L
L
P
L
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 base layer events: (0,4)(0,-1)(0,7)(2,-5)(1,2)(0,1)(1,-2)(0,4) + PBP the bits: 8 + 5 + 11 + 20 + 7 + 5 + 7 + 8 + 6 = 77 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
base:
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.
Mbit/s, 28.8/8 = 3.5, thus three TV programmes.
for B = 1, α = 1 and β = 10-5
for B = 5, α=0.2 and β = 2 × 10-6
34 980 Mbits = 4.3725 Gbytes
mean bit rate = 34 980/(90 x 60) = 6.478 Mbit/s
CBR = 90 × 60 × 20/8 = 13.5 Gbytes, peak-to-mean = 3.09
if error in any video bits, packet is in error; P = 47 × 8 × 10-7 = 3.76 × 10-7
if error in the header, packet is lost; P = 5 × 8 × 10-7 = 4 × 10-6
available link rate = 50 × 30 × 10-2 = 15 Mbit/s
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
25 × 4 = 100 Mbits/s, load and the error rate is P = .
With SNR scalability, assume 30 per cent more load, then total load = 100 x 1.3 = 130 Mbit/s, of which 65 Mbit/s is assigned to the base layer. Since the base layer has an absolute priority, then network load for the base layer:
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.
In data partitioning, with 4 per cent extra bits over one layer and 50 per cent to the base layer, the base layer load will be:
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.
With spatial scalability, assuming 50 per cent more bits over one layer and 50 per cent assigned to the base layer, then the allocated bits to the base layer will be 75 Mbit/s. Base layer load is:
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!