Why isn't my quantization noise in my code matching the SNR=6.02(number of bits) + 1.76

I have written a code for an oversampled quantizer and the SNR which i get is not following SNR = 6.02n + 1.76 value. It differs slightly for every bit. Here :

    clear all
    close all
    fs=1024e3; %Very high sampling rate 500 kHz
    N=8
    m=16
    f=m*fs/(2^N); %Frequency of sinusoid
    nCyl=16; %generate sixteen cycles of sinusoid
    t=0:1/fs:nCyl*1/f; %time index

    x=cos(2*pi*f*t);

    ra=2%range of signal
    n=20%number of bits
    q=ra/((2^n)-1)

    t=0:1/fs:nCyl*1/f;
    y=1+cos(2*pi*f*t);
    x0=round(y/q)
    y1=x0*q
    y2=y1-1

    nfft=length(y2);
    nfft2=2^nextpow2(nfft);
    ff=fft(y2,256);%FFT of the quantized signal

    ff0=fft(x,256)%FFT of the unquantized signal

    fabs=abs(ff)
    s=((fabs(17))^2 )*2/256 %signal power (17th peak of FFT)
    t=sum((fabs).^2)/256%complete signal spectrum power
    noise=t-s%noise power

    snrc =s/noise%SNR calculated
    snrcf=10*log10(snrc)%SNR in dB
    isnr=(6.02*n)+1.76%ideal SNR by formula

    s=snr(y1);
    s0=snr(x);

    qe=x-y2;%quantization error
    deltaSNR= snrcf-isnr%error in SNR

Please help!

Answer 1

[Kambiz Shoarinejad]

Because that equation assumes uniformly distributed quantization noise. And with a pure tone, that may not be quite true. To see that, in your code, replace your actual quantization by the following line:

y2 = x + (-q/2) + q.*rand(size(x));

In other words, you would just be adding uniformly distributed quantization noise, in (-q/2,q/2) range, to your signal. And you will see that your calculated SNR would closely match the equation.

Also if you just change the number of points in your FFT from 256 to 1024 (i.e., change your N from 8 to 10), again you will see that your simulated SNR would be in line with the equation.

You may find this article useful (for example, see the plots on Page 6): https://www.analog.com/media/en/training-seminars/tutorials/MT-001.pdf