John Atkinson: Yes, MQA IS Elegant...

[Shadders]

10 minutes ago, Fokus said:

Moreover, mentioning the envelope in this story is quite irrelevant.

Hi,

It is not irrelevant. If the envelope was different, then the frequency spectrum would be different.

 

The envelope is essentially critical to the signal being analysed.

 

Regards,

Shadders.

[Shadders]

27 minutes ago, Fokus said:

 

If the envelope were different then the filters would be different. And in discussing filters we talk about transition frequency, transition band(width), steepness/order, ... There is no need to single out the envelope of the (impulse) response.

 

Hi,

You are confusing what you know about the filter with the analysis of a waveform.

 

If you were presented with the impulse response without prior knowledge of the system, you would implement the frequency transform, and see the result.

 

Just because you "know" about the system does NOT stop the envelope being critical to the frequency spectrum of the signal.

 

Anyway, it is the examples given, such as the ramp and the sine wave with discontinuity at the 0.5volts signal value relaxed to a tangent, which shows you that there is a variation in the output of the filter which is not ringing (where ringing always been attributed to energy outside the passband).

 

Regards,

Shadders.

[Shadders]

1 minute ago, mansr said:

You're looking at this backwards. The envelope is a result of the spectrum.

Hi,

No, the envelope is part of the response of the system - where the response just happens to be the filter coefficients.

 

Assume you were given the signal as a sample from someone and told to investigate, and you did not know that it was the impulse response of a filter.

 

Prior knowledge skews peoples approach.

 

Regards,

Shadders.

[Shadders]

5 minutes ago, Fokus said:

Deep sigh ...

Hi,

Which part do you not agree with ?.

 

Why not move the discussion forward instead of condescending responses.

 

If you lack the capability to discuss, then just state you don't understand. Will not be an issue for me.

 

Regards,

Shadders.

[Shadders]

On 8/23/2018 at 7:34 PM, John_Atkinson said:

 

You can see from the article of mine that triggered this thread an example where a perfectly legal, band-limited impulse nevertheless excites the DAC reconstruction filter's sinc-function ringing.

 

John Atkinson

Editor, Stereophile

 

Hi,

I agree, that you DO NOT require the input signal into the filter to have any out of band frequencies to cause ringing. Here is the signal analysed – this is the output of the 512tap linear phase filter. The input signal is a 1kHz sine wave, sampled at 192kHz.

 

Ringing of the filter is purely a result of the input signal second order behaviour. For example, assume the signal is at rest into the filter (0volts), and then you excite it with a sine wave starting at 0volts. At the time = 0 where the sine wave starts, you see the filter ringing behaviour. You have gone from a dV/dt=0,  to a dV/dt=Cos(phi).  This is shown in the graphic below.

As the sine wave continues to enter the filter, the ringing, which is a transient behaviour caused by the change in dV/dt, subsides to nil, and the filter outputs the sine wave. There is no ringing.

 

If at the zenith of the sine wave, where Sin(pi/2)=1.0, the signal remains at 1.0volts, here again is a change in the signal rate of change, from dV/dt=Cos(phi) to dV/dt=0. This change in the rate of change causes  transient response in the filter. This second order effect is less in the second example, and hence the amplitude of the ringing is therefore smaller. This is shown in the graphic below.

Regarding the article text “the more you constrain the data in the frequency domain, the less you can do so in the time domain, and a sinc-function filter smears the transient's energy in an extreme manner. ”

 

I would not state that the filter is smearing the transient energy. I would state it is added noise which is effectively the filter coefficients, whose amplitude is directly proportional to the second order derivative magnitude of the transients in the music.

 

Using a smaller tap filter reduces the length of the noise added to the signal, but it does not reduce the noise energy magnitude.

 

So, is all MQA proposing is the use of smaller length filters which will add less noise in terms of the length of the noise, but still adds relatively the same magnitude of noise as a longer filter ???

 

Regards,

Shadders.

[Shadders]

6 minutes ago, Fokus said:

 

You claim something, and then you try to demonstrate it with a signal that happens to have a lot of 'out of band frequencies' ...

 

Hi,

No, the signal does not have any out of band energy.

 

Can you explain why it does have out of band energy ?.

 

I recall something about the "boxcar" function ?, but cannot find the post.

 

Regards,

Shadders.

