 10−6(−2× 1016f2 + 10−2)

5

for |f | ≤ 106 and zero otherwise. The power content of the in-phase and quadra-ture component is

PYc = PYs = 10−6∫ 106

−106(−2× 10−16f2 + 10−2)df

= 10−6(

2× 104 − 4

3102)

= PY .

6