John Atkinson: Yes, MQA IS Elegant...

[mansr]

Just now, ddetaey said:

Sorry, cannot do as it is now working !

Good enough for me.

[mansr]

Just now, Shadders said:

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.

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

 

Just now, Shadders said:

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.

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

[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.

[mansr]

21 minutes ago, Shadders said:

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) 

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.

[Jud]

1 hour ago, mansr said:

Strange. It's using Cloudflare certificates. What browsers do you use?

 

On iPhone, Opera and Safari.

 

Edit:  Also, connecting through OpenDNS.

 

Further edit: Works here now also. 

[mansr]

4 minutes ago, Jud said:

On iPhone, Opera and Safari.

Is it still broken? I changed a Cloudflare setting that might be related.

 

4 minutes ago, Jud said:

Edit:  Also, connecting through OpenDNS.

That shouldn't make a difference.

[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.

[Jud]

17 minutes ago, mansr said:

Is it still broken? I changed a Cloudflare setting that might be related.

 

It does work now, thanks.

[Jud]

Way beyond my first year high school calculus (which unfortunately I don't remember anyway - 46 years ago), but on a layperson's level, I was just wondering: Is the straightline waveform resulting from the rectangular gating/windowing anything like those square waves at e.g. 10kHz you sometimes see "proving" only DSD is capable of correct reproduction of audible frequencies?  In other words, the straight vertical rise and fall times show you there's "illegal" (above Nyquist) bandwidth, regardless of frequency.

[psjug]

6 minutes ago, Jud said:

Way beyond my first year high school calculus (which unfortunately I don't remember anyway - 46 years ago), but on a layperson's level, I was just wondering: Is the straightline waveform resulting from the rectangular gating/windowing anything like those square waves at e.g. 10kHz you sometimes see "proving" only DSD is capable of correct reproduction of audible frequencies?  In other words, the straight vertical rise and fall times show you there's "illegal" (above Nyquist) bandwidth, regardless of frequency.

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.

[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.

[mansr]

1 hour ago, Shadders said:

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

Yes, very useful.

 

Quote

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.

Of course it doesn't. That statement has nothing to do with Fourier transforms.

 

Quote

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

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

 

Quote

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.

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.

[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.

[mansr]

4 minutes ago, Shadders said:

I agree that it is smooth, but it is not a smooth function.

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

 

6 minutes ago, Shadders said:

I understand the principle you are illustrating,

No, you do not.

 

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

[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.

[mansr]

2 minutes ago, Shadders said:

Look at the equation defining the function i created :

 

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

Good. Now calculate the Fourier transform.

[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.

[psjug]

This really doesn't have to be very complicated, at least for the gated sine example.  As @mansr pointed out, you have a step in the dV/dt.  So this example is not band limited. 

[mansr]

7 minutes ago, Shadders said:

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

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

[psjug]

9 minutes ago, Shadders said:

Hi,

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

Regards,

Shadders.

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

[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.

[psjug]

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

[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.

[mansr]

4 minutes ago, Shadders said:

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

Yes, your function is continuous. Nobody disagrees with that. If you calculate the first and second derivatives, you will find that they are not continuous, and this means the function cannot be band-limited. Alternatively, calculate the Fourier transform directly and notice that it never reaches zero.