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Why is 50 Î© often chosen as the input impedance of antennas, whereas the free space impedance is 377 Î©?
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In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched. Free space can be regarded as a further element, since a transmitting antenna eventually should radiate all power from the transmission line into it.
Now, if the impedances in the transmission line and in the antenna are matched at 50 Ω, but the impedance of free space is 377 Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
EDIT:
As far as I gathered from the answers, literature and discussions online, the antenna acts as a impedance transformer between the feed line and free space. The argument goes: no power from the feed line is reflected and must go to the antenna. The antenna can be assumed to be resonant and therefore radiates all its power into free space (disregarding heat losses etc). This means that there is no reflected power between antenna and free space, and the transition between antenna and free space is therefore matched.
The same should be true in the reverse direction for a receiving antenna (Reciprocity Principle): a wave in free space ($Z_0$) impinges onto an antenna, and the received power is fed into the transmission line (again through impedance transformation). At least in one paper (Devi et al., Design of a wideband 377 Ω E-shaped patch antenna for RF energy harvesting, Microwave and Optical Letters (2012) Vol. 54, No. 3, 10.1002/mop.26607) it was mentioned that a 377 Ω antenna with a separate circuit to match it to 50 Ω was used to "achieve a wide impedance bandwidth" with a high power level. If the antenna normally is already the impedance transformer, what is the matching circuit needed for then? Or alternatively, under what circumstances is the antenna not also the impedance transformer?
Some helpful sources and discussions I found:
- Klaus Kark, Antenne und Strahlungsfelder (in German)
- Impedance Matching (http://www.phys.ufl.edu/~majewski/nqr/reference2015/nqr_detection_educational/Impedance_matching_networks.pdf)
- Forum discussion that mentions impedance transformation for an inverted-F antenna (http://www.antenna-theory.com/phpbb2/viewtopic.php?t=776&sid=dede0d4127170d16cc3a583ab0929f3e)
- Some general notes about antennas 8http://fab.cba.mit.edu/classes/862.16/notes/antennas.pdf)
antenna impedance-matching high-frequency
edited 9 mins ago
Loong
1094
asked 10 hours ago
ahemmetter
1738
 |Â
show 4 more comments
up vote
14
down vote
favorite
In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched. Free space can be regarded as a further element, since a transmitting antenna eventually should radiate all power from the transmission line into it.
Now, if the impedances in the transmission line and in the antenna are matched at 50 Ω, but the impedance of free space is 377 Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
EDIT:
As far as I gathered from the answers, literature and discussions online, the antenna acts as a impedance transformer between the feed line and free space. The argument goes: no power from the feed line is reflected and must go to the antenna. The antenna can be assumed to be resonant and therefore radiates all its power into free space (disregarding heat losses etc). This means that there is no reflected power between antenna and free space, and the transition between antenna and free space is therefore matched.
The same should be true in the reverse direction for a receiving antenna (Reciprocity Principle): a wave in free space ($Z_0$) impinges onto an antenna, and the received power is fed into the transmission line (again through impedance transformation). At least in one paper (Devi et al., Design of a wideband 377 Ω E-shaped patch antenna for RF energy harvesting, Microwave and Optical Letters (2012) Vol. 54, No. 3, 10.1002/mop.26607) it was mentioned that a 377 Ω antenna with a separate circuit to match it to 50 Ω was used to "achieve a wide impedance bandwidth" with a high power level. If the antenna normally is already the impedance transformer, what is the matching circuit needed for then? Or alternatively, under what circumstances is the antenna not also the impedance transformer?
Some helpful sources and discussions I found:
- Klaus Kark, Antenne und Strahlungsfelder (in German)
- Impedance Matching (http://www.phys.ufl.edu/~majewski/nqr/reference2015/nqr_detection_educational/Impedance_matching_networks.pdf)
- Forum discussion that mentions impedance transformation for an inverted-F antenna (http://www.antenna-theory.com/phpbb2/viewtopic.php?t=776&sid=dede0d4127170d16cc3a583ab0929f3e)
- Some general notes about antennas 8http://fab.cba.mit.edu/classes/862.16/notes/antennas.pdf)
antenna impedance-matching high-frequency
edited 9 mins ago
Loong
1094
asked 10 hours ago
ahemmetter
1738
For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise
– PlasmaHH
10 hours ago
In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily.
