Why does a sine bandwidth decrease with 1/dur, dur beeing the signal duration

Dear community, recently i was interested in the bandwidth of spectra. That's why i did some plots for signals with a single frequency at once but for different amplitude, duration and frequency. I saw, that the amplitude and frequency and amplitude does not influence the bandwidth in this case but signal duration does. can someone tell me why and why it is like a constant divided by the duration (~1/x). The constant changes a bit if a analyze the power spectrum instead of PSD, but the behavior stays the same. the behavior also stays the same if take the way over fft function and a linspaced frequency vector here a sample script; for anyone who is confused about the many loops can just set the ampNr and f0 value to one number only.

close all;
fs=8000; 

dur=0.1:0.1:2;
cl=lines(6);
amp=1:6;

tiledlayout('flow');
for ampNr=amp
    nexttile();
    for f0=100:100:1000

          myBW=nan(size(dur));
          for durNr=1:numel(dur)
              t1=0:1/fs:dur(durNr)-1/fs;
              s1=sin(2*pi*f0*t1);
              s1=s1+rand(size(s1))-0.5;
              [spec,fSpec,tSpec,p]=spectrogram(s1,length(s1),[],[],fs);
              [val,~,bw]=findpeaks(abs(spec),fSpec,'MinPeakHeight',0.9*max(p));
              [~,idx]=max(val);
              bw=bw(idx);
              myBW(durNr)=bw;
          end

      end
  plot(dur,myBW,'Color',cl(ampNr,:)); hold on;

xlabel('sine duration (s)');
ylabel('bandwidth (Hz)');
title(['sine amplitude ' num2str(ampNr)])

plot(dur,1.76./dur,'--k');

end
linkaxes()
best regards
Jonas

Answer 1

[David Goodmanson]

Hi Jonas,

I believe what you are seeing is just the natural reciprocal relation between width in the time domain and width in the frequency domain of the transformed signal. In physics this is nothing more than the Heisenberg uncertainly principle.

Suppose you have a signal f(t), with a characteristic width (let's say it is determined by the standard deviation) of width t0. Now suppose the same signal is stretched in time by a factor of b. Without getting into details of normalization and so forth, it looks like f(t/b) and its width is b*t0. The transform of the original function is

Int f(t) exp(iwt) dt = g(w)

Suppose it has a width w0. The transform of the stretched function is

Int f(t/b) exp(iwt) dt = Int f(t/b) exp(i(bw)(t/b)) (dt/b) b
and denoting t/b by s,
= Int f(s) exp(i(bw)s) ds b
= g(bw) b

The hanging b is due to not paying attention to normalization and would go away with proper normalization of all this. That doesn't change the fact that the frequency domain function, g(bw) has width w0/b and so is shrunk by the same factor that f(t/b) was expanded by. So the product of widths, width(f)*width(g), is a constant. It's also dimensionless.

As you have alluded to, ploting dur.*myBW gives a nice nearly flat line.

[Jonas]

oh dear, i just realized hiw dumb i am. i had in mind that fft frequency Resolution also increases linearly with signal length, followed by the thought that the width if the one peak spectrum triangle must decrease in similar manner. so far so good, but then my head did the mistake, i had a line width positive slooe in mind for resolution of the spectrun followed by the thought that the inverse beahvior is a line with negative slope and that's of course the mistake, the inverse behavior of a*x is a/x and not -a*x

thanks for shaking my head until it got the thoughts right again!

[Paul]

I agree with everything @David Goodmanson said. However, I'm not sure that the code actually illustrates the effect that he explains so well.