[Shadders]

1 hour ago, Fokus said:

Your function is the multiplication of a sine and a rectangle function. The spectrum of the

result is the convolution of the part spectra. The spectrum of the rectangle function, while

falling with frequency, still extends to infinity.

Hi,

Apologies for the delay in responding, it is a bank holiday here in the UK - so it is eat, drink and be merry.

 

OK - so, in my slightly inebriated state, are you are stating that at the sin(phi)=0 (start of the sine wave in the example) that there is a discontinuity ?

 

Or, that the function applied to the filter which you state is the multiplication of the sine which is continuous (-infinity to +infinity) is gated by the rectangular function, and this gating function of the rectangular pulse causes the ringing ?

 

Sorry, may not be making sense - too much drink for merriment.

 

Regards,

Shadders.

[Shadders]

2 hours ago, mansr said:

Exactly. It's obvious if you think about it, but probably unintuitive to the average person.

Hi,

OK - i am assuming that you completely agree with the statements from Fokus. Is this correct ???.

 

I need to investigate the rectangular function a bit more. hic.

 

Regards,

Shadders.

[Shadders]

Just now, mansr said:

Those statements are equivalent.

Hi,

OK - will await for the response from Fokus.

 

Sorry for labouring the point, i am only average engineer - if an engineer at all.

 

Regards,

Shadders.

[Shadders]

2 hours ago, Fokus said:

There certainly is a discontinuity. Switching something on or off in our ideal mathematical world implies a discontinuity. Your particular test signal is simple enough to be separated into its component functions, namely an eternal sine, a rectangle (keying the sine on and off), and a step (keeping the signal at high after the sine went off). For all of these functions we have closed-form Fourier transforms, two of which stretch to infinity. This allows us to see/know quickly that the gated sine also stretches to infinity in the frequency domain.

Hi,

I can see how you can for the gated sine state that it is a rectangular function multiplied by the eternal sine.

 

This issue i have with your response is that you state it is "implied". The filter has no knowledge of what the signal is.

 

As such, are you stating that when the sine wave starts that there is definitely a discontinuity, or no discontinuity ?

 

Please do not use the word implied, there either is or there isn't. The filter has to respond to the incoming data, not an "implied" signal.

 

Regards,

Shadders.

[Shadders]

2 hours ago, Fokus said:

and a step (keeping the signal at high after the sine went off)

Hi,

Regarding your statement above. Let us assume that your "implied" signal theory is correct.

 

For the sine to stop, the rectangular pulse has a negative going discontinuity at (pi/2).

 

For the step to start, the step has a positive going discontinuity at (pi/2).

 

The two discontinuities cancel out at (pi/2).

 

Therefore, how can there be a discontinuity at (pi/2) causing the ringing ?

 

Regards,

Shadders.

[Shadders]

1 minute ago, mansr said:

There's still a corner, or a discontinuity in the first derivative, where the sine wave switches to a straight line. That also has unbounded bandwidth.

Hi,

No, there is no discontinuity, neither in the sine waveform starting, nor the sine wave levelling to value 1 (constant).

 

For the sine wave starting, approaching from the negative axis, the value is zero. Approaching from the positive axis, the value tends to zero. NO discontinuity.

 

For the sine wave levelling to 1. Approach from the negative direction, the value tends to wards 1. Approach from the positive direction the value is 1. NO discontinuity.

 

You state that there is a discontinuity in the first derivative. Approach from the negative direction, the first derivative value tends to zero (cos(pi/2) = 0). Approach from the positive direction the derivative is zero. NO discontinuity.

 

Regards,

Shadders.

[Shadders]

27 minutes ago, mansr said:

The first derivative is zero in the negative direction, one in the positive. That's a discontinuity.

 

Here the second derivative is discontinuous, -1 from the negative direction, zero continuing on.

Hi,

Not sure what you are saying with the first derivative. Why have you implemented the first derivative for the sine starting ?

 

Are you testing for smoothness ?

 

Same for the where the sine levels off - again, are you taking it further to the second derivative to test for smoothness ?

 

What mathematical rules are you using to state a discontinuity ? (wiki reference please)

 

Regards,

Shadders.

[Shadders]

4 minutes ago, mansr said:

See https://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms. Entry 106 in the table of transforms has the implication that if any derivative of f(x) has a discontinuity, then the Fourier transform F(ξ) can be decomposed as a convolution involving the transform of the step function, which extends to infinity (entry 313). In other words, any band-limited function is smooth.

