Decay

Smoke detectors are fascinating. They contain a tiny radioactive sample of Americium-241, the atoms of which are continuously decaying into Neptunium-237. Alpha decay is when the nucleus of an atom emits an alpha particle, which is made of 2 protons and 2 neutrons. The half-life of \(Am\) is approximately 432.2 years, which means that if you start off with \(N_0\) atoms at \(t=0\) then at \(t=1\) you'll have \(\frac{1}{2}N_0\) and at \(t=2\) you'll have \(\frac{1}{4}N_0\) left. Generally:

$$N_t = N_0(\frac{1}{2})^t\tag{1}$$

This is an exponential function, so if you wanted to, you could rewrite it as:

$$N_t = N_0e^{t\ln\frac{1}{2}}\tag{2}$$

\(\ln\frac{1}{2}\) is an interesting constant that will reappear later. It's roughly equivalent to \(-0.69314718\).

If you plotted the above graph, you might note that time is on the x-axis and the number of \(Am\) atoms is on the y-axis. This number is always decreasing as time increases. For a given time interval \(t=a, t=b\), the decrease in \(N\) from \(N_{a}\) to \(N_{b}\) is a direct consequence of the activity. 

Activity, measured in Becquerels \(Bq\), is the number of decays per second, but as our graph's time axis is currently measured in half-lives we can't just read the activity directly from the slope. 

Approach #1

$$432.2 years\times365.25\times24\times3600 = 13639194720 secs\tag{3}$$

If we already had \(N_0\) we could plug \(\frac{1}{13639194720}\) into our first equation and see \(N\) one second later. This would be the number of Becquerels.

Approach #2

Modify the original equation so that \(t\) is no longer measured in half-lives but instead in seconds:

$$N_t = N_0(\frac{1}{2})^\frac{t}{13639194720}\tag{4}$$

So actually, both approaches are identical, but the second approach is more reusable: we've taken the general equation \(1\) and altered it to be specific to the half-life of Americium-241. 

Let's find a typical value of \(N_0\) for our smoke detector example, where a Google search reveals about 0.3μg is common. How do we find the number of Americium-241 atoms in 0.3μg? Avogadro's number: \(6.02214076\times10^{23}\) is the number of particles in one mole.

$$N_0 = 0.3\times10^{-6} / 241 \times 6.02214076\times10^{23} = 7.49644\times10^{14}\tag{5}$$

Let's plug this number into equation \(4\). Unsurprisingly, if you follow along in Excel, you'll see the same number for \(N_0\) as \(N_1\).

$$N_1 = 7.49644\times10^{14}\tag{6}$$

However, if you rearrange it, you'll find that Excel didn't throw away the bits, it just hid them away with some formatting that prioritises the most significant digits:

$$N_0 - N_1 = 38097.125\tag{7}$$

Here you have it. In a smoke detector with just 0.3μg of Americium 241, there are approximately 38k decays per second. Alpha particles are being emitted at 38kHz! Note that in equation \(7\), \(N_1 - N_0\) would give us a negative slope of \(-38097.125\) indicating that our sample contained fewer atoms at \(t = 1\).

Bonus section:

The slope of the exponential function \(1\) is always directly proportional to the value on the y-axis! In this case, the \(N\) value is scaled by \(ln\frac{1}{2}\) while exponential function \(4\) is scaled by a different number \(\frac{\ln\frac{1}{2}}{13639194720}\). Intuitively this should make sense: when the amount of radioactive material is halved, so is the activity.

Heterodyne

Frequency modulation (FM) is one of the techniques used to transmit audio signals at radio frequencies. If we "multiply" a sinusoidal signal wave with a sinusoidal carrier wave, the result is two waves at frequencies just above and just below the original carrier.

By way of an example let's try and transmit a 440 Hz tone using a 10 kHz carrier.

\(signal(t) * carrier(t) = mixed(t)\)

The mixed result will then be the superposition of a 10,000 Hz - 440 Hz and a 10,000 Hz + 440 Hz wave.



We can un-mix the mixed wave by applying the same function to our pair of tones at 9,560 Hz and 10,440 Hz; doing so will create another pair of tones: one for each input. Two of those tones will be very high frequency and almost inaudible; the other two will be centered around 0 Hz. Originally, the tone at 9,560 Hz would be a kind of "mirror image" of our original tone (though this distinction isn't readily apparent for a single frequency), but after the second multiplication, it's now the "mirror image of a mirror image" so it's the right way around. The tone that was at 10,440 Hz when modulated will now be at -440 Hz and inaudible.

As a second example, let's say the signal is an audio file containing many frequencies between 20 Hz and 20 kHz. We know from the superposition property of waves that we can decompose our song into 19,980 different signals (yes, there will be some rounding errors if we only allow integer values, but hear me out). And let's chose a slightly higher frequency carrier of 100 kHz. Our lowest frequency of 20 Hz will now be mixed to 99,980 Hz and 100,020 Hz while our highest frequency will be mixed to 80,000 Hz and 120,000 Hz. The signal frequencies in between the lowest and highest will then fall into the two bands when mixed. One band will appear to have the "mirror image" of our original signal: that is to say the higher signal frequencies will now occupy lower mixed frequencies and the lower signal frequencies will now occupy higher mixed frequencies. The other band will contain the same information as the signal but simply shifted to a higher base frequency.

