Difference between revisions of "Aufgaben:Exercise 5.8Z: Falsification of BMP Images"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Anwendungen bei Multimedia–Dateien
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Applications_for_Multimedia_Files
 
}}
 
}}
  
[[File:P_ID1856__Dig_Z_5_8.png|right|frame|Verfälschte BMP–Dateien]]
+
[[File:P_ID1856__Dig_Z_5_8.png|right|frame|Falsified BMP files <br>$($"Erde" &nbsp; &rArr; &nbsp;  "earth"$)$]]
Wir gehen hier von den folgenden Bildern im Format 160x120 aus:
+
We assume here the following images in the format 160x120 (pixels):
* dem Bild &bdquo;Weiß&rdquo; mit der Farbtiefe 1 BPP (ein Bit per Pixel) und
+
* the image "White" with the color depth "1 BPP" (one bit per pixel) and
* dem Bild &bdquo;Erde&rdquo; mit 24 BPP, auch wenn hier nur wenige der $2^{24}$ möglichen Farben genutzt werden.
+
* the image "Earth" with "24 BPP", even if only a few of the $2^{24}$ possible colors are used here.
  
  
Das Bild &bdquo;W1&rdquo; ist durch Verfälschung mit einem Gilbert&ndash;Elliott&ndash;Modell unter Verwendung folgender Parameter entstanden:
+
The image "W1" was created by falsification with a Gilbert&ndash;Elliott model using the following parameters:
 
:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,
 
:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,
\hspace{0.2cm}p_{\rm B} = 0.1,$$
+
\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}
:$${\rm Pr}(\rm
+
{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \
G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \
 
 
\hspace{-0.1cm}  0.1, \hspace{0.2cm} {\rm Pr}(\rm
 
\hspace{-0.1cm}  0.1, \hspace{0.2cm} {\rm Pr}(\rm
 
B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$
 
B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$
  
Damit erhält man für die mittlere Fehlerwahrscheinlichkeit
+
Thus, we obtain for the mean error probability
:$$p_{\rm M} =  \frac{p_{\rm G} \cdot {\rm Pr}({\rm
+
:$$p_{\rm M} =  \frac{p_{\rm G} \cdot {\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}+ p_{\rm B} \cdot {\rm Pr}(\rm
G\hspace{0.05cm}|\hspace{0.05cm} B)}+ p_{\rm B} \cdot {\rm Pr}(\rm
+
B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rm
B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm
 
G\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rm
 
 
B\hspace{0.05cm}|\hspace{0.05cm} G)} = 0.01 \hspace{0.05cm},$$
 
B\hspace{0.05cm}|\hspace{0.05cm} G)} = 0.01 \hspace{0.05cm},$$
  
und für die Fehlerkorrelationsdauer
+
and for the error correlation duration
 
:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm}
 
:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm}
 
B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1 \approx
 
B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1 \approx
 
8 \hspace{0.05cm}.$$
 
8 \hspace{0.05cm}.$$
  
Das Bild &bdquo;W2&rdquo; entstand nach Verfälschung mit den GE&ndash;Parametern
+
The image "W2" was obtained after falsification with the GE parameters
 
:$$p_{\rm B} = 0.2\hspace{0.05cm},\hspace{0.2cm}
 
:$$p_{\rm B} = 0.2\hspace{0.05cm},\hspace{0.2cm}
 
  {\rm Pr}({\rm
 
  {\rm Pr}({\rm
Line 36: Line 33:
 
0.0005\hspace{0.05cm}.$$
 
0.0005\hspace{0.05cm}.$$
  
Die Fehlerwahrscheinlichkeit im Zustand &bdquo;G&rdquo; wurde so gewählt, dass sich die mittlere Fehlerwahrscheinlichkeit ebenfalls zu $p_{\rm M} = 0.01$ ergibt.
+
The error probability in the state "$\rm G$" was chosen so that the average error probability is &nbsp;$p_{\rm M} = 0.01$.&nbsp;
  
Die beiden unteren Bilder &bdquo;E3&rdquo; und &bdquo;E4&rdquo; können entstanden sein durch Verfälschung mit
+
The two lower images "E3" and "E4" may have been created by falsification with
* dem BSC&ndash;Modell $(p = 0.01)$,
+
* the BSC model&nbsp; $(p = 0.01)$,
* dem gleichen GE&ndash;Modell, das zu &bdquo;W1&rdquo; geführt hat,
+
* that GE model which led to "W1",
* dem gleichen GE&ndash;Modell, das zu &bdquo;W2&rdquo; geführt hat.
+
* the GE model that led to "W2".
  
Dies zu klären, ist Ihre Aufgabe. Eine der Antworten ist jeweils richtig.
 
