# Exercise 4.6: Quantization Characteristics

Non-linear quantization is considered.  The system model according to  Exercise 4.5 still applies.

The graph shows two compressor characteristics  $q_{\rm K}(q_{\rm A})$:

• Drawn in red is the so-called  "A-characteristic" recommended by the CCITT  ("Comité Consultatif International Téléphonique et Télégraphique")  for the standard system PCM 30/32.  For  $0 ≤ q_{\rm A} ≤ 1$  applies here:
$$q_{\rm K}(q_{\rm A}) = \left\{ \begin{array}{l} \frac{1 \hspace{0.05cm}+\hspace{0.05cm} {\rm ln}(A \hspace{0.05cm}\cdot \hspace{0.05cm}q_{\rm A})} {1 \hspace{0.05cm}+ \hspace{0.05cm}{\rm ln}(A )} \\ \\ \frac{A \hspace{0.05cm}\cdot \hspace{0.05cm}q_{\rm A}} {1 \hspace{0.05cm}+ \hspace{0.05cm}{\rm ln}(A )} \\ \end{array} \right.\quad \begin{array}{*{10}c} {{1}/{A} \le q_{\rm A} \le 1} \hspace{0.05cm}, \\ \\ {q_{\rm A} < {1}/{A}} \hspace{0.05cm}. \\ \end{array}$$
• The blue-dashed curve applies to the so-called  "13-segment characteristic".  This is obtained from the A-characteristic by piecewise linearization;  it is treated in detail in the  Exercise 4.5 .

Hints:

• The Exercise belongs to the chapter  "Pulse Code Modulation".
• Reference is made in particular to the page  "Compression and Expansion".
• For the A-characteristic drawn in solid red,  the quantization parameter  $A = 100$  is chosen.  With the value  $A = 87.56$  suggested by CCITT,  a similar curve is obtained.
• For the other two curves,
•  $A = A_1$  (dash–dotted curve) and
•  $A = A_2$  (dotted curve),

where for  $A_1$  and  $A_2$  the two possible numerical values  $50$  and  $200$  are given.  In the subtask  (3)  you are to decide which curve belongs to which numerical value.

### Questions

1

What are the arguments for non-linear quantization?

 The larger SNR – even with equally likely amplitudes. For audio,  small amplitudes are more likely than large ones. The distortion of small amplitudes is subjectively more disturbing.

2

What are the differences between the  "A-characteristic" and the  "13-segment characteristic"?

 The A-characteristic curve describes a continuous course. The 13-segment curve approximates the A-characteristic linearly piece by piece. In the realization,  the A-characteristic shows significant advantages.

3

Can the parameter  $A$  be derived from  $q_{\rm A} = 1$   ⇒   $q_{\rm K} = 1$  alone?

 Yes. No.

4

Can the parameter  $A$  be determined if we specify that the transition between the two domains should be continuous?

 Yes. No.

5

Determine the parameter  $A$  from the condition  $q_{\rm K}(q_{\rm A} = 1/2) = 0.8756$.

 $A \ = \$

6

What parameter values were used for the other curves?

 It holds  $A_1 = 50$  and  $A_2 = 200$. It holds  $A_1 = 200$  and  $A_2 = 50$.

### Solution

#### Solution

(1)  Correct are the  statements 2 and 3:

• Signal distortion of soft sounds or in speech pauses is subjectively perceived as more disturbing than e.g. additional noise in heavy metal.
• In terms of quantization noise or SNR,  however,  there is no improvement due to non-linear quantization if an uniformly distribution of the amplitude values is assumed.
• However,  if one considers that in speech and music signals smaller amplitudes occur much more frequently than large   ⇒   "Laplace distribution",  non-linear quantization also results in a better SNR.

(2)  Correct are the statements 1 and 2:

• Due to the linearization in the individual segments,  the interval width of the various quantization levels is constant in these for the  "13-segment characteristic",  which has a favorable effect in realization.
• In contrast,  with the non-linear quantization according to the  "A-characteristic",  there are no quantization intervals of equal width.
This means:   The statement 3 is false.

(3)  Correct is  "NO":

• For  $q_{\rm A} = 1$  one obtains independently of  $A$  the value  $q_{\rm K} = 1$.
• So with this specification alone  $A$  cannot be determined.

(4)  Correct is again  "NO":

• For  $q_{\rm A} = 1/A$  both range equations yield the same value  $q_{\rm K}= 1/[1 + \ln(A)]$.
• Also with this  $A$  cannot be determined.

(5)  With this requirement  $A$  is now computable:

$$0.875 = \frac{1 \hspace{0.05cm}+\hspace{0.05cm} {\rm ln}(A/2)} {1 \hspace{0.05cm}+ \hspace{0.05cm}{\rm ln}(A )} = \frac{1\hspace{0.05cm}-\hspace{0.05cm} {\rm ln}(2) \hspace{0.05cm}+\hspace{0.05cm} {\rm ln}(A)} {1 \hspace{0.05cm}+ \hspace{0.05cm}{\rm ln}(A )}\approx \frac{1-0.693 \hspace{0.05cm}+\hspace{0.05cm} {\rm ln}(A)} {1 \hspace{0.05cm}+ \hspace{0.05cm}{\rm ln}(A )}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm ln}(A) = \frac{0.875 - 0.307 } {1 -0.875 }= 4.544 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} A \hspace{0.15cm}\underline {\approx 94} \hspace{0.05cm}.$$

(6)  Correct is  statement 2:

• The curve for  $A_1 = 200$  lies above the curve with  $A = 100$,  the curve with  $A_2 = 50$  below.
• This is shown by the following calculation for  $q_{\rm A} = 0.5$:
$$A= 100\text{:}\hspace{0.2cm} q_{\rm K}= \frac{1 + \ln(100) - \ln(2)}{1 + \ln(100)}= \frac{1+4.605- 0.693} {1 +4.605}\approx 0.876 \hspace{0.05cm},$$
$$A= 200\text{:}\hspace{0.2cm} q_{\rm K}= \frac{1+5.298- 0.693} {1 +5.298}\approx 0.890 \hspace{0.05cm},$$
$$A= 50\text{:}\hspace{0.4cm} q_{\rm K}= \frac{1+3.912- 0.693} {1 +3.912}\approx 0.859 \hspace{0.05cm}.$$