Binary vs Decimal, Part 2

Microsoft's .NET framework gives developers options for representing floating point numbers as well as integers. The same is true of Microsoft's SQL Server, though the options differ subtly.

Remembering from my earlier post that we can convert between base 10 (decimal) and base 2 (binary) integers with no loss of information, it shouldn't surprise you that .NET and SQL Server offer similar integer types with varying number of bits. Even though we're storing numbers that we entered as a sequence of digits (i.e. literals) in C# source or SQL script, they are converted into base 2 integers. However, this post is not about integers.

If, instead of integers, we wish to store fractions (e.g. negative powers of our chosen bases) we must ask a few questions, the first being:

Are we trying to represent a fractional number that has a base 10 representation? If so, we already know that the conversion from base 10 to base 2 cannot necessarily be done without losing information. We can very easily represent the decimal number 0.625 in binary because it's simply the sum of 0.500 (or 2^-1) and 0.125 (or 2^-3), but we cannot exactly represent the decimal number 0.99 because the number of base 2 powers required to add up to 0.99 would quickly exceed our chosen precision (the number of bits available to represent the number.) Instead, if wanted to store our decimal number in a binary field with a fixed number of bits, we would need to approximate the decimal.

0.989999949932098388671875
0.9900000095367431640625
0.990000069141387939453125

The three numbers above are actually sums of powers of 2 and can be stored in a 24 bit binary field without losing information^[1]. Notice that the middle number is closest to 0.99.

But what if we wanted to represent the price of our chocolate: 34.99? In order to fit into a field of the same width we would need to adjust the largest exponent and make a corresponding adjustment to
smallest exponent. Our closest "sum of powers of 2" approximation might be:

34.990001678466796875

In binary this same number would look like

100010.111111010111000011

25	24	23	22	21	20	2-1	2-2	2-3	2-4	2-5	2-6	2-7	2-8	2-9	2-10	2-11	2-12	2-13	2-14	2-15	2-16	2-17	2-18
1	0	0	0	1	0	1	1	1	1	1	1	0	1	0	1	1	1	0	0	0	0	1	1

We're ready to ask ourselves the second question: can our application tolerate the approximation in converting from base 10 to base 2? In applications where the inputs, intermediate values and outputs are not exact numbers to start with, (e.g. in the fields of science and statistics, to name a few) the answer is probably yes, but certain applications with exact inputs (e.g. in the field of finance) might strictly require us not to approximate the values. The next section presents a way to circumvent the approximation.

In this series' first post, we learned that we can convert numbers between all bases without loss of information, but only when the lowest power for both bases is b^0 or 1. Some smart engineers realised that if we pre-scale our number by some chosen scale so that it becomes an integer, then we can post-scale it when we're done operating on it. Going back to our example number of 0.99, we could pre-scale it by 10^2 (or 100) giving us 99, which converts exactly into the binary number 0110 0011. Now we're able to store the exact value in 7 bits when previously we could only store the approximation in 24 bits.

At this point SQL Server and .NET diverge:

SQL Server offers the DECIMAL data type, which has a fixed precision and scale. In other words, when you define a value of this type you explicitly provide the precision and scale (e.g. DECIMAL(26, 2). If we wanted to store the number 31.42 in this field, SQL Server would pre-scale it by 10^2, convert the number 3142 to binary, and store the binary representation.

.NET, on the other hand, offers the System.Decimal struct type which is a floating point number. The scale that's chosen to pre-scale any given decimal number is packed and stored along with the binary integer representation, and can be accessed to post-scale the integer back into a fraction when needed. The struct is 128 bits wide, and stores the integer part in 96 of those bits. 2^96 is approximately 7.92282E+28. We can create a System.Decimal with anywhere between 0 and 27 (or 28) digits and we choose where the floating decimal point will go.

It might be worth emphasizing that SQL Server's DECIMAL is not floating point. The distinction between fixed and floating point is probably best made by looking at where the scale value is stored.

Converting between SQL Server's DECIMAL and .NET's System.Decimal can take a little forethought.

[1] https://www.h-schmidt.net/FloatConverter/IEEE754.html
[2] https://docs.microsoft.com/en-us/sql/t-sql/data-types/decimal-and-numeric-transact-sql

Binary vs Decimal, Part 1

We commonly represent numbers in base 10, or decimal. The number three thousand, one hundred and forty-two, for example, can then be represented by the composite symbol 3142.

Power	103	102	101	100
Digit	3	1	4	2
Product	3000	100	40	2

The arithmetical sum of products (of powers and digits) gives us back our number: 3142.

Computers represent numbers in base 2, or binary. The same number three thousand, one hundred and forty-two has a binary representation of 1100 0100 0110 (I added some spaces to make the long sequence of bits easier to process visually.)

Power	211	210	29	28	27	26	25	24	23	22	21	20
Bit	1	1	0	0	0	1	0	0	0	1	1	0
Product	2048	1024	0	0	0	64	0	0	0	4	2	0

Again, the arithmetical sum of products (of powers and bits) gives us back our number: 3142.

Notice the similarity of the mechanical operations between the two tables above: a pattern (or algorithm) should emerge. Once it does, you may continue reading.

