#### Transcript MS PowerPoint

```Sampling
and
Pulse Code Modulation
ECE460
Spring, 2012
How to Measure Information Flow
• Signals of interest are band-limited which lend themselves to
sampling (e.g., discrete values)
• Study the simplest model: Discrete Memoryless Source (DMS)
• Alphabet: A  {a1 , a2 , , aN }
• Probability Mass Function:
pi  P( X  ai )
The DMS is fully defined given its Alphabet and PMF.
• How to measure information flow where a1 is the most likely
and aN is the least likely:
1. Information content of output aj depends only on the
probability of aj and not on the value.
Denote this as I(pj) – called self-information
2. Self-information is a continuous function of pj
3. Self-information is increases as pj decreases.
j1
j2
4. If pj = p(j1) p(j2), then I  p j   I p    I p  

 

Only function that satisfies all these properties:
I  x    log  x 
Unit of measure:
•
•
Log2 Loge -
bits (b)
nats
2
Entropy
The mean value of I(xi) over the alphabet of source X with N
different symbols is given by
H  X   E  I  xi  
N
  pi log  pi 
i 1
 1 
  pi log  
i 1
 pi 
N
Note: 0 log 0 = 0
H(X) is called entropy and is a measure of the average
information content per source symbol and is measured in
b/symbol
Source entropy H(X) is bounded:
0  H  X   log 2 N
3
Entropy Example
A DMS X has an alphabet of four symbols, {x1, x2, x3, x4} with
probabilities P(x1) = 0.4, P(x1) = 0.3, P(x1) = 0.2, and P(x1) = 0.1.
a) Calculate the average bits/symbol for this system.
b) Find the amount of information contained in the messages
x1, x2, x1, x3 and x4, x3, x3, x2 and compare it to the average.
c) If the source has a bandwidth of 4000 Hz and it is sampled
at the Nyquist rate, determine the average rate of the
source in bits/sec.
4
Joint & Conditional Entropy
Definitions
• Joint entropy of two discrete random variables (X, Y)
H  X , Y    p  xi , y j  log p  xi , y j 
n
m
i 1 j 1
• Conditional entropy of the random variable X given Y
H  X | Y    p  xi , y j  log p  xi | y j 
n
m
i 1 j 1
• Relationships
H  X , Y   H  X | Y   H Y 
H  X , Y   H Y | X   H  X 
• Entropy rate of a stationary discrete-time random process is
defined by
Memoryless:
Memory:
H  X   lim H  X n | X 1 , X 2 ,
n 
1
H  X   lim H  X n | X 1 , X 2 ,
n  n
, X n 1 
, X n 1 
5
Example
Two binary random variables X and Y are distributed according
to the joint distribution
p  X  Y  0   p  X  0, Y  1  p  X  Y  1 
1
3
Compute:
• H(X)
• H(Y)
• H(X,Y)
• H(X|Y)
• H(Y|X)
6
7
Source Coding
Which of these codes are viable?
Symbol Code 1 Code 2 Code 3
a
00
00
0
b
00
01
1
c
11
10
11
Code 4
1
10
100
Code 5
1
00
01
Code 6
1
01
11
Which have uniquely decodable properties?
Which are prefix-free?
A sufficient (but not necessary) condition to assure that a
code is uniquely decodable is that no code word be the prefix of any
other code words.
Which are instantaneously decodable?
Those for which the boundary of the present code word can be
identified by the end of the present code word rather than by the
beginning of the next code word.
Theorem: A source with entropy H can be encoded with arbitrarily
small error probability at any rate R (bits/source
output)as long as R > H. Conversely if
R < H, the error probability will be bounded away from
zero, independent of the complexity of the encoder and
the decoder employed.
R: the average code word length per source symbol
R
 p  x l  x
where l  x  is the length of the code word
x X
8
Huffman Coding
Steps to Huffman Codes:
1. List the source symbols in order of decreasing probability
2. Combine the probabilities of the two symbols having the
lowest probabilities, and reorder the resultant probabilities;
this step is called reduction 1. The same procedure is
repeated until there are two ordered probabilities
remaining.
3. Start encoding with the last reduction, which consist of
exactly two ordered probabilities. Assign 0 as the first digit
in the code words for all the source symbols associated with
the first probability; assign 1 to the second probability.
4. Now go back and assign 0 and 1 to the second digit for the
two probabilities that were combined in the previous
reduction step, retaining all assignments made in step 3.
5. Keep regressing this way until the first column is reached.
Huffman code: H  X   R  H  X   1
9
Example
Sym Prob
x1 0.30
x2
0.25
x3
0.20
x4
0.12
x5
0.08
x6
0.05
Code
H  X :
R:
10
Quantization
Scalar Quantization
x  ¡ is partitioned into N disjoint subsets k ,1  k  N .
If at time i, xi belongs to k , then xi is represented by xˆk .
