View source for Aufgaben:Exercise 4.6: Quantization Characteristics


{{quiz-Header|Buchseite=Modulation_Methods/Pulse_Code_Modulation
}}

[[File:P_ID1623__Mod_Z_4_5.png|right|frame|Non-linear quantization characteristics]]
Non-linear quantization is considered.&nbsp; The system model according to&nbsp; [[Aufgaben:Exercise_4.5:_Non-Linear_Quantization| Exercise 4.5]] still applies. 

The graph shows two compressor characteristics &nbsp;$q_{\rm K}(q_{\rm A})$:
* Drawn in red is the so-called&nbsp; "'''A-characteristic'''" recommended by the CCITT&nbsp; ("Comité Consultatif International Téléphonique et Télégraphique")&nbsp; for the standard system PCM 30/32.&nbsp; For &nbsp;$0 ≤ q_{\rm A} ≤ 1$&nbsp; 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&nbsp; "'''13-segment characteristic'''".&nbsp; This is obtained from the A-characteristic by piecewise linearization;&nbsp; it is treated in detail in the&nbsp; [[Aufgaben:Exercise_4.5:_Non-Linear_Quantization| Exercise 4.5]]&nbsp;.







Hints: 
*The Exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|"Pulse Code Modulation"]].
*Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Compression_and_expansion|"Compression and Expansion"]]. 
*For the A-characteristic drawn in solid red,&nbsp; the quantization parameter &nbsp;$A = 100$&nbsp; is chosen.&nbsp; With the value &nbsp;$A = 87.56$&nbsp; suggested by CCITT,&nbsp; a similar curve is obtained. 
*For the other two curves, 
:*&nbsp;$A = A_1$&nbsp; (dash&ndash;dotted curve) and 
:*&nbsp;$A = A_2$&nbsp; (dotted curve),&nbsp; 

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




===Questions===

<quiz display=simple>
{What are the arguments for non-linear quantization?
|type="[]"}
- The larger SNR &ndash; even with equally likely amplitudes. 
+ For audio,&nbsp; small amplitudes are more likely than large ones.
+ The distortion of small amplitudes is subjectively more disturbing.

{What are the differences between the&nbsp; "A-characteristic" and the&nbsp; "13-segment characteristic"?
|type="[]"}
+ The A-characteristic curve describes a continuous course.
+ The 13-segment curve approximates the A-characteristic linearly piece by piece.
- In the realization,&nbsp; the A-characteristic shows significant advantages.

{Can the parameter &nbsp;$A$&nbsp; be derived from &nbsp;$q_{\rm A} = 1$ &nbsp; ⇒ &nbsp;  $q_{\rm K} = 1$&nbsp; alone?
|type="()"}
- Yes. 
+ No. 

{Can the parameter &nbsp;$A$&nbsp; be determined if we specify that the transition between the two domains should be continuous? 
|type="()"}	
- Yes. 
+ No. 

{Determine the parameter &nbsp;$A$&nbsp; from the condition &nbsp;$q_{\rm K}(q_{\rm A} = 1/2) = 0.8756$.
|type="{}"}
$A \ = \ $ { 94 3% } 

{What parameter values were used for the other curves?
|type="()"}
- It holds &nbsp;$A_1 = 50$&nbsp; and &nbsp;$A_2 = 200$.
+ It holds &nbsp;$A_1 = 200$&nbsp; and &nbsp;$A_2 = 50$.




</quiz>

===Solution===
{{ML-Kopf}}

'''(1)'''&nbsp; Correct are the&nbsp; <u>statements 2 and 3</u>: 
*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,&nbsp; however,&nbsp; there is no improvement due to non-linear quantization if an uniformly distribution of the amplitude values is assumed. 
*However,&nbsp; if one considers that in speech and music signals smaller amplitudes occur much more frequently than large &nbsp; &rArr; &nbsp; "Laplace distribution",&nbsp; non-linear quantization also results in a better SNR.



'''(2)'''&nbsp; Correct are the&nbsp;<u>statements 1 and 2</u>: 
*Due to the linearization in the individual segments,&nbsp; the interval width of the various quantization levels is constant in these for the&nbsp; "13-segment characteristic",&nbsp; which has a favorable effect in realization. 
*In contrast,&nbsp; with the non-linear quantization according to the&nbsp; "A-characteristic",&nbsp; there are no quantization intervals of equal width.&nbsp; <br>This means: &nbsp; The statement 3 is false.



'''(3)'''&nbsp; Correct is&nbsp; "<u>NO</u>":
*For&nbsp; $q_{\rm A} = 1$&nbsp; one obtains independently of&nbsp; $A$&nbsp; the value&nbsp; $q_{\rm K} = 1$. 
*So with this specification alone&nbsp; $A$&nbsp; cannot be determined.
 


'''(4)'''&nbsp; Correct is again&nbsp; "<u>NO</u>":
*For&nbsp; $q_{\rm A} = 1/A$&nbsp; both range equations yield the same value&nbsp; $q_{\rm K}= 1/[1 + \ln(A)]$. 
*Also with this&nbsp; $A$&nbsp; cannot be determined. 



'''(5)'''&nbsp; With this requirement&nbsp; $A$&nbsp; 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)'''&nbsp; Correct is&nbsp; <u>statement 2</u>:
*The curve for &nbsp;$A_1 = 200$&nbsp; lies above the curve with &nbsp;$A = 100$,&nbsp; the curve with &nbsp;$A_2 = 50$&nbsp; below. 
*This is shown by the following calculation for &nbsp;$q_{\rm A} = 0.5$:
:$$A= 100\text{:}\hspace{0.2cm} q_{\rm K}= \frac{1 + \ln(100) - \ln(2)}{1 + \ln(100)}=
\frac{1+4.605- 0.693} {1 +4.605}\approx
0.876 \hspace{0.05cm},$$
:$$A= 200\text{:}\hspace{0.2cm} q_{\rm K}= \frac{1+5.298- 0.693} {1 +5.298}\approx
0.890 \hspace{0.05cm},$$
:$$A= 50\text{:}\hspace{0.4cm} q_{\rm K}= \frac{1+3.912- 0.693} {1 +3.912}\approx
0.859 \hspace{0.05cm}.$$





{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^4.1 Pulse Code Modulation^]]