With representations like these, it's instructive to notice that we can represent any number that is a multiple of the rightmost power, up to a maximum that is determined by the leftmost power. Once we choose (fix) the rightmost and leftmost power, we restrict our representation to only those real numbers that can be represented subject to those constraints.

Consider a real world example: the price of a bar of chocolate at your local grocery store is R 34.99. If the shop has one of those electronic price displays on the shelf, that display is likely to be physically limited to displaying only 6 digits.

To enable us to represent any number that is a multiple of 1 cent (R 0.01) we must set the rightmost exponent to -2. Consequently, because we also defined the number of digits as 6, we are also constrained to represent numbers less than or equal to 9999.99. So, we can encode any number from 0.01 to 9999.99 inclusive, but we cannot encode smaller numbers (such as 0.001) or larger numbers (such as 10000.00) as they exceed our chosen constraints.

Power	103	102	101	100	10-1	10-2
Digit	0	0	3	4	9	9
Product	0	0	30	4	0.9	0.09

Once we restrict our representations to a limited number of digits (in the case of decimal) or bits (in the case of binary) we essentially restrict the set of numbers that we can represent. Remember: every number must be some multiple of the rightmost power. It's not coincidental that I chose an exponent of 0 for the first two examples; b⁰ is always 1 whatever base b is. That has the consequence that we can represent all integers in both decimal and binary (as long as we increase the number of digits or bits sufficiently.) As long as our smallest representable value is common to both schemes, we can represent the same numbers in each of them. If the following equation holds true then we can convert between two schemes d and b, with integer exponents m and n respectively, without loss of information:

\({d}^{m} = {b}^{n}\)

When d is 10 and b is 2 the only solution is where m and n are both zero. We can only convert integers (i.e. multiples of b⁰, or 1) between base 10 and base 2 without losing information.

If we chose a different pair of bases 16 (hexadecimal) and 2 (binary, again) then we see an additional repeating solution each time the larger base power is an integer multiple of the smaller base power, but we still wouldn't be able to represent the other powers of 2.

Base 16			...	161	160	16-1	16-2	...
Base 2	24	23	22	21	20	2-1	2-2	2-3	2-4	2-5	2-6	2-7	2-8

Multiples of

16

1

0.0625

0.00390625

Wake On Lan (WOL)

Wake On Lan (WOL) is possible on the ASUS M3A78-CM board, but it requires a level of finesse to get it working. First, the cryptically named BIOS option "Power > APM Configuration > Power on From S5 by PMEs" must be enabled. Then, it's important to note that WOL won't work when the power switch is off (i.e. SB_PWR LED is still lit, but the computer has been shut down). Instead, the machine needs to be merely suspended in order to pick up the "magic" WOL ethernet packet. You can suspend it with the command:

sudo systemctl suspend

You'll also need to decide what to do with the BIOS setting "Power > APM Configuration > Restore on AC Power Loss"... [Last State] won't bring it back to suspended after an AC power loss, but it will bring it back to running if the server was running when the power went off. If the power goes out when you're suspended, then you're up a certain creek without a paddle.

On waking, some network drivers need to be reloaded. Check my github's private ubuntu/WOL repository for the script that does this, or follow the steps in this accepted answer: ubuntu 18.04 - ethernet disconnected after suspend

To send a WOL packet, install and run etherwake from some other machine in the same subnet:

sudo apt install etherwake
sudo etherwake -i enp1s0 00:22:15:DE:31:42

Hypothesis Testing Errors

Hypothesis testing allows us to quantify relationships between known samples and the unknown population from which they were taken. What do I mean by that? Let's say I am interested in the returns from investing in a particular stock. The daily return might be calculated as the ratio of today's closing price over yesterday's closing price. Whether I was to take a sample of 30 returns (over 31 days) or 60 returns (over 61 days), I still couldn't know the population mean return, but I could hypothesize about it... so I do.

I choose a "null hypothesis" that the population mean is 4.5%. Given that my sample mean was 5% and there was a 2% standard deviation, 30 observations would produce a test statistic of \(\frac{0.05 - 0.045}{\frac{0.02}{\sqrt{30}}} = 1.37\) standard deviations. In other words, the p-values would be 8.5% and 91.5%. For an 80% confidence two-tailed test (10% in the left tail and 10% in the right tail) we would reject the hypothesis, but at 90% confidence we would accept it. Note how we've already accepted or rejected the hypothesis regardless of its truth.

Now, imagine an all-seeing and all-knowing supernatural bystander watching the same events unfurl... they could know the population mean... and even though they wouldn't be obliged to share the exact value with me, let's say that they'd at least tell me if my hypothesis was true or false; that is to say: if I hypothesized that the population mean was 4.5% and it actually was 4.5% then the hypothesis would be true, otherwise if would be false (it could be 4.2% or 4.3% or even -5% or 63%; the point is we don't know).

If we take our two test results and combine them with the two possible truth values, it produces this 2X2 matrix of outcomes.

	Accept	Reject
True	Correct	Type 1 Error
False	Type 2 Error	Correct

• True/False: does the actual population mean match the hypothesized mean?
• Reject/Accept: does our statistic fall outside/inside the confidence interval?
• Correct/Incorrect: did we accept a true (or reject a false) hypothesis or did we commit an error?