– DonFusili
10 hours ago
3
"My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna".
– Bruce Abbott
10 hours ago
1
Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines.
– Curd
9 hours ago
1
@ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need.
– Curd
9 hours ago
 |Â
show 4 more comments
up vote
14
down vote
favorite
up vote
14
down vote
favorite
In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched. Free space can be regarded as a further element, since a transmitting antenna eventually should radiate all power from the transmission line into it.
Now, if the impedances in the transmission line and in the antenna are matched at 50 Ω, but the impedance of free space is 377 Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
EDIT:
As far as I gathered from the answers, literature and discussions online, the antenna acts as a impedance transformer between the feed line and free space. The argument goes: no power from the feed line is reflected and must go to the antenna. The antenna can be assumed to be resonant and therefore radiates all its power into free space (disregarding heat losses etc). This means that there is no reflected power between antenna and free space, and the transition between antenna and free space is therefore matched.
The same should be true in the reverse direction for a receiving antenna (Reciprocity Principle): a wave in free space ($Z_0$) impinges onto an antenna, and the received power is fed into the transmission line (again through impedance transformation). At least in one paper (Devi et al., Design of a wideband 377 Ω E-shaped patch antenna for RF energy harvesting, Microwave and Optical Letters (2012) Vol. 54, No. 3, 10.1002/mop.26607) it was mentioned that a 377 Ω antenna with a separate circuit to match it to 50 Ω was used to "achieve a wide impedance bandwidth" with a high power level. If the antenna normally is already the impedance transformer, what is the matching circuit needed for then? Or alternatively, under what circumstances is the antenna not also the impedance transformer?
Some helpful sources and discussions I found:
- Klaus Kark, Antenne und Strahlungsfelder (in German)
- Impedance Matching (http://www.phys.ufl.edu/~majewski/nqr/reference2015/nqr_detection_educational/Impedance_matching_networks.pdf)
- Forum discussion that mentions impedance transformation for an inverted-F antenna (http://www.antenna-theory.com/phpbb2/viewtopic.php?t=776&sid=dede0d4127170d16cc3a583ab0929f3e)
- Some general notes about antennas 8http://fab.cba.mit.edu/classes/862.16/notes/antennas.pdf)
antenna impedance-matching high-frequency
edited 9 mins ago
Loong
1094
asked 10 hours ago
ahemmetter
1738
In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched. Free space can be regarded as a further element, since a transmitting antenna eventually should radiate all power from the transmission line into it.
Now, if the impedances in the transmission line and in the antenna are matched at 50 Ω, but the impedance of free space is 377 Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
EDIT:
As far as I gathered from the answers, literature and discussions online, the antenna acts as a impedance transformer between the feed line and free space. The argument goes: no power from the feed line is reflected and must go to the antenna. The antenna can be assumed to be resonant and therefore radiates all its power into free space (disregarding heat losses etc). This means that there is no reflected power between antenna and free space, and the transition between antenna and free space is therefore matched.
The same should be true in the reverse direction for a receiving antenna (Reciprocity Principle): a wave in free space ($Z_0$) impinges onto an antenna, and the received power is fed into the transmission line (again through impedance transformation). At least in one paper (Devi et al., Design of a wideband 377 Ω E-shaped patch antenna for RF energy harvesting, Microwave and Optical Letters (2012) Vol. 54, No. 3, 10.1002/mop.26607) it was mentioned that a 377 Ω antenna with a separate circuit to match it to 50 Ω was used to "achieve a wide impedance bandwidth" with a high power level. If the antenna normally is already the impedance transformer, what is the matching circuit needed for then? Or alternatively, under what circumstances is the antenna not also the impedance transformer?
