Have fun! At some point I gave up on it because it was too much like a job.

Yeah this just causes those cars to be registered out of state without any regulation at all and the registration revenue goes elsewhere.

You can still lessen assembly, it’s not that hard actually, just tedious. It can play TIS-100 if you want a “fun” way to learn it.

What about Cyberpunk?

Not all QA is destructive. For example, X-rays can be used to detect internal cracks without damaging the sample.

You can perform QA on parts.

Wait which Red Alert was this?

Except it has?

Geosynchronous means the orbital period is a day and relatively circular. Satellites with these orbits basically look down at the same part of the Earth.

Lagrange points are locations along an orbit where gravitational forces balance out with the centrifugal force of the orbit. This does allow less fuel expenditure to maintain.

For the Earth-Sun or Earth-Moon systems, the Lagrange points do not occur at the altitude of the geosynchronous orbit. The Lagrange points are either significantly further away or at a different orbital phase.

any system that communicates information is apparently a digital communications system, so long as you can imagine an arbitrary scheme to interpret at least one bit of information from the signal,

This has always been my point since the beginning! There exist very low bandwidth digital communication systems in real life, with less than one bit per second. The bandwidth available should be defined where something is digital or not.

regardless of whether that was the message intended to be communicated.

Seeing the bird in the spectrogram is quite intentional and sufficient to consider this a communications system.

It seems if instead of a bird picture, a random set of bits were encoded and then detected In the spectrogram, you’d consider this more of a digital system since instead of a human doing the check you use an algorithm?

If we’re being pedantic, shouldn’t we consider that it can be a one bit signal? Otherwise you should be specific about what bandwidth you’d consider digital.

I thought we were being pedantic here?

Yes, eventually a signal may degrade or be corrupted, but prior to that point the reproduction is literally and exactly perfect.

Modulation schemes are characterized via a probabilistic tolerance, so even when you are within the tolerances, you can get an incorrect value at some expected rate. Note that you can even define a modulation scheme with a high error rate and be ok with that.

That’s why I take issue with the concept of an exactly perfect reproduction. Usually there are layers above the digital modulation to handle these possibility to decrease the error rates even lower.

And no, I don’t consider the PNG to be the data carried. I think the way the author does the bandwidth calculations is incorrect.

How do you get destructor behavior in C?

The problem here is that people will just register those cars in other states. So that means you still get the same smog issues without the registration revenues.

The bird drawing is just a proxy for arbitrary data. In your example, you could convert bitstream into a pattern of black and white squares into a YouTube Video. Send it through the VHS channel, and when you digitize it, you would get back the exact bitstream.

If your argument is that the bandwidth calculation is incorrect, then sure I think that’s fair.

But I don’t think it’s correct to say it’s not a digital channel juts because it doesn’t have optimal bandwidth.

The entire point is that the modulated signal can be reconstructed exactly,

But this isn’t true. Just because a signal is modulated doesn’t mean it can’t be distorted.

A spectrogram is just showing that arbitrary data can be sent though this channel. It’s literally a form of modulation.

My point is that it doesn’t have to be optimal to be considered digital. Which in the general case means basically any communication channel can be digital.

If the argument is that they didn’t correctly calculate the bandwidth, then sure.

Why couldn’t you have a likelihood function for the bird?

As a trivial case, you can just say: Does the spectrum look like a bird? Then you’d have a digital channel by your definition for a single bit.

The actual channel bandwidth is obviously higher than that.

Isn’t this for C++?