Thread Subject: Effect of sampling frequency for FFT

Hi,

I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
Highter sampling frequency then, of course, reduces the time window size.
This should simulate measurements of some sinusoid with different sampling frequencies

Do you know what would cause better SNG values, when sampling frequency increases?

Thanks already for answers,

-Juho

"juho salminen" <jssalmi3@cc.hut.fi> wrote in message <hcev3g$lej$1@fred.mathworks.com>...
> Hi,
>
> I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> Highter sampling frequency then, of course, reduces the time window size.
> This should simulate measurements of some sinusoid with different sampling frequencies
>
> Do you know what would cause better SNG values, when sampling frequency increases?
>
> Thanks already for answers,
>
> -Juho

Note: I am talking about SNR (!) for detecting amplitde of signal. Amplitudes seem to be always closer to right value, when sampling frequency is increased

"juho salminen" <jssalmi3@cc.hut.fi> wrote in message <hcev3g$lej$1@fred.mathworks.com>...

> Highter sampling frequency then, of course, reduces the time window size.

By time-frequency duality, windowing in the time domain is equivalent to low-pass filtering the frequency spectrum. By decreasing the window-size you are applying a low pass filter to your spectral data of lower and lower cut-offs and hence greater and greater smoothing.

"Matt " <xys@whatever.com> wrote in message <hcf0h0$kqk$1@fred.mathworks.com>...
> "juho salminen" <jssalmi3@cc.hut.fi> wrote in message <hcev3g$lej$1@fred.mathworks.com>...
>
> > Highter sampling frequency then, of course, reduces the time window size.
>
> By time-frequency duality, windowing in the time domain is equivalent to low-pass filtering the frequency spectrum. By decreasing the window-size you are applying a low pass filter to your spectral data of lower and lower cut-offs and hence greater and greater smoothing.

Sorry for inconvenience, what I meaned was, that actual FFT window size is always the same, let's say 500 points. But, when I sample artificial signal with highter sampling frequency, Amplitude of sine wave in FFT is much more accurate than with lower sampling frequency.

For example, 1. I sample some artificial noisy sine wave with 50Hz, and I collect 5000samples, 2. I sample some artificial sine wave with 200Hz, and I collect 5000samples, the second case gives more accurate amplitude results in FFT

juho salminen wrote:
...

> For example, 1. I sample some artificial noisy sine wave with 50Hz,
> and I collect 5000samples, 2. I sample some artificial sine wave with
> 200Hz, and I collect 5000samples, the second case gives more accurate
> amplitude results in FFT

That is, in essence, the same effect as fitting any function--you've got
more information to work with that defines the resolution of the shape
of the curve more distinctly.

--

On Oct 30, 11:01 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
> Hi,
>
> I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> Highter sampling frequency then, of course, reduces the time window size.
> This should simulate measurements of some sinusoid with  different sampling frequencies
>
> Do you know what would cause better SNG values, when sampling frequency increases?

SNR is independent of Fs.

close all, clear all, clc

f0 = 30
N = 500
S = 5
Ps = 5^2/2 % = 12.5 Signal Power
N0 = 1
Pn = 1 % = 1 Noise Power
SNR0 = Ps/Pn % 12.5
M = 30 % No. of trials

rand('state',0)
randn('state',0)

for i = 1:M

    Fs(i) = f0*i; % Sampling Frequency
    dt = 1/Fs(i)
    t = dt*(0:N-1);
    s = S*cos(2*pi*(f0*t+rand(1,N)));
    n = N0*randn(1,N);

    vars(i) = var(s);
    varn(i) = var(n);
    SNR(i) = var(s)/var(n);

end

summary = [Fs' vars' varn' SNR']
meansummary = mean(summary(:,2:4))
stdsummary = std(summary(:,2:4))
meanSNR = meansummary(3)
stdSNR = stdsummary(3)

figure,hold on
plot(Fs,SNR0*ones(1,M),'k')
plot(Fs,meanSNR*ones(1,M),'r')
plot(Fs,(meanSNR-stdSNR)*ones(1,M),'r--')
plot(Fs,(meanSNR+stdSNR)*ones(1,M),'r--')
plot(Fs,SNR)
axis([0 M*f0 0 20])
xlabel('Sampling Frequency (Hz)')
ylabel('Signal-to-Noise Ratio')
title('SNR Estimate vs Sampling Frequency')