Some helpful sources and discussions I found:
- Klaus Kark, Antenne und Strahlungsfelder (in German)
- Impedance Matching (http://www.phys.ufl.edu/~majewski/nqr/reference2015/nqr_detection_educational/Impedance_matching_networks.pdf)
- Forum discussion that mentions impedance transformation for an inverted-F antenna (http://www.antenna-theory.com/phpbb2/viewtopic.php?t=776&sid=dede0d4127170d16cc3a583ab0929f3e)
- Some general notes about antennas 8http://fab.cba.mit.edu/classes/862.16/notes/antennas.pdf)
antenna impedance-matching high-frequency
antenna impedance-matching high-frequency
edited 9 mins ago
Loong
1094
asked 10 hours ago
ahemmetter
1738
edited 9 mins ago
Loong
1094
asked 10 hours ago
ahemmetter
1738
edited 9 mins ago
Loong
1094
edited 9 mins ago
Loong
1094
edited 9 mins ago
Loong
1094
1094
asked 10 hours ago
ahemmetter
1738
asked 10 hours ago
ahemmetter
1738
asked 10 hours ago
ahemmetter
1738
1738
For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise
– PlasmaHH
10 hours ago
In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily.
– DonFusili
10 hours ago
3
"My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna".
– Bruce Abbott
10 hours ago
1
Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines.
– Curd
9 hours ago
1
@ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need.
– Curd
9 hours ago
 |Â
show 4 more comments
For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise
– PlasmaHH
10 hours ago
In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily.
– DonFusili
10 hours ago
3
"My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna".
– Bruce Abbott
10 hours ago
1
Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines.
– Curd
9 hours ago
1
@ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need.
– Curd
9 hours ago
For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise
– PlasmaHH
10 hours ago
For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise
– PlasmaHH
10 hours ago
In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily.
– DonFusili
10 hours ago
In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily.
– DonFusili
10 hours ago
3
3
"My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna".
– Bruce Abbott
10 hours ago
"My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna".
– Bruce Abbott
10 hours ago
1
1
Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines.
– Curd
9 hours ago
Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines.
– Curd
9 hours ago
1
1
@ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need.
– Curd
9 hours ago
@ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need.
– Curd
9 hours ago
 |Â
show 4 more comments
3 Answers
3
active
oldest
votes
up vote
9
down vote
The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance.
Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side).
The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed.
So the antenna does match the impedance of free space (on one side); and ideally also the impedance of the circuit (on the other side).
Since the space side's impedance is always the same (for all kinds of antennas operated in vacuum or air), it doesn't need to be mentioned.
Only the wire side is what you need and can care about.
The reason 50Ω or 75Ω or 300Ω or ... is choosen as antenna impedances is because of practical reasons to construct particular antennas/transmission lines/amplifiers with that impedance.
A possible ansatz for calculating the radiation resistance $R$ of an antenna is:
Find an answer to the question:
"How much power $P$ (average over one period) is radiated if a sinusoidal signal of given voltage (or current) amplitude $V_0$ (or $I_0$) is applied to the antenna?"
Then you get
$R = fracV_0^22P$ (or $=frac2PI_0^2$)
You get radiated power $P$ by integrating the Poynting vector $mathbfS$ (=radiated power per area) over the sphere enclosing the antenna.
The Poynting vector is $mathbfS = frac1mu_0 mathbfE times mathbfB$ where $mathbfE$ and $mathbfB$ are electric/magnetic fields caused by the voltages and currents in your antenna.
You can find an example for such a calculation in the Wikipedia acticle about "Dipole antenna", in paragraph Short Dipole.
answered 10 hours ago
Curd
12.2k2031
5
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
1
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
1
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
1
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
 |Â
show 4 more comments
up vote
2
down vote
The radiation resistance, $small R_r$, of a half-wave dipole is $small 73Omega$. This relates directly to the feedpoint impedance, i.e. this is the impedance presented to the transmission line by the antenna at the design frequency.
$small R_r$ is related to the impedance of free space (i.e. the impedance seen by an E-M wave travelling in free-space), but is not equal to it.
answered 9 hours ago
Chu
4,8762611
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
add a comment |Â
up vote
2
down vote
"...In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched...."
