我不知道，我可能以前发过这个。但这是一个二阶 sigma-delta 量化器，它不作为过采样率运行。所以你可以听到量化噪声，但你也可以听到它下面的音乐。并且输出是二进制的：。y[n]=±A
如果您的教授认为他们可以在没有反馈路径中的或者忽略比较器的增益的情况下对此进行建模，那么他/她是错误的。z−1
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% simulated 1 bit sigma-delta converter:
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% x(n)-y(n-1) w(n) v(n) ( mean(y^2) = A^2 )
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% x ---->(+)--->[1/(z-1)]--->(+)--->[1/(z-1)]--->[Quantizer]----.---> y = +/- A = quantized value
% ^ ^ |
% | | |
% | '----[-fbg]<----. |
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% '------[-1]<------------------------'------[1/z]<-----'
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% "linearized" model:
% .---- q = quantization noise ( mean(q) = 0 )
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% x - y/z w v | ( mean(y^2) = G^2*mean(v^2) + mean(q^2) )
% v
% x ---->(+)--->[1/(z-1)]--->(+)--->[1/(z-1)]--->[G]--->(+)-----.---> y = G*v + q
% ^ ^ |
% | | |
% | '----[-fbg]<----. |
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% '------[-1]<------------------------'------[1/z]<-----'
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% W = 1/(z-1)*(X - Y/z)
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% V = 1/(z-1)*(W - fbg*Y/z)
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% = (X - Y/z - fbg*Y*(z-1)/z)/(z-1)^2
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% = (X*z - Y*(1+fbg*(z-1))) / (z*(z-1)^2)
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% Y = G*V + Q = G*(X*z - Y*(1+fbg*(z-1)))/(z*(z-1)^2) + Q
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% = G*X/(z-1)^2 - G*Y*(1+fbg*(z-1))/(z*(z-1)^2) + Q
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% Y + G*Y*(1-fbg + fbg*z)/(z*(z-1)^2) = G*X/(z-1)^2 + Q
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% Y = (G*X/(z-1)^2 + Q)/(1 + G*(1-fbg + fbg*z)/(z*(z-1)^2))
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% = (G*X/(z-1)^2 + Q)*(z*(z-1)^2)/((z*(z-1)^2) + G*(1-fbg + fbg*z))
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% = z*(G*X + Q*(z-1)^2)/(z^3 - 2*z^2 + (G*fbg+1)*z + G*(1-fbg))
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% = z*(G*X + Q*(z-1)^2)/(z*(z-1)^2 + G*fbg*z + G*(1-fbga))
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% as z -> 1 (DC)
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% Y -> z*X/(fbg*z + (1-fbg)) = X/(fbg + (1-fbg)/z) --> X
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%
if ~exist('mean_vv', 'var')
linearized_model = 0 % run this with 0 the first time to define G and mean(q^2)
end
if ~exist('A', 'var')
A = 1.0 % comparator output magnitude
end
if ~exist('fbg', 'var')
fbg = 2.0 % feedback gain to internal integrator
end
%
% if there is an input soundfile specified, use it. else, create a sin wave
%
if exist('inputFile', 'var')
[inputBuffer, Fs] = audioread(inputFile);
fileSize = length(inputBuffer);
numSamples = 2.^(ceil(log2(fileSize(1)))); % round up to nearest power of 2
x = zeros(numSamples, 1); % zero pad if necessary
x(1:fileSize) = inputBuffer(:,1); % if multi-channel, use left channel only
clear inputBuffer; % free this memory
clear fileSize;
t = linspace(0.0, (numSamples-1)/Fs, numSamples); % time
else
if ~exist('numSamples', 'var')
numSamples = 65536 % number of samples in simulation
end
if ~exist('Fs', 'var')
Fs = 44100 % (oversampled) sample rate
end
if ~exist('f0', 'var')
f0 = 261.6255653 % input freq (middle C)
end
if ~exist('Amplitude', 'var')
Amplitude = 0.25 % input amplitude
end
t = linspace(0.0, (numSamples-1)/Fs, numSamples); % time
x = Amplitude*cos(2*pi*f0*t); % the input
end
sound(x, Fs); % listen to input sound
pause;
y = zeros(1, numSamples); % the output (created and initialized for speed later)
if linearized_model
% artificial quantization noise for linearized model
% mean(q) = 0, var(q) = mean(q^2) = mean(y^2) - G^2*mean(v^2)
% does not have to be uniform or triangle p.d.f.
q = sqrt(6.0*(A^2 - G^2*mean_vv))*( rand(1, numSamples) - rand(1, numSamples) );
else
q = zeros(1, numSamples);
end
sum_yv = 0.0;
sum_vv = 0.0;
w = 0;
v = 0;
for n = 1:numSamples
if linearized_model
y(n) = G*v + q(n); % here the comparator is modelled as a little gain with additive noise
else
if (v >= 0) % the comparator
y(n) = +A;
else
y(n) = -A;
end
q(n) = y(n) - (sum_vv+1e-20)/(sum_yv+1e-20)*v;
end
sum_yv = sum_yv + y(n)*v; % collect some statistics on v
sum_vv = sum_vv + v*v;
v = v + w - fbg*y(n); % second integrator
w = w + x(n) - y(n); % first integrator
end
if ~linearized_model % don't recalculate this if using the linearized model
mean_yv = sum_yv/numSamples;
mean_vv = sum_vv/numSamples;
G = mean_yv/mean_vv; % the apparent comparator gain (assuming stationary input)
end
%
%
%
% Y = ((G*z)*X + (z^3 - 2*z^2 + z)*Q) / (z^3 - 2*z^2 + (G*a+1)*z + G*(1-a))
%
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%
Hx = freqz([0 0 G 0], [1 -2 G*fbg+1 G*(1-fbg)], numSamples/2);
Hq = freqz([1 -2 1 0], [1 -2 G*fbg+1 G*(1-fbg)], numSamples/2);
plot(t, y, 'b');
sound(y, Fs); % this could sound pretty bad
pause;
Y = fft(fftshift(y .* kaiser(numSamples, 5.0)'));
Q = fft(fftshift(q .* kaiser(numSamples, 5.0)'));
f = linspace(0.0, (numSamples/2-1)/numSamples*Fs, numSamples/2);
plot(f, 20*log10(abs(Y(1:numSamples/2)) + 1e-10), 'b');
hold on;
plot(f, 20*log10(abs(Q(1:numSamples/2)) + 1e-10), 'r');
plot(f, 20*log10(abs(Hq) + 1e-10), 'g');
axis([0 Fs/2 -50 100]);
hold off;
pause;
semilogx(f(2:numSamples/2), 20*log10(abs(Y(2:numSamples/2)) + 1e-10), 'b');
hold on;
semilogx(f(2:numSamples/2), 20*log10(abs(Q(2:numSamples/2)) + 1e-10), 'r');
semilogx(f(2:numSamples/2), 20*log10(abs(Hq(2:numSamples/2)) + 1e-10), 'g');
axis([Fs/numSamples Fs/2 -50 100]);
hold off;
pause;
semilogx(f(2:numSamples/2), 20*log10(abs(Y(2:numSamples/2)) + 1e-10), 'b');
hold on;
semilogx(f(2:numSamples/2), 20*log10(abs(Hq(2:numSamples/2)) + 1e-10), 'r');
semilogx(f(2:numSamples/2), 20*log10(abs(Hx(2:numSamples/2)) + 1e-10), 'g');
axis([Fs/numSamples Fs/2 -50 110]);
hold off;
对于所有三个求和节点，第二张图片中都缺少减号。那是一个错误。反馈确实应该是负面的