On Nov 1, 2:27 am, Greg Heath <he...@alumni.brown.edu> wrote:
> On Oct 30, 11:01 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
>
> > Hi,
>
> > I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> > Highter sampling frequency then, of course, reduces the time window size.
> > This should simulate measurements of some sinusoid with  different sampling frequencies
>
> > Do you know what would cause better SNG values, when sampling frequency increases?
>
> SNR is independent of Fs.
-----SNIP

Whoops. You mean SNR of the FFT.
I'll look at that now.
Greg

On Nov 1, 2:27 am, Greg Heath <he...@alumni.brown.edu> wrote:
> On Oct 30, 11:01 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
>
> > Hi,
>
> > I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> > Highter sampling frequency then, of course, reduces the time window size.
> > This should simulate measurements of some sinusoid with different sampling frequencies
>
> > Do you know what would cause better SNG values, when sampling frequency increases?
>
> SNR is independent of Fs.
>
> close all, clear all, clc
>
> f0 = 30
> N = 500
> S = 5
> Ps = 5^2/2 % = 12.5 Signal Power
> N0 = 1
> Pn = 1 % = 1 Noise Power
> SNR0 = Ps/Pn % 12.5
> M = 30 % No. of trials
>
> rand('state',0)
> randn('state',0)
>
> for i = 1:M
>
> Fs(i) = f0*i; % Sampling Frequency
> dt = 1/Fs(i)
> t = dt*(0:N-1);
> s = S*cos(2*pi*(f0*t+rand(1,N)));

      s = S*cos(2*pi*(f0*t+rand));

> n = N0*randn(1,N);
>
> vars(i) = var(s);
> varn(i) = var(n);
> SNR(i) = var(s)/var(n);
>
> end
>
> summary = [Fs' vars' varn' SNR']
-----SNIP

% Exclude cases Fs < = 2*f0
meansummary = mean(summary(3:M,2:4))
stdsummary = std(summary(3:M,,2:4))
> meanSNR = meansummary(3)
> stdSNR = stdsummary(3)
-----SNIP

 % Restricting to Fs > 2*f0.

Fs1 = Fs(3:M);
M1 = M-2

figure,hold on
plot(Fs1,SNR0*ones(1,M1),'k')
plot(Fs1,meanSNR*ones(1,M1),'r')
plot(Fs1,(meanSNR-stdSNR)*ones(1,M1),'r--')
plot(Fs1,(meanSNR+stdSNR)*ones(1,M1),'r--')
plot(Fs1,SNR(3:M))

> axis([0 M*f0 0 20])
> xlabel('Sampling Frequency (Hz)')
> ylabel('Signal-to-Noise Ratio')
> title('SNR Estimate vs Sampling Frequency')

Hope this helps.

Greg

Greg Heath <heath@alumni.brown.edu> wrote in message <d65291df-9d35-4375-abf6-1915e52fc5c6@m1g2000vbi.googlegroups.com>...
>
> On Nov 1, 2:27 am, Greg Heath <he...@alumni.brown.edu> wrote:
> > On Oct 30, 11:01 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
> >
> > > Hi,
> >
> > > I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> > > Highter sampling frequency then, of course, reduces the time window size.
> > > This should simulate measurements of some sinusoid with different sampling frequencies
> >
> > > Do you know what would cause better SNG values, when sampling frequency increases?
> >
> > SNR is independent of Fs.
> >
> > close all, clear all, clc
> >
> > f0 = 30
> > N = 500
> > S = 5
> > Ps = 5^2/2 % = 12.5 Signal Power
> > N0 = 1
> > Pn = 1 % = 1 Noise Power
> > SNR0 = Ps/Pn % 12.5
> > M = 30 % No. of trials
> >
> > rand('state',0)
> > randn('state',0)
> >
> > for i = 1:M
> >
> > Fs(i) = f0*i; % Sampling Frequency
> > dt = 1/Fs(i)
> > t = dt*(0:N-1);
> > s = S*cos(2*pi*(f0*t+rand(1,N)));
>
> s = S*cos(2*pi*(f0*t+rand));
>
> > n = N0*randn(1,N);
> >
> > vars(i) = var(s);
> > varn(i) = var(n);
> > SNR(i) = var(s)/var(n);
> >
> > end
> >
> > summary = [Fs' vars' varn' SNR']
> -----SNIP
>
> % Exclude cases Fs < = 2*f0
> meansummary = mean(summary(3:M,2:4))
> stdsummary = std(summary(3:M,,2:4))
> > meanSNR = meansummary(3)
> > stdSNR = stdsummary(3)
> -----SNIP
>
> % Restricting to Fs > 2*f0.
>
> Fs1 = Fs(3:M);
> M1 = M-2
>
> figure,hold on
> plot(Fs1,SNR0*ones(1,M1),'k')
> plot(Fs1,meanSNR*ones(1,M1),'r')
> plot(Fs1,(meanSNR-stdSNR)*ones(1,M1),'r--')
> plot(Fs1,(meanSNR+stdSNR)*ones(1,M1),'r--')
> plot(Fs1,SNR(3:M))
>
> > axis([0 M*f0 0 20])
> > xlabel('Sampling Frequency (Hz)')
> > ylabel('Signal-to-Noise Ratio')
> > title('SNR Estimate vs Sampling Frequency')
>
> Hope this helps.
>
> Greg


