%% Description of Z1
First, we recognize that loop is equal to sample number.
So L_level(loop) is the quantization level of the current sample.
Because MATLAB indexes from 1 to n, instead of 0 to n-1, we would never
reach level 10 quantization unless we subtract 1 from the index.
The level L=10-(index-1), for index=1 = 10.
Also, they are inverting the levels, bacause the test is "amplitude-index...
In other words, the first index is a test of Level 10, so index=1 gives
Level=10-(index-1) = 10-(1-1) = Level_10.
To continue, index 10 = (Level=10)-(index-1) = (Level=10) - (index=10 - 1)
= 10 - (10-1) = 10-9 = 1.
Therefore, indexes 1:10 test for Levels 10:1 (this could have been programmed
better, without this inverse relationship)
%% Description of Z2
Because the index loop (for index = 1:L) tests for A-index being less than xs,
and also tests whether A-(index-1) is greater than xs, it won't catch the
case where A-(index=1 -1) = A-0 = A is equal to the max amplitude A.
Therefore, the code developer followed up with a test of.... if xs(loop) = A...
then xq(loop) = A - A/L = 4 - 4/10 = 3.6
Note that this test is unnecessary if the code developer changes the second
half of the first test to be (A-(index-1)*(2*A)/L >= xs(loop)
Maybe to be clear, the L_levels are... Level 10 >= 3.2(to 4), Level 9 >=2.4(to 3.5999999...),
Level 8 >= 1.6 to 2.4-, Level 7 >= 0.8.., Level 6 >= 0.0-0.799999999, Level 5 >=
-0.8(to 0-), Level 4 >= -1.6(to just under -0.8), Level 3 >= -2.4..., Level
2 >= -3.2..., Level 1 >= -4 to just under -3.2)
Note that the L_levels are not the final answer. Eventually the final
answer will be the midpoint of the L_levels for each bin.... So Level 10 =
3.2 to 4.0, and will eventually be encoded 3.6, the midpoint of the band.
%% Description of Z3
Because we have 10 levels, and want to express them in binary...
log2 of a number tells what power of 2 = 10
NN = ceil(log2(L))
NN = ceil(log2(10))
since log2(10) = 3.3219
ceil(log2(10) = ceil(3.3219), and ceil rounds up to next integer
so NN = 4(bits needed) when this statement is finished,
implying 4 bits to hold 1:10 in binary representation.
Level Binary(encoded)
1 '0001
2 '0010
3 '0011
4 '0100
5 '0101
6 '0110
7 '0111
8 '1000
9 '1001
10 '1010
%% Description of Z4
The dec2bin function converts the base_10 level numbers to base_2 (binary).
In addition, since NN was specified, we will fully pupulate 4 bit binary
text representatons of the 10 levels.
'encoded' contains the binary representation of the level at each of the 21
signal samples.
Level Binary(encoded)
1 '0001
2 '0010
3 '0011
4 '0100
5 '0101
6 '0110
7 '0111
8 '1000
9 '1001
10 '1010
