Is it possible that a radio wave could be direct current only? I know that dc fields exist , but could a radio be recitifed at it's output to produce a signal with no negative or inverting component to the em wave?
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NO.Is it possible that a radio wave could be direct current only?
Hi,I've spent quite a few months experimenting with small antennas connected to the output pins of microcontrollers to transmit the edges of pulses up to several meters. These were strings of pulses at rates ranging up to a few tens of kHz. They were pulsating DC when then went into the antenna, but they were AC when the arrived at the receiver because the coupling was capacitive and the DC component was lost. You can transmit "almost DC".
"Pulsating DC" is not DC. The actual spectrum of "pulsating DC" is humongous.I've spent quite a few months experimenting with small antennas connected to the output pins of microcontrollers to transmit the edges of pulses up to several meters. These were strings of pulses at rates ranging up to a few tens of kHz. They were pulsating DC when then went into the antenna, but they were AC when the arrived at the receiver because the coupling was capacitive and the DC component was lost. You can transmit "almost DC".
'Nuff said!
It would seem that if radiated waves depended on acceleration of electrons traveling as a continuous (non-pulsating) current in a circular piece of wire that drift velocity would be important.@MrAl said:Here is something to think about though...
If we have a round (circle) antenna 1 meter in diameter and pump a 1 amp DC current though it, do we get radiation?
The electrons are traveling at constant speed, however, they are also turning constantly and that means that they have angular acceleration.
My guess is that there could be radiation however it is probably very tiny because of the size of the electrons relative to the diameter of the antenna.
Very good points about using wire. With accelerated charged in vacuum the effect of Cyclotron radiation is detectable down to one electron.It would seem that if radiated waves depended on acceleration of electrons traveling as a continuous (non-pulsating) current in a circular piece of wire that drift velocity would be important.
Unfortunately, that velocity is quite slow (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html) compared to reversals caused by AC of a few kHz (or sharp edges).
Using that calculator for drift velocity, 10A in 24 awg (0.81 mm dia) copper wire has a drift velocity of 1.425 ^10-3 m/s. A ring with an effective path length* of 1.425 mm and would produce 2 reversals/loop, i.e., each second. Acceleration (v^2/r) = 5 mm/sec^2. That would require a very small loop for anything that was not mHz.**
*I am assuming the mean path that should be used would be something like the magnetic path length of a toroid and not the simple average of OD and ID x π. The formula for magnetic path length is here: http://www.mhw-intl.com/assets/CSC/CSC Design Formulas 2011.pdf However, regardless of the model, the path would be short.
**Assume the wire is 0.8 mm diameter and one can make a loop with an OD of 2.4 mm and ID of 0.8 mm. (I am ignoring deformation of the wire as it is formed.) The mean path is 4.58 mm. A 1.425 mm mean path would require an ID on the order of 45 microns.
Yes and we see why DC is not used for RF transmissionIt would seem that if radiated waves depended on acceleration of electrons traveling as a continuous (non-pulsating) current in a circular piece of wire that drift velocity would be important.
Unfortunately, that velocity is quite slow (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html) compared to reversals caused by AC of a few kHz (or sharp edges).
Using that calculator for drift velocity, 10A in 24 awg (0.81 mm dia) copper wire has a drift velocity of 1.425 ^10-3 m/s. A ring with an effective path length* of 1.425 mm and would produce 2 reversals/loop, i.e., each second. Acceleration (v^2/r) = 5 mm/sec^2. That would require a very small loop for anything that was not mHz.**
*I am assuming the mean path that should be used would be something like the magnetic path length of a toroid and not the simple average of OD and ID x π. The formula for magnetic path length is here: http://www.mhw-intl.com/assets/CSC/CSC Design Formulas 2011.pdf However, regardless of the model, the path would be short.
**Assume the wire is 0.8 mm diameter and one can make a loop with an OD of 2.4 mm and ID of 0.8 mm. (I am ignoring deformation of the wire as it is formed.) The mean path is 4.58 mm. A 1.425 mm mean path would require an ID on the order of 45 microns.