Hi,

doesn't this basically mean, that there is no SNR improvement, when sampling frequency is increased?

-Juho

Btw, thanks for all the effort!

On Nov 2, 3:51 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
> Greg Heath <he...@alumni.brown.edu> wrote in message <d65291df-9d35-4375-abf6-1915e52fc...@m1g2000vbi.googlegroups.com>...
>
> > On Nov 1, 2:27 am, Greg Heath <he...@alumni.brown.edu> wrote:
> > > On Oct 30, 11:01 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
>
> > > > Hi,
>
> > > > I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> > > > Highter sampling frequency then, of course, reduces the time window size.
> > > > This should simulate measurements of some sinusoid with different sampling frequencies
>
> > > > Do you know what would cause better SNG values, when sampling frequency increases?
>
> > > SNR is independent of Fs.

-----SNIP (Revised code below)

close all, clear all, clc

f0 = 30
N = 500
S0 = 5 % Sinusoidal amplitude
SNR0 = 15 % Signal-to-Noise Ratio
M = 100 % No. of trials

Ps = S0^2/2 % 12.5 Signal Power
Pn = Ps/SNR0 % 0.8333 Noise Power
N0 = sqrt(Pn) % S0/sqrt(2*SNR0) = 0.9129

rand('state',0)
randn('state',0)

for i = 1:M

    Fs(i) = f0*i; % Sampling Frequency
    dt = 1/Fs(i);
    t = dt*(0:N-1);
    s = S0*cos(2*pi*(f0*t+rand));
    n = N0*randn(1,N);

    vars(i) = var(s);
    varn(i) = var(n);
    SNR(i) = var(s)/var(n);

end

summary = [Fs' vars' varn' SNR'];

% Restricting to Fs > 2*f0.

meansummary = mean(summary(3:M,2:4))
% 12.5174 0.8428 14.9103

stdsummary = std(summary(3:M,2:4))
% 0.1117 0.0519 0.9746

meanSNR = meansummary(3)
stdSNR = stdsummary(3)
Fs1 = Fs(3:M);
M1 = M-2

figure,hold on
plot(Fs1,SNR0*ones(1,M1),'k')
plot(Fs1,SNR(3:M))
plot(Fs1,meanSNR*ones(1,M1),'r')
plot(Fs1,(meanSNR-stdSNR)*ones(1,M1),'r--')
plot(Fs1,(meanSNR+stdSNR)*ones(1,M1),'r--')
legend('True SNR','SNR Estimate',...
       ['meanSNR = ',num2str(meanSNR)], ...
       ['stdSNR = ',num2str(stdSNR)],4)
axis([0 M*f0 0 20])
xlabel('Sampling Frequency (Hz)')
ylabel('Signal-to-Noise Ratio')
title('SNR Estimate vs Sampling Frequency')

Hope this helps.

Greg

> Hi,
>
> doesn't this basically mean, that there is no SNR improvement, when sampling
> frequency is increased?

Yes. I repeat:

"SNR is independent of Fs"

The example indicates that Power based estimates of SNR are ~14.9103
+/- 0.9746
independently of Fs, when the true value is SNR = 15.

