Mixing
The right antenna size
Clash Royale CLAN TAG#URR8PPP
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I'm preparing for the USA Technician exam and I was reading in the manual that the ideal size for a dipole is following this strange formula: 1/2-lambda. But based on what I read after, it seems it means 1/2*lambda. Is it right ? And does it mean that each dipole spoke is 1/4*lambda ? Then that would mean you can only emit frequencies that are harmonics to that lambda ?
Thanks,
Nicolas
antenna dipole math
asked 22 hours ago
Nicolas Dufour
113
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Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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down vote
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I'm preparing for the USA Technician exam and I was reading in the manual that the ideal size for a dipole is following this strange formula: 1/2-lambda. But based on what I read after, it seems it means 1/2*lambda. Is it right ? And does it mean that each dipole spoke is 1/4*lambda ? Then that would mean you can only emit frequencies that are harmonics to that lambda ?
Thanks,
Nicolas
antenna dipole math
asked 22 hours ago
Nicolas Dufour
113
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Welcome to Amateur Radio stackexchange.
– mike65535
22 hours ago
Thank you! Lots to learn and experiment!
– Nicolas Dufour
22 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm preparing for the USA Technician exam and I was reading in the manual that the ideal size for a dipole is following this strange formula: 1/2-lambda. But based on what I read after, it seems it means 1/2*lambda. Is it right ? And does it mean that each dipole spoke is 1/4*lambda ? Then that would mean you can only emit frequencies that are harmonics to that lambda ?
Thanks,
Nicolas
antenna dipole math
asked 22 hours ago
Nicolas Dufour
113
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm preparing for the USA Technician exam and I was reading in the manual that the ideal size for a dipole is following this strange formula: 1/2-lambda. But based on what I read after, it seems it means 1/2*lambda. Is it right ? And does it mean that each dipole spoke is 1/4*lambda ? Then that would mean you can only emit frequencies that are harmonics to that lambda ?
Thanks,
Nicolas
antenna dipole math
antenna dipole math
asked 22 hours ago
Nicolas Dufour
113
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 22 hours ago
Nicolas Dufour
113
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 49 mins ago
asked 22 hours ago
Nicolas Dufour
113
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 22 hours ago
Nicolas Dufour
113
asked 22 hours ago
Nicolas Dufour
113
113
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nicolas Dufour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Welcome to Amateur Radio stackexchange.
– mike65535
22 hours ago
Thank you! Lots to learn and experiment!
– Nicolas Dufour
22 hours ago
add a comment |Â
3
Welcome to Amateur Radio stackexchange.
– mike65535
22 hours ago
Thank you! Lots to learn and experiment!
– Nicolas Dufour
22 hours ago
3
3
Welcome to Amateur Radio stackexchange.
– mike65535
22 hours ago
Welcome to Amateur Radio stackexchange.
– mike65535
22 hours ago
Thank you! Lots to learn and experiment!
– Nicolas Dufour
22 hours ago
Thank you! Lots to learn and experiment!
– Nicolas Dufour
22 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
The classic dipole is a half-wave antenna. This means that the total length of the antenna is lambda/2. So writing it as 1/2-lambda is OK from an English language point of view, but not IMO as a rigorous mathematical formula.
For a half-wave dipole, each side of the feedpoint is one-half of that or a quarter-wave.
answered 22 hours ago
mike65535
419216
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
add a comment |Â
up vote
3
down vote
If you make a dipole exactly a half-wavelength long, then it will be too long and out of resonance.
The formula for determining the length of a half-wavelength dipole in feet is 468÷frequency in MHz. That's shorter than the actual wavelength in free space which is 492÷MHz.
answered 21 hours ago
Mike Waters♦
2,5522531
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
1
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
2
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
 |Â
show 1 more comment
up vote
1
down vote
There are already a couple of nice answers to your question. I thought I would add a little more context - no need to vote for this as it is tangential to the question.
