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This is a follow up question from my question answered here:

Why does a loop antenna not measure the differential magnetic field?

Essentially I just found out today that we are actually measuring the derivative of the time varying electric field with an antenna, rather than its magnitude at a given point in time. So my questions is if that is the case, why do we not normally need to integrate the output (measured) of a receive antenna? I know the integral of a sin is a cos etc. but there will usually be an amplitude term as well which seems important?

To clarify the question as there seems to be confusion in the comments, I am referring to time varying fields not static ones.

Christian
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Most communication is done using signals that are sufficiently bandlimited that we simply don't have any reason to care. The derivative of the signal is the same thing as the signal, except for a term that's proportional to the freqency (d/dt sin(ωt) = ω cos(ωt)). As long as the fractional bandwidth of the signal is reasonably small, this transform doesn't matter at all. As for the extra ω in the magnitude, we just say that the antenna has a response that depends on frequency — but all realistic antennas have a response that depends on frequency, so again, whatever.

hobbs
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    This is the answer I was looking for. Thank you. – Christian May 15 '23 at 20:23
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    (As for the difference between sin and cos, that's the same thing that happens if you move the antenna slightly closer to or farther away from the transmitter so nobody cares) – user253751 May 16 '23 at 16:32
  • @user253751 Sure for a typical communications type signal (CW) it's as you say unimportant but for a transient signal I would imagine the phase information becomes quite important – Christian May 17 '23 at 08:59
  • @Christian no, it really isn't. A *change* in phase would potentially matter, but a constant phase offset is undetectable. It's the same, as user253751 says, as moving the receiver a foot closer or further from the transmitter. For GPS, you would have to calibrate it out because measuring that phase distance is the purpose of the system. For almost everything else, you would never notice. – hobbs May 17 '23 at 13:33
  • @hobbs if I am measuring an impulse (lets say with the shape of sinc(t)) then the shape of the derivative of that waveform is very different to the shape of the sinc(t) itself. I don't understand how we can say that this is not important, if you are interested in the shape of a waveform? There are application for example, lets say you want to measure the electric field from a lightning strike – Christian May 17 '23 at 14:22
  • @Christian Did you precisely measure the distance between your antenna and the signal source? – user253751 May 17 '23 at 15:46
  • @user253751 I'm interested in the theory, so would be interested in your response to both yes and no to that question. I have not made any actual measurements – Christian May 17 '23 at 15:50
  • @Christian everyone is wondering why the integral is useful to you, that's why, and if your need is based on phase being so important, this excuse only makes sense in a precisely measured lab setup... of course, if you are doing a science experiment, then maybe you *do* need the integral for some reason because science is often like that, but nobody else does – user253751 May 17 '23 at 15:51
  • @user253751 yes it is for a science experiment – Christian May 17 '23 at 15:59
  • @Christian then maybe don't apply advice used for communication antennas. Design your experiment however you need to to get the measurements you want. If that means integrating an antenna, so be it, although I've never heard of that being a thing, so maybe double-check it actually works. Note that integration still will not tell you the DC value of something because that is the unknown "constant of integration". – user253751 May 17 '23 at 16:02
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why do we not normally need to integrate the output of an antenna

We don't care about the integral.

This sounds a bit banal, but it's really the case.

A transmitter needs to change a field to cause a signal to radiate – so the whole phenomenon is always about fields' changes, never about static fields.

Luckily, for very long parts, electric and magnetic field in air and free space are linear – so that the changing fields actually carrying a signal can overlay linearly with the static magnetic field of earth and the static electric field that typically exists between ground (as in: dirt) and upper atmospheric layers. We don't care about these static fields.

A static field would "always have been there", and contains no information. No power can be extracted from it, either, until it changes (and no longer is static). (This is generally true; of course, if the amount of energy you extract is very small compared to the energy in the field, you won't change the "pseudo-static" field enough to really matter – turning the needle of a compass doesn't demagnetize the earth in any significant way.)

