How the chain rule applies to integral equations.

In Calculus, one of the most basic things that can be solved for, is that a principal function receives a parameter, multiplies it by a multiplier, and then passes the product to a nested function, of which either the derivative or the integral can subsequently be found. But what needs to be done over the multiplier, is opposite for integration, from what it was for differentiation. The following two work-sheets illustrate:

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Please pardon the poor typesetting of the EPUB File. It’s the result of some compatibility issues (with EPUB readers which do not support EPUB3 that uses MathML.)

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Exploring the Discrete Sine Transform…

I can sometimes describe a way of using certain tools – such as in this case, one of the Discrete Cosine Transforms – which is correct in principle, but which has an underlying flaw, that needs to be corrected, from my first approximation of how it can be applied.

One of the things which I had said was possible was, to take a series of frequency-domain ‘equalizer settings’, which be at one per unit of frequency, not, at so many per octave, compute whichever DCT was relevant, such that the result had the lowest frequency as its first element, and then to apply that result as a convolution, in order finally to apply the computed equalizer to a signal.

One of the facts which I’m only realizing recently is, that if the DCT is computed in a one-sided way, the results are ‘completely non-ideal’, because it gives no control over what the phase-shifts will be, at any frequency! Similarly, such a one-sided convolution can also not be applied as the sinc function, because the amount of sine-wave output, in response to a cosine-wave input, will approach infinity, when the frequency is actually at the cutoff frequency.

What I have found instead is, that if such a cosine transform is mirrored around a centre-point, the amount of sine response, to an input cosine-wave, will cancel out and become zero, thus giving phase-shifts of zero.

But a result which some people might like is, to be able to apply controlled phase-shifts, differently for each frequency, such that those people specify a cosine as well as a sine component, for an assumed input cosine-wave.

The way to accomplish that is, to add-in the corresponding (normalized) sine-transform, of the series of phase-shifted response values, and to observe that the sine-transform will actually be zero at the centre-point. Then, the thing to do is, to apply the results negatively on the other side of the centre-point, which were to be applied positively on one side.


 

 

I have carried out a certain experiment with the Computer Algebra System named “wxMaxima”, in order first to observe what happens if a set of equal, discrete frequency-coefficients belonging to a series is summed. And then, I plotted the result of the definite integral, of the sine function, over a short interval. Just as with the sinc function, The integral of the cosine function was (sin(x) – sin(0)) / x, the definite integral of the sine function will be (1 – cos(x)) / x, and, Because the derivative of cos(x) is zero at (x = 0), the limit equation based on the divide by zero, will actually approach zero, and be well-behaved.


 

 

(Update 1/31/2021, 13h35: )

There is an underlying truth about Integral Equations in general, which people who studied Calculus 2 generally know, but, I have no right just to assume that any reader of my blog did so. There exist certain standard Integrals, which behave in the reverse way of how the standard Derivatives behave, just because ‘Integrals’ are ‘Antiderivatives’…

When one solves the Derivatives of certain trig functions repeatedly, one obtains the sequence:

sin(x) -> cos(x) -> -sin(x) -> -cos(x) -> sin(x)

Solving the Indefinite Integrals of the same trig functions yields the result:

sin(x) -> -cos(x) -> -sin(x) -> cos(x) -> sin(x)

Hence, the Indefinite Integral of sin(x) is in fact -cos(x), and:

( -(-cos(0)) = +1 )

(End of Update, 1/31/2021, 13h35.)

 

(Updated 2/04/2021, 17h10…)

Continue reading Exploring the Discrete Sine Transform…

There has been some confusion about the Sinc-Filter.

I have read descriptions about the Sinc-Filter somewhere, which predicted that it would become unstable, if the frequency of the input stream, happened to correspond to the spacing, between its non-zero coefficients. As far as I can tell, this prediction was based on a casual inspection of the Sinc Function, but overlooks something which is easy to overlook about it. This case also happens to correspond, to the input stream having a frequency equal to the Nyquist Frequency, of certain practical applications, such as over-sampling.

The Sinc Function has zero-crossings at regular intervals, including the center-point, where its coefficient is stated as being equal to (1.0) . This happens because the value at the center-point, is the solution to a limit equation, that corresponds to (0/0) .

This center coefficient is symmetrically flanked by two positive ones, one of which is only positive, because it forms as a division of the sine of x by the corresponding negative value of x. At frequencies below the Nyquist Frequency, the sum of their products starts to reinforce the center element. Above Nyquist, they start to cancel the product with the center coefficient.

sincplot_2

This can be complicated to plot using Computer Algebra Systems, because plotting functions are always numerical, and at (x=0), there is no numerical solution (only the Algebraic solution given lHôpitals Rule). So, a CAS typically needs to have the Sinc Function defined as a special case, to be able to plot it, otherwise requiring a complex workaround.

So it is possible that the frequency of the incoming stream aligns to the spacing between the maxima and minima of the Sinc Function. If that happens, there are two behaviors to bear in mind:

  1. The peak of the input stream could be aligned with the center-point. In that case, all the other waves will have zero-crossings, where the Sinc Function has maxima. The fact that the single input-sample seems to produce (1.0) as the output amplitude, is due to how the function is frequently normalized for practical use. According to that, maximum output should reach (2.0) at a frequency of zero…
  2. The input stream could have a zero-crossing, at the center-point of the Sinc Function, so that its product from there should equal (0.0) . In that case, the input stream will have positive peaks on one side of the center-point, that all correspond to negative peaks on the other side of the center-point. According to that, the instantaneous output should equal (0.0) .

All of this would suggest to me, that the Sinc-Filter will work properly.

sincplot_3

One way in which people can misinterpret the plot of the curve, would be to notice it has a positive peak in the center, to notice that after a zero-crossing, it forms two negative peaks, and then to conclude that those negative peaks are also the two closest non-zero coefficients to the center.

Continue reading There has been some confusion about the Sinc-Filter.

Guessing at the Number of Coefficients Filters Might Need

There probably exist Mathematically-more-rigorous ways to derive the following information. But just in order to be able to understand concepts clearly, I often find that I need to do some estimating, that will give some idea, of how many zero-crossings, for example, a Sinc Filter should realistically have, on each side of its center sample. Or, of what kind of cutoff-performance the low-pass part of a Daubechies Wavelet will have, If it only has 8 coefficients…

If the idea is accepted that a low-pass filter is supposed to be of some type, based on the ‘Sinc Function’, including filters that only have 2x / 1-octave over-sampling, then a question which Electronics Experts will face, is what number of zero-crossings is appropriate. This question is especially difficult to find a precise answer to, because the series does not converge. It is a modified series of the form Infinite Sum (1/n) .

Just to orient ourselves within the Sinc Function when applied this way, the center sample is technically one of the zero-crossings, but is equal to 1, because it has the only coefficient of the form (0/0). After that, each coefficient twice removed is a zero-crossing, and the coefficients displaced from those are the standard non-zero examples.

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