MATLAB: FFT signal power is 3dB higher than definition

fft

It seems the signal power is 3dB higher than actual power. Does anyone can explain me why?
For example, I have f(t)=A*sin(2*pi*f*t), its power is f(t)^2 integrated over a cycle then divide the cycle period, (1/T*int(f(t)^2)) which is equal to A^2/2 as expected. (match with RMS calculation as well =(A/sqrt(2))^2)
However, both Cadence Virtuoso dft and Matlab FFT give me signal power = A^2.
The way how I calculate the power is as below: I assume the amplitude of the single tone in one bin of FFT results (coherent sampling) is magx. I calculate the power as magx^2.
I have to calculate the power in this way instead of (magx/sqrt(2))^2 because the noise from all other bins are calculated as sum squared. Otherwise, my SNR will be wrong.
From another perspective, it means the noise is also 3dB higher than actual gaussion noise variance value. So the final SNR is correct if I calculate the signal power as magx^2, which is 3dB higher than actual signal power.
I just wonder why there is 3dB difference in fft results compared with actual power, especially for noise? It seems to me the FFT report peak magnitude instead of rms magnitude.
I have the code attached.
%%whether fft give noise in rms or peak, what is the correct snr?
    %%SNR/FFT/SFDR/SNDR/noise simulation
    %%https://www.mathworks.com/help/signal/ref/snr.html#bt3er4n
    %%https://www.mathworks.com/help/matlab/ref/fft.html
    clear;close all;
    rng default
    %rng(1);
    FlagTwoTone = false;
    tone1 = 53/128*800e6;
    Amp = 0.532 * 10^(1.37/20); %input amplitude * PGA gain = 0.623
    fs = 800e6;
    N = 128;
    t = (0:N-1)*1/fs;
    freq = fs*(0:(N/2))/N;
    noisestd = 300e-6; % assume noise std = 300 uV
    sig = Amp*sin(2*pi*tone1*t);
    noise = noisestd*randn(size(sig)); %zero mean, variance = noisestd^2
    myx = sig + noise;
    Tone1dB = 20*log10(Amp/sqrt(2))
    noisedB = 20*log10(noisestd)
    %deltadB= Tone1dB - noisedB
    mySNRviasnr = snr(sig, noise)
    Y = fft(myx, N); %length(myx) = N
    P2 = abs(Y/N);  %double sided fft spectrum, convert complex to mag and then divided by N=128, why divide 128?
    P1 = P2(1:N/2+1); %matlab starts with 1
    P1(2:end-1) = 2*P1(2:end-1);
    modP1 = P1;
    figure(1);
    plot(freq,P1) 
    title('Single-Sided Amplitude Spectrum of myx(t)')
    xlabel('f (Hz)')
    ylabel('|P1(f)|')
    grid on;
    simTone1amp = 0;
    simTone1index = 0;
    simTone1freq = 0;
    for i = 1:length(modP1)
        if simTone1amp < modP1(i)
            simTone1amp = modP1(i);
            simTone1index = i;        
        end
    end
    %simTone1index
    %simTone1amp =P1(simTone1index)
    simTone1freq = freq(simTone1index);
    modP1(simTone1index) = 0;
    simTone1dB = 20*log10(simTone1amp)
    if FlagTwoTone %two tones only 
        simTone2amp = 0;
        simTone2index = 0;
        simTone2freq = 0;
        for i = 1:length(modP1)
            if simTone2amp < modP1(i)
                simTone2amp = modP1(i);
                simTone2index = i;        
            end
        end
        simTone2freq = freq(simTone2index);
        modP1(simTone2index) = 0;
    end
        simHarmonicAmp = 0;
        simHarmonicIndex = 0;
        simHarmonicFreq = 0;
        %test = 0;
        for i = 1:length(modP1)
            %test = test + modP1(i)^2;
            if simHarmonicAmp < modP1(i)
                simHarmonicAmp = modP1(i);
                simHarmonicIndex = i;        
            end
        end
        simHarmonicFreq = freq(simHarmonicIndex);  
        fftTotalNoise = sum(modP1.^2);%sum squared total noises
        simNoisedB = 10*log10(fftTotalNoise)
        %test
    if FlagTwoTone
    else
        SNDR = 10*log10(simTone1amp^2/fftTotalNoise)
    end

Best Answer

The normalization is all in your code--
P2 = abs(Y/N);
P1 = P2(1:N/2+1);
P1(2:end-1) = 2*P1(2:end-1);
This is the way to normalize a one-sided PSD such that the peak amplitude for a pure signal matches the peak of the signal in time domain; if you want RMS, then that's up to you.
NB: However, this normalization for a single bin works(*) only if the frequency sampling is such the tone of the input signal exactly matches a frequency bin; otherwise the energy will be spread over adjacent bins...discussion/illustration just a few days ago at Answer_319928
(*) That is, the normalization is correct; the maximum amplitude of a single bin in the PSD only matches the time domain peak if the energy is all at the precise bin frequency.
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