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Why is 50Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377Ω?
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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 50Ω, but the impedance of free space is 377Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
antenna impedance-matching high-frequency
asked 2 hours ago
ahemmetter
1186
 |Â
show 2 more comments
up vote
3
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 50Ω, but the impedance of free space is 377Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
antenna impedance-matching high-frequency
asked 2 hours ago
ahemmetter
1186
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
2 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
1 hour 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
1 hour ago
This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses?
– ahemmetter
1 hour 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
1 hour ago
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
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 50Ω, but the impedance of free space is 377Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
antenna impedance-matching high-frequency
asked 2 hours ago
ahemmetter
1186
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 50Ω, but the impedance of free space is 377Ω, won't there be a impedance mismatch and consequently a less-than-optimal radiation from the antenna?
antenna impedance-matching high-frequency
antenna impedance-matching high-frequency
asked 2 hours ago
ahemmetter
1186
asked 2 hours ago
ahemmetter
1186
edited 1 hour ago
asked 2 hours ago
ahemmetter
1186
asked 2 hours ago
ahemmetter
1186
asked 2 hours ago
ahemmetter
1186
1186
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
2 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
1 hour 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
1 hour ago
This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses?
– ahemmetter
1 hour 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
1 hour ago
 |Â
show 2 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
2 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
1 hour 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
1 hour ago
This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses?
– ahemmetter
1 hour 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
1 hour 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
2 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
2 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses?
– ahemmetter
1 hour ago
This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses?
– ahemmetter
1 hour 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
1 hour 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
1 hour ago
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
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 antenne 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), its 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.
answered 1 hour ago
Curd
12.1k2031
3
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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour ago
 |Â
show 2 more comments
up vote
0
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 25 mins ago
Chu
4,8382611
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
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 antenne 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), its 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.
answered 1 hour ago
Curd
12.1k2031
3
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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour ago
 |Â
show 2 more comments
up vote
2
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 antenne 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), its 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.
answered 1 hour ago
Curd
12.1k2031
3
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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour ago
 |Â
show 2 more comments
up vote
2
down vote
up vote
2
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 antenne 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), its 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.
answered 1 hour ago
Curd
12.1k2031
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 antenne 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), its 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.
answered 1 hour ago
Curd
12.1k2031
edited 1 hour ago
answered 1 hour ago
Curd
12.1k2031
answered 1 hour ago
Curd
12.1k2031
answered 1 hour ago
Curd
12.1k2031
12.1k2031
3
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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour ago
 |Â
show 2 more comments
3
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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour 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? There isn't an obvious 2:15 ratio there.
– Puffafish
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour ago
Yes, it's not trivial to see. Nobody said it would be. Maybe I can find a simple example... not within the next few hours however.
– Curd
1 hour ago
 |Â
show 2 more comments
up vote
0
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 25 mins ago
Chu
4,8382611
add a comment |Â
up vote
0
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 25 mins ago
Chu
4,8382611
add a comment |Â
up vote
0
down vote
up vote
0
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 25 mins ago
Chu
4,8382611
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 25 mins ago
Chu
4,8382611
edited 14 mins ago
answered 25 mins ago
Chu
4,8382611
answered 25 mins ago
Chu
4,8382611
answered 25 mins ago
Chu
4,8382611
4,8382611
add a comment |Â
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
2 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
1 hour 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
1 hour ago
This seems to be in contradiction to @Curd's answer. Does an antenna act as a impedance transformer (50 to 377Ω) or do we just chose 50Ω because the cables and transmitters are easier to build that way and we just accept the radiation losses?
– ahemmetter
1 hour 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
1 hour ago