  
''Hinweise:''
+
It is your task to clarify this. One of the answers is correct in each case.
* Die Aufgabe gehört zum Themengebiet des Kapitels [[Digitalsignal%C3%BCbertragung/Anwendungen_bei_Multimedia%E2%80%93Dateien| Kapitel 5.4]].
 
* Alle Bilder wurden mit dem Windows&ndash;Programm [[Digitale Kanalmodelle & Multimedia]] erzeugt. Der angegebene Link verweist auf die Zip&ndash;Version dieses Programms.
 
  
  
  
===Fragebogen===
+
 
 +
 
 +
 
 +
Notes:&nbsp;
 +
*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Applications_for_Multimedia_Files| "Applications for Multimedia Files"]].
 +
 
 +
* All images were created with the Windows program&nbsp; "Digital Channel Models & Multimedia".
 +
 +
 
 +
 
 +
 
 +
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice
+
{For the image "W2" falsified with the Gilbert&ndash;Elliott model, determine the error probability in the state "GOOD", resulting in&nbsp; $p_{\rm M} = 1\%$?&nbsp;
|type="[]"}
+
|type="{}"}
+ correct
+
$p_{\rm G} \ = \ ${ 0.05 3% } $\ \%$
- false
+
 
 +
{What is the correlation period of the errors in the image "W2"?
 +
|type="{}"}
 +
$D_{\rm K} \ = \ ${ 94.2 3% }
  
{Input-Box Frage
+
{How many bit errors &nbsp;$(N_{\rm W})$&nbsp; occur (statistically) in image "W1" (or "W2") at &nbsp;$p_{\rm M} = 1\%$?&nbsp;
 
|type="{}"}
 
|type="{}"}
$xyz \ = \ ${ 5.4 3% } $ab$
+
$N_{\rm W} \ = \ ${ 192 3% }
 +
 
 +
{How many bit errors &nbsp;$(N_{\rm E})$&nbsp; occur (statistically) in image "E3" (or "E4") at &nbsp;$p_{\rm M} = 1\%$?&nbsp;
 +
|type="{}"}
 +
$N_{\rm E} \ = \ ${ 4608 3% }
 +
 
 +
{Which error model is the image "E3" based on?
 +
|type="()"}
 +
+ The BSC model with&nbsp; $p = 1\%$,
 +
- the same GE model as for "W1",
 +
- the same GE model as for "W2"
 +
 
 +
{Which error model is the image "E4" based on?
 +
|type="()"}
 +
- The BSC model with&nbsp; $p = 1\%$,
 +
- the same GE model as for "W1",
 +
+ the same GE model as for "W2".
 +
 
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
 
 +
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  
+
'''(1)'''&nbsp; Rearranging the given $p_{\rm M}$ equation leads to the result we are looking for:
'''(2)'''&nbsp;  
+
:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \frac{p_{\rm M}
'''(3)'''&nbsp;  
+
\cdot \big[{\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}+  {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)\big] -
'''(4)'''&nbsp;  
+
p_{\rm B} \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm}
'''(5)'''&nbsp;  
+
G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) } = \frac{  0.01 \cdot [0.01+0.0005] - 0.2 \cdot
 +
0.0005}{0.01} \hspace{0.15cm}\underline {= 0.05\%}\hspace{0.05cm}.$$
 +
 
 +
 
 +
'''(2)'''&nbsp; Using the given equation, we obtain:
 +
:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm}
 +
B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1
 +
=\frac{1}{0.0105}-1\hspace{0.15cm}\underline {\approx 94.2}\hspace{0.05cm}.$$
 +
 
 +
 
 +
'''(3)'''&nbsp; The image "White" consists of $160 \cdot 120 = 19200 \ \rm pixels$ and is also described by $19200 \ \rm bits$ because of the color depth $1 \ \rm BPP$.
 +
*With the average bit error probability $p_{\rm M} = 0.01$,  $N_{\rm W}  \underline{= 192}$ bit errors are expected in both images ("W1" and "W2").
 +
 
 +
 
 +
'''(4)'''&nbsp; With the same image size and error probability, there are now significantly more bit errors because of the color depth $24 \ \rm BPP$, i.e.
 +
:$$N_{\rm E} = 24 \cdot 192 \ \underline{= 4608}.$$
 +
 
 +
 
 +
'''(5)'''&nbsp; <u>Answer 1</u> is correct:
 +
*Image "E3" shows the typical structure of statistically independent errors.
 +
 
