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.
CookieOfFortune
joined 2 years ago
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 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++?
- Played through a DAC and speaker to produce an analogue signal (lossy)
- Analogue modulation of bit stream played through DAC (lossy)
These steps are literally the same thing. You’re converting some data into sound for the bird to hear.
Edit: Actually, most physical modulation schemes use sinusoids anyways. So that’s exactly the same as playing a spectrum.
That’s not really how it works in the real world. Usually you have both bandwidth and noise constraints.
Sure you can send something like a square wave but this isn’t practical for real communication channels. Typically you’re sending many sine waves in parallel with multiple amplitudes and phase offsets to represent a sequence of bits (QAM). Then on top of that you’d encode the original data with both a randomizer (to prevent long runs from looking like nothing) and error correction. So usually the system can handle some level of distortion.
What you’re hoping is that by the time the data reaches the user (really, Layer 3), all the errors have already been handled and you never see any issues.
The bird is just another type of noisy channel with its own distortion characteristics.
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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.
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?