Exam 5: Lossy Compression Algorithms, Image Compression Standards and Basic Video Compression Techniques

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Given the following two consecutive frames in an H.261 video, if the first image is treated as an Iframe and the second a P-frame. Show in detail how the second image is encoded. (Work out the numbers at each step. You can assume any quantizer and stop after the RLC (Run-length Coding).) To simplify the data, we assume that the size of macroblocks is 8×88 \times 8 instead of 16×1616 \times 16 . Intensity Values for the First Image:  Given the following two consecutive frames in an H.261 video, if the first image is treated as an Iframe and the second a P-frame. Show in detail how the second image is encoded. (Work out the numbers at each step. You can assume any quantizer and stop after the RLC (Run-length Coding).) To simplify the data, we assume that the size of macroblocks is  8 \times 8  instead of  16 \times 16 . Intensity Values for the First Image:     Intensity Values for the Second Image:    Intensity Values for the Second Image:  Given the following two consecutive frames in an H.261 video, if the first image is treated as an Iframe and the second a P-frame. Show in detail how the second image is encoded. (Work out the numbers at each step. You can assume any quantizer and stop after the RLC (Run-length Coding).) To simplify the data, we assume that the size of macroblocks is  8 \times 8  instead of  16 \times 16 . Intensity Values for the First Image:     Intensity Values for the Second Image:

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Another name for zig-zag coding is "zonal coding". Suppose you invent a new zonal coding scheme for JPEG that simply discards diagonals above the first few. Suppose we keep the first six zig-zag lines. (a) How many coefficients are we keeping? (b) How will we do, compared to keeping all the zig-zags (still using run-length encoding)? Comment on both compression capability and image quality.

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(a) 21
(b) Better compression, marginally, but might have discernable quality reduction.

Explain why the factorization steps in Eqs. (8.24) and (8.25) save CPU cycles. About how much faster is the factorized approach to the DCT than the straightforward definition in (8.17)? (In both approaches, assume that of course the cosine matrix values are pre-stored outside any loop.)

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The factorization steps in Eqs. (8.24) and (8.25) save CPU cycles because they reduce the number of multiplications and additions required to compute the DCT. In the straightforward definition in Eq. (8.17), each element of the DCT matrix is computed using a series of multiplications and additions, which can be computationally expensive.

On the other hand, the factorized approach in Eqs. (8.24) and (8.25) breaks down the DCT computation into smaller, more efficient steps. This reduces the overall number of arithmetic operations required, leading to a faster computation of the DCT.

The exact speedup achieved by the factorized approach will depend on the specific implementation and hardware, but in general, it can be significantly faster than the straightforward definition in Eq. (8.17). In some cases, the factorized approach can be several times faster, especially for larger DCT matrices. This makes it a more efficient and practical choice for real-time or high-performance applications.

The top part of Figure 8.1 shows the DCT F(u)F(u) for a 1-D signal f(i)f(i) . i. What does F(5)F(5) tell you in terms of the image content in f(i)f(i) ? ii. What is the reason that F(5)F(5) is less than 0 ? (b) The Inverse Discrete Cosine Transform (IDCT) can be viewed as a reconstruction process in which the Cosine basis functions are scaled and added one by one to reconstruct f~(i)\tilde{f}(i) . Assume F(0)=500F(0)=500 and F(1)=70F(1)=-70 ; in the space provided at the bottom part of the figure, i. draw f~(i)\tilde{f}(i) after only the F(0)F(0) is used, ii. draw f~(i)\tilde{f}(i) after F(0)F(0) and F(1)F(1) are used. Show all details including magnitudes, and explain why they look like that.

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Describe how H.261 deals with temporal and spatial redundancies in video.

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Explain why the block DCT is preferred to taking the whole image DCT in JPEG compression.

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We have seen that a logarithmic-based block search strategy is fast: it is "suboptimal but still usually effective." That being the case, why should we be bothered carrying out a full search in any situation? Why not just stick to the fast method - what is the advantage to be gained, if any, from a comprehensive, optimal search strategy?

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Briefly explain why DCT was chosen for JPEG compression.

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Suppose we have 32 samples from a mono audio file. (a) If the audio was sampled at 8kHz8 \mathrm{kHz} , how many seconds does this represent? (b) If we carry out a DCT with blocksize equal to 32, how many DCT coefficients do we end up with? (c) Numbering the DCT coefficients from 0, i) What frequency, in Hz\mathrm{Hz} , does coefficient number 1 represent? ii) What frequency, in Hz\mathrm{Hz} , does coefficient number 16 represent?

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Why does JPEG give compression? At which points in the algorithm does compression occur? Please estimate (as a general guess) how much compression occurs at each of these points.

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