Consider the code from the OQ, but modified to not add noise to the signal, and modified arguments to spectrogram so that it applies a rectangular window and does no overlap

fs=8000;
dur=0.1:0.1:2;
cl=lines(6);
amp=1:6;
tiledlayout('flow')
for ampNr=amp
    nexttile();
    for f0 = 100:100:1000
        myBW=nan(size(dur));
        for durNr=1:numel(dur)
            t1=0:1/fs:dur(durNr)-1/fs;
            s1=sin(2*pi*f0*t1);
%            s1=s1 + (rand(size(s1))-0.5)*0;
%            [spec,fSpec,tSpec,p]=spectrogram(s1,length(s1),[],[],fs);
            [spec,fSpec,tSpec,p]=spectrogram(s1,0*s1+1,0,numel(s1),fs);
            [val,~,bw,prom]=findpeaks(abs(spec),fSpec,'MinPeakHeight',0.9*max(p));
            [~,idx]=max(val);
            bw=bw(idx);
            myBW(durNr)=bw;
        end
    end
    plot(dur,myBW,'Color',cl(ampNr,:)); hold on;
    xlabel('sine duration (s)');
    ylabel('bandwidth (Hz)');
    title(['sine amplitude ' num2str(ampNr)])
    plot(dur,1.76./dur,'--k');
end
linkaxes()

Now, let's look closely at how myBW(end) was caculated.

Fist, plot abs(spec)

figure
plot(fSpec,abs(spec),'-o'),grid
xlim([997 1003])

As expected, we see a single peak at 1000 Hz and 0 everywhere else. In other words, looking at the samples of spec one could (rightly?) conclude the signal has zero bandwidth. I suspect that this result obtains because spectrogram() is fft-based and the fft inherently assumes an infinite duration, periodic signal.

The prominence of the peak computed from findpeaks() is

prom

prom = 8.0000e+03

and the width of the peak computed by findpeaks() is the width at 0.5*prom using linear interpolation, which is the width at 4000, which is 0.5

myBW(end)

ans = 0.5000

It seems to me that the elements of myBW are really related to the frequency resolution in the processing. Because fs is fixed, as the duration increases the frequency resolution decreases and it appears that's the effect the code illustrates.

[David Goodmanson]

Hi Paul,

You may well be correct. It could boil down to a difference of point of view, though. I took a physics point of view, where scaling arguments apply quite pervasively. In this particular problem, there is an initial idea that the bandwidth is zero. However, some at-the-moment unspecified process has come up with a nonzero bandwidth. With scaling as a general principle and some admitted handwaving, one can argue that no matter what (linear) process was involved, frequency resolution or anything else, the end result will still have to come out width(f)*width(g) = constant. So I think a frequency resolution argument could be tracked back to the scaling law. However, looking at the details of a specific process as you are doing will provide better understanding of what is going on.

Answer 2

[Jon]

I don't have enough time at the moment to look at your code in depth, but I'm suspecting that you are experiencing "leakage" effects by not having an integral number of periods of the signal. So the truncation of your signal causes some extra energy outside of the pure sine frequency.

I suggest that you try changing your duration, but always keep the total duration an integral multiple of the sine wave's period and see if the behavior changes.

[Jonas]

actually the number of periods is always exact, the signals are also always 0 and the start and the beginning. the behavuir also stays the same if i use wholenseconds duration like 1 to 10 seconds

[Jonas]

i saw that the behavior stays also the same if i alter the window type eg to recangular, only the constant of c/durations changes a bit, like when i change to power spectrum instead of PSD

[Jon]

Sorry, I have not used the spectrogram function, and am not that familiar with the theory behind it In general though, it is aimed at giving you some time resolution of how the frequencies in a time varying signal evolve with time.

This doesn't seem to be directly applicable to your situation where you have pure sine waves.

Obviously a pure sine wave should have only a single frequency in its spectrum. I think if you plot the magnitude of the fft of these signals you will see that.

I think the fact that the peak of the spectrogram, with the parameters that you specified has some finite bandwidth is an artifact of the overlap and add used to compute the spectrogram. It is probably correct, I just don't know how to interpret it without digging deeper.

Hopefully someone else with more expertise on spectrograms can add more on this.

[Jonas]

the use of the function spectrogram may be confusing here, but in this case i choose the length of the analysis window as the length of the whole signal, this way i have one 1 time bin and dont have to bother with creation of the fft frequency vector, normalization or fftshifting