Actually Madison used it in 1788:

https://founders.archives.gov/documents/Hamilton/01-05-02-0012-0060

This isn't due to any bug. Merrill's study uses normalized utilities, so in the 2-candidate case, the majority vote always goes to the utilitarian winner.

AES17 doesn't define "dynamic range". It defines "Signal-to-noise ratio or noise in the presence of signal" which is what includes the test tone:

The test signal for the measurement shall be a 997-Hz sine wave producing – 60 dB FS at the output of the EUT.

The hex spells lint not list, also in a real WAV file it appears to be capitalized LIST = 0x4C495354

Utter nonsense. To paraphrase:

Any election system that favors unrepresentative candidates [like IRV] would be considered unreasonable; the same rationale must be applied to one that favors representative candidates.

Uh, no. That doesn't follow.

FairVote starts from the conclusion that IRV is the best voting method, and then works backwards to try to justify it, in this case arguing that a flaw of IRV is actually a feature, by making a false equivalent between a voting system that favors unrepresentative candidates and one that favors representative candidates.

The whole point of an election is to find the most-representative candidate.

Yes, candidates who are moderate relative to the voters should always win.

The goal of an election is to find the candidate who best represents the electorate. If the electorate is left-wing on average, the winner should be too. If the electorate is "strong on both personal freedoms and economic freedoms", then the winner should be too.

Anything else is undemocratic.

This isn't true. It's possible for a Condorcet candidate to be extremist relative to the other candidates or the electorate, since weak preferences are given equal weight to strong preferences. Simple example here:

This is true, but IRV doesn't choose the candidate who is most-liked (the "utilitarian winner"), either.

Under range/score, the best strategy to promote the election of a preferred candidate is always to give that candidate the maximum score and then give every other competitor the minimum score.

Yeah, this is false.

If you have perfect knowledge of how everyone else is voting (and you usually don't), then the best strategy is to give a maximum score to the frontrunner that you prefer, and also to everyone you like more than them, and to likewise give a minimum score to the other frontrunner, and to everyone you dislike more.

This is not bullet voting; it's equivalent to Approval Voting, and leads to more moderate winners who are good representatives of the electorate.

Real-world Score elections don't show this behavior, anyway, because polls are imprecise and the consequences of voting honestly under Score aren't as dire as they are under FPTP or IRV.

Rebuttal to [the original version of] this page at https://www.equal.vote/fv

This is nonsense. All voting systems are subject to tactical voting.

"Failing the majority criterion" is actually a good thing, because these systems find the candidate who is most acceptable to the entire population, rather than the candidate who is most acceptable to half of the population.

Majoritarianism that ignores the desires of half of the population leads to adversarial politics, polarization, and eventually even to violence and civil war.

...which is why it subsequently gets repealed?

If the SM58 noise floor is calculated at room temperature, the voltage output is 0.00000032 volts.

This is equal to -130 dBV. The SM58 Vocal Microphone Specification Sheet says:

Sensitivity (at 1,000 Hz Open Circuit Voltage) –54.5 dBV/Pa (1.85 mV)

At this sensitivity, the self-noise would be 94 - (-54.5 - -130) = 19 dB SPL, which is pretty typical and certainly not "lower than can be typically measured".

This is incorrect. If R1 is infinite and R2 is 0, the parallel impedance is 0 ohms, but the input impedance is much higher than the input impedance of the op-amp itself, due to feedback making the inputs very similar in voltage.

The input impedance is actually

$$(1 + A_0 B)\cdot Z_\mathrm{ino}$$

where

• \(A_0\) is the gain without feedback (the open loop gain)
• \(B\) is the fraction of the output which feeds back as a negative voltage at the input
• \(Z_\mathrm{ino}\) is input impedance without negative feedback

For the above buffer example, it would be close to \(A_0 Z_\mathrm{ino}\)

See https://electronics.stackexchange.com/q/177007/142

Simpson - Introductory electronics for scientists and engineers section 7.2 Negative Voltage Feedback explains this clearly

Actually, he first used π in 1727, and it meant 6.28.., not 3.14..

Equation 10-27 is wrong:

• Resistors are in parallel, their noise does not sum.

This is 3 terms (10 MΩ noise, 100 kΩ noise, and amplifier noise), but the equation only includes 2.

Equation 10-25 has several errors:

• The left and right side are not equal.
• Resistors are in parallel, so their noise does not sum.
• Resistor noise needs to be multiplied by gain to be combined with 113.1 μV output noise of op-amp.

Equation 10-23 is incorrect:

• The noise of the resistors does not sum, since the resistors are in parallel and load each other. It would be equal to a single 50 kΩ resistor producing 4.0 μV.
• The 113.1 value is output noise. The resistor noise needs to be multiplied by the gain to get the output noise, too.

Should be 0.1 μF

Should be 0.1 μF

The noise calculations have many errors. See annotations on https://via.hypothes.is/http://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf for details

actually should be multiplied by the non-inverting noise gain, which is 101

Should be TLC2201

The noise calculations have many errors. See annotations on https://via.hypothes.is/http://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf for details

It's been 3 years! Where is it??

This is incorrect:

• The poles are not located on a circle
• The imaginary parts of the poles are not equally spaced

To generate the poles of a Bessel filter you need to use root-finding methods on the reverse Bessel polynomials. There's no other shortcut that I'm aware of.

This is incorrect: There is a small amount of overshoot in Bessel filters.

This isn't quite correct: Bessel filters have a small amount of overshoot.

Link is dead :D