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How to calculate the time domain amplitude of noise added Sine Gaussian signal from corresponding frequency domain. Here time domain amplitude is 2. How to get this from frequency domain.

N = 4096;
Samplingfrequency= 1024;
timelength=N/Samplingfrequency;

% Generate random
Noise=rand(1,N)*.1;

t    = linspace(-timelength/2,timelength/2,N);
Freq=(1/timelength)*[0:N/2-1 -N/2:-1];

signalFrequency = 64 ;
amplitude=2;
Q=10;
tau  = Q /(2 * pi * signalFrequency);

%Sine Gaussian
Gaussian=((amplitude*exp(-1 .* ((t)/tau).^2) .* cos(2 * pi * signalFrequency .* t))+Noise);

%Time Domain
subplot(2,1 ,1)
plot(t,Gaussian);    

%Frequency Domain 
FFTGaussian=abs(fft(Gaussian));

subplot(2,1 ,2)
% One sided frequency domain plot
plot(Freq(2:N/2),FFTGaussian(2:N/2));
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  • $\begingroup$ The noise in your example is uniform. I assume you want to use randn for Gaussian noise. $\endgroup$ – Harris Sep 8 '15 at 19:53
  • $\begingroup$ The way you've used uniformly distributed noise, it will have both a non-zero mean and variance. The mean will lead to a influence the DC value in the frequency domain. You'll need to estimate the mean before estimating the variance. $\endgroup$ – David Dec 7 '15 at 18:03
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Random noise doesn't have an amplitude, it has statistical properties such as standard deviation and variance. Variance (standard dev. squared) is interpreted as the AC power of the noise. In the freq domain you can estimate the total power of a signal and the total 'power' (variance) of the noise. The ratio of the signal power over the noise power is your signal-to-noise ratio. Multiplying the log-base-ten of that ratio by ten gives you the signal-to-noise ratio measured in dB. [-Rick-]

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  • $\begingroup$ @Rick: Thanks, here I set the noise level 0 to 0.1 not an amplitude, but my interest is amplitude of the true signal not interested to calculate SNR. In other case I have SNR say 'x1' and one sided power spectral density. I would like to generate a sine Gaussian signal with noise which should give SNR value of x1. Frequencies of both signals are same. $\endgroup$ – ma.aero Apr 11 '15 at 14:04
  • $\begingroup$ @ma.areo: Sorry. I misunderstood your words. As it turns out, estimating such an amplitude is a rather tough problem. There are no simple math equations that will answer your question. Some smart people have worked on this problem with moderate success. Estimating such an amplitude is affected by (1) normal spectral leakage, (2) spec leakage on positive-freq spec components from negative-freq-components, and (3) the signal's SNR. ma.aero, if you really want to tackle this problem I can give you some web pages that will help you get started. $\endgroup$ – Richard Lyons Apr 13 '15 at 10:32
  • $\begingroup$ @ma.aero: I wonder if the material at <dsprelated.com/showarticle/155.php> would be helpful. [-Rick-] $\endgroup$ – Richard Lyons Apr 14 '15 at 10:01
  • $\begingroup$ Thanks a lot Richard. It (<dsprelated.com/showarticle/155.php>) gave me some idea to solve this. $\endgroup$ – ma.aero Apr 17 '15 at 13:00

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