A 1/2 wavelength (1/2 $lambda$) long dipole is one of the most basic and easy to deploy antennas. It is popular because when it is cut and fed in the center, it very closely matches the impedance of the most popular transmission line - 50 ohm coax. This helps to minimize power lost in the coax as a result of a mismatch. Such a mismatch is also the cause of elevated SWR (standing wave ratio).
The wire on each side of the feed point of a dipole is called a "leg". In the case of a 1/2 $lambda$, center fed dipole each leg is 1/4 $lambda$.
The length of a dipole is not limited to 1/2 $lambda$. If the dipole becomes significantly shorter than 1/2 $lambda$, the mismatch to the transmission line becomes high and so the losses in the transmission line go up. The antenna also becomes inefficient in itself which results in more power converted to heat and less radiated as electromagnetic energy.
A dipole longer than 1/2 $lambda$ can also be used. In fact, a dipole that is 10/8 $lambda$ will actually have more gain than a 1/2 $lambda$ dipole. The disadvantage is that it takes a little more work to make it a match for 50 ohm coax. Dipoles longer than 10/8 $lambda$ can be used as well but they will generally not offer any more gain.
In recent years, there has been a great deal of interest in feeding a dipole at a spot other than its center. So called OCF (off center fed) dipoles offer the possibility of a multi-band dipole with reasonable matching to 50 ohm coax. Typically it is a 1/2 $lambda$ or slightly longer dipole for the lowest frequency of interest and then at higher frequencies, it simply acts as a >1/2 $lambda$ dipole.
Other dipoles get specific names due to their historical references such as the double Zepp (from its Zeplin origins), extended double Zepp, a doublet, the G5RV, etc. Some of these designations may also imply the use of an open wire feedline instead of coax. But in the end, they are simply another form of a dipole.
Then there is the so called "End Fed Half Wave" (EFHW) antenna. This is a piece of wire that is the length of a 1/2 $lambda$ dipole but instead of being split somewhere to create a feed point, it is left as one solid piece of wire and the coax is attached to the end. This is a very misunderstood antenna by many hams. It takes careful construction and installation techniques to achieve performance comparable to a center fed 1/2 $lambda$ antenna. Many of these antennas are used in a multiband configuration. Their popularity is due to the convenience of feeding it from one end.
answered 7 hours ago
Glenn W9IQ
11.6k1738
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The classic dipole is a half-wave antenna. This means that the total length of the antenna is lambda/2. So writing it as 1/2-lambda is OK from an English language point of view, but not IMO as a rigorous mathematical formula.
For a half-wave dipole, each side of the feedpoint is one-half of that or a quarter-wave.
answered 22 hours ago
mike65535
419216
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
add a comment |Â
up vote
4
down vote
The classic dipole is a half-wave antenna. This means that the total length of the antenna is lambda/2. So writing it as 1/2-lambda is OK from an English language point of view, but not IMO as a rigorous mathematical formula.
For a half-wave dipole, each side of the feedpoint is one-half of that or a quarter-wave.
answered 22 hours ago
mike65535
419216
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The classic dipole is a half-wave antenna. This means that the total length of the antenna is lambda/2. So writing it as 1/2-lambda is OK from an English language point of view, but not IMO as a rigorous mathematical formula.
For a half-wave dipole, each side of the feedpoint is one-half of that or a quarter-wave.
answered 22 hours ago
mike65535
419216
The classic dipole is a half-wave antenna. This means that the total length of the antenna is lambda/2. So writing it as 1/2-lambda is OK from an English language point of view, but not IMO as a rigorous mathematical formula.
For a half-wave dipole, each side of the feedpoint is one-half of that or a quarter-wave.
answered 22 hours ago
mike65535
419216
edited 19 hours ago
answered 22 hours ago
mike65535
419216
answered 22 hours ago
mike65535
419216
answered 22 hours ago
mike65535
419216
419216
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
add a comment |Â
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
Oh I see! It's a notation. Thanks!
– Nicolas Dufour
22 hours ago
add a comment |Â
up vote
3
down vote
If you make a dipole exactly a half-wavelength long, then it will be too long and out of resonance.
