Quantization of a standard Cosine signal

Hello, could someone please explain the function of the folllowing ( Z1 to Z4 )lines in the MATLAB script along with the summary of what the code does. A slightly detailed explanation is much appreciated.

Thank you in advance.

close all; clear all; clc

A = 4; fm = 1; M = 20;
t = 0:(1/(M*fm)):1;
xs = A*cos(2*pi*fm*t);  %sampled cosine signal
%Quantization
L = 10;
Length_xs = length(xs);
xq = zeros(1, Length_xs);
L_level = zeros(1, Length_xs);
for loop = 1:Length_xs
     for index = 1:L
       if (A-index*(2*A)/L <=xs(loop)) && (A-(index-1)*(2*A)/L > xs(loop)) 
          xq(loop) = A - index*(2*A)/L + A/L;
          L_level(loop) = L-(index-1); %-------------------------------------Z1
       end
     end
     if xs(loop) == A
        xq(loop) = A - A/L;
        L_level(loop) = L;  %-----------------------------------------Z2
     end
end
%xq is the quantized cosine signal

NN = ceil(log2(L));  %---------------------------------------------Z3
encoded = dec2bin(L_level, NN) %-----------------------------------Z4
plot(t, xs, 'r-')
hold on
plot(t, xq, 'bo')
xlabel('time')
ylabel('signals')
legend('Original signal','quantized signal')

Answer 1

[Robert Brown]

%% 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

[John D'Errico]

PLEASE don't add answers for every comment you will make! All of your multiple answers could have been ONE single response. ONE answer.

I would point out that if the OP wanted to mark any of these multiple answers as the correct one, which one should they mark? You cannot accept more than one answer.

[John D'Errico]

I just merged all of your answers into one.

Note there is NO need to enclose all of your comments in %{ and }%. That does nothing in Answers. So I deleted those lines in the merge.

Remember that you edit any response of yours by clicking on the icon to the top right of it that looks like a pencil. It is between the flag icon and the trash can icon.

[Robert Brown]

Thanks for the edit pencil advice...

I first wrote the comments into a MATLAB code copy of the example, hence the comment % marks in the answers. They did nothing here, but they did something (commented those answer comments) in the code. I'm not ADDING % comment markers to my answers here, they were already added when I was writing the answers into the code while debugging this code question. When I extracted the answers from the code and pasted here, the % markers tagged along...

You MVP's sure are particular about your answers. Perhaps I'll just leave the answering to the MVPs. I don't seem to be getting any thanks for my attempts, just complaints that my answers aren't up to "standards". Disappointing...