Let's ask the bystander if our hypothesis was indeed true or if it was false.

• Yes, it's true:
  • At 90% we accepted it
  • At 80% we rejected it (Type 1 Error)
• No, it's false:
  • At 90% we accepted it (Type 2 Error)
  • At 80% we rejected it.

This last pair of possibilities deserves more analysis: when the bystander tells us our hypothesis was false, it doesn't seem to matter why we were correct to reject the hypothesis; all that matters is that we did.

This video tries to explain it but I am not confident the author is correct. I'd prefer to side with authors of the CFA curriculum who say - in short - "it's complicated".

Fourier Transform in C#

I am fascinated by the Fourier transform; specifically: its ability to decompose a wave into a series of fundamental waves. Mathematicians might prefer to see it described it more formally, but here I'll be trying to make it understandable.

Obviously, for a wave to have motion, time needs to pass. If we freeze time then our wave cannot oscillate. To say a wave has frequency, we need first to observe a period in which the wave has motion. The longer we observe its movement the more we can mathematically extract from it, but for now let's simply say that we need to acquire a range of samples measured as the wave oscillates. From this point, armed with our array of floating point numerical samples, we menacingly approach the rabbit hole.

The remainder of this post will focus on a single practical aspect of the Fourier transform: how to interpret the result, especially its magnitude. I'm using the MathNet.Numerics package from Nuget, which provides the handy method Fourier.Forward(samples, FourierOptions.NoScaling).
Note: you need to specifically pass FourierOptions.NoScaling if you want control of the output.

Since I grew up in an age when Compact Discs were seen as technological marvels, I am going to use the standard 44.1 kHz as my sample rate for the following example. A full second's worth of samples (44100) might seem a reasonable place to start, but as we'll see later, it's by no means a requirement. The chosen sample rate theoretically limits the upper frequency to 22050 Hz, and even though we pass in 44100 samples to Fourier.Forward we get back a symmetrical result with the first 22050 data points carrying the frequencies up to 22050 Hz, and the second 22050 carrying them all in again reverse. We can safely ignore the second half of the output for this post. If our input signal (encoded in the 44100 samples) was a pure sinusoidal signal (of any frequency up to 22050 Hz) oscillating between +1 and -1, it will contribute 22050 units to the frequency bin associated with the signal. Those units can be found in the Complex32.Magnitude property. Multiplying the amplitude of the input signal by any constant A will scale the magnitude to A x 22050. However, if we pass in a fraction of the full second's worth of samples - let's say \(\frac{4}{5}\), which is 35280 instead of 44100 - then we see the number of units fall to \(\sqrt{\frac{4}{5}}A\frac{samplerate}{2}\). And yes, the number would rise if the fraction was greater than one.

It's probably important to note that we haven't changed the sample rate: that's still 44100 and it still means we cannot get any higher frequencies than 22050 Hz. What we've done is alter the resolution. With \(\frac{4}{5}\) of the original resolution each frequency bin would be \(\frac{5}{4}\) as wide so that we could still accommodate all 22050 Hz, but in fewer distinct bins. It's not a net loss, though; it's a trade off: we've lost resolution in the frequency domain but we've gained resolution in the time domain (there will be \(\frac{5}{4}\) as many Fourier transforms in any given time period). Of course, this now means we cannot simply read off the frequency from the bin number; now we need to scale the bin number by our resolution factor. Staying with the \(\frac{4}{5}\) example: 800 Hz would now be found in bin 640, which is \(\frac{4}{5}\) of 800.

So... why did we choose to use the Complex32.Magnitude property, and what does it mean? Each of the data points we get as output from Fourier.Forward is a complex number, having both a Complex32.Real and an Complex32.Imaginary part. Besides these two orthogonal vectors pointing in the real and imaginary axes, another way to interpret them is as a single vector that has two properties: Complex32.Magnitude and Complex32.Phase. Together, these allow us to imagine the frequency bin as a clock with a single spinning hand of variable length. If the hand is long: there's more signal; if it's short: there is less. And the angle of the hand shows us the offset of the signal (relative to convention).

Input Shift	Re	Im	Mag	Phase
+0	0	-22050	22050	\(-\frac{π}{2}\)
+\(\frac{π}{2}\)	22050	0	22050	0
+π	0	22050	22050	\(\frac{π}{2}\)
+\(\frac{3π}{2}\)	-22050	0	22050	π

Finally, how did we get 22050 for the maximum Complex32.Magnitude given a maximum amplitude of 1 (and minimum amplitude of -1) for our input signal? Well, that's how the Fourier transform works: it essentially adds up the 22050 vector results of pairing the input signal with the known pure sinusoidal signal.

Demand and Supply: Part 2

In the previous post, there was a table of coefficients for average cost (AC), total cost (TC) and marginal cost (MC). Let's decompose TC into two components: one that doesn't vary as quantity (Q) varies and another that does. We'll call the components total fixed cost (TFC) and total variable cost (TVC). We can also decompose AC into average fixed cost (AFC) and average variable cost (AVC)

	AC	AFC	AVC	TC	TFC	TVC	MC
Q-1	a	a	0				0
Q0	b	0	b	a	a	0	b
Q1	c	0	c	b	0	b	2c
Q2	d	0	d	c	0	c	3d
Q3				d	0	d

AFC: \(p = \frac{a}{q}\)
AVC: \(p = b + cq + dq^2\)
TFC: \(p = a\)
TVC: \(p = bq + cq^2 + dq^3\)

Note how \(AC = AFC + AVC\) and \(TC = TFC + TVC\)
Note too how MR is unrelated to TFC, but has the same relation to both TC and TVC.