This is your assumption. And it is correct, but not in the case of antennas.
Because in antennas, we have "reflection". Power applied to the feed point (in a dipole, for example) travels down to the end of the wire, and is reflected back to the feed point, where (if resonant) it will meet a voltage or current 180 degrees out of phase, thus canceling, and represented by the (so-called) standing wave.
So, the applied power bounces back and forth in the antenna wire until all is radiated or lost as heat. So it does not matter if the antenna impedance is different than free space. What really matters, practically speaking, if the energy is reflected back into the transmitter and warms the final amp device, thus wasting the power/energy appliled. This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna). But once the antenna system is matched to the transmitter, almost all the energy will be transmitted to free space (except for resistance in the wire, which is usually negligible. Or so I am told.
answered 2 hours ago
Baruch Atta
411
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance.
Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side).
The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed.
So the antenna does match the impedance of free space (on one side); and ideally also the impedance of the circuit (on the other side).
Since the space side's impedance is always the same (for all kinds of antennas operated in vacuum or air), it doesn't need to be mentioned.
Only the wire side is what you need and can care about.
The reason 50Ω or 75Ω or 300Ω or ... is choosen as antenna impedances is because of practical reasons to construct particular antennas/transmission lines/amplifiers with that impedance.
A possible ansatz for calculating the radiation resistance $R$ of an antenna is:
Find an answer to the question:
"How much power $P$ (average over one period) is radiated if a sinusoidal signal of given voltage (or current) amplitude $V_0$ (or $I_0$) is applied to the antenna?"
Then you get
$R = fracV_0^22P$ (or $=frac2PI_0^2$)
You get radiated power $P$ by integrating the Poynting vector $mathbfS$ (=radiated power per area) over the sphere enclosing the antenna.
The Poynting vector is $mathbfS = frac1mu_0 mathbfE times mathbfB$ where $mathbfE$ and $mathbfB$ are electric/magnetic fields caused by the voltages and currents in your antenna.
You can find an example for such a calculation in the Wikipedia acticle about "Dipole antenna", in paragraph Short Dipole.
answered 10 hours ago
Curd
12.2k2031
5
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
1
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
1
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
1
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
 |Â
show 4 more comments
up vote
9
down vote
The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance.
Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side).
The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed.
So the antenna does match the impedance of free space (on one side); and ideally also the impedance of the circuit (on the other side).
Since the space side's impedance is always the same (for all kinds of antennas operated in vacuum or air), it doesn't need to be mentioned.
Only the wire side is what you need and can care about.
The reason 50Ω or 75Ω or 300Ω or ... is choosen as antenna impedances is because of practical reasons to construct particular antennas/transmission lines/amplifiers with that impedance.
A possible ansatz for calculating the radiation resistance $R$ of an antenna is:
Find an answer to the question:
"How much power $P$ (average over one period) is radiated if a sinusoidal signal of given voltage (or current) amplitude $V_0$ (or $I_0$) is applied to the antenna?"
Then you get
$R = fracV_0^22P$ (or $=frac2PI_0^2$)
You get radiated power $P$ by integrating the Poynting vector $mathbfS$ (=radiated power per area) over the sphere enclosing the antenna.
The Poynting vector is $mathbfS = frac1mu_0 mathbfE times mathbfB$ where $mathbfE$ and $mathbfB$ are electric/magnetic fields caused by the voltages and currents in your antenna.
You can find an example for such a calculation in the Wikipedia acticle about "Dipole antenna", in paragraph Short Dipole.
answered 10 hours ago
Curd
12.2k2031
5
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
1
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
1
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
1
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
 |Â
show 4 more comments
up vote
9
down vote
up vote
9
down vote
The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance.
Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side).
The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed.
So the antenna does match the impedance of free space (on one side); and ideally also the impedance of the circuit (on the other side).
Since the space side's impedance is always the same (for all kinds of antennas operated in vacuum or air), it doesn't need to be mentioned.
Only the wire side is what you need and can care about.
The reason 50Ω or 75Ω or 300Ω or ... is choosen as antenna impedances is because of practical reasons to construct particular antennas/transmission lines/amplifiers with that impedance.