The formula for determining the length of a half-wavelength dipole in feet is 468÷frequency in MHz. That's shorter than the actual wavelength in free space which is 492÷MHz.
answered 21 hours ago
Mike Waters♦
2,5522531
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
1
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
2
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
 |Â
show 1 more comment
up vote
3
down vote
If you make a dipole exactly a half-wavelength long, then it will be too long and out of resonance.
The formula for determining the length of a half-wavelength dipole in feet is 468÷frequency in MHz. That's shorter than the actual wavelength in free space which is 492÷MHz.
answered 21 hours ago
Mike Waters♦
2,5522531
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
1
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
2
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
 |Â
show 1 more comment
up vote
3
down vote
up vote
3
down vote
If you make a dipole exactly a half-wavelength long, then it will be too long and out of resonance.
The formula for determining the length of a half-wavelength dipole in feet is 468÷frequency in MHz. That's shorter than the actual wavelength in free space which is 492÷MHz.
answered 21 hours ago
Mike Waters♦
2,5522531
If you make a dipole exactly a half-wavelength long, then it will be too long and out of resonance.
The formula for determining the length of a half-wavelength dipole in feet is 468÷frequency in MHz. That's shorter than the actual wavelength in free space which is 492÷MHz.
answered 21 hours ago
Mike Waters♦
2,5522531
answered 21 hours ago
Mike Waters♦
2,5522531
answered 21 hours ago
Mike Waters♦
2,5522531
answered 21 hours ago
Mike Waters♦
2,5522531
2,5522531
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
1
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
2
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
 |Â
show 1 more comment
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
1
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
2
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Ah I see, is it related to the 5/8 of lambda ?
– Nicolas Dufour
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
Or divide frequency by 150 to get the half-wave dipole length in metres (or divide by 1.5 to get it in centimetres).
– Jim MacKenzie VE5EV
20 hours ago
1
1
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
Ah I see, is it related to the 5/8 of lambda Not at all. The shortening of the actual antenna is due to the practical nature of its construction. kb6nu.com/dipoles-shorter-half-wavelength
– mike65535
19 hours ago
2
2
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
A 1/2 wave dipole is not a resonant antenna as Mike said. We shorten the antenna slightly from a 1/2λ length in order to make it resonant. There is also a small amount of capacitance on the ends of each wire that affects the tuning.
– Glenn W9IQ
16 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
The metric formula for wavelength = 300 / f (where f is in MHz, result in meters)
– Edwin van Mierlo
6 hours ago
 |Â
show 1 more comment
up vote
1
down vote
There are already a couple of nice answers to your question. I thought I would add a little more context - no need to vote for this as it is tangential to the question.
A 1/2 wavelength (1/2 $lambda$) long dipole is one of the most basic and easy to deploy antennas. It is popular because when it is cut and fed in the center, it very closely matches the impedance of the most popular transmission line - 50 ohm coax. This helps to minimize power lost in the coax as a result of a mismatch. Such a mismatch is also the cause of elevated SWR (standing wave ratio).
The wire on each side of the feed point of a dipole is called a "leg". In the case of a 1/2 $lambda$, center fed dipole each leg is 1/4 $lambda$.
The length of a dipole is not limited to 1/2 $lambda$. If the dipole becomes significantly shorter than 1/2 $lambda$, the mismatch to the transmission line becomes high and so the losses in the transmission line go up. The antenna also becomes inefficient in itself which results in more power converted to heat and less radiated as electromagnetic energy.
A dipole longer than 1/2 $lambda$ can also be used. In fact, a dipole that is 10/8 $lambda$ will actually have more gain than a 1/2 $lambda$ dipole. The disadvantage is that it takes a little more work to make it a match for 50 ohm coax. Dipoles longer than 10/8 $lambda$ can be used as well but they will generally not offer any more gain.
In recent years, there has been a great deal of interest in feeding a dipole at a spot other than its center. So called OCF (off center fed) dipoles offer the possibility of a multi-band dipole with reasonable matching to 50 ohm coax. Typically it is a 1/2 $lambda$ or slightly longer dipole for the lowest frequency of interest and then at higher frequencies, it simply acts as a >1/2 $lambda$ dipole.