Demand and Supply: Part 1

Let's use a little bit of mathematics to assist our understanding of demand and supply, focusing here on the slightly more complicated concept of supply (demand functions identically, but we don't often decompose revenue into fixed and variable components). In the study of economics we often encounter three "curves" that describe how the price of an item is related to the quantity (Q) a firm is willing to supply: average cost (AC), total cost (TC) and marginal cost (MC). These three functions are mathematically related by two equations: \(AC = \frac{TC}{Q}\) (which also means that \(TC = ACxQ\)) and \(\frac{dTC}{dQ} = MC\) (which also means that \(TC_q = \int_0^q MC dQ\)). It's too soon to explain, but these are not quantities the firm is willing to supply, but rather the quantities the firm is able to supply.

A third degree polynomial function should be sufficient for this demonstration. In the table below, I've displayed the coefficients of each power of Q, highlighting the "fixed" component of TC (at Q⁰) and AC (at Q^-1) noting that MC has no such "fixed" component.

	AC	TC	MC
Q-1	a		0
Q0	b	a	b
Q1	c	b	2c
Q2	d	c	3d
Q3		d

AC: \(p = \frac{a}{q} + b + cq + dq^2\)
TC: \(p = a + bq + cq^2 + dq^3\)
MC: \(p = b + 2cq + 3dq^2\)

If the firm has any fixed costs (e.g. monthly rental payments on commercial property), they would be represented by the coefficient \(a\); if it has variable costs (e.g. monthly wages payable to labour) they would be represented by the coefficients \(b\),\(c\) and \(d\) (these are the costs that vary as Q is varied.

Eigenv{ector;alue}s

There are some nice instructional videos on YouTube that explain eigenvectors and eigenvalues. What follows is a brief summary as well as a caveat or two.

We start with a square transformation matrix M, which can have one or more eigenvectors. If you restrict eigenvalues to real numbers (i.e. if you exclude imaginary numbers) then it's possible to have a transformation matrix without any real eigenvalues (e.g. in a pure rotation). It's also possible to have just one eigenvalue but an infinite number of eigenvectors (e.g. common scale in all axes) ... but I've probably distracted you too much already...

Anyway:
What's an eigenvector? It's a vector that only changes (stretches/squeezes) in scale along its span when transformed by M.
What's a span? That's a radial line to any vector from the origin of the coordinate system.
Each eigenvector has a corresponding scalar eigenvalue λ that, when the two are multiplied, has the same result as M being multiplied by that eigenvector.
It's a rather circular definition so far:
\(\matrix{M}\times\vec{v} = λ\times\vec{v}\)
We already know about a transformation matrix that represents a uniform scale of λ, and that's \(λ\times\matrix{I}\) (it's the identity matrix, but instead of 1s running down the diagonal, it's got λs). Now we have a starting point for our calculation:
\(\matrix{M}\times\vec{v} = (λ\times\matrix{I})\times\vec{v}\)
As tempting as it looks, we should never divide by a vector. Instead, what we want next is the value of λ that results in:
\((\matrix{M} - (λ\times\matrix{I}))\times\vec{v} = \vec{0}\)
Let's let:
\(\matrix{T} = \matrix{M} - (λ\times\matrix{I})\)
Ordinarily, it would now be trivial to solve for the eigenvector using:
\(\matrix{T}\times\vec{v} = \vec{0}\)
\(=> \matrix{T}^{-1}\times\vec{0} = \vec{v}\)
But we cannot do that in this instance, because:
\(det(\matrix{T}) = 0\)
And that means that T is not invertible. But surely that means we've missed a step? Nobody said we need the determinant to be zero. What's happening? I'll explain: if \(\vec{v} = \vec{0}\) then anything multiplied by \(\vec{v}\) will also equal \(\vec{0}\).
If we restrict our eigenvector to non-zero vectors then the only way we can get a zero vector after transformation is if the transformation matrix collapses a dimension: i.e. the determinant (or scale) of the transform must be zero. That's how we get our constraint.

Note: Matrices are either invertible or singular. A matrix is invertible iff the determinant is non-zero, and it is singular iff the determinant is zero. Our transform T must be singular. That means: it collapses down by a dimension.

So, if the determinant of our 2X2 transformation matrix is zero:
\(det\begin{bmatrix}a-λ & b \\ c & d-λ \\\end{bmatrix} = 0\)
Then we know (from the Leibniz formula of a determinant) \((a-λ) \times (d-λ) - c \times b = 0\)
And we can expand to:
\(λ^2 - λ(a+d) + (a \times d) - (c \times b) = 0\)
That's just a quadratic, easily solved (but take care with complex numbers, i does appear in rotation matrices). So we have our eigenvalue(s) λ and (by substitution) also T. But how do we find the eigenvectors if we cannot take the inverse of T? Well, you can try a technique like row reduction. But don't expect a unique solution... in fact, because the transform has a determinant of zero, the eigenvector is strictly a ratio of X:Y in two dimensions (or X:Y:Z in three dimensions, etc.) and you're going to find the lines overlap; in jargon: they are linearly dependent.