Hi,

The table you refer to, are the fourier transforms of functions.

 

It does not state or imply that if you continue to take the derivative of a function to the nth order, that if that nth order derivative has a different value between two curves approaching a point on the curve, that the point is a discontinuity.

 

The starting sine and levelling sine BOTH fail the criteria for a discontinuity at the relevant points. As such, they are continuous.

 

The page you referenced has a lot of discussion on smoothness. You cannot use discontinuity in the discussion of the waveforms presented. Of course, you can use the definition of discontinuity in the discussion of the generating function which is the rectangular pulse.

 

Regards,

Shadders.

[Shadders]

2 minutes ago, psjug said:

It has to be smooth to be band limited.  Maybe it is easier to think in mechanical terms?  Say Y axis is transducer position, for example. A function that is not smooth has infinite + or - acceleration.

Hi,

Smoothness reference here :

https://en.wikipedia.org/wiki/Smoothness

 

In mathematical analysis, the smoothness of a function is a property measured by the number of derivatives it has that are continuous. A smooth function is a function that has derivatives of all orders everywhere in its domain.

 

So, a function can be smooth for a specific number of derivatives, but for it to be a smooth function, every order of derivative has to exist.

 

The levelling sine is smooth in the first derivative only. BUT - the relevant text does not exist in the link : "Smoothness of piecewise defined curves and surfaces" has to be written on the wiki.

 

The question is - is music in the band 0Hz to 22.5kHz a smooth function ? or just a curve that has some derivatives that are smooth ?

 

Regards,

Shadders.

[Shadders]

5 minutes ago, mansr said:

Your function is continuous but not smooth. Do you at least agree with this?

Hi,

I agree that it is smooth, but it is not a smooth function. See the wiki i referenced.

 

6 minutes ago, mansr said:

A non-smooth function f(x) has a discontinuous nth derivative for some n. This means said derivative can be written as g(x) + a · u(x), where u(x) is the step function. Since the Fourier transform of the nth derivate of f(x) equals (2πiξ)^n · F(ξ) and the Fourier transform of the step function extends to infinity, it follows that F(ξ) does as well. In other words, f(x) is not band-limited.

The problem here is that your representative function g(x) should stop when the unit step is applied. Therefore your equation fails to take other unit steps into account - the positive unit step rise for the sine start, the negative unit step for when the sine stops, and the positive unit step for the levelling.

 

Your function has a discontinuity at x=0, where as my function does not at 0 and pi/2.

 

I understand the principle you are illustrating, but two opposing unit steps occurring at the same time, the discontinuities of these functions will cancel out - no spectrum, and the result is just the sine and the levelling, at pi/2.

 

Regards,

Shadders.

[Shadders]

Just now, mansr said:

A function can't be both smooth and not smooth.

Hi,

As per my wiki reference :

https://en.wikipedia.org/wiki/Smoothness#Smoothness_of_piecewise_defined_curves_and_surfaces

 

A function can be smooth, but does not mean it is a smooth function. There is a difference.

 

2 minutes ago, mansr said:

No, you do not.

 

There seems to be some miscommunication at a fundamental level going on here. How much maths education do you have

Look at the equation defining the function i created :

 

y(x) = [u(x)-u(x-5pi/2)].sin(x) + u(x-5pi/2)

 

At (5pi/2) where the sine is at its zenith, the unit steps which are both delayed to (5pi/2) CANCEL out. No spectrum from the unit steps.

 

Regards,

Shadders.

[Shadders]

1 minute ago, mansr said:

Good. Now calculate the Fourier transform.

Hi,

If you disagree with the analysis, state why the two unit steps do not cancel.

Regards,

Shadders.

[Shadders]

Just now, psjug said:

Why don't the steps cancel in a square wave?

Hi,

At the function (5pi/2), the positive and negative unit steps cancel.

 

y(x) = [u(x)-u(x-5pi/2)].sin(x) + u(x-5pi/2)

 

Regards,

Shadders.

[Shadders]

4 minutes ago, mansr said:

That's not the point. Your function is not infinitely differentiable and thus also not band-limited.

Hi,

It is the point if the result of the contribution of that specific point is nil (zero), from the unit steps which cancel.

 

Does band limited music have infinitely differentiable functions ???. There are transients etc.

 

Regards,

Shadders.