For example:
  x  a1
a1  x  a2
a2  x  a3
a3  x  a4
a4  x  a5
a5  x  a6
a6  x  a7
a7  x  
xˆ1
xˆ2
xˆ3
xˆ4
xˆ5
xˆ6
xˆ7
xˆ8
N is usually chosen to be a power of 2
 R  log 2 N
and, for this example, R  log 2 8  3
11
Quantization
Define a Quantization Function Q as
Q  x   xˆi
for all x i
Is the quantization function invertible?
Define the squared-error distortion for a single measurement:
d  x, xˆ    x  Q  x    x%2
2
Since X is a random variable, so are Xˆ and X%, and the distortion
D for the source is the expected value of this random variable


2
D  E  d X , Xˆ   E  X  Q  X  


12
Example
The source X(t) is a stationary Gaussian source with mean zero
and power-spectral density
2,
SX  f   
0,
f  100 Hz
otherwise
The source is sampled at the Nyquest rate, and each sample is
quantized using the 8-level quantizer shown below.
What is the rate R and the distortion D?
  x  60
60  x  40
40  x  20
20  x  0
0  x  20
20  x  40
40  x  60
60  x  
xˆ1  70
xˆ2  50
xˆ3  30
xˆ4  10
xˆ5  10
xˆ6  30
xˆ7  50
xˆ8  70
13
Signal-to-Quantized Noise Ratio
In the example, we used the mean-squared distortion, or
quantization noise, as the measure of performance. SQNR
provides a better measure:
Definition: If the random variable X is quantized Q(x), the
signal-to-quantized noise ration (SQNR) is defined by
SQNR 
E  X 2 
E  X  Q  X  
2
For signals, the quantization-noise power is
T /2
2
1


PX  lim
E
X
t

Q
X
t
    dt
  
T  T 
T /2
And the signal power is
T /2
1
PX  lim
E  X 2  t   dt

T  T
T /2
Therefore:
SQNR 
PX
PX
If X(t) is a strictly stationary random process, how is the SQNR
related to the autocorrelation functions of X(t) and X% t  ?
14
Uniform Quantization
The example before was for a uniform quantization where the
first and last ranges were (-∞,-70] and (70, ∞). The remaining
ranges went in steps of 20 (e.g., (-70, -50], … , (50, 70] ).
We can generalize for a zero-mean, unit variance Gaussian by
breaking the ranges into N symmetric segments of width Δ
about the origin. The distortion would then be given by:
Even N:
 N 
  1 
 2 

D  2
N
1
2
 x  xˆ1 
 N

   i 1 
2


 N 
  i 
 2 
2  
i 1
Odd N:
 N 
  1 
 2 

D  2
N 3
2
2  
i 1
f X  x  dx
 x  xˆi 1 
 x  xˆ1 
 N

   i 1 
 2

 N 
  i 
 2 
2
2
2
f X  x  dx
f X  x  dx
 x  xˆi 1 
2
f X  x  dx

2


2
  x 2 f X  x  dx
How do you select the optimized value for Δ give N ?
15
Optimal Uniform Quantizer
16
Non-Uniform Quantizers
The Lloyd-Max conditions for optimal quantization:
1. The boundaries of the quantization regions are the
midpoints of the corresponding quantized values (nearest
neighbors)
2. The quantized values are the centroids of the quantization
regions.
Typically determined by trial and error! Note the optimal nonuniform quantizer table for a zero-mean, unit variance Gaussian
source.
Example (Continued):
How would the results of the previous example change if instead
of the uniform quantizer we used an optimal non-uniform
quantizer with the same number of levels?
17
Optimized Non-Uniform Quantizer
18
Waveform Coding
Our focus will be on Pulse-Code Modulation (PCM)
x t 
xn
xˆn
Uniform PCM
• Input range for x(t): [ xmax ,  xmax ]
• Length of each quantization region:

2 xmax xmax
  1
N
2
• xˆn is chosen as the midpoint of each quantization level
• Quantization error x% x  Q  x  is a random variable in the
interval   / 2,  / 2
19
Uniform PCM
• Typically, N is very high and the variations of input low so
that the error can be approximated as uniformly distributed
on the interval   / 2,  / 2
• This give the quantization noise as
 /2 1
E  x%2   
x%2 dx%
 /2 
2

12
2
xmax

3N 2
2
xmax

3  4
• Signal-to-quantization noise ration (SQNR)
(2
X 2 3  N 2 X 2 3  4 X 2

SQNR 


 3 4 X
2
2
2
%
x
x
X
max
max
(
X
where X 
xmax
20
Uniform PCM
Example: What is the resulting SQNR for a signal uniformly
distributed on [-1,1] when uniform PCM with 256
levels is employed? What is the minimum bandwidth
required?
21
How to Improve Results
A more effective method would be to make use a non-uniform
PCM:
g  x

i
xn 
x
xmax
Speech coding in the U.S. uses the µ-law compander with a
typical value of µ = 255:

x 
ln 1  

x
max
 sgn m
g  x  
 
ln 1   
x
xmax
1
22
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