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I read the following in a book:

"When the transmitted signal is passed through the air using electromagnetic waves, it must take the form of a continuous (analog) waveform."

Why is this so? Why can't the signal take the form of a digital waveform?

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    \$\begingroup\$ Calling something digital is done for convenience. A digital voltage signal can still be perfectly treated as an analogue signal. \$\endgroup\$ – Andy aka Mar 4 '17 at 11:17
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    \$\begingroup\$ A step change in a waveform requires infinite bandwidth. \$\endgroup\$ – Tom Carpenter Mar 4 '17 at 11:33
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    \$\begingroup\$ Since the author states that analog is continuous, he must be assuming that digital is discontinuous. An ideal, discontinuous digital signal CANNOT be transmitted because it would need infinite bandwidth as Tom Carpenter already said. \$\endgroup\$ – Claudio Avi Chami Mar 4 '17 at 11:47
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    \$\begingroup\$ Note that once you get underneath all the coding and thresholds, etc., all signals are actually analog on the wire or in the air. \$\endgroup\$ – Todd Wilcox Mar 4 '17 at 16:36
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    \$\begingroup\$ @ClaudioAviChami, But, the text is explicitly differentiating between transmission though the air and (presumably) a wire. An ideal, discontinuous digital signal cannot exist, due to requiring infinite bandwidth (regardless of being on a wire, in air, etc.). Thus, trying to differentiate between transmitting it over a wire vs. in air, as the text appears to do, is nonsensical. \$\endgroup\$ – Makyen Mar 5 '17 at 6:11
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Adding to Tom's answer:

The wording is not very clear, but what this means is that digital signals do not actually exist in reality. All signals are analog.

When we decide that a voltage above a certain threshold is a "1", a voltage below a certain threshold is a "0", and the space in between is "undefined", then we interpret an analog signal as a digital value. However, it is only a very convenient approximation that greatly simplifies the job of the designer.

Digital is abstract information. It is a meaning we choose to assign to physical values. This is why you cannot send a digital signal over the air as radio waves. It must be converted first into something that exists outside of abstraction, like an analog signal which represents the information to be transmitted.

The real signal is made of physical analog values: voltage, light, current, fields, acoustic pressure, whatever.

For your radio application, you could encode your digital bits into the frequency of a carrier, or its phase, or any other encoding, of which they are many. Now, you have an analog signal which carries your information, and you can transmit it, then receive it and recover your bits.

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  • \$\begingroup\$ How about I take a simple circuit consisting of a pair of wires, a battery, a switch and a resistor. Now if I take the output as the voltage of the resistor at periodic intervals, and at the input I switch the key on/off. Won't I get a perfectly digital signal? \$\endgroup\$ – Vishal Sharma Mar 4 '17 at 11:59
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    \$\begingroup\$ You'll get a voltage, not bits. It only becomes digital bits when you compare it with a threshold you decided... \$\endgroup\$ – peufeu Mar 4 '17 at 12:26
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    \$\begingroup\$ In your example, when you press the button you'll get random garbage for a few milliseconds as the switch bounces, then RC settling to the battery voltage (which is variable) plus noise depending on the environment. If you want a digital bit, you need to add specific decision criteria regarding debouncing, filtering, voltage thresholds, etc. \$\endgroup\$ – peufeu Mar 4 '17 at 12:47
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    \$\begingroup\$ @VishalSharma A real, physical switch does not instantly change from zero resistance to infinite resistance. And every time you close it, you will get a slightly different resistance as different parts of the switch make different levels of physical contact. You only get digital in the real world by abstracting away information. (With perhaps some exceptions from quantum mechanics.) \$\endgroup\$ – David Schwartz Mar 4 '17 at 19:44
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    \$\begingroup\$ @VishalSharma Take your example circuit and replace the resistor with a resistor that glows when current flows through it, like a light bulb. Record video of that light bulb with a very high frame rate. When you operate the switch, you'll see on the video that the bulb does not go from off to fully on instantly. The inflow of current and the onset of the EMF is not instantaneous. The "signal" from the switch to the bulb is not digital, it's analog. A way to understand why is all real world circuits have nonzero resistance, capacitance, and inductance, so no signal can change instantly. \$\endgroup\$ – Todd Wilcox Mar 4 '17 at 23:52
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The important take away point is that you need a continuous waveform if you are using electromagnetic waves. That is not to say you can't have a signal which represents digital data, just that the signal itself must be continuous.

