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From: Greg Heath on 2 Aug 2010 13:30
On Jul 31, 9:31 pm, "Dean " <wj.antis...(a)gmail.com> wrote:
> Hi guys,
>
>     I've a signal consists of a frequency 3/32 Hz, sampled at 1 time per second (1 Hz). Since the signal amplitude is 1 volt, I am expecting the signal power is 0.5 v^2.
>
> Although I can obtain the frequency estimation perfectly, I don't know how to convert  Periodogram PSD to the signal amplitude.
>
> If anyone knows the answer, please kindly let me know.
>
> Thank you all.
>
> Dean
>
> ____ Code as follows ________
-----SNIP

You can deduce the relationship from the following demo


clear all, close all, clc, p = 0
N = 24 % Not a power of 2
Nfft = 2^nextpow2(N) % 32
T = 3*pi
dt = T/N
t = dt*(0:N-1)';
t = (0:dt:T-dt)';
t = linspace(0,T-dt,N)';

Fs = 1/dt
df = Fs/N
f = df*(0:N-1)';
f = (0:df:Fs-dt)';
f = linspace(0,Fs-df,N)';

% Choose f0/df = integer, f0 ~ 0.5*(Fs/2)

f0 = round(N/4)*df
A0 = 1
x0 = A0*cos(2*pi*f0*t);

X0 = fft(x0);
absX0 = abs(X0);

p=p+1, figure(p) % Note the stem scaling
subplot(211)
plot(t,x0,'-o')
title( 'x0 = A0 *cos( 2 *pi *f0 *t )' )
subplot(212)
stem(f,absX0/N,'o')
title( ' abs( fft( x0 ) / N )' )

%Power Spectral Density
PSD = absX0.^2/(N*Fs);

Q = ceil((N+1)/2) % 13 = No. of unique components

% Single-Sided PSD

fss = f(1:Q);
if mod(N,2)==0 % Neven
PSDss = [ PSD(1); 2*PSD(2:Q-1); PSD(Q) ];

else % Nodd
PSDss = [ PSD(1); 2*PSD(2:Q) ];
end

p=p+1, figure(p)
subplot(211)
stem(f,PSD,'-o')
title( 'Doublesided Power Spectral Density' )
subplot(212)
stem(fss,PSDss,'-o')
title( 'Singlesided Power Spectral Density' )

minmax(PSD') % [1.662e-031 2.3562 ]
minmax(PSDss') % [1.8907e-031 4.7124 ]

win = rectwin(N);
[Pxx,fxx] = periodogram(x0,win,N,Fs);
length(fxx) % 13 = 1 + N/2
minmax(fxx') % [0 1.2732 ]
minmax(Pxx') % [1.8907e-031 4.7124 ]

p=p+1, figure(p) , hold on
plot(fxx,Pxx,'k-o')
stem(fss,PSDss,'r')
legend('periodogram (Nfft = 24)','fft (N = 24)')
xlabel('Frequency, Hz');
ylabel('PSD, v^2/Hz')
title('Single Sided Power Spectral Density')

win = rectwin(N);
[Pxx,fxx] = periodogram(x0,win,Nfft,Fs);
length(fxx) % 17 = 1 + Nfft/2
minmax(fxx') % [0 1.2732 ]
minmax(Pxx') % [1.8907e-031 4.7124 ]

p=p+1, figure(p) , hold on
plot(fxx,Pxx,'k-o')
stem(fss,PSDss,'r')
legend('periodogram (Nfft = 32)','fft (N = 24)')
xlabel('Frequency, Hz');
ylabel('PSD, v^2/Hz')
title('Single Sided Power Spectral Density')

for j=p:-1:1
figure(j)
end

Hope this helps,

Greg


From: Dean on 5 Aug 2010 00:33
Thank you all for your comments and code. Highly appreciated.
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