Databases Reference
In-Depth Information
T A B L E 13 . 11
The quantized values of the
coefficients.
32
11
0
0
0
0
0
0
−
108
0
0
0
0
0
0
0
42
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
T A B L E 13 . 12
The reconstructed block.
123
122
122
121
120
120
119
119
121
121
121
120
119
118
118
118
121
121
120
119
119
118
117
117
124
124
123
122
122
121
120
120
130
130
129
129
128
128
128
127
141
141
140
140
139
138
138
137
152
152
151
151
150
149
149
148
159
159
158
157
157
156
155
155
is the seventh element in Category 4. Therefore, we send the binary string 1011, which is
the Huffman code for 0/4, followed by 0110 to indicate that
9 is the seventh element in
Category 4. The next label is 3, which belongs to Category 2, so we send the Huffman code
01 corresponding to 0/2, followed by the 2 bits 11. All the labels after this point are 0, so we
send the EOB Huffman code, which in this case is 1010. If we assume that the Huffman code
for the DC coefficient is 2 bits long, we have sent a grand total of 21 bits to represent this 8
−
×
8
21
block. This translates to an average of
64
bits per pixel.
To obtain a reconstruction of the original block, we perform the dequantization, which
simply consists of multiplying the labels in Table
13.8
with the corresponding values in
Table
13.7
. Taking the inverse transform of the quantized coefficients shown in Table
13.11
and adding 128, we get the reconstructed block shown in Table
13.12
. We can see that in spite
of going from 8 bits per pixel to
9
32
bits per pixel, the reproduction is remarkably close to the
original.
If we wanted an even more accurate reproduction, we could do so at the cost of an increased
bit rate by multiplying the step sizes in the quantization table by one-half and using these values
as the new step sizes. Using the same assumptions as before, we can show that this will result
in an increase in the number of bits transmitted. We can go in the other direction by multiplying
the step sizes with a number greater than one. This will result in a reduction in bit rate at the
cost of increased distortion.