Other dipoles get specific names due to their historical references such as the double Zepp (from its Zeplin origins), extended double Zepp, a doublet, the G5RV, etc. Some of these designations may also imply the use of an open wire feedline instead of coax. But in the end, they are simply another form of a dipole.
Then there is the so called "End Fed Half Wave" (EFHW) antenna. This is a piece of wire that is the length of a 1/2 $lambda$ dipole but instead of being split somewhere to create a feed point, it is left as one solid piece of wire and the coax is attached to the end. This is a very misunderstood antenna by many hams. It takes careful construction and installation techniques to achieve performance comparable to a center fed 1/2 $lambda$ antenna. Many of these antennas are used in a multiband configuration. Their popularity is due to the convenience of feeding it from one end.
answered 7 hours ago
Glenn W9IQ
11.6k1738
add a comment |Â
up vote
1
down vote
There are already a couple of nice answers to your question. I thought I would add a little more context - no need to vote for this as it is tangential to the question.
A 1/2 wavelength (1/2 $lambda$) long dipole is one of the most basic and easy to deploy antennas. It is popular because when it is cut and fed in the center, it very closely matches the impedance of the most popular transmission line - 50 ohm coax. This helps to minimize power lost in the coax as a result of a mismatch. Such a mismatch is also the cause of elevated SWR (standing wave ratio).
The wire on each side of the feed point of a dipole is called a "leg". In the case of a 1/2 $lambda$, center fed dipole each leg is 1/4 $lambda$.
The length of a dipole is not limited to 1/2 $lambda$. If the dipole becomes significantly shorter than 1/2 $lambda$, the mismatch to the transmission line becomes high and so the losses in the transmission line go up. The antenna also becomes inefficient in itself which results in more power converted to heat and less radiated as electromagnetic energy.
A dipole longer than 1/2 $lambda$ can also be used. In fact, a dipole that is 10/8 $lambda$ will actually have more gain than a 1/2 $lambda$ dipole. The disadvantage is that it takes a little more work to make it a match for 50 ohm coax. Dipoles longer than 10/8 $lambda$ can be used as well but they will generally not offer any more gain.
In recent years, there has been a great deal of interest in feeding a dipole at a spot other than its center. So called OCF (off center fed) dipoles offer the possibility of a multi-band dipole with reasonable matching to 50 ohm coax. Typically it is a 1/2 $lambda$ or slightly longer dipole for the lowest frequency of interest and then at higher frequencies, it simply acts as a >1/2 $lambda$ dipole.
Other dipoles get specific names due to their historical references such as the double Zepp (from its Zeplin origins), extended double Zepp, a doublet, the G5RV, etc. Some of these designations may also imply the use of an open wire feedline instead of coax. But in the end, they are simply another form of a dipole.
Then there is the so called "End Fed Half Wave" (EFHW) antenna. This is a piece of wire that is the length of a 1/2 $lambda$ dipole but instead of being split somewhere to create a feed point, it is left as one solid piece of wire and the coax is attached to the end. This is a very misunderstood antenna by many hams. It takes careful construction and installation techniques to achieve performance comparable to a center fed 1/2 $lambda$ antenna. Many of these antennas are used in a multiband configuration. Their popularity is due to the convenience of feeding it from one end.
answered 7 hours ago
Glenn W9IQ
11.6k1738
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are already a couple of nice answers to your question. I thought I would add a little more context - no need to vote for this as it is tangential to the question.
A 1/2 wavelength (1/2 $lambda$) long dipole is one of the most basic and easy to deploy antennas. It is popular because when it is cut and fed in the center, it very closely matches the impedance of the most popular transmission line - 50 ohm coax. This helps to minimize power lost in the coax as a result of a mismatch. Such a mismatch is also the cause of elevated SWR (standing wave ratio).
The wire on each side of the feed point of a dipole is called a "leg". In the case of a 1/2 $lambda$, center fed dipole each leg is 1/4 $lambda$.