Vinyl LP Record

In this post I will detail one 12" stereophonic 33⅓ vinyl LP record. Initially I was hoping to be able to image the grooves cut into the surface and recover the audio track visually, but the resolution of a borrowed microscope demonstrated that the task was beyond its capabilities. Unfortunately that's left me to issue a caveat from the outset: reader beware; no experimental fact checking has taken place...

First, some mathematical properties:
Rotations per minute: 33⅓ = 100/3
Rotations per second: => 5/9ths
Outermost groove radius: 146.05mm
Innermost groove radius: 60.3mm
This gives us about 85.75mm of playable radius.
We're told that each side can contain up to 23 minutes (1380 seconds) of audio, the playback of which would require the disc to rotate 767 times (5/9 x 1380).
Fitting 767 rotations into 85.75mm gives us roughly 0.112mm separation between adjacent edges of the groove.
Finally, we define the spiral travelled by the stylus at time t as some initial offset less 0.112mm for each 2πθ of rotation as it moves from the outermost to the innermost radius.
let r(t) = 146.05 - 0.112 * (5 / 9) * t;
Let's switch to using µm, as things are starting to get tiny!

Note: the groove on a stereophonic record is a 90 degree cut rotated 45 degrees so that it looks like a V cut into the vinyl. The inner surface (closest to the center of the record) modulates one of the stereo audio tracks, while the outer surface (closest to the edge of the record) modulates the other. To increase likelihood of the stylus tracking the groove the V should never be shallower (or, equivalently, narrower) than 25µm.

We'll divide the 112µm groove into a handful of regions:

1. First, our mandatory 25µm notch, which we later split into an inner 12.5µm and an outer 12.5µm.
2. An optional (but recommended) gap
3. Inner and outer channels for modulating the left and right stereo data

Mathematically we say the groove width is made up of these components: 112µm - 25µm - gap = 2 * audio;
This tells us: once we've picked a size for the gap, we will know the maximum width for each of our audio channels; we don't need to refer to the gap size again once we've worked out the audio channel widths.

Now it's time to scale our input audio signals... For simplicity, let's ignore the bitness of a digital audio signal and assume that the signal amplitude fits into the range from +1 to -1. Note that the 0 point maps exactly onto the half-channel.
*Another assumption: positive samples have shallower grooves*
Remember: at time t, the stylus can be found at radius = r(t) [defined above].
let inner and outer be the left and right audio samples (numbers between -1 and +1) at time t. Combined with the rotation of the disc (giving us θ at time t) these four equations below define the groove geometry...
Surface: inner radial offset = (audio / 2) * (inner - 1) - 12.5µm
Surface: outer radial offset = (audio / 2) * (1 - outer) + 12.5µm
Groove: radial offset = (audio / 2) * (inner - outer) / 2
Groove: vertical offset = (audio / 2) * ((inner + outer) / 2 - 1) - 12.5µm

The image below is a stylized aid for identifying the described parts of a groove, as seen looking down the groove from head on. The image shows the inner channel modulated to a positive audio sample and the outer channel modulated to a negative audio sample. The stylus is offset to the right as a result of this stereo modulation.

• Green: maximum and minimum audio modulation points (where audio signal is +1 or -1)
• Orange: the imaginary "zero" line
• Light blue: surface of the inner modulation area
• Red: surface of the outer modulation area
• Black: stylus position
• Dark gray: uncut portion of vinyl LP record
• Light gray: mandatory unusable 25µm notch

Questions:

• do we map left to inner and right to outer?
• do we map positive to high (shallower V) and negative to low (deeper V)
• what's an appropriate audio channel width? if we know the maximum modulation in the 45 degree vector, we can scale it by SQRT(2) to find the channel width.
• is it appropriate to map the zero sample point to half of the audio channel width? it might

Numerical Integration Basics

Consider the equation \(y = 4\). If we plot this on two axes, x and y, it has no slope; it's just a horizontal line. We could also write the equation as \(y = 0x + 4\) because the value of y is unaffected by the value of x. We could also write it as \(y = f(x)\).

The integral of a function can be used to find the area between the function and the x-axis and is written \(I = \int_a^b f(x) dx\). It's essentially the sum of tiny changes in x (\(dx\)) multiplied by their corresponding y values (\(f(x)\)). The equation reminds us that we need to restrict ourselves to a lower and an upper value of x (otherwise the area would be infinite).

Let's choose a = x = 0 (the y-axis) and b = x = 3 as those two bounds.

\(I = \int_0^3 4 dx\)
\(I = \left[4 x\right]_0^3\)
\(I = \left(4\times3\right) - \left(4\times0\right)\)
\(I = 12\)

We're not restricted to 0 as the lower bound, but it exposes an interesting property: the area under a-to-b is equal to the area under 0-to-b less the area under 0-to-a. Let's try the same integral as above, but from 0-to-1.

