- There can only be four (4) states with 2 bits; 00, 01, 10, 11 but your quantization process produces 5 levels. NB: there are (N-1) differences between N levels.
- Half the elements of y2 are zero; precisely every other one, in fact. Therefore, half of the values in y2 are -Inf
- The plot doesn't show anything because of 2. above there is no line between consecutive points and the size of the dots is small enough just not visible. Try stem(f,y2) or plot(f,y2,'*k')
FFT of quantized signal
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I am having difficulty in plotting FFT of quantized signal of an ideal qunatizer in MATLAB . When I plot the FFT in dB Scale , I am getting infinity as the value & figure is blank . Please help me, I have posted the Code below
fs = 1e6;
fx = 50e3;
Afs = 1;
N = 2^11;
% time vector
t = linspace(0, (N-1)/fs, N);
% input signal
y = Afs * cos(2*pi*fx*t);
% spectrum
s = 20 * log10(abs(fft(y)/N/Afs*2));
% drop redundant half
s = s(1:N/2);
% frequency vector (normalized to fs)
f = (0:length(s)-1);
figure(1)
plot(y);
title('input signal ')
ylabel('y');
xlabel('frequency bins');
B = 2; % no of bits
delta = 2/2^B;
fs = 1e6;
% Number of cycles in test
cycles = 67;
%Make N/cycles non-integer! accomplished by choosing cyclesÆprime #
%N=power of 2 speeds up analysis
% N = 2^10;
%signal frequency
fx = fs*cycles/N
y = Afs * cos(2*pi*fx*t);
s = 20 * log10(abs(fft(y)/N/Afs*2));
f = (0:length(s)-1);
figure(2)
plot(f,s);
title('spectrum of input signal ');
ylabel('y')
xlabel('frequency bins')
y1=round(y/delta)*delta;
figure(3)
plot(y,'b');
hold on
stem(y1,'r')
hold on
plot(y-y1,'g')
title(sprintf('Signal, Quantized signal and Error for %g bits, %g quantization levels',B,2^B));
xlabel('frequency');
hold off
y2=20*log10((abs(fft(y1)/N*2)));
figure(10)
f1 = (0:length(y2)-1);
plot(f1 ,y2)
댓글 수: 4
Sanjaya Senavirathne
2021년 8월 11일
problem in this case is some values contains 0, so the FFT value of 0 is which is un imagineble that's why figure is blank. so i would suggest please add really very low noise like e-30 , then you will be able to see an output.
Chunru
2021년 8월 11일
In theory, quantization can have any integer number of levels (not necessarily 2^B, though 2^B is prefered). The quantization can be non-uniform as well.
답변 (2개)
Chunru
2021년 8월 11일
Quantizing the ideal sinusoidal signal will produce a perodical signal, which may have some weird spectrum with 0 at some frequencies causing the display problem. To study the quatization effect, you can use some other signals, for example chirp, pink noise. The following code add in an offset so that the spectrum is no longer "weird".
fs = 1e6;
fx = 50e3;
Afs = 1;
N = 2^11;
% time vector
t = linspace(0, (N-1)/fs, N);
% input signal
y = Afs * cos(2*pi*fx*t);
% spectrum
s = 20 * log10(abs(fft(y)/N/Afs*2));
% drop redundant half
s = s(1:N/2);
% frequency vector (normalized to fs)
f = (0:length(s)-1);
figure(1)
plot(y);
title('input signal ')
ylabel('y');
xlabel('frequency bins');
B = 2; % no of bits
delta = 2/2^B;
fs = 1e6;
% Number of cycles in test
cycles = 167;
%Make N/cycles non-integer! accomplished by choosing cyclesÆprime #
%N=power of 2 speeds up analysis
% N = 2^10;
%signal frequency
fx = fs*cycles/N
y = Afs * cos(2*pi*fx*t) + 0.123; % add a small offset
s = 20 * log10(abs(fft(y)/N/Afs*2));
f = (0:length(s)-1);
figure(2)
plot(f,s);
title('spectrum of input signal ');
ylabel('y')
xlabel('frequency bins')
y1=round(y/delta)*delta; y1'
figure(3)
plot(y,'b');
hold on
stem(y1,'r')
hold on
plot(y-y1,'g')
title(sprintf('Signal, Quantized signal and Error for %g bits, %g quantization levels',B,2^B));
xlabel('frequency');
hold off
y2=20*log10((abs(fft(y1)/N*2)))'
%y2=((abs(fft(y1)/N*2)))'
figure(10)
f1 = (0:length(y2)-1);
plot(f1 ,y2)
댓글 수: 0
Yazan
2021년 8월 11일
There are some issues with your code. See my modifications and comments below. You cannot calculate SNR, because you did not add noise to any signal (unless I'm missing something here).
