Devoir 2

Sébastien Roy
February 20, 2012

GEL7060 (Communications numériques avancées)

Hiver 2012
à remettre avant jeudi le 1er mars, 23h00

1. (2 points) Random variables

(a) (1 point) The covariance of 2 random variables X and Y is given
by
cov(X, Y ) = h[X − hX i] [Y − hY i]i
Show that cov (aX + b, cY + d) = ac cov (X, Y ).

cov(aX + b, cY + d) = h(aX + b − a hXi − b)(cY + d − c hY i − d)i

= ha(X − hXi)c(Y − hY i)i
= ac h(X − hXi)(Y − hY i)i
= accov(X, Y ).

(b) (1 point) If X and Y are independent and Y follows a uniform law

between 0 and 1, show that the PDF of Z = X + Y is

fZ (z) = FX (z) − FX (z − 1)

where FX (x) is the CDF of X.

1
GEL7060 – (Communications numériques avancées) 2

The PDF of z is given by

Z ∞
fZ (z) = fY (α)fX (z − α)dα,
−∞

but
fY (y) = u(y) − u(y − 1),
leading to Z 1
fZ (z) = fX (z − α)dα.
0
Letting u = z − α and therefore du = −dα, we have
Z z−1
fZ (z) = − fX (u)du
Z zz
= fX (u)du
z−1
Z z Z z−1
= fX (u)du − fX (u)du
−∞ −∞
= FX (z) − FX (z − 1).

2. (3 points) Discrete stochastic processes

Given a discrete stochastic process X[k] having the following power
spectral density :
SXX (f )

− 14 1 f
4

(a) (1 point) Derive the corresponding autocorrelation function.

Since this is a discrete process, the autocorrelation function is expressed
Z
1
RXX [k] = SXX (f )ej2πf k df
− 21 2
Z 1
4
= ej2πf k df
− 14
GEL7060 – (Communications numériques avancées) 3

#1
ej2πf k
"
4
=
j2πk − 14
jπk jπk
e− e− 2
2
=
j2πk
!
sin(πk/2) 1 k
= = sinc .
πk 2 2

(b) (1 point) If this process is presented at the input of a filter whose

impulse response is h[k] = δ[k] + δ[k − 4], what is the power spectral
density of the output process Y [k] ?
The frequency response of the filter in discrete time is given by
∞
h[k]e−j2πf k
X
H(f ) =
k=−∞
∞
(δ[k] + δ[k + 4]) e−j2πf k
X
=
k=−∞
−j8πf
= 1−e  
= e−j4πf ej4πf + e−j4πf
= 2e−j4πf cos(4πf ),

which naturally leads to

|H(f )|2 = 4 cos2 (4πf ),

and we have

SY Y (f ) = |H(f )|2 SXX (f )

= 4 cos2 (4πf )SXX (f )
1 1
    
= 4 cos2 (4πf ) u f + +u f − .
4 4
GEL7060 – (Communications numériques avancées) 4

(c) (1 point) Derive the output discrete autocorrelation function RY Y [k].

∞
X ∞
X
RY Y [k] = h[m]h[l]RXX [k + m − l]
m=−∞ l=−∞
∞ ∞
1 k+m−l
X X  
= (δ[m] + δ[m − 4]) (δ[l] + δ[l − 4]) sinc
m=−∞ l=−∞
2 2
∞
1 k−l k+4−l
X     
= (δ[l] + δ[l − 4]) sinc + sinc
2 l=−∞ 2 2
1 k k+4 k−4 k
        
= sinc + sinc + sinc + sinc
2 2 2 2 2
k 1 k−4 k+4
      
= sinc + sinc + sinc .
2 2 2 2

3. (2 points) Cyclostationary processes

Given a process of the form
∞
X
S(t) = A[k]g(t − kT ),
k=−∞

where g(t) is a square wave of unit amplitude and duration T2 , i.e.

g(t) = u(t) − u(t − T /2), and the discrete stochastic process A[k] has
the following autocorrelation function :
 1
 2,
 k = 0,
RAA [k] =  − 14 , k = ±1,

0, ailleurs.

(a) (0.5 point) What is the mean of this process ?

GEL7060 – (Communications numériques avancées) 5

* ∞ +
X
µS = A[k]g(t − kT )
k=−∞
∞
X
= hA[k]i g(t − kT )
k=−∞
X∞
= µA g(t − kT ).
k=−∞

In the strict sense, we cannot reduce more than this given the information
that we have.
Observing RAA [k], we see that it is possible that A[k] is bipolar of ampli-
tude √12 with equally-likely pulses. However, we cannot confirm this. If it
were true, then we would have µS = µA = 0.

(b) (0.5 point) What is its autocorrelation function ?

1 ∗
RSS (t, t + τ ) = hS (t)S(t + τ )i
2*
∞ ∞
+
1 X ∗
X
= A [k]g(t − kT ) A[l]g(t + τ − lT )
2 k=−∞ l=−∞
∞ ∞
1 X
hA∗ [k]A[l]i g(t − kT )g(t + τ − lT )
X
=
2 k=−∞ l=−∞
∞
X ∞
X
= RAA [l − k]g(t − kT )g(t + τ − lT )
k=−∞ l=−∞
∞ 
1 1
X 
= g(t) − (g(t − T ) + g(t + T )) g(t + τ − lT )
l=−∞ 2 4
GEL7060 – (Communications numériques avancées) 6

(c) (0.5 point) Compute its period-averaged autocorrelation function.

