This the result, and the code is below,

clear

L=200;

deltax=0.01;

fs=1/deltax;

x = 0:deltax:L;

y = cos(x);

nfft=2*round(numel(x)/4);

%---pwelch

[px2,fx2]=pwelch(y,hanning(nfft),nfft/2,nfft,fs);

umin=0;

umax=1e3;

u=umin:deltax:umax;

psd_temp=(2*sin(L/2)*cos(pi*L*u)-4*pi*u*cos(L/2).*sin(pi*L*u))./(1-4*pi^2*u.^2);

psd_temp=psd_temp.^2/L;

figure

loglog(u,psd_temp)

hold on

loglog(fx2,px2,'r')

legend({'Analytical by defination','Numerical result by matlab function'})

set(gca,'fontsize',12)

figsize_only(12,10)

However, in the analytical solution, the denominator will be zero when u equals 1/2/pi, Can any body give explaination from math view?

Aparna Sathya Murthy

Dear Aparna Sathya Murthy ,

I also tried the psd function in matlab which uses the fft to calculate the PSD, the result is the same with pwelch()

Aparna Sathya Murthy

When I decreace the interval of u in the analytical expression of PSD for finite length，I got strange figure！

Code belw:

clear

L=1000;

deltax=0.0001;

fs=1/deltax;%---------sample frequency

x = 0:deltax:L;

y = cos(x);

nfft=2*round(numel(x)/4);%-------fft num

%---psd

[px,fx]=psd(y,nfft,fs,hanning(nfft),nfft/2);

px=px/fs*2;

px([1,end])=px([1,end])/2;

%---pwelch

[px2,fx2]=pwelch(y,hanning(nfft),nfft/2,nfft,fs);

%---plot

figure

loglog(fx,px);

hold on

loglog(fx2,px2,'r')

legend('psd','pwelch')

title('psd与pwelch method by matlab')

set(gca,'fontsize',12)

figsize_only(12,10)

figure

plot(fx,px);

hold on

plot(fx2,px2)

legend('psd','pwelch')

title('psd与pwelch method by matlab')

set(gca,'fontsize',12)

umin=0;

umax=1e3;

u=umin:deltax:umax;

psd_temp=(2*sin(L/2)*cos(pi*L*u)-4*pi*u*cos(L/2).*sin(pi*L*u))./(1-4*pi^2*u.^2);

psd_temp=psd_temp.^2/L;

figure

loglog(u,psd_temp)

hold on

loglog(fx2,px2,'r')

legend({'Analytical by defination','Numerical result by matlab function'})

set(gca,'fontsize',12)

Aparna Sathya Murthy

For finite length PSD analytical solution, can I use the Window function to make the head and taile of the sample length zeros?