clc, clear, close all
fs = 1e6;
fx = 50e3;
Afs = 1;
N = 2^11;
% time vector
t = linspace(0, (N-1)/fs, N);
% input signal
y = Afs * cos(2*pi*fx*t);
figure('Units', 'normalized', 'Position', [0.05 0.15 0.85 0.7])
subplot(3,2,1)
plot(t, y), title('Signal y'); axis tight
xlabel('Time - sec');
ylabel('Amplitude');
% spectrum
Nfft = N;
s = 1/Nfft * abs(fft(y, Nfft));
f = linspace(0, fs/2, Nfft/2+1);
subplot(3,2,2)
s = s(1:Nfft/2+1).^2;
sy = [s(1), 2*s(2:end-1), s(end)];
plot(f, pow2db(sy)), grid minor
title('One-sided power spectrum of y'), xlabel('Frequency - Hz');
ylabel('dB')
% or you can use the periodogram function with rect window
% you'll get exactly the same result
% [s2, f2] = periodogram(y, rectwin(length(y)), Nfft, fs, 'onesided', 'power');
% hold on, plot(f2, pow2db(s2))
% signal 2
B = 2;
delta = 2/2^B;
fs = 1e6;
cycles = 67;
fx = fs*cycles/N;
yy = Afs * cos(2*pi*fx*t);
subplot(3,2,3)
plot(t, yy), title('Signal yy'); axis tight
xlabel('Time - sec');
ylabel('Amplitude');
% spectrum
Nfft = N;
s = 1/Nfft * abs(fft(yy, Nfft));
s = s(1:Nfft/2+1).^2;
syy = [s(1), 2*s(2:end-1), s(end)];
subplot(3,2,4)
plot(f, pow2db(syy)), grid minor
title('One-sided power spectrum of yy'), xlabel('Frequency - Hz');
ylabel('dB')
% quantized signal
y1 = round(y/delta)*delta;
subplot(3,2,5)
n = 100;
plot(t(1:n), y1(1:n)), title('Signal y1'); axis tight
xlabel('Time - sec');
ylabel('Amplitude'); grid minor
hold on,
plot(t(1:n), y(1:n), '--')
err = y-y1;
plot(t(1:n), err(1:n))
legend({'Quantized', 'Original', 'Error'})
s = 1/Nfft * abs(fft(y1, Nfft));
s = s(1:Nfft/2+1).^2;
sy1 = [s(1), 2*s(2:end-1), s(end)];
subplot(3,2,6)
plot(f, pow2db(sy1)), grid minor
title('One-sided power spectrum of y1'), xlabel('Frequency - Hz');
ylabel('dB')
% estimate power
pRMSy = mag2db(rms(y));
pRMSyy = mag2db(rms(yy));
pRMSy1 = mag2db(rms(y1));
fprintf('Power of signal y = %g dBW\n', pRMSy)
fprintf('Power of signal yy = %g dBW\n', pRMSyy)
fprintf('Power of signal y1 = %g dBW\n', pRMSy1)
댓글 수: 1
Addy
2022년 8월 17일
Hey in your code, 2 bit quantization, there are 5 levels.
you can see when you do "unique(y1)"
But for 2 bit quantization, there is only 4 levels. 00, 01, 10, 11.
So, dividing your signal into 4 parts is necessary (At least that is what I am understanding. Please correct me if I am wrong. I am still learning.
Otherwise, your code looks great.
I was working on that 4 level quantization.
clear;clc
A = 4;
f = 10e3;
% fs = 25e6; ts = 1/fs;
ts = 0.2/10/f(end);
t_end = 1/f;
t = 0:ts:t_end;
x = A*sin(2*pi*f*t)';
% x = awgn(x,100);
t = t*1e6;
%%
q_bits = 2;
q_num_levels = 2^q_bits;
q_levels = linspace(min(x),max(x),q_num_levels);
bins = interp1(1:numel(q_levels),q_levels,0.5 + (1:numel(q_levels)-1));
y = discretize(x,[-Inf bins Inf]);
y = normalize(y,'range',[min(q_levels),max(q_levels)]);
y = y-mean(y);
u = unique(y);
%%
bit_levels = 0:q_num_levels-1;
bit_levels = dec2bin(bit_levels,q_bits);
q = diff(q_levels); q = q(1);
q_error = (max(x)-min(x))/(2*q_num_levels);
q_noise_pwr = q/sqrt(12);
%%
figure(1);clf
hold on;box on
set(gcf,'color','k')
set(gca,'color','k')
set(gca, 'YGrid', 'on', 'XGrid', 'off')
set(gca,'GridColor','w')
set(gca,'XColor','w')
set(gca,'YColor','w')
set(gca,'GridAlpha',0.5)
plot(t,x,'c')
plot(t,y,'m');
plot(t,x-y,'-.g')
yticks(q_levels);
yticklabels(bit_levels);
yticklabels([]);
ylim([min(x)+min(x)/10,max(x)+max(x)/10])
xlim([min(t),max(t)])
tit = title(['Example with ',num2str(q_bits),' Bits[',num2str(q_num_levels),' levels] Linear Quantization']);
lgnd = legend( {'\color{cyan} Analog signal',...
'\color{magenta} Quantized signal',...
['\color{green} Quantized Error signal.',newline,...
' Max.Quant.Error = ',num2str(q_error)]});
set(lgnd, 'Color', 'k');
set(tit, 'Color', 'w');
ylabel('Quantization levels')
xlabel('Time - [µs]')
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