A possible ansatz for calculating the radiation resistance $R$ of an antenna is:
Find an answer to the question:
"How much power $P$ (average over one period) is radiated if a sinusoidal signal of given voltage (or current) amplitude $V_0$ (or $I_0$) is applied to the antenna?"
Then you get
$R = fracV_0^22P$ (or $=frac2PI_0^2$)
You get radiated power $P$ by integrating the Poynting vector $mathbfS$ (=radiated power per area) over the sphere enclosing the antenna.
The Poynting vector is $mathbfS = frac1mu_0 mathbfE times mathbfB$ where $mathbfE$ and $mathbfB$ are electric/magnetic fields caused by the voltages and currents in your antenna.
You can find an example for such a calculation in the Wikipedia acticle about "Dipole antenna", in paragraph Short Dipole.
answered 10 hours ago
Curd
12.2k2031
The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance.
Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side).
The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed.
So the antenna does match the impedance of free space (on one side); and ideally also the impedance of the circuit (on the other side).
Since the space side's impedance is always the same (for all kinds of antennas operated in vacuum or air), it doesn't need to be mentioned.
Only the wire side is what you need and can care about.
The reason 50Ω or 75Ω or 300Ω or ... is choosen as antenna impedances is because of practical reasons to construct particular antennas/transmission lines/amplifiers with that impedance.
A possible ansatz for calculating the radiation resistance $R$ of an antenna is:
Find an answer to the question:
"How much power $P$ (average over one period) is radiated if a sinusoidal signal of given voltage (or current) amplitude $V_0$ (or $I_0$) is applied to the antenna?"
Then you get
$R = fracV_0^22P$ (or $=frac2PI_0^2$)
You get radiated power $P$ by integrating the Poynting vector $mathbfS$ (=radiated power per area) over the sphere enclosing the antenna.
The Poynting vector is $mathbfS = frac1mu_0 mathbfE times mathbfB$ where $mathbfE$ and $mathbfB$ are electric/magnetic fields caused by the voltages and currents in your antenna.
You can find an example for such a calculation in the Wikipedia acticle about "Dipole antenna", in paragraph Short Dipole.
answered 10 hours ago
Curd
12.2k2031
edited 1 hour ago
answered 10 hours ago
Curd
12.2k2031
answered 10 hours ago
Curd
12.2k2031
answered 10 hours ago
Curd
12.2k2031
12.2k2031
5
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
1
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
1
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
1
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
 |Â
show 4 more comments
5
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
1
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
1
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
1
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
5
5
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377? There isn't an obvious 2:15 ratio there.
– Puffafish
10 hours ago
1
1
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
"Just" apply Maxwell's equations to your antenna geometry and you will find, that it will turn out that it does (not exactly but about). Your expectation to immediately "see" the 50/377 ratio in wire or wave length ratios is not justified; but you will get the result if you do the integrations etc.
– Curd
10 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
I'd appreciate to see at least the ansatz to this calculation for a dipole antenna. It's not trivial to see where the impedances appear in such a case and how Maxwell's equations can be applied to find them. Would it be possible to add this to the answer or link to an article explaining it?
– ahemmetter
9 hours ago
1
1
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
At best you are arguing that the feedpoint impedance is what it is because that is what works. That's not an answer. An answer would explain why the feedpoint impedance is what it is. And no, it isn't too match the feedline, if anything the other way around, the feedline is designed with the antenna impedance as one of the goals.
– Chris Stratton
9 hours ago
1
1
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
Thanks for adding the ansatz. So, to clarify: the input impedance (especially the radiation resistance $R$) is the impedance 'seen' by the transmission line, whereas power radiated into free space depends on the free space impedance in the Poynting vector $S = fracE^2Z_0$. And the antenna just transforms between both impedances. Is that more or less correct?
– ahemmetter
4 hours ago
 |Â
show 4 more comments
up vote
2
down vote
The radiation resistance, $small R_r$, of a half-wave dipole is $small 73Omega$. This relates directly to the feedpoint impedance, i.e. this is the impedance presented to the transmission line by the antenna at the design frequency.