However, although your OP posed the question in terms of estimating
SNR vs Fs, your
followup post indicated that you are really interested in using the
FFT to estimate the
amplitude of a sine wave corrupted by noise.

In the latter case the simulation in my next post shows that peak-
picking FFT estimates
of both the frequency and amplitude degrade as Fs increases.

The basic reason appears to be that the FFT frequency spacing df = Fs/
N increases
as Fs increases with N = constant.

Hope this helps.

Greg

On Nov 1, 2:36 am, Greg Heath <he...@alumni.brown.edu> wrote:
> On Nov 1, 2:27 am, Greg Heath <he...@alumni.brown.edu> wrote:> On Oct 30, 11:01 am, "juho salminen" <jssal...@cc.hut.fi> wrote:
>
> > > Hi,
>
> > > I have done some tests with artificial sinusoidial data with white noise added on it. It seems, that the higher sampling frequency I have, the better is signal to noise ratio in FFT. With this I mean, that I take FFT with constant window length, let's say 500 samples. Signal itself is kept constant, let's say 30Hz.
> > > Highter sampling frequency then, of course, reduces the time window size.
> > > This should simulate measurements of some sinusoid with  different sampling frequencies
>
> > > Do you know what would cause better SNG values, when sampling frequency increases?
>
> > SNR is independent of Fs.
>
> -----SNIP
>
> Whoops. You mean SNR of the FFT.
> I'll look at that now.

Although your OP posed the question in terms of estimating SNR vs Fs,
I
interpret your above comments to mean that you are really interested
in
using the FFT to estimate the amplitude of a sine wave corrupted by
noise.

In the latter case the simulation below shows that peak-picking FFT
estimates
of both the frequency and amplitude degrade as Fs increases.

The basic reason appears to be the FFT frequency spacing df = Fs/N
increases
as Fs increases with N = constant.

close all, clear all, clc, k = 0

f0 = 30
N = 500
S0 = 5 % Sinusoidal amplitude
SNR0 = 15 % Signal-to-Noise Ratio
M = 100 % No. of trials

Ps = S0^2/2 % 12.5 Signal Power
Pn = Ps/SNR0 % 0.8333 Noise Power
N0 = sqrt(Pn) % S0/sqrt(2*SNR0) = 0.9129

rand('state',0)
randn('state',0)

for i = 1:M

    i = i
    Fs(i) = f0*i; % Sampling Frequency
    dt = 1/Fs(i);
    t = dt*(0:N-1);
    s = S0*cos(2*pi*(f0*t+rand));
    n = N0*randn(1,N);
    z = s+n;
    meanz = mean(z)
    check1 = meanz - mean(n)
    z = z - mean(z); % Exclude DC

% k = k+1,figure(k),hold on
% plot(t,s)
% plot(t,n,'g')
% plot(t,z,'r')
% xlabel('Time')
% ylabel('Signal and Noise')

    df = Fs(i)/N; % Frequency Resolution
    Q = ceil((N+1)/2); % No. unique points
    f = df*(0:Q-1); % One-sided spectrum
    Z = fft(z-mean(z))/N ; % Aliased for Fs <= 2*f0
    absZ = abs(Z);
    PSD = N*(absZ.^2);
    check2 = sum(PSD)-sumsqr(z)

    absZ1 = [absZ(1),sqrt(2)*absZ(2:Q-1),absZ(Q)];%Single Sided
    check2 = max(abs(sum(absZ1.^2)-sum(absZ.^2)))

    [S(i) I(i)] = max(sqrt(2)*absZ1); %Dominant Amplitude
    fmax(i) = f(I(i));

    if(i==1),k=k+1; figure(k), end

    plot(f(1:Q),sqrt(2)*absZ1),hold on
    plot(fmax(i),S(i),'or'),hold off
    xlim([0 60])
    xlabel('Frequency')
    ylabel('Amplitude')
    title(['Dominant Peak Estimate for Fs = ',num2str(Fs(i))])
 end

k=k+1; figure(k)
subplot(2,1,1)
plot(Fs(3:M),fmax(3:M))
ylabel('Dominant Frequency')
subplot(2,1,2)
plot(Fs(3:M),S(3:M))
xlabel('Sampling Frequency')
ylabel('Peak Amplitude')

Hope this helps.

Greg