Marcus Müller
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  • Surely if you wanted to measure the amplitude of the electric field then the integral is important? Even the power? – Christian May 15 '23 at 19:35
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    but you don't measure the static amplitude of any field with an antenna. Antennas typically don't even *have* paths for DC to flow. Finite-sized antennas are infinitely inefficient at receiving 0-frequency fields. – Marcus Müller May 15 '23 at 19:36
  • I'm referring to the maximum amplitude ot the time varying signal – Christian May 15 '23 at 19:37
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    Electronic equipment probably has plenty of other sources of static or low-frequency E and B fields, besides the Earth's magnetic poles and atmospheric charge. It's desirable NOT to have these included in the output of the radio receiver. It isn't that they don't contain information, it's that the information they contain is, from the perspective of the communication, interfering noise not signal – Ben Voigt May 15 '23 at 19:37
  • Exactly, that has brought us full circle to my question: why don't we integrate this change of amplitude magnitude to get the actual amplitude? – Christian May 15 '23 at 19:38
  • @Christian sorry, overread the "maximum" in "maximum amplitude": For harmonic signals, the maximum of the oscillation is the same as the maximum of its derivative. – Marcus Müller May 15 '23 at 19:39
  • @Christian how often do we need to repeat: we don't *care* about the integral. – Marcus Müller May 15 '23 at 19:39
  • It's a very boring integral, too. It's composed of the sum of harmonic oscillation, and a constant. So, the anti-derivative of the harmonic oscillation is the harmonic oscillation again, shifted in phase, and the integral of a constant just diverges. – Marcus Müller May 15 '23 at 19:40
  • @MarcusMüller the integral of cos ax = 1/a sin ax. So the amplitude term in front changes when you integrate, which is why I thought there would be a difference in amplitude – Christian May 15 '23 at 19:43
  • then that's not a propagating wave. You really need to refresh your Maxwell's Equations, which why I asked you to clarify how familiar you are with them under your last question. – Marcus Müller May 15 '23 at 19:44
  • @Christian are you looking for an [Envelope Detector](https://en.wikipedia.org/wiki/Envelope_detector)? – brhans May 15 '23 at 19:46
  • @brhans what I was originally looking for is how a so-called differential field probe (b-dot or d-dot) differs from a conventional antenna. I was always lead to believe the difference was that they measure the differential of the field but as I found out today, so too does the conventional antenna! So I'm a bit lost. Appreciate everyone trying to help me – Christian May 15 '23 at 19:50
  • @MarcusMüller I am familiar of maxwells equations but no part of them that I'm aware of suggests that the time derivative of either the B or E field should have the same max amplitude as the B or E field itself. So perhaps I need to focus on reading up more on maxwells equations – Christian May 15 '23 at 19:55
  • @Christian In communication applications, we are interested in the peak change of the derivative, and its RMS, and its power, but rarely the waveform. So the derivative contains all we need to know. When we are interested in the integral, for instance let's say we want the waveform of it, then we integrate. Consider the [Rogowski Coil](https://en.wikipedia.org/wiki/Rogowski_coil) – Neil_UK May 15 '23 at 20:02
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    @Christian you'll notice that the time derivative of the B-field happens to be proportional (through material constants and vacuum speed of light) to the amplitude of the curl of the E-field – and the derivative of the E-field (plus the current density) proportional to the B-field (again, material constants and speed of light). So you're right, the amplitude does change – but what we measure with the antenna is always the derivative of the "other" field. – Marcus Müller May 15 '23 at 20:03
  • @Neil_UK thank you that pretty much answers my question. I realise I should have been more specific in the original wording. – Christian May 15 '23 at 20:10
  • @Christian integrating a derivative of any constant gives you 0+C and you get to choose what C is - it doesn't tell you the constant. – user253751 May 16 '23 at 14:47
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I have come to believe that some of the responses to this question are misleading or perhaps just incorrect. I have discovered a paper which sheds some light on this question:

Chapter 4. Time domain characterisation of antennas by normalised impulse response

Essentially in the case of an antenna designed to measure the electric field component, we are not measuring the time differential of the electric field. However in the case of a pair of identical antennas, the transmit antenna will radiate the time differential of the input signal, so that what we measure on the receive antenna is the time differential of the original signal. It is not however measuring the time differential of the radiated waveform itself.

I found it helpful to think about the relative phase of current and electric field to understand this better. In the case of the (resonant) transmit antenna, the current drives a potential difference which is out of phase with the driving current, and this potential difference is directly proportional to the electric field transmitted. In the case of the receive antenna, the electric field drives a current directly, and it is this that we measure.

Christianidis Vasileios
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Christian
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