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Communication Systems Engineering Episode 1 Part 4 docx

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Nội dung

– Non-uniform quantization regions Finer regions around more likely values – Optimal quantization values not necessarily the region midpoints – Use uniform quantizer anyway Optimal c

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Sampling provides a discrete-time representation of a continuous waveform

– Sample points are real-valued numbers – In order to transmit over a digital system we must first convert into

discrete valued numbers

Quantization levels

Q3

Q2

Q1 � �

� � � � � �

Sample points

What are the quantization regions

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∆ ∆ 3∆

−3∆ −2∆ −∆

∆ 2

– Except first and last regions if samples are not finite valued

quantized value

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Quantization Error

e(x) = Q(x) - x

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E X

Example

f(x) = 1/2A, -A<=x<=A and 0 otherwise

– Q(x) = quantization level = midpoint of quantization region in which x lies

D = E e x [ ( ) | x2 ∈ Ri ] = ∫−∆ ∆ / 2 / 2 x2 f (x)dx = 1 ∆ / 2 ∆2

∆ ∫−∆ / 2 x2 dx =

12

1 A

2 A2

[ ] =

2 A ∫− Ax dx =

3

∆2

/12 =

(2 A N)2 /12 = N 2, ( ∆ = 2 A / N)

/

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– Non-uniform quantization regions

Finer regions around more likely values

– Optimal quantization values not necessarily the region midpoints

– Use uniform quantizer anyway

Optimal choice of

– Use non-uniform quantizer

Choice of quantization regions and values

– Transform signal into one that looks uniform and use uniform

quantizer

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∆∆∆

– Find the optimal value of

– Find the optimal quantization values within each region – Optimization over N+2 variables

quantization regions (except first and last regions, when input not finite valued)

– Solution depends on input pdf and can be done numerically for

commonly used pdfs (e.g., Gaussian pdf, table 6.2, p 296 of text)

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f x

Uniform quantizer example

x ( ) =

2 πσ

1

ex 2 / 2σ2, σ2 = 1

– Notice that H(Q) = the entropy of the quantized source is < 2 – Two bits can be used to represent 4quantization levels

– Soon we will learn that you only need H(Q) bits

− ∆

q1 = -3∆/2 q2 =∆/2 q3 =∆/2 q4 = 3∆/2

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Quantization regions need not be of same length

– Given quantization regions, what should the quantization levels be? – What should the quantization regions be?

– Minimize distortion

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Optimal quantization levels

a

a i−

– Optimal value affects distortion

only within its region dDR

= ∫a i

2(x x√ i )2 fx (x)dx =

a

x√ i = ∫ xfx ( | ai −1 ≤ x ai

a i

[ | ai −1 ≤ x ai

– The conditional expected value of that region

uniform quantizer as well

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Optimal quantization regions

– Take derivative with respect to integral boundaries

2

dD

= f ax ( i )[(ai x√ i )2 − (ai x√ i +1) ] = 0

dai

√ x

xi + i +1

ai =

2

– Boundaries of the quantization regions are the midpoint of the

quantization values

1 Quantization values are the “centroid” of their region

2 Boundaries of the quantization regions are the midpoint of the quantization values

3 Clearly 1 depends on 2 and visa-versa The two can be solved iteratively to obtain optimal quantizer

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A) Find optimal quantization values (“centroids”)

B) Use quantization values to get new regions (“midpoints”)

– Repeat A & B until convergence is achieved

– Table 6.3 (p 299) gives optimal quantizer for Gaussian source

– D = 0.1175, H(x) = 1.911 – Recall: uniform quantizer, D= 0.1188, H(x) = 1.904 (slight improvement)

1.51 0.4528

-0.9816

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µµµ µµµ

g x

Companders

– Requires knowledge of source statistics – Different quantizers for different input types

then use uniform quantizer

( ) = Log(1 + µ | x |)

sgn(x)

Log(1 + µ )

– µ controls the level of compression

– µ = 255 typically used for voice

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× ×

Pulse code modulation

µ -law Uniform Q

– N = 2 V quantization levels, each level encoded using v bits

– SQNR: same as uniform quantizer

v

E X 2

XMAX

– Notice that increasing the number of bits by 1 decreases SQNR by a

factor of 4 (6 dB)

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µµµ

– Sample at 8KHZ => Ts = 1/8000 = 125 µs

8000 samples per second at 7 bits per sample => 56 Kbps

– Speech samples are typically correlated

– Instead of coding samples independently, code the difference

between samples

– Result: improved performance, lower bit rate speech

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