$small R_r$ is related to the impedance of free space (i.e. the impedance seen by an E-M wave travelling in free-space), but is not equal to it.
answered 9 hours ago
Chu
4,8762611
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
add a comment |Â
up vote
2
down vote
The radiation resistance, $small R_r$, of a half-wave dipole is $small 73Omega$. This relates directly to the feedpoint impedance, i.e. this is the impedance presented to the transmission line by the antenna at the design frequency.
$small R_r$ is related to the impedance of free space (i.e. the impedance seen by an E-M wave travelling in free-space), but is not equal to it.
answered 9 hours ago
Chu
4,8762611
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The radiation resistance, $small R_r$, of a half-wave dipole is $small 73Omega$. This relates directly to the feedpoint impedance, i.e. this is the impedance presented to the transmission line by the antenna at the design frequency.
$small R_r$ is related to the impedance of free space (i.e. the impedance seen by an E-M wave travelling in free-space), but is not equal to it.
answered 9 hours ago
Chu
4,8762611
The radiation resistance, $small R_r$, of a half-wave dipole is $small 73Omega$. This relates directly to the feedpoint impedance, i.e. this is the impedance presented to the transmission line by the antenna at the design frequency.
$small R_r$ is related to the impedance of free space (i.e. the impedance seen by an E-M wave travelling in free-space), but is not equal to it.
answered 9 hours ago
Chu
4,8762611
edited 9 hours ago
answered 9 hours ago
Chu
4,8762611
answered 9 hours ago
Chu
4,8762611
answered 9 hours ago
Chu
4,8762611
4,8762611
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
add a comment |Â
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
That's the point though: how is the radiation resistance related to the free space impedance? Alternatively, can the antenna be changed so that it is matched to the feed line but doesn't radiate its power into free space (and is lost as heat instead)?
– ahemmetter
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
@ahemmeter a non radiating antenna is called a dummy load. Typically it is constructed of a resistor, at larger power capacities with careful measures to achieve cooling and manage the impedance across geometry of the element so that the SWR remains close to ideal even at higher frequencies. You can of course add resistors in series or parallel with a real antenna, but you would probably not want to.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
What this answer is missing is a statement of why the feedpoint impedance of a dipole is what it is.
– Chris Stratton
4 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
@ChrisStratton Ah, I completely forgot about the dummy load, right. So this would be an example of something that is matched to the input but not to free space anymore, since it doesn't transform any impedances.
– ahemmetter
3 hours ago
add a comment |Â
up vote
2
down vote
"...In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched...."
This is your assumption. And it is correct, but not in the case of antennas.
Because in antennas, we have "reflection". Power applied to the feed point (in a dipole, for example) travels down to the end of the wire, and is reflected back to the feed point, where (if resonant) it will meet a voltage or current 180 degrees out of phase, thus canceling, and represented by the (so-called) standing wave.
So, the applied power bounces back and forth in the antenna wire until all is radiated or lost as heat. So it does not matter if the antenna impedance is different than free space. What really matters, practically speaking, if the energy is reflected back into the transmitter and warms the final amp device, thus wasting the power/energy appliled. This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna). But once the antenna system is matched to the transmitter, almost all the energy will be transmitted to free space (except for resistance in the wire, which is usually negligible. Or so I am told.
answered 2 hours ago
Baruch Atta
411
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
add a comment |Â
up vote
2
down vote
"...In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched...."
This is your assumption. And it is correct, but not in the case of antennas.
Because in antennas, we have "reflection". Power applied to the feed point (in a dipole, for example) travels down to the end of the wire, and is reflected back to the feed point, where (if resonant) it will meet a voltage or current 180 degrees out of phase, thus canceling, and represented by the (so-called) standing wave.
So, the applied power bounces back and forth in the antenna wire until all is radiated or lost as heat. So it does not matter if the antenna impedance is different than free space. What really matters, practically speaking, if the energy is reflected back into the transmitter and warms the final amp device, thus wasting the power/energy appliled. This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna). But once the antenna system is matched to the transmitter, almost all the energy will be transmitted to free space (except for resistance in the wire, which is usually negligible. Or so I am told.
answered 2 hours ago
Baruch Atta
411
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
"...In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched...."
