Example 1:
Consider a signal X that follows a Gaussian distribution with mean b and variance a. The signal X is to be
quantized by an optimal scalar 2-level scalar Quantizer where the representation levels are y1<y2
i.
Design the optimal scalar quantizer for the signal fX(x) as a function of ‘a’ and ‘b’
Let work on a Gaussian Random Variable with mean 0 and variance ‘a’
Gaussian PDF is symmetric  y1 = -y2 and x1 = 0
[ ]
∫
[ ]
∫
∫
√
∫
√
√
√
(
)|
(
)|
( )|
√
√
√
Quantizer Levels are:
√
,
√
ii. Calculate the SQNR for the corresponding optimal 4-level transform coding quantizer
[
[
]
]
√
∫(
[(
√
(
)
∫
√
∫
(√
)
√
∫(
√
(
√
∫(
√
[ ]
)
)
(√
√
∫
)
)
∫
√
)
∫
√
)
[
]
[
]
[
[
]
]
√
√
()
(
√
∫
)]
Example 2:
Consider a random variable T that follows a Pareto distribution where the pdf fT(t) is expressed as
 βα
α
fT t    t α 1
 0

i.
t β
t β
Derive the mean for the random variable T (What is the restriction on the value of α? Explain your
answer)


β
β
E T   tfT t  dt  tα


E T   αβα

1
t
α
βα
dt
t α 1
dt
β
E T   αβ
α


t α dt  αβα
β
t  α 1
α  1

β
 α 1


β  α 1 
α β
E T   αβα 0 
  αβ 

 α  1 
 α 1 
 αβ 
E T   

α 1
α > 1 Otherwise E[T] would be infinite and then negative
ii. Derive the variance for T (What is the restriction on the value of α? Explain your answer)


β
β
E T 2   t 2 fT t  dt  t 2α

E T 2   αβα


t
1
α 1
βα
dt
t α 1
dt
β

t  α 2
E T 2   αβα t α 1dt  αβα
α  2
β


β
 α 2


β  α 2 
α β
E T 2   αβα 0 
  αβ 

 α  2 
 α 2 
 αβ2 
E T 2   

α 2
 αβ2   αβ 
2
Var T   E T 2   E T   


α 2  α 1
2
 αβ2  α  12   αβ 2  α  2  

Var T   


 α  2 α  12


 αβ2 α2  2α  1   αβ 2  α  2    3 2
2 2
2
3 2
2 2
   α β  2α β  αβ  α β  2α β 
Var T   

 

 α  2 α  12
 α  2 α  12

 


αβ2

Var T   
2
  α  2  α  1 
α > 2 Otherwise Var[T] would be infinite and then negative
iii. Internet traffic packet sizes L could be modeled using a truncated Pareto distribution where the
parameter β depicts the minimum allowable packet size. The maximum allowable packet size is γ =
15000 Bytes. Suppose we would like to develop a model that approximates Internet packets into
two possible sizes L1, L2. Determine the values of L1, L2 that will minimize the mean square error of
the approximation given that the minimum packet size is 40 Bytes and α =1.5.
βα
α 1
fT t 
1
1
fL t   fT t β  t  γ  
γ t
 α 1 γ
α
P β  t  γ 
t
β
α α 1 dt
t α 1dt
t
β
β
α

fL t  
1
t

α 1
α
t α
γ
β


1

α 1   α
α
t
 β γ
α
X1
 tf t dt
L
L1  E L β  t  X1  
β
X1
 f  t  dt
L
β


1  αγ α βα 

 α 1   α α 
 t
 γ β 
X1
 αγ α βα  1
t  α α  α  1 dt
 γ β t
L1  βX

1
 αγ α βα  1
 α α  α  1 dt
γ β t
β 
X1



β
1
dt
tα
X1
X1

1
t
α
dt
β
X1
t
dt
α 1
β
t  α  1 X1
1α
β
 α t

α

1
L1  α


t X1
 α  1  t α
β
α
t
 α 1
dt
β
X1
β
X1
β
1α
1α
 α   X1  β 
L1  


α
α 
 α  1   X1  β 
γ
 tf t dt
L
L2  E L X1  t  γ  
X1
γ
 f  t  dt
L
X1
γ
L2 
 αγ α βα
t α α
γ β
X1 

γ
 αγ β
α
 βα
  γ
X1
α α
 1
 α 1 dt
t
 1
 α 1 dt
t
γ


X1
γ
t
1
dt
tα
1
α 1
X1
t  α 1 γ
1α
X1
 α t

α

1
L2  α


t γ
 α  1  t α
X
α 1
dt
γ

t
α
dt
X1
γ
t
 α 1
dt
X1
γ
X1
γ
X1
1α
1α
 α   γ  X1 
L2  

 α
α 
 α  1   γ  X1 
L1  L2  2X1
For α = 1.5 and β=40, γ = 15000 and solving by trail and error
X1  2239.5 Bytes
L1  104 Bytes
L2  4375 Bytes
Example 3:
Assume that the CDFs shown below represent the call duration CDFs for the three mobile network
providers in EGYPT. For any given mobile subscriber assume the probability that this user is an Operator
A subscriber is equal to 0.4 and the probability that this user is an Operator B subscriber is equal to 0.35
and the probability that this user is an Operator C subscriber is equal to 0.25.
1
1 2 3 4 5 6 7 8
Call Duration in minutes
CDF of call durations in
Operator A
i.
CDF
1
CDF
CDF
1
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
Call Duration in minutes
Call Duration in minutes
CDF of call durations in
Operator B
CDF of call durations in
Operator C
Write down the pdf of the call duration for subscribers of Operator B
t 0
0

CDFOp2 t   0.2t 0  t  5
1
t 5

t 0
0

PDFOp2 t   0.2 0  t  5
0
t 5

ii. What is the average call duration in Egypt
ET   E T OP1 POP1  E T OP2 POP2  E T OP3 P OP3
ET   1.5  0.4  2.5  0.35  4  0.25  0.6  0.875  1
ET   2.475
iii. Use the figure below to draw the pdf for the call duration in EGYPT
PDF
f T   f T OP1 POP1  f T OP2 P OP2  f T OP3 P OP3
0
0.333  0.4  0.2  0.35  0.125  0.25

PDF t   0.2  0.35  0.125  0.25
0.125  0.25

0
0
0.234583

PDF t   0.10125
0.03125

0
t 0
0t 3
3t 5
5t 8
t 8
t 0
0t 3
3t 5
5t 8
t 8
iv. What is the probability that the user making a call is an Operator C user given
that the call duration is less than 4 minutes
P T  4 OP3 P OP3
P OP3 T  4   
P T  4 
P OP3 T  4  
0.125  4 0.25
0.234583  3  0.10125
 0.1553
v. If Operator C decides to charges 20 Pt per call for any call that is less than two
minutes and charges and 40 Pt per call for any call that is more than two
minutes what would be the average call charge.
EC   0.25  20  0.75  40  35