 +
 
 +
'''(6)'''&nbsp; <u>Answer 3</u> is correct:
 +
*Image "E4" shows a typical burst error structure.
 +
*Here, the GE model with $D_{\rm K} \approx 94$, was used, which was also used for "W2".
 +
*However, since now every single pixel is represented by $24 \ \rm bits$, the average error correlation duration (related to pixels) is only about ${D_{\rm K}}' = 4$.
 +
*The GE model with $D_{\rm K} \approx 8$ (in terms of bits) for a $24 \ \rm BPP$ image would look approximately like the image "E3" based on the BSC model.
 +
*In terms of pixels, this would result in rather statistically independent errors.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
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[[Category:Aufgaben zu Digitalsignalübertragung|^5.4 Multimedia-Dateien^]]
+
[[Category:Digital Signal Transmission: Exercises|^5.4 Multimedia Files^]]

Latest revision as of 14:49, 18 October 2022

Falsified BMP files
$($"Erde"   ⇒   "earth"$)$

We assume here the following images in the format 160x120 (pixels):

  • the image "White" with the color depth "1 BPP" (one bit per pixel) and
  • the image "Earth" with "24 BPP", even if only a few of the $2^{24}$ possible colors are used here.


The image "W1" was created by falsification with a Gilbert–Elliott model using the following parameters:

$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001, \hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm} {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.1, \hspace{0.2cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

Thus, we obtain for the mean error probability

$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}+ p_{\rm B} \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)} = 0.01 \hspace{0.05cm},$$

and for the error correlation duration

$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1 \approx 8 \hspace{0.05cm}.$$

The image "W2" was obtained after falsification with the GE parameters

$$p_{\rm B} = 0.2\hspace{0.05cm},\hspace{0.2cm} {\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B})= 0.01, \hspace{0.2cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.0005\hspace{0.05cm}.$$

The error probability in the state "$\rm G$" was chosen so that the average error probability is  $p_{\rm M} = 0.01$. 

The two lower images "E3" and "E4" may have been created by falsification with

  • the BSC model  $(p = 0.01)$,
  • that GE model which led to "W1",
  • the GE model that led to "W2".


It is your task to clarify this. One of the answers is correct in each case.




Notes: 

  • All images were created with the Windows program  "Digital Channel Models & Multimedia".



Questions

1

For the image "W2" falsified with the Gilbert–Elliott model, determine the error probability in the state "GOOD", resulting in  $p_{\rm M} = 1\%$? 

$p_{\rm G} \ = \ $

$\ \%$

2

What is the correlation period of the errors in the image "W2"?

$D_{\rm K} \ = \ $

3

How many bit errors  $(N_{\rm W})$  occur (statistically) in image "W1" (or "W2") at  $p_{\rm M} = 1\%$? 

$N_{\rm W} \ = \ $

4

How many bit errors  $(N_{\rm E})$  occur (statistically) in image "E3" (or "E4") at  $p_{\rm M} = 1\%$? 

$N_{\rm E} \ = \ $

5

Which error model is the image "E3" based on?

The BSC model with  $p = 1\%$,
the same GE model as for "W1",
the same GE model as for "W2"

6

Which error model is the image "E4" based on?

The BSC model with  $p = 1\%$,
the same GE model as for "W1",
the same GE model as for "W2".


Solution

(1)  Rearranging the given $p_{\rm M}$ equation leads to the result we are looking for:

$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{p_{\rm M} \cdot \big[{\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}+ {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)\big] - p_{\rm B} \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) } = \frac{ 0.01 \cdot [0.01+0.0005] - 0.2 \cdot 0.0005}{0.01} \hspace{0.15cm}\underline {= 0.05\%}\hspace{0.05cm}.$$


(2)  Using the given equation, we obtain:

$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1 =\frac{1}{0.0105}-1\hspace{0.15cm}\underline {\approx 94.2}\hspace{0.05cm}.$$


(3)  The image "White" consists of $160 \cdot 120 = 19200 \ \rm pixels$ and is also described by $19200 \ \rm bits$ because of the color depth $1 \ \rm BPP$.

  • With the average bit error probability $p_{\rm M} = 0.01$, $N_{\rm W} \underline{= 192}$ bit errors are expected in both images ("W1" and "W2").


(4)  With the same image size and error probability, there are now significantly more bit errors because of the color depth $24 \ \rm BPP$, i.e.

$$N_{\rm E} = 24 \cdot 192 \ \underline{= 4608}.$$


(5)  Answer 1 is correct:

  • Image "E3" shows the typical structure of statistically independent errors.


(6)  Answer 3 is correct:

  • Image "E4" shows a typical burst error structure.
  • Here, the GE model with $D_{\rm K} \approx 94$, was used, which was also used for "W2".
  • However, since now every single pixel is represented by $24 \ \rm bits$, the average error correlation duration (related to pixels) is only about ${D_{\rm K}}' = 4$.
  • The GE model with $D_{\rm K} \approx 8$ (in terms of bits) for a $24 \ \rm BPP$ image would look approximately like the image "E3" based on the BSC model.
  • In terms of pixels, this would result in rather statistically independent errors.