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 + z1)2 = 1 + 2z1 + z2, that gives the threetap highpass filter of H1(z) = 1  2z1 + z2 and the remaining parts give the ninetap 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= 3150 = 19, symbol_1 = 5, symbol_2 = 191 =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 Ipictures:
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 
2Devents: (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 onelayer 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 × 108.
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 × 109. The enhancement layer has a load of 2 x 3.783 = 0.7566 that leads to an error rate of P = 5.3 × 105.
leading to P = 9.5 × 108. 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.