Exercise 4.6: Quantization Characteristics

From LNTwww

Non-linear 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 .


  • 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.



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.


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.


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



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



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

$A \ = \ $


What parameter values were used for the other curves?

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


(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  $q_{\rm A} = 0.5$:
$$A= 100\text{:}\hspace{0.2cm} q_{\rm K}= \frac{1 + \ln(100) - \ln(2)}{1 + \ln(100)}= \frac{1+4.605- 0.693} {1 +4.605}\approx 0.876 \hspace{0.05cm},$$
$$A= 200\text{:}\hspace{0.2cm} q_{\rm K}= \frac{1+5.298- 0.693} {1 +5.298}\approx 0.890 \hspace{0.05cm},$$
$$A= 50\text{:}\hspace{0.4cm} q_{\rm K}= \frac{1+3.912- 0.693} {1 +3.912}\approx 0.859 \hspace{0.05cm}.$$