r |
g |
b |
y |
c |
m |
w |
|
R |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
G |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
B |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
r |
g |
b |
y |
c |
m |
w |
|
Y |
82 |
145 |
41 |
210 |
170 |
107 |
235 |
cb |
90 |
54 |
240 |
16 |
166 |
202 |
128 |
cr |
240 |
34 |
110 |
146 |
16 |
222 |
128 |
higher than its mean power ⇒ PSNR = 10.78 + 6n
quantised coefficients: |
360 |
0 |
-216 |
-24 |
0 |
40 |
0 |
0 |
reconstructed pixels: PSNR = 30 dB |
30 |
70 |
185 |
250 |
204 |
153 |
104 |
27 |
|
128 |
128 |
128 |
128 |
128 |
128 |
128 |
128 |
|
28 |
87 |
169 |
227 |
227 |
169 |
87 |
28 |
type |
operations |
Multiplications |
additions |
---|---|---|---|
TDL |
23 |
23 × 256 |
23 × 511 |
TSS |
25 |
25 × 256 |
25 × 511 |
CSA |
17 |
17 × 256 |
17 × 511 |
OSA |
13 |
13 × 256 |
13 × 511 |
Thus in (i) the lowpass analysis filter will be:
and the high pass analysis filter is:
In (ii) and the highpass:
which are the (5,3) subband filter pairs.
In (iii) and the highpass
which gives the second set of (4,4) subband filters.
In (iv) and
Any other combinations may be used, as desired.
retaining H1(-z) = (1 + z-1)2 = 1 + 2z-1 + z-2, that gives the three-tap highpass filter of H1(z) = 1 - 2z-1 + z-2 and the remaining parts give the nine-tap lowpass filter
or
Had we divided the highpass filter coefficients, H1(z), by , and hence multiplying those of lowpass H0(z) by this amount, we get the 9/3 tap filter coefficients of Table 4.2.
62 |
0 |
3 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
-1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
-1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
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 |
3 |
31 |
0 |
1 |
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 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
(3,1)(0,2)(0,1)(0,1)(1,1)(6,1)(6,1)(1,1)(1,1)(15,0)(15,0)(3,3)
for 25% quality: DIFF= 31-50 = -19, symbol_1 = 5, symbol_2 = -19-1 =-20
scanned pairs: (4, 1)(57, 1)
events: (4,1)(15,0)(15,0)(15,0)(9,1)
VLC for CAT = 5 is 110 and -20 in binary is 11101100, hence the VLC for the DC coefficient is 11001100
for AC, using the AC VLC tables:
for each (15,0) the VLC is 11111111001
and for (9, 1) the VLC is 111111001
total number of bits: 8 + 3 x 11 + 9 = 50 bits.
83 |
0 |
2 |
1 |
0 |
0 |
4 |
0 |
0 |
0 |
-1 |
0 |
3 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
3 |
-1 |
0 |
0 |
-1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
3 |
0 |
2 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
-1 |
4 |
0 |
0 |
0 |
0 |
0 |
0 |
2 |
0 |
0 |
0 |
0 |
0 |
0 |
31 |
events: (0, 83)(4, 2)(0,1)(0, -1)(1, -1)(1, 1)(4, 3)(4, -1)(1,3)(1,3)(1,4)(2, -1) (3, 4)(0, 2) (3, 2)(20,1)(2, 31)
number of bits (including the sign bit): 20 + 9 + 5 + 5 + 6 + 6 + 11 + 8 + 8 + 8 + 9 + 7 + 11 + 5 + 8 + 20 + 20 = 166
no EOB is used, as the last coefficient is coded
159 |
149 |
170 |
105 |
113 |
133 |
for I-pictures:
for P = 54.3 kbits and for B = 38 kbits.
(8 × 5) + (3 × 10) + 20 = 90, and bits for I = 128 kbits, for P = 64 and for B = 32 kbits
for P, the average index of (10 + 10 + 20)/3 = 13.3 should be used, hence the target bit rates for I = 99.3 kbits, for P = 66.2 kbits and for B = 34.75 kbits.