T
1
Z
2
R̄SS (τ ) = RSS (t, t + τ )dt
T − T2
∞ ∞ Z T
1 X X 2
= RAA [l − k]g(t − kT )g(t + τ − lT )dt
T k=−∞ l=−∞ − T2
∞ ∞ Z T
1 X X 2
= g(t − kT )g(t + τ − lT )dt
T k=−∞ l=−∞ − T2
∞ ∞ Z T 
1 X T
 
X 2
= RAA [l − k] u(t − kT ) − u t − kT − ×
T k=−∞ l=−∞ − T2 2
T
  
u(t + τ − lT ) − u t + τ − lT − dt
2
∞ Z T 
1 X T
 
2
= RAA [l] u(t + τ − lT ) − u t + τ − lT − dt
T l=−∞ 0 2
 T h i
∞  R2 T T
1 X lT −τ dt = 2 − lT + τ, if lT − τ ∈ 0, 2 ,
= T
 lT −τ − 2 dt = lT − τ − T , if lT − τ ∈ − T , 0 .
h i
T l=−∞  R
0 2 2
 h i
1 X ∞  T − lT + τ, if τ ∈ lT − T , lT ,
2 2
= RAA [l] h i
T l=−∞  lT − τ − T , if τ ∈ lT, lT + T .
2 2
1 T T T
    
= − τ+ sign(τ )W − , +
2T 2 2 2
1 T T 3T
   
τ+ sign(τ − T )W , +
2 2 2 2
1 3T 3T T
   
τ+ sign(τ + T )W − , ,
2 2 2 2
where
W (a, b) = u(t − a) − u(t − b). (1)
GEL7060 – (Communications numériques avancées) 7

(d) (0.5 point) Compute its period-averaged mean.

∞
Z T X
1
µ̄S = µA g(t − kT )dt
T 0 k=−∞
∞ Z T
1 X
= µA g(t − kT )dt
T k=−∞ 0

1
= µA .
2

4. (3 points) Random variables Consider two χ2 random variables Z1 et Z2

respectively having PDFs
1
fZ1 (x) = xn1 −1 e−x ,
Γ(n1 )
1
fZ2 (x) = xn2 −1 e−x/2.5 ,
Γ(n2 )2.5n2
Z1
(a) (2 points) What is the PDF of the ratio X = Z2
?
We proceed using the Mellin convolution. Thus, we need to express the problem
as a product of random variables.
First, we apply the transformation Y = Z12 , yielding the PDF

1 1
−1−n2 − 2.5y
fY (y) = y e .
Γ(n2 )2.5n2
Z1
Then, using the Mellin convolution, we compute the PDF of X = Z2 = Z1 Y :
Z ∞
x 1
 
fX (x) = fY fZ1 (a) da
0 a a
x −1−n2
an1 −1 e−a
Z ∞

a
= a
e− 2.5x da
0 Γ(n2 )2.5n2 Γ(n1 )a
1 ∞ Z
1
= an1 +n2 −1 e−a(1+ 2.5x ) da
Γ(n1 )Γ(n2 )2.5n2 x1+n2 0
n1 +n2
1 5x

= Γ(n1 + n2 )
Γ(n1 )Γ(n2 )2.5n2 x1+n2 5x + 2
xn1 −1
=  n1 +n2 u(x).
B(n1 , n2 )2.5n2 x + 25
GEL7060 – (Communications numériques avancées) 8

(b) (1 point) If n1 = 2 and n2 = 2, derive the PDF of Y = Z1 + Z2 .

Hint : use characteristic functions and partial fractions expansion.
If n1 = n2 = 2, this corresponds to 2 complex degrees of freedom, or 4 real
degrees of freedom. Hence, we have

fZ1 (x) = xe−x u(x),

x
xe− 2.5
fZ2 (x) = u(x),
2.52
and the corresponding characteristic functions are
1
φZ1 (jt) = ,
(1 − jt)2
1
φZ2 (jt) = .
1 − 2.5jt)2

The characteristic function of Y = Z1 + Z2 is given by

1
φY (jt) =
(1 − jt)2 (1
− 2.5jt)2
A1 A2 A3 A4
= + + + .
(1 − jt)2 1 − jt (1 − 2.5jt)2 1 − 2.5jt

To resolve the partial fractions expansion, we can apply Heaviside’s formula or

just re-express the sum with a common denominator. The numerator must be
equal to 1 and we have

A1 (1 − 2.5jt)2 + A2 (1 − jt)(1 − 2.5jt)2 +

A3 (1 − jt)2 + A4 (1 − jt)2 (1 − 2.5jt) = 1,

from which we can deduce

A1 + A2 + A3 + A4 = 1,
−5A1 − 6A2 − 2A3 − 4.5A4 = 0,
−6.25A1 − (6.25 + 5)A2 − A3 − (1 + 5)A4 = 0,
6.25A2 + 2.5A4 = 0,
GEL7060 – (Communications numériques avancées) 9

which leads to the system of equations

    
1 1 1 1 A1
 −5 −6 −2 −4.5 
  A2
   0 
= .
   
−6.25 −11.25 −1 −6   A3   0 
 

0 6.25 0 2.5 A4 0

Resolving the system, we find that

A1 = 0.4444,
A2 = 1.4815,
A3 = 2.7778,
A4 = −3.7037.

Hence, we have
A3 − x A4 − x
fY (y) = A1 xe−x + A2 e−x + 2
xe 2.5 + e 2.5
2.5 2.5
A3 A4
 
x
= (A1 x + A2 ) e−x + 2
x + e− 2.5 .
2.5 2.5