Difference between revisions of "Aufgaben:Exercise 5.8Z: Falsification of BMP Images"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Applications_for_Multimedia_Files |
}} | }} | ||
− | [[File:P_ID1856__Dig_Z_5_8.png|right|frame| | + | [[File:P_ID1856__Dig_Z_5_8.png|right|frame|Falsified BMP files <br>$($"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: | |
:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001, | :$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001, | ||
\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm} | \hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm} | ||
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B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$ | B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$ | ||
− | + | Thus, we obtain for the mean error probability | |
:$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}+ p_{\rm B} \cdot {\rm Pr}(\rm | :$$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)}{{\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},$$ | B\hspace{0.05cm}|\hspace{0.05cm} G)} = 0.01 \hspace{0.05cm},$$ | ||
− | + | 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}.$$ | 8 \hspace{0.05cm}.$$ | ||
− | + | 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 | ||
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0.0005\hspace{0.05cm}.$$ | 0.0005\hspace{0.05cm}.$$ | ||
− | + | 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. | |
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− | + | Notes: | |
− | * | + | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Applications_for_Multimedia_Files| "Applications for Multimedia Files"]]. |
− | * | + | |
+ | * All images were created with the Windows program "Digital Channel Models & Multimedia". | ||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {For the image "W2" falsified with the Gilbert–Elliott model, determine the error probability in the state "GOOD", resulting in $p_{\rm M} = 1\%$? |
|type="{}"} | |type="{}"} | ||
$p_{\rm G} \ = \ ${ 0.05 3% } $\ \%$ | $p_{\rm G} \ = \ ${ 0.05 3% } $\ \%$ | ||
− | { | + | {What is the correlation period of the errors in the image "W2"? |
|type="{}"} | |type="{}"} | ||
$D_{\rm K} \ = \ ${ 94.2 3% } | $D_{\rm K} \ = \ ${ 94.2 3% } | ||
− | { | + | {How many bit errors $(N_{\rm W})$ occur (statistically) in image "W1" (or "W2") at $p_{\rm M} = 1\%$? |
|type="{}"} | |type="{}"} | ||
$N_{\rm W} \ = \ ${ 192 3% } | $N_{\rm W} \ = \ ${ 192 3% } | ||
− | { | + | {How many bit errors $(N_{\rm E})$ occur (statistically) in image "E3" (or "E4") at $p_{\rm M} = 1\%$? |
|type="{}"} | |type="{}"} | ||
$N_{\rm E} \ = \ ${ 4608 3% } | $N_{\rm E} \ = \ ${ 4608 3% } | ||
− | { | + | {Which error model is the image "E3" based on? |
|type="()"} | |type="()"} | ||
− | + | + | + The BSC model with $p = 1\%$, |
− | - | + | - the same GE model as for "W1", |
− | - | + | - the same GE model as for "W2" |
− | { | + | {Which error model is the image "E4" based on? |
|type="()"} | |type="()"} | ||
− | - | + | - The BSC model with $p = 1\%$, |
− | - | + | - the same GE model as for "W1", |
− | + | + | + the same GE model as for "W2". |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Rearranging the given $p_{\rm M}$ equation leads to the result we are looking for: |
:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{p_{\rm M} | :$$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] - | \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] - | ||
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− | '''(2)''' | + | '''(2)''' Using the given equation, we obtain: |
:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} | :$$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 | B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1 | ||
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− | '''(3)''' | + | '''(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)''' | + | '''(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. |
:$$N_{\rm E} = 24 \cdot 192 \ \underline{= 4608}.$$ | :$$N_{\rm E} = 24 \cdot 192 \ \underline{= 4608}.$$ | ||
− | '''(5)''' | + | '''(5)''' <u>Answer 1</u> is correct: |
− | * | + | *Image "E3" shows the typical structure of statistically independent errors. |
− | '''(6)''' | + | '''(6)''' <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:Digital Signal Transmission: Exercises|^5.4 Multimedia | + | [[Category:Digital Signal Transmission: Exercises|^5.4 Multimedia Files^]] |
Latest revision as of 13:49, 18 October 2022
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:
- The exercise belongs to the chapter "Applications for Multimedia Files".
- All images were created with the Windows program "Digital Channel Models & Multimedia".
Questions
Solution
- $$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{p_{\rm M} \cdot \big[{\rm Pr}({\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}+ {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)\big] - p_{\rm B} \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) } = \frac{ 0.01 \cdot [0.01+0.0005] - 0.2 \cdot 0.0005}{0.01} \hspace{0.15cm}\underline {= 0.05\%}\hspace{0.05cm}.$$
(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.