\(I = \int_0^1 4 dx\)
\(I = \left[4 x\right]_0^1\)
\(I = \left(4\times1\right) - \left(4\times0\right)\)
\(I = 4\)

We might then subtract 4 from 12 to arrive at 8, but why not do this in one go?

\(I = \int_1^3 4 dx\)
\(I = \left[4 x\right]_1^3\)
\(I = \left(4\times3\right) - \left(4\times1\right)\)
\(I = 8\)

Singleton Pattern vs. C# Static Class

I ran into an old foe the other day when we were discussing interview questions; which to choose: standard singleton pattern or static class? At the most basic level, both constructs are useful in limiting the number of instances of an object that can be created. There are numerous subtle differences at the implementation level, but the single biggest difference is at an object-oriented level: static classes do not support polymorphism. This stems from a combination of factors at the language level; in short, a static class cannot inherit from a base class (abstract or concrete) nor can it implement any interface methods. Conversely, you can neither define static virtual methods on a class, nor can you define them on an interface. So, if you're ever using dependency injection - a technique that relies heavily on polymorphism - you will likely find that static classes will be inadequate. There are a number of other runtime level differences, but these are all secondary to the polymorphism issue. The simplest is that a static class can never be instantiated, and all methods are executed against the type instance. Contrast this with the singleton, of which exactly one instance is created, and where methods are executed against that instance. Next time I choose, I'm going anti-static for all but the simplest "utility" classes.

Cisco 7960 SIP Phone

I bought a Cisco 7960 IP phone (preloaded with SIP firmware) to connect with my experimental asterisk server running on a Raspberry Pi. Setting it up wasn't that hard, once I stopped following the bad advice that abounded on Google search results. I'm partially documenting my own experience here - not to help, but to hinder. You don't need a TFTP server (and therefore you don't need to set any hard-to-reach DHCP options) to distribute configuration files, you can do everything you need through the phone's keypad. If you're going to make changes, you need to perform the "Unlock Configuration" steps each time.

Unlock the Configuration

1. Press Settings
2. Scroll down to 9: Unlock
3. Enter the password "cisco"

Set the TFTP server IP address manually

1. Press Settings
2. Scroll down to 3: Network Configuration
3. Scroll down to TFTP Server
4. Press Edit
5. Enter the IP address of the TFTP server

Configure SIP on the Phone

1. Press Settings
2. Scroll down to SIP Configuration
3. Select the line you want to edit
4. Confirm the following entries exist (or add them) [They should match what you've got setup in sip.config on the Asterisk server).

Setup SIP on the Asterisk Server

1. Add a new block to sip.conf. Make sure the extension and secret match the values you provided when configuring SIP on the phone. [1234] context=home secret=secret etc.
2. Add a dialplan rule or two to extensions.conf. Make sure the context in sip.conf matches the section name in extensions.conf. exten => 1234,1,Dial(SIP/1234)
asterisk -r then sip reload and dialplan reload

ADSL2+ Sync Rate

If you have ever called up your ISP support team to complain, you may have heard them quote a number called SNR to describe the quality of your connection.

It's not quite as simple as a single number, though.

ADSL2+ uses multiple "channels" at once to transfer data between the ISP and your home modem. In the case of my ASUS modem, there are 512 channels: there is 2.208 MHz of available spectrum on the copper wire; these are split into equally spaced 4.3125 kHz wide channels.

However, not every channel (or bin) is created equally: there is almost certainly some electrical "noise" or congestion on the cables. Some bins are reserved for normal telephone connections, others for upstream and downstream communication with the server, with a few reserved as guards to keep sections separated.

I pay for a 4 Mb/s connection, so let's assume the ISP allows me 1672 bits' worth of downstream sync rate*. It is up to the modem and ISP to negotiate how to allocate the sync rate across the available bins. My modem surveys the ratio of signal to noise (SNR) across each bin in the downstream set of channels. A rule of thumb is that each bit requires about 3dB of SNR to reliably decode the signal without errors. Bins with higher SNR will therefore have higher bit capacity for bits.

My modem allows me to connect with SSH and look at three files, which are ordinarily used by the administrative web server interface to present me a way to easily change modem settings and view diagnostics:

• /var/tmp/spectrum-bpc-ds
• /var/tmp/spectrum-bpc-us
• /var/tmp/spectrum-snr

Let's do some very simple mathematics with the data in these files that won't be too difficult to follow:

Data taken from the SNR file:
1672 bits to allocate
268 channels into which they must be allocated
10,523.27 dB (the sum of SNR across the available reception channels)

Assumptions:
3 dB/bit required to convert received signal without errors

1) 1672 bits * 3 dB/bit = 5016 dB SNR required to decode the signal

2) 10,523.27 dB - 5016 dB = 5507.27 dB unused SNR

3) 5507.27 dB / 268 channels = 20.54951 dB unused SNR per channel

4) Now, we iterate each of the 268 channels in turn. I'll demonstrate with the first channel, where the SNR was 24.21 dB.

a) 24.21 dB - 20.54951 dB = ~3.66 dB to be used for data reception

b) 3.66 dB / 3 dB/bit = ~1.22 bits

c) round 1.22 bits to a whole number: 1

It turns out that this simple algorithm gets us remarkably close to the numbers of bits per channel (BPC) in the other files, which are arrived at by the modem's own algorithms. The difference could even be accounted for by (lack of, or outdated) "bitswap", where bins are periodically re-checked for SNR and reallocated to make the best use of the spectrum as it evolves over time.


