6.7 Problems

6.7 Problems

1. 

In a CIF picture, find the number of macroblocks and blocks per

1. GOB

2. picture

2. 

Calculate the macroblock interval in CIF pictures. How large is this value for a QCIF video at 10 Hz?

3. 

The absolute value of the motion compensated frame difference per macroblock, |mfd|, is normally smaller than that without motion compensation, |fd|. In a search for motion compensation, the following values for |mfd| and |fd| have been calculated. Using Figure 6.4, determine in which of the following cases motion compensation should be used:

1. |fd| = 1200 and |mfd| = 1000

2. |fd| = 600 and |mfd| = 500

3. |fd| = 200 and |mfd| = 50

4. 

Why is nonmotion compensation preferred to motion compensation for very small frame difference images?

5. 

To decide whether a macroblock should be interframe or intraframe coded, the variances of intra and motion compensated interframe macroblocks are compared, according to Figure 6.5. Find in each of the following whether a macroblock should be intra or interframe coded:

6. 

In macroblocks with small energy (inter or intra), inter macroblock is preferred to intra, why?

7. 

Calculate the coded block pattern (CBP) indices of the following macroblocks in a H.261 codec if:

1. all the blocks in the macroblock are coded

2. only the luminance blocks are coded

3. only the chrominance blocks are coded

8. 

Assume the transform coefficients of the block given in problem 3 of Chapter 5 belong to an H.261 codec. These coefficients are linearly quantised with the quantiser of Figure 6.9, with th = 16 and q = 12. Using Figure 6.12, calculate the number of bits required to code this block.

9. 

Pixels on the top left corner of a picture have values of:

200	135	180	210
75	110	134	230
62	89	52	14

and are filtered with the loop filter of Figure 6.16. Find the filtered values of these pixels.

10. 

The maximum quantiser step size in H.261 is 62 (quantiser parameter is 31, defined with five bits). A 384 kbit/s H.261 encoder with an RM8 type rate control has a smoothing buffer of 5 kbytes. Find the spare capacity of the buffer, when the quantiser step size is at its maximum value.

Answers

1. 

1. 33, 198

2. 396, 2376

2. 

; QCIF:

3. 

1. MC

2. NOMC

3. NOMC

4. 

due to motion vector overhead

5. 

1. inter

2. intra

3. inter

6. 

For small values in intra mode DC still needs eight bits, while in inter mode it is less.

7. 

1. 63

2. 60

3. 3

8. 

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

9. 

159	149	170
105	113	133

10. 

@384 kbit/s P = 6 and with q = 62 and P = 6, the buffer content is 36 kbits, left over capacity = 5000 x 8 - 36000 = 4000 bits