ࡱ> 463|S@ #bjbj WT,$)))PL*l*T9+^v+v+v++///9999999$:R<A9-/'////A9v++n9g6g6g6/v++9g6/9g64g66Z66+ + V) 3699096=5=6=6//g6/////A9A9$$?6(Power Spectral Density - the basics Power Spectral Densfty (PSD) is the frequency response of a random or periodic signal. It tells us where the average power is distributed as a function of frequency. ( The PSD is deterministic, and for certain types of random signals is independent of time1. This is useful because the Fourier transform of a random time signal is itself random, and therefore of little use calculating transfer relationships (i.e., finding the output of a filter when the input is random). ( The PSD of a random time signal x(t) can be expressed in one of two ways that are equivalent to each other 1. The PSD is the average of the Fourier transform magnitude squared, over a large time interval  EMBED Equation.3  2. The PSD is the Fourier transform of the auto-correlation function.  EMBED Equation.3   EMBED Equation.3  ( The power can be calculated from a random signal over a given band of frequencies as follows: 1. Total Power in x(t):  EMBED Equation.3  2. Power in x(t) in range f1 - f2:  EMBED Equation.3  1The signal has to be stationary, which means that us statistics do not change as a function of time. ( If a random signal x(t) is passed through a time-invariant filter with frequency response H(f), the resulting signal y(t) has a PSD as follows:    Example: Random Gaussian noise signals. Two random signals x1(t) and x2(t) of 10 seconds duration are created as follows: xl=randn(1,10000); % signal 1 x2=randn(1,10000); % signal 2 t=linspace(0,10,10000); % time series dt = t(2)-t(1); % time increment subplot(2,2,1),plot(t,x1) %plot of x1 subplot(2,2,2),plot(t,x2) %plot of x2 The sample frequency is 1kHz. The FFT of each of these signals is calculated from X1=fftshift(fft(x1)); % FFT of signal 1 X2=fftshift(fft(x2)); % FFT of signal 2 Df=1/dt; % fequency span f=linspace(Df/2,Df/2,10000); % frequency series subplot(2,2,3),plot(f,dB(X1)); %plot of FFT of x1 subplot(2,2,4),plot(f,dB(X2)); %plot FFT of x2 The plot below shows a 0.1 second snapshot of two random time functions x1(t) and x2(t) and the first 10 Hz of the fast Fourier transform of these signals. This shows that the frequency responses of these random signals are generally different, although they seem to have a common average level, and have similar overall randomness, which would be exhibited in the rest of the spectrum if we were to show it. Perhaps what we should be doing is looking at the average Fourier transform instead of just a single sample of the Fourier transform this is, more or less, what the PSD is; it is the average Fourier transform squared taken over a very long time interval.  We now calculate and plot the PSD of the original time series x1(t) and x2(t). x1=randn(1,10000); % signal 1 x2=randn(1,10000); % signal 2 t=linspace(0,10,10000); % time series dt = t(2)-t(1); % time increment Df = 1/dt % fequency span subplot(2,2,1),plot(t,x1) %plot of x1 subplot(2,2,2),plot(t,x2) %plot of x2 Sx1=psd(x1,1024,1000); %PSD of x1 Sx2=psd(x2,1024,1000); %PSD of x2 f=linspace0,Df/2,length(Sx1)); % frequency series subplot(2,2,3),plot(f,db(Sx1)); %plot of PSD of x1 subplot(2,2,4),plot(f,db(Sx2)); %plot of PSD of x2  The PSD for each signal looks more or less flat across the frequency band. This type of noise is referred to as white, and if we had taken an infinitesimally small time increment, we would see this flatness across the entire frequency band. The reason that there is some variation about the constant level is that we didnt take a large enough (i.e., infinite) time sequence of random numbers to calculate the PSD from. The estimate of the PSD (as calculated in MATLAB) becomes more accurate as the sample size becomes infinite. The mean of the PSDs of xl and x2 turn out to be very close to 1. mean(Sxl) %= 1.0137 mean(Sx2) %= 1.0241 This is telling us that the average value of the MATLAB PSD, which is the variance, is close to unity. Now lets consider filtering the Gaussian noise. First design a high order Butterworth filter that cuts off at half the Nyquist frequency (500 Hz) [b,a]=butter(40,0.5); %n=40th order,fc=0.5 nyquist Now plot the frequency response, normalized to the nyquist frequency (this just makes the maximum frequency be 1) freqz(b,a) % plot the frequency and phase response  Now produce new noise samples y1(t) and y2(t) by filtering x1(t) and x2(t). y1=filter(b,a,x1); %filter x1 y2=filter(b,a,x2); %filter x2 Y1=fftshift(fft(y1)); %calculate frequency response of y1 Y2=fftshift(fft(y2)); %calculate frequency response of y2 subplot(2,2,1),plot(t,x1) %plot of y1 subplot(2,2,2),plot(t,x2) %plot of y2 subplot(2,2,3),plot(f,dB(X1)); %plot of FFT of y1 subplot(2,2,4),plot(f,dB(X2)); %plot FFT of y2 We see that the time series is smoother (it has been filtered), but the FFT remains random. The variance of the filtered noise is reduced to roughly 0.5. var(y1) %= 0.4895 var(y2) %= 0.4956  Taking the PSD of the filtered noise Sy1=psd(y1,1024,1000); %PSD of y1 Sy2=psd(y2,1024,1000); %PSD of y2 f=linspace(0,Df/2,length(Sx1)); % frequency series subplot(2,2,1),plot(t,x1) %plot of y1 subplot(2,2,2),plot(t,x2) %plot of y2 subplot(2,2,3),plot(f,db(Sy1)); %plot of PSD of x1 subplot(2,2,4),plot(f,db(Sy2)); %plot of PSD of x2 Basically what we have done here is filtered white noise. The PSD of the filtered noise takes on the shape of the filter frequency response, and is the same (within experimental variation) for each independent noise signal.  Calculating the variance of the filtered noise results in mean(Syl) %= 0.4851 mean(Sy2) %= 0.4910 The means are both close to 0.5. Not surprisingly, cutting out half the noise with a steep high order Butterworth filter has reduced the total variance by approximately 0.5. The variance for a flat PSD with amplitude Sx (W/Hz) and total positive and negative frequency extent Df is found by (sx)2 = Sx Df. In MATLABs calculation of PSD, Df is normalized to 1, so (sx)2 = Sy = 1. For the filtered signal, Sx is 1 for half of Df and 0 for the other half of Df, so (sx)2 = Sy/2= 1/2. In general, when the PSD is not simple shaped as in the above examples, you will need to integrate the PSD in order to find the variance. In MATLAB, this is equivalent to simply finding the mean of the PSD. What can you conclude about the PSD and white Gaussian Noise? x(t) H(f) y(t)  EMBED Equation.3  $&@A\}' (   5 9 : M N O P U V i j k l p 뽰뚍~qjhDhDEHUj!LC hDCJUVaJjEhDhDEHUjLC hDCJUVaJ hDhDjhDhDEHUjLC hDCJUVaJjhDUh @hDH* jh @hD hD>*hD h @hDh @hD5\*$%&  o p 6 7 Q $u$u$$ $ޅ$$$$І$$$$$1Ud^U`gdDdgdDdgd d^gdD gdD$ d^`a$gdD d^gdDdgdD""#Q R S m n o p   A B C < = > ? 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