how to make a 4 bits quantizer with 15 levels

im working in project that i should make a 4 bits quantizers with 15 levels and encoding each levels with a binary sequence of 4 bits:

can i have some idea please. thank you

Answer 1

[Walter Roberson]

discretize() and dec2bin

Or is this a Simulink question?

[mohamed chahma]

no it's a script, i'm just need some idea, about the binary sequence, i dont wnt matlab give the sequence binary , i want to ecode the levels some specefique binary sequence.

[Image Analyst]

For your 15 levels, what value do you want mapped to 0 and what value do you want mapped to 14? Or do you want 16 values between and including 0 to 15? Anyway, I don't see any additional question so I imagine you're looking at discretize() right now. So if that works for you, which it should, then please "Accept this answer" to give Walter credit.

[mohamed chahma]

i want 15 levels between 0 and 14, 0 is "0000" and 14 "1110", my signal beetween -1 and 1

[Walter Roberson]

So 1111 represents all values that are negative?

discretize to get a bin number. Use the bin numbers as row indices into a 2d array with 16 rows and 4 columns.

[mohamed chahma]

i want to present -1 with 0000 and 1 with 1000 i will not use 1111

[Walter Roberson]

We are going to need a list of voltages and associated bit patterns. Please be specific about whether any particular voltage < or <= .

You have indicated that your signal is -1 to +1 . You have indicated that you want 0 exactly to be 0000 and that you want 1 exactly to be 1000 and that 1111 will not be used. What about values between -1 (exactly) and 0 (excluded)? What will be represented by 1001 1010 1011 1100 1101 1110 ?

[Walter Roberson]

format long g
bitpatterns = dec2bin(0:15,4) - '0';
bitpatterns = bitpatterns(randperm(15),:)

bitpatterns = 15×4

1 1 0 1 0 1 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1

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boundaries = [-1 sort(rand(1,14)*2-1) 1];
mat2str(boundaries)

ans = '[-1 -0.884897455427377 -0.805515680810448 -0.52755891710828 -0.457066695586775 -0.421367409673541 -0.413658951403185 -0.256435083908991 -0.255358677307649 -0.187772339939494 -0.079464055460132 0.15663450872117 0.281287350041566 0.588341090826803 0.786210928528617 1]'

t = sort(rand(1,10));
mat2str(t)

ans = '[0.021032837054374 0.105139509835107 0.16540077837713 0.18325666397305 0.552700488919609 0.658471471291779 0.659400298626749 0.661681959007068 0.662279200025889 0.780426713723395]'

signal = cos(3*t*pi);
mat2str(signal)

ans = '[0.980416722769212 0.547923383075194 0.011930433013734 -0.15572072886789 0.476518476686621 0.99701863625309 0.997655893732092 0.998896653894506 0.999145173395954 0.478225560797132]'

plot(t, signal, '*-')

bin = discretize(signal, boundaries)

bin = 1×10

15 13 11 10 13 15 15 15 15 13

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m_arry = bitpatterns(bin,:)

m_arry = 10×4

1 0 1 1 0 1 0 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 0 1

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