* Ignore - for now - that those numbers are not equivalent, they are effectively the same thing quoted for different units of time
 

OpenVPN

This article on Ars Technica inspired me to try and setup a VPN, but it seemed to lack a couple of extra steps that were required to run the server on a Raspberry Pi behind a router. The official OpenVPN documentation was helpful but long-winded, and I resorted to a few askubuntu answers in the end.

1. Certificate names

   client-no-pass doesn't have to be replaced with the client host name, but can be (pretty much) any identifier for the client. The important thing is to sign all certificates with the same CA key.

2. VPN topology

   The default topology of net30 is perhaps not as easy to comprehend as subnet, and subnet works with Ubuntu, Windows and iPhone clients.

3. Default gateway

   On the server, I needed to push redirect-gateway, push "route xxx.xxx.xxx.0 255.255.255.0" and consequently push "dhcp-option DNS xxx.xxx.xxx.1" so that the client would be able to route to the server's local network, and the tun0 device would be able to do DNS lookups.

4. Firewall rules

   Besides enabling the forwarding of IPv4 so that the Raspberry Pi acted as a router, I need to add three firewall rules in iptables: one for NAT masquerading, and two for accepting new and related established connections.

5. DNS setup for clients

   The dhcp-options that were pushed to the client were ignored, which resulted in the client being unable to resolve any DNS names to IP addresses. This was only a problem on Ubuntu clients, but the script /etc/openvpn/update-resolv-conf was provided when I installed openvpn; all I had to do was reference it from client-name.config as a pair of lines:

   up /etc/openvpn/update-resolv-conf
   down /etc/openvpn/update-resolv-conf
   

Grub

It's nice when the Grub bootloader remembers your last choice and reuses it as the default value next time your machine boots. Using Linux, put the following in /etc/default/grub:

GRUB_DEFAULT=saved
GRUB_SAVEDEFAULT=true

Then run:

sudo update-grub

SubscribeOn / ObserveOn

Once you have an instance of an IObservable<T> object, Reactive Extensions (Rx) provides at least two points where you can choose an IScheduler which will determine the threading context under which the later operations will run. Those operations can be broken down into two phases: subscription and observation.

Subscription refers to the operations that modify the behaviours of an IObservable<T> (e.g. Throttle, Merge, Concat) The IScheduler for the subscription phase is assigned by the SubscribeOn() method.

Observation refers to the callback operations that are invoked as the result of observing an IObservable<T> (e.g. OnNext, OnCompleted, OnError) The IScheduler for the observation phase is assigned by the ObserveOn() method.

While not the subject of this post, you can also choose an IScheduler implementation to convert any IEnumerable<T> to an IObservable<T> with the ToObservable method. This is relevant when the IEnumerable<T> has the potential to take a long time to be enumerated.

6 Simple Rules for Async/Await

1. If a method A calls another method B that returns a Task (or Task<T>) then the calling method does not block on the completion of that task, unless it either:
1. awaits that task, or
2. waits on the result of that task
2. The calling method cannot await a task unless it is declared with the async modifier, which causes the compiler to build a state machine (similar to the IEnumerable iterator) for that awaitable method.
3. In point 1 above, the behaviour is independent of whether or not B is marked with the async keyword
4. It is this non-blocking behaviour of A that allows the calling thread to proceed past the point where B is invoked. It's even possible for the thread's call stack to unwind, and for the thread to go back to reading the Windows message queue if it was the UI thread.
5. The compiler will only warn you "because this call is not awaited ..." if B was marked with async.
6. It is expected that A will do something with the Task returned by B; at the very least there should be some code to check that the Task did not throw any exceptions. If - instead of B - we have an async method C that returns void then we do not present A with any opportunity to monitor the completion of C. Unobserved exceptions thrown during the execution of C could indicate corrupted program state and can be configured to terminate the application in much the same way that an unhandled exception in synchronous code can unwind a stack fully and terminate the process.

iSCSI Enterprise Target

In the presence of conflicting information (lots of it) this is how I got iSCSI Enterprise Target running on Ubuntu 16.04 (LTS). First, install the required software with

sudo apt-get install iscsitarget

man ietd indicates that the configuration file is located at /etc/ietd.conf but it isn't, so instead:

sudo nano /etc/iet/ietd.conf

If you're following the "Creating an Open Source SAN" chapter of "Pro Ubuntu Server Administration" by Sander van Vugt, be careful when adding your target not to include a space between the comma and the type

Target iqn.2008-08.com.sandervanvugt:mytarget
    Lun 0 Path=/dev/sdb,Type=fileio
    Lun 1 Path=/dev/sdc, Type=fileio

One step that's missing from this chapter is to enable it (why isn't it enabled?).

sudo nano /etc/default/iscsitarget

Ensure this line is present in the configuration

ISCSITARGET_ENABLE=true

Then exit nano, saving changes, and restart the iscsitarget service

sudo /etc/init.d/iscsitarget restart

If everything has gone to plan, you should be able to see these files

cat /proc/net/iet/session
cat /proc/net/iet/volume

Otherwise, have a look at the log

tail /var /log/syslog