Consider a square wave, or even a sequence of binary voltages (1 0 1 1 0 etc.). If you take the FFT of such a signal you will find that it has spectral content over an infinite bandwidth. In other words, to produce a perfect step change you need a channel with infinite bandwidth.

There is no such thing as a channel with infinite bandwidth. In the case of sending electromagnetic wave based signals wirelessly we have a massive limitation on bandwidth which prevents a non-continuous waveform (i.e. ones with step changes) being sent.

However just because you can't send a signal which is non-continuous, doesn't mean you can't send a signal which represents one. All of the digital modulation schemes do just that. OOK is the most basic example - a zero is represented by no signal, a one is represented by a simple tone.

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    \$\begingroup\$ Your description of OOK also requires infinite bandwidth... \$\endgroup\$ – R.. Mar 4 '17 at 15:59
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    \$\begingroup\$ No, because the switching on and off need not occur in near-zero time. Good OOK'd transmitters ramp the transmitter power up and down to minimize bandwidth. See for ex. ivarc.org.uk/uploads/1/2/3/8/12380834/keyclicks_version_1.pdf \$\endgroup\$ – Jamie Hanrahan Mar 5 '17 at 11:47
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Digital signals are an abstraction that humans use to describe and understand things by omitting information that we are not interested in.

For example, consider the following: 1001010101000101010

Is that a digital signal or an analog one? If you only care about the pattern of ones and zeroes, then it's a digital signal. But the actual physical thing you are looking at is entirely analog because each digit is in a slightly different physical position and has a slightly different level of brightness and so on.

There might be exceptions in quantum mechanics, but that's not relevant here.

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Consider an electromagnetic channel such that you can send and receive, say, voltage within -5 V and +5 V (wire, radio with different modulations, etc.) If you give different relevance to all voltage levels within this range (e.g. to operate a speaker) then you are in an analog regime. If you choose two voltage intervals, say (-4, -2) and (2, 4), and you only care if the voltage falls within one or the other — into one could mean “0” and into the other could mean “1”, otherwise it would mean nothing — then you are in a digital regime.

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    \$\begingroup\$ So in the second case, the signal being used is still analog but the way we are interpreting it is digital. So an analog signal can be interpreted in both ways i.e. analog as well as digital data can be extracted from it. This part I'm able to understand. What I'm confused about right now is if there really exists a perfectly digital waveform because as Tom had mentioned, it would require infinite bandwidth. \$\endgroup\$ – Vishal Sharma Mar 4 '17 at 12:15
  • \$\begingroup\$ Any electromagnetic channel is open to (random) noise, interference. So, the output cannot be exactly determined from the input. Hence whatever waveform you put at the entrance will never be the same at the exit. Now, within the framework of Quantum Physics, maybe it is possible to send (pure) binary coded information, but not like in classical eletromagnetism, through EM waves. (I cannot add to this.) \$\endgroup\$ – Pedro J. V. Mendes Mar 4 '17 at 14:41
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    \$\begingroup\$ @VishalSharma the answer is no, there can be no perfect digital waveform in real world electronics because infinite bandwidth is impossible because zero resistance, capacitance, and inductance are all impossible. Digital signaling works because you only have to make the actual signal close to digital for the receiver of the signal to be able to interpret it as discrete values, but the underlying waveform of the signal will still be continuous and analog. \$\endgroup\$ – Todd Wilcox Mar 4 '17 at 23:56
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    \$\begingroup\$ What I'm going to take away from all the discussion is this: All signals in this world are Analog signal. We can't produce a perfectly digital signal. We can only generate closer approximations of a digital signal. Therefore whether the medium is wired or wireless, it doesn't matter. The transmitted signal's waveform will always be Analog. \$\endgroup\$ – Vishal Sharma Mar 5 '17 at 4:29
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Electromagnetic waves, or signals that pass through space or air, exist due to the energy exchange between the electric-wave and the magnetic-wave.

This energy exchange happens at a fixed frequency, or colour for light-waves. The frequency is determined by the source radiating the energy.

Waves of different frequencies can be mixed, but they exist as separate colours that cannot be changed as they trave through space. This is why white sunlight can be seperated into the colours in the rainbow.

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