For for P = 48 kbits and for B = 36 kbits
4 |
-1 |
-5 |
0 |
2 |
0 |
0 |
-1 |
7 |
0 |
2 |
1 |
-1 |
0 |
0 |
-1 |
0 |
1 |
0 |
1 |
-1 |
0 |
0 |
0 |
0 |
-4 |
0 |
0 |
0 |
0 |
0 |
0 |
-2 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
2D-events: (0,4)(0,-1)(0,7)(2, -5)(1, 2) (0, 1)(1, -2) (0, -4)(1, 1)(0, 2)(1, -1) (0, 1)(7, -1)(2, -1)(4, 1)(5, 1) (2, -1)EOB (using Figure 6.12, the bits including the sign bit)
8 + 5 + 11 + 20 + 7 + 5 + 7 + 8 + 6 + 5 + 6 + 5 + 9 + 6 + 8 + 8 + 6 + 2 =132 bits
The enhancement layer events: (1,1)(0,2)(1,-1)(0,1)(7,-1)(2,-1)(4,1)(5,1) (2,-1) + EOB the bits: 6 + 5 + 6 + 5 + 9 + 6 + 8 + 8 + 6 + 2 = 61
Total bits 77 + 61 = 138 bits, about 4.5 per cent extra over one-layer coding
2 |
0 |
2 |
0 |
1 |
0 |
0 |
0 |
4 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
-1 |
0 |
0 |
0 |
0 |
-2 |
0 |
0 |
0 |
0 |
0 |
0 |
-1 |
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 |
enhancement:
0 |
-1 |
0 |
0 |
0 |
0 |
0 |
-1 |
0 |
0 |
0 |
1 |
-1 |
0 |
0 |
-1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
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 |
Base layer events: (0, 2)(1, 4)(2, -2)(1, 1)(2, -1)(0, -2)(2, 1)(10, -1)EOB
Bits: 5 + 9 + 7 + 6 + 6 + 5 + 6 + 10 + 2 = 56 bits
Enhancement layer events: (1, -1)(6, 1)(4, 1)(2, -1)(0, 1)(10, -1)(10, 1)(2, -1)
EOB
Bits: 6 + 9 + 8 + 6 + 5 + 10 +10 + 6 + 2 = 62 bits
Total bits = 56 + 62 = 118
Note: the overall bit rate is less than the one layer, but the distortion will be larger.
total data to be sent = 34 980 × 53/47 = 39 445.5 Mbits
time required = 39 445.5/15 Mbits/s = 43 min 50 s
and at this load P = 1.72 × 10-8.
For the enhancement layer, it sees the whole load, of 130 Mbits/s, thus the load will be:
and the error rate will be P = 0.088.
and the error rate P = 2.7 × 10-9. The enhancement layer has a load of 2 x 3.783 = 0.7566 that leads to an error rate of P = 5.3 × 10-5.
leading to P = 9.5 × 10-8. For the enhancement layer, the load will be more than 100 per cent and the loss probability will be P = 1!
thus B1 = 150 - 8=142 and C1 = 115 + 8 = 123
Thus the forward 4 × 4 integer transform becomes
And the inverse transform is its transpose
As can be tested, this inverse transform is orthornormal, e.g.:
431 |
-156 |
91 |
-15 |
43 |
52 |
30 |
1 |
-6 |
-46 |
-26 |
-7 |
-13 |
28 |
-19 |
14 |
The reconstructed pixels with the retained coefficients are; for N = 10:
105 |
121 |
69 |
21 |
69 |
85 |
62 |
44 |
102 |
100 |
98 |
119 |
196 |
175 |
164 |
195 |
which gives an MSE error of 128.75, or PSNR of 27.03 dB. The reconstructed pixels with the retained 6 and 3 coefficients give PSNR of 22.90 and 18 dB, respectively.
With 4 × 4 DCT, these values are 26.7, 23.05 and 17.24 dB, respectively.
As we see the integer transform has the same performance as the DCT. If we see it is even better for some, this is due to the approximation of cosine elements.
At higher indices, the largest quantiser parameter for H.263 is 31QP, but that of H.26L is 88 QP, hence at larger indices H.26L has a coarser quantiser.
For B: c0 = c1 = c2 = c3 = c4 = 1, and index = 1 + 2 + 4 + 8 + 16 = 31, prob(0) = 6554. As we see this odd pixel of 0 among the 1s has a lower probability.
For C: c1 = c2 = c3 = c4 = c5 = c7 = 1, and the index becomes 190. Like pixel A the prob(0) = 91, but its prob(1) = 65535 - 91 = 65444 out of 65535, which is as expected.
127 |
-40 |
1 |
10 |
-15 |
|
23 |
-7 |
|
-6 |
confined to the top left corner, while for those of the normal DCT with padded zeros, the significant coefficients are scattered all over the 8 × 8 area.
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