How to resolve between @spectrum/psd.m and psd.m?

Question

Bob Li 2011-12-29

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Hi,

In Spectral Analysis http://www.mathworks.com/help/toolbox/signal/ug/f12-6587.html, under Nonparametric Methods>> Periodogram, the following code snippet is shown

fs  =  1000; %  Sampling  frequency
t  =  (0:fs)/fs; %  One  second  worth  of  samples
A  =  [1  2]; %  Sinusoid  amplitudes  (row  vector)
f  =  [150;140]; %  Sinusoid  frequencies  (column  vector)
xn  =  A*sin(2*pi*f*t)  +  0.1*randn(size(t));
psd(Hs,xn,'Fs',fs,'NFFT',1024,'SpectrumType','onesided')

I am curious on which function psd is, so I used which to locate it:

>> which psd

C:\Program Files\MATLAB\R2008a\toolbox\signal\signal\psd.m

>> which spectrum.psd

C:\Program Files\MATLAB\R2008a\toolbox\signal\signal\@spectrum\psd.m % static method or package function

It seems that “which psd” returns a standalone function, whereas “which spectrum.psd” returns a static function of the spectrum class.

In standard Object-oriented programming language like C++, compiler examines argument list to resolve overloaded functions. But here,

Standalone psd.m;

function [Pxx, Pxxc, f] = psd(varargin)

@spectrum\psd.m:

function varargout = psd(this,x,varargin)

both take variable input argument lists, and the calling statement

psd(Hs,xn,'Fs',fs,'NFFT',1024,'SpectrumType','onesided')

has no return argument which, if length ≥4, could possibly be used to resolve ambiguity since standalone psd.m has maximum of 3 outputs.

So how could the compiler determine which psd.m is executed, the standalone one or the spectrum class method under @spectrum?

Bob

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Walter Roberson 2011-12-29

Bob, when you are adding tags, please separate the tags with comma instead of semi-colon. I have fixed this for you here.

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Answer 1

Daniel Shub 2011-12-29

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https://ww2.mathworks.cn/matlabcentral/answers/24917-how-to-resolve-between-spectrum-psd-m-and-psd-m#answer_32744

在 MATLAB Online 中打开

For

...
Hs = spectrum.periodogram;
psd(Hs,xn,'Fs',fs,'NFFT',1024,'SpectrumType','onesided')

MATLAB knows to call the spectrum.psd method sicne the first argument is of class spectrum.

I don't have access to MATLAB right now, but if instead you do

...
eval('Hs = spectrum.periodogram;');
psd(Hs,xn,'Fs',fs,'NFFT',1024,'SpectrumType','onesided')

I am pretty sure you confuse the JIT and you get the psd function.

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Bob Li 2012-1-2

Daniel,

I tested it. The 1st argument was indeed the case.

Thanks,

Bob

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Answer 2

Wayne King 2011-12-29

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/24917-how-to-resolve-between-spectrum-psd-m-and-psd-m#answer_32731

在 MATLAB Online 中打开

Hi Bob, in

fs  =  1000; %  Sampling  frequency
t  =  (0:fs)/fs; %  One  second  worth  of  samples
A  =  [1  2]; %  Sinusoid  amplitudes  (row  vector)
f  =  [150;140]; %  Sinusoid  frequencies  (column  vector)
xn  =  A*sin(2*pi*f*t)  +  0.1*randn(size(t));
Hs = spectrum.periodogram;
psd(Hs,xn,'Fs',fs,'NFFT',1024,'SpectrumType','onesided')

You are using the psd() method for spectrum objects. C:\Program Files\MATLAB\R2008a\toolbox\signal\signal\@spectrum\psd.m

Note that I have added the line: Hs = spectrum.periodogram above. This was not present in your original post, but it must have been somewhere in the documentation previous to the call

psd(Hs,xn,'Fs',fs,'NFFT',1024,'SpectrumType','onesided')

Hs is the spectrum object. Note the following use of the psd method for spectrum objects.

    t = 0:0.001:1-0.001;
    x = cos(2*pi*100*t)+randn(size(t));
    psdest = psd(spectrum.periodogram,x,'NFFT',length(x),'Fs',1/0.001);
    plot(psdest);

Note that psd() the function is being deprecated.

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Wayne King 2011-12-29

psd() is a method of the spectrum object. Hs is a spectrum object. MATLAB knows because you have a spectrum object as the first input to the method. this (is the spectrum object)