The length of a dipole is not limited to 1/2 $lambda$. If the dipole becomes significantly shorter than 1/2 $lambda$, the mismatch to the transmission line becomes high and so the losses in the transmission line go up. The antenna also becomes inefficient in itself which results in more power converted to heat and less radiated as electromagnetic energy.
A dipole longer than 1/2 $lambda$ can also be used. In fact, a dipole that is 10/8 $lambda$ will actually have more gain than a 1/2 $lambda$ dipole. The disadvantage is that it takes a little more work to make it a match for 50 ohm coax. Dipoles longer than 10/8 $lambda$ can be used as well but they will generally not offer any more gain.
In recent years, there has been a great deal of interest in feeding a dipole at a spot other than its center. So called OCF (off center fed) dipoles offer the possibility of a multi-band dipole with reasonable matching to 50 ohm coax. Typically it is a 1/2 $lambda$ or slightly longer dipole for the lowest frequency of interest and then at higher frequencies, it simply acts as a >1/2 $lambda$ dipole.
Other dipoles get specific names due to their historical references such as the double Zepp (from its Zeplin origins), extended double Zepp, a doublet, the G5RV, etc. Some of these designations may also imply the use of an open wire feedline instead of coax. But in the end, they are simply another form of a dipole.
Then there is the so called "End Fed Half Wave" (EFHW) antenna. This is a piece of wire that is the length of a 1/2 $lambda$ dipole but instead of being split somewhere to create a feed point, it is left as one solid piece of wire and the coax is attached to the end. This is a very misunderstood antenna by many hams. It takes careful construction and installation techniques to achieve performance comparable to a center fed 1/2 $lambda$ antenna. Many of these antennas are used in a multiband configuration. Their popularity is due to the convenience of feeding it from one end.
answered 7 hours ago
Glenn W9IQ
11.6k1738
There are already a couple of nice answers to your question. I thought I would add a little more context - no need to vote for this as it is tangential to the question.
A 1/2 wavelength (1/2 $lambda$) long dipole is one of the most basic and easy to deploy antennas. It is popular because when it is cut and fed in the center, it very closely matches the impedance of the most popular transmission line - 50 ohm coax. This helps to minimize power lost in the coax as a result of a mismatch. Such a mismatch is also the cause of elevated SWR (standing wave ratio).
The wire on each side of the feed point of a dipole is called a "leg". In the case of a 1/2 $lambda$, center fed dipole each leg is 1/4 $lambda$.
The length of a dipole is not limited to 1/2 $lambda$. If the dipole becomes significantly shorter than 1/2 $lambda$, the mismatch to the transmission line becomes high and so the losses in the transmission line go up. The antenna also becomes inefficient in itself which results in more power converted to heat and less radiated as electromagnetic energy.
A dipole longer than 1/2 $lambda$ can also be used. In fact, a dipole that is 10/8 $lambda$ will actually have more gain than a 1/2 $lambda$ dipole. The disadvantage is that it takes a little more work to make it a match for 50 ohm coax. Dipoles longer than 10/8 $lambda$ can be used as well but they will generally not offer any more gain.
In recent years, there has been a great deal of interest in feeding a dipole at a spot other than its center. So called OCF (off center fed) dipoles offer the possibility of a multi-band dipole with reasonable matching to 50 ohm coax. Typically it is a 1/2 $lambda$ or slightly longer dipole for the lowest frequency of interest and then at higher frequencies, it simply acts as a >1/2 $lambda$ dipole.
Other dipoles get specific names due to their historical references such as the double Zepp (from its Zeplin origins), extended double Zepp, a doublet, the G5RV, etc. Some of these designations may also imply the use of an open wire feedline instead of coax. But in the end, they are simply another form of a dipole.
Then there is the so called "End Fed Half Wave" (EFHW) antenna. This is a piece of wire that is the length of a 1/2 $lambda$ dipole but instead of being split somewhere to create a feed point, it is left as one solid piece of wire and the coax is attached to the end. This is a very misunderstood antenna by many hams. It takes careful construction and installation techniques to achieve performance comparable to a center fed 1/2 