MATLAB Answers

Quantization of a standard Cosine signal

5 views (last 30 days)
Andre Gates
Andre Gates on 25 Mar 2021
Commented: Robert Brown on 5 Apr 2021
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')

Answers (1)

Robert Brown
Robert Brown on 30 Mar 2021
Edited: John D'Errico on 4 Apr 2021
%% 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
  3 Comments
Robert Brown
Robert Brown on 5 Apr 2021
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...

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!