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Does antenna radiate DC?

  1. Apr 29, 2010 #1
    Dear all,

    This question is related to the radiation of a UWB antenna.

    If I feed a Gaussian pulse to a UWB antenna, would a differentiated gaussian pulse radiated? It seems so as in the tutorial paper (attached) by Wiesbeck says that "any antenna differentiates any signal because antenna do not radiate any DC".

    Is it true? If I send a diferentiated gaussian signal, would a twice-differentiated gaussian (vicker's) signal being radiated?

    Thank you for any discussion.


    Attached Files:

  2. jcsd
  3. Apr 29, 2010 #2
    From my understanding, for like dipole/monopole antenna, the signal line and the ground line are separated. For DC, it will sees the transducer (antenna) as an open circuit, hence no current flow results no radiation.
  4. Apr 29, 2010 #3


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    DC itself cannot radiate, but turning the DC on or off produces a rising or falling edge to the DC and this contains harmonics which can radiate if the antenna is suitable.
  5. Apr 29, 2010 #4
    Thank you for the reply.

    Will I observe a step function in the radiated field?

    I am puzzled that if I send a signal with energy at frequency=0, e.g. Gaussian pulse g(f) to an antenna and measure the radiate field, how would antenna handle the energy at frequency=0? will I observe the field being a Gaussian function or a differentiated Gaussian function?

    My feeling is that the radiated field will be Gaussian, since it is a wave being radiated.

    However, I cannot reconcile with the fact that if an antenna has a transfer function h(f) with 0 gain at 0 and unity everywhere else, then the linear model says the radiated signal is g(f)h(f). Inverse fourier transform of g(f)h(f) will give a Gaussian with is bias removed.
  6. Apr 29, 2010 #5


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    You will observe a radiated "splatt" of energy when the DC is turned on and when it is turned off. This will not be a step function, but a burst of radiation on many frequencies, depending on the shape and rise or fall time of the rising or falling edge.

    You can verify this. Just turn a flashlight on and off near a radio tuned between broadcast stations. You may hear a distinct "click" when the flashlight is turned on and another when it is turned off.

    Applying a DC voltage to an antenna that is open circuited to DC is not applying power to it. If there is nowhere for current to flow and it can't be radiated, the voltage will just stay there until the voltage source is removed. No current will flow, so there is no power being used.
  7. Apr 30, 2010 #6
    Dear vk6kro,

    Thank you for the reply. I understand your interpretation of the physics of the phenomenon at DC.

    I find it challenging to understand the mathematical equivalent of the problem.

    If I transmit a Gaussian pulse with a finite time-span, this Gaussian has some value at f=0 in the frequency domain. Could I interpret this value at f=0 corresponding some kind of DC?

  8. Apr 30, 2010 #7


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    There are loop antennae as well as electric dipole antennae, remember! Loads of current but no volts, this time.
  9. May 1, 2010 #8
    Well, I think Vk6kro says "open circuited to DC". So no current.

    On a second thought, I think the confusion comes from the physical meaning of DC and what Fourier transform gives you when f=0. When f=0, Fourier transform gives the algebraic mean of the signal.
  10. May 1, 2010 #9


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    My point is that there needs to be both current and volts at a frequency that you want to radiate power from. In a dipole or loop antenna (i.e. any antenna) there will only be volts OR current at DC.
    During a step change, there will be both current and volts and this will cause a pulse of power to radiate.
    If the rise time of the pulse is long then the LF component of the pulse will extend to a lower and lower frequency - but approaches zero at DC.
  11. Jan 19, 2011 #10
    The pulse radiated by an ultra-wideband antenna is the time-derivative of the pulse fed to the the antenna. The fed pulse may have DC content, e.g. a Gaussian pulse, but the radiated pulse has 0 DC content. See the reference below.

    author = {Werner Wiesbeck and Grzegorz Adamiuk and Christian Sturm},
    title = {Basic properties and design principles of {UWB} antennas},
    journal = IEEE_J_PROC
  12. Jan 19, 2011 #11
    I think you need to back up a few steps. DC implicitly means constant voltage or constant current. "Pulsed DC" isn't constant voltage so it isn't really direct current, and neither is a Gaussian voltage pulse. "Pulsed DC" has a DC component in it's spectrum, but a host of other frequencies, as noted earlier.

    Having said all that, the easiest analysis I can think for you is consider sending a Gaussian or square pulse voltage through a single turn loop and ask about the resulting electric and magnetic fields that develop. It's a simpler geometry than a dipole.

    As to your question,

    "Is it true? If I send a diferentiated gaussian signal, would a twice-differentiated gaussian (vicker's) signal being radiated?",

    This seems to be a mathematically obvious if you assume parameters such as permittivity and permeability are invariant with frequency and that the speed of light is large compared to your antenna size.
    Last edited: Jan 19, 2011
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