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Say I have the following 4x4 block:

[ 120  121  125  119
  120  120  118  117
  122  121  117  118
  122  122  120  120]

If this block is intra coded the resulting DCT coefficients are the following and rounding to nearest integer:

[ 480.8  3.8  -1.5  0
  -0.1  -1.8  -2.7  2.7
    4   -2.4  -2    2
   1.1   0.2   0    0.3]

Assuming I used QP size of 2 for quantization:

[ 240  2  -1  0
   0  -1  -1  1
   2  -1  -1  1
   1   0   0  0]

How do I calculate the compression ration for the following assuming a pixel is encoded using 8bits/pixel and the DCT coefficient uses 10 bits

I normally calculate the compression by first calculating the

uncompressed bit rate = image width × image height × colour depth × bits per pixels 

and then solve uncompressed size/compressed size

But I am not sure how to calculate the compression ratio in this case where DCT coefficient is given!!

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First, motivations for an engineer solution, which might be the initial reason of such a question:

The compression ratio of a JPEG encoded image result from a sort of average from different image blocks "compressibility". The JPEG principles rely on two combined aspects:

  • the average luminosity from block to block varies slowly, hence it can be nicely predicted from preceding blocks. This is the differential prediction used on DC coefficients (here your 240);
  • local stationarized parts of images can be faithfully approximated by a few quantized 2D cosines.

The scheme you provide is a simplified version, here are a few. First, the DCT is applied on $8\times8$ blocks, not $4\times4$. Hence, it is more difficult to generate long runs of zeros after quantization. Second, the quantization scheme is linear (dividing by 2). Third, with only one block, you cannot gain from inter-block prediction on DC. Fourth, with only 15 AC coefficients, the (value,run) Huffman coding might be a little inefficient. To add a few, non-adaptive tables, lack of header, may highly impact the estimated compression ratio on such a small file.

Last, the "DCT coefficient uses 10 bits" is a bit awkward. So, I interpret it as "The DC coefficient uses 10 bits". With that, you are left with 4 values and their occurrences:

  • $2$ (2 times)
  • $1$ (3 times)
  • $-1$ (5 times)
  • $0$ (5 times)

from with you can build a Huffman tree, and a non-prefix code-word for each of them. Adding their overall length to the above 10 bits can give you a rough estimate of the total number of bytes actually needed. You can use for instance an Online Huffman coding tool.

Second, you can work out your own JPEG format, on $4\times4$ blocks, and input an sufficiency large image with those repeated $4\times4$ blocks, to compute actual compression ratios.

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