giải 10 bài thêm trong Đề cương
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8/11/2019 Giải 10 Bài Thêm Trong Đề Cương
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Bài 3. Audio signals...
-CD quality audio, the maximum frequency: 44,100Hz / 2 = 22,050Hz.
-Telephone quality audio, the maximum frequency: 8kHz / 2 = 4kHz.
-This is based on Nyquist theorem: the sampling frequency for a signal must be at least
twice the highest frequency component in the signal.
If an arbitrary input signal is directly sampled, what artefact may result and how
to solve this?
-This may result in aliasing artefact.
-To solve this, add an analog low pass filter before sampling to eliminate highfrequency components.
Bài 4.
a)Different colour models...
The CMYK colour model use Cyan, Magenta, Yellow and Black as primaries
(components).
The CMYK colour model is mostly used in printing because the colour pigments on
the paper absorb certain colours thus a subtractive model is suitable; black is used to
produce darker black than simply mixing CMY.
b)Give a colour
c)Give three colour models other than...
CIE L*a*b* : relate more closely to human perception, useful for image processing
such as Photoshop
YUV(YCbCr) : separate colour from luminance component, useful for PAL
video/MPEG
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YIQ : separate colour from luminance component, useful for NTSC video/JPEG. Other
sensible answers are acceptable as well.
d)What is chrome subsampling...
-Chroma subsampling is a method that stores colour information at lower resolution than intensity information.
-Chroma subsampling is meaningful because human visual system is less sensitive tovariations in colour than brightness.
-Chroma subsampling can reduce the bandwidth for colour detail in almost no
perceivable visual difference.
e)For the following array
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Bài 5: Huffman coding algorithm
Bài 6.
a)Encode the following stream ... decimal...
Sort Data into largest probabilities first and make cumulative probabilities:
0 - E - 0.3 - D - 0.6 – I – 0.8 – M - 0.9 – A – 1.0
There are only 5 Characters so there are 5 segments of width determined by the
probability of the related character.
The first character to encoded is M which is in the range 0.8 – 0.9, therefore the
range of the final codeword is in the range 0.8 to 0.89999…..
Each subsequent character subdivides the range 0.8 – 0.9
SO after coding M we get
0.8 - E - 0.83 - D - 0.86 – I – 0.88 – M - 0.89 – A – 0.9
So to code E we get range 0.8 – 0.83 SO we subdivide this range
0 - E - 0.809 - D - 0.818 – I – 0.824 – M - 0.827 – A – 0.83
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