This is your assumption. And it is correct, but not in the case of antennas.
Because in antennas, we have "reflection". Power applied to the feed point (in a dipole, for example) travels down to the end of the wire, and is reflected back to the feed point, where (if resonant) it will meet a voltage or current 180 degrees out of phase, thus canceling, and represented by the (so-called) standing wave.
So, the applied power bounces back and forth in the antenna wire until all is radiated or lost as heat. So it does not matter if the antenna impedance is different than free space. What really matters, practically speaking, if the energy is reflected back into the transmitter and warms the final amp device, thus wasting the power/energy appliled. This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna). But once the antenna system is matched to the transmitter, almost all the energy will be transmitted to free space (except for resistance in the wire, which is usually negligible. Or so I am told.
answered 2 hours ago
Baruch Atta
411
"...In order to efficiently deliver power to a different part of a circuit without reflection, the impedances of all circuit elements need to be matched...."
This is your assumption. And it is correct, but not in the case of antennas.
Because in antennas, we have "reflection". Power applied to the feed point (in a dipole, for example) travels down to the end of the wire, and is reflected back to the feed point, where (if resonant) it will meet a voltage or current 180 degrees out of phase, thus canceling, and represented by the (so-called) standing wave.
So, the applied power bounces back and forth in the antenna wire until all is radiated or lost as heat. So it does not matter if the antenna impedance is different than free space. What really matters, practically speaking, if the energy is reflected back into the transmitter and warms the final amp device, thus wasting the power/energy appliled. This happens when the impedance of the final amp does not match the antenna system (transmission line plus antenna). But once the antenna system is matched to the transmitter, almost all the energy will be transmitted to free space (except for resistance in the wire, which is usually negligible. Or so I am told.
answered 2 hours ago
Baruch Atta
411
answered 2 hours ago
Baruch Atta
411
answered 2 hours ago
Baruch Atta
411
answered 2 hours ago
Baruch Atta
411
411
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
add a comment |Â
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
That's an interesting thought! How does it account for the fact that the same antenna in different surrounding media (different $Z_0$) behaves differently? As in optics, there is still an interface that creates some kind of reflection if the impedances of both media are not equal. And it seems to me that the constructive interference (standing wave) is only determined by the properties of the antenna: material and length.
– ahemmetter
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
ahemmetter: that's also a good question - and my thought is to consider a Yagi antenna - the driven element has power applied, but the E fields affect the reflector and director elements and affect the total impedance and radiation pattern.
– Baruch Atta
2 hours ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
Hm, in a Yagi antenna the different induced waves from the passive elements are just superimposed in the far field, but not in the active part of the antenna itself. They change the radiation pattern no doubt, but is the output impedance also different?
– ahemmetter
1 hour ago
add a comment |Â
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For TV I see more often 75Ω and you need to consider the impedance of the feedline, and then you look up where the best power transfer lies (wikipedia has a chart) and other parameters and then you find a compromise
– PlasmaHH
10 hours ago
In short: 50 ohms is nice compromise between power transmission towards the antenna and dielectric losses inside cables we can make easily. It's nice to be able to make stuff easily.
– DonFusili
10 hours ago
3
"My question to this is: how does a single wire, (1/4 or 1/2 wavelength long) convert form 50 to 377?" - you mean how does the antenna transform from 50 to 377 Ohms? If that is what you want to know then it should be in your question. Otherwise the answer is simply "because that is the impedance of that type of antenna".
– Bruce Abbott
10 hours ago
1
Both is true. That's no contradiction. Anennas act as transmores and you can build them in ways to transform to high or low impedance depending on the antenna design. The same is true for amplifiers or transmission lines.
– Curd
9 hours ago
1
@ahemmetter: ...because it is just a transmission line. It simply does not have the special property of antennas: efficiently transmitting energy to/picking energy up from space. Just matching impedance is not all you need.
– Curd
9 hours ago