[Shadders]

2 minutes ago, psjug said:

I am not following you... are you maintaining that your example is band limited?

Hi,

I am stating that the positive and negative unit steps discontinuities at the point (5pi/2) cancel, and contribute no spectrum.

 

Regards,

Shadders.

[Shadders]

Just now, psjug said:

still not following... are you saying there are + and - steps in the dV/dt of the gated sine that occur at the same time?

Hi,

For the equation :

y(x) = [u(x)-u(x-5pi/2)].sin(x) + u(x-5pi/2)

 

The negative unit step multiplied by the sin function occurs at (5pi/2). The positive unit step at (5pi/2) cancels this negative unit step.

 

The positive unit step [u(x)] multiplied by the sin(x) occurs at x=0, not at (5pi/2).

 

Regards,

Shadders.

[Shadders]

2 minutes ago, SoundAndMotion said:

Hi Shadders,

Given that your function does violate the bandwidth requirements, you can look at this in 3 ways (or more?):

 

1- Why? As explained, your function is a rectangle multiplied by a sine added to a step. Both the rectangle and step have infinite frequency content. And the sine is a single frequency. In the frequency domain, you convolve the spectrums of the rectangle(infinite bandwidth) and sine and add the spectrum of the step (infinite bandwidth). The cancellation you seek (and see in the time domain) doesn't occur in the frequency domain.

 

2- Prove it. Just look at the Fourier Transform of your function. @mansr suggests doing it analytically, but you can also just do it numerically on your time domain data:

3- Work backwards. Look at your filtered result:

This shows that that transition contains higher frequencies than you expect.

Hi,

Thanks. I will examine the start of the gated sine more closely later.

 

To explain, the filter has a finite memory - 512 taps. So, any spectrum generated by the start of the gated sine will have decayed to zero - the transient will have stopped and the steady state condition remains - the sine wave.

 

At the transition of the sine to the constant 1, there is no discontinuity. This transition can be seen as a negative unit step multiplied by the sine (the unit step will essentially be time reversed and delayed to 5pi/2), and the positive unit step delayed to (5pi/2). It is this transition to the constant level 1, which has no discontinuity, where the unit steps discontinuities cancel out, is of interest.

 

My point is - there are transients in music, and the transient response of the filter can be seen graphically, and they do not need to have out of band energy for the transient response to be seen. Every case of "ringing" is seen as the result of out of band energy, whereas this ringing occurs without the need for out of band energy. Every filter has the steady state response, and the transient response.

 

Regards,

Shadders.

[Shadders]

5 hours ago, mansr said:

Ringing is the result of energy at the cut-off frequency. What happens at either side is irrelevant. Please see the examples on the website I linked earlier.

Hi,

I ran the sine with constant level through Octave - indeed there is out of band energy - see plot :

 

The signal is sampled at 192kHz, filter is set to cut off at 48kHz. Energy at 48kHz is -108.21dB. This energy causes the following "ringing" :

 

The peak value in the waveform is 1.0003 - this is 0.0003 greater than the input waveform - where the 0.0003 is -70.46dB (compared to 1.0volt). The maximum value of the filter coefficient is 0.24905. To be analysed, is how the power at 48kHz which is -108.21dB causes the -70dB peak (+38dB), when the peak coefficient is 0.25.

 

What does need to be investigated is the hump in the plot - which is not part of the filter ringing.

 

Regards,

Shadders.

[Shadders]

1 minute ago, mansr said:

Could you post the Octave commands you used to create those plots.

Hi,

As follows - should produce the plots :

fs = 192000;
ts = 1/fs;
n = 0:192000;               % Variable used for indexing
x = zeros(1, length(n));  % Input signal initialisation
b = fir1(500, 0.25);        % Create the FIR response

f = 1000;
t = n(1:fs);
x = sin(2*pi*f*t/fs);
gate = zeros(1,fs);
gate(9409:10370) = 1;  %Select 5 sine waves
z = gate.*x;
z(9841:192000) = 1;      %Set the remaining samples to 1 after 2 1/4 sine waves
zf = filter(b, 1, z);       %filter the input signal
 

xftfilt=abs(fft(filter(b,1, z)));
xftfilt = xftfilt./max(xftfilt);
figure(1);
plot(1:length(xftfilt), 20*log10(xftfilt));

 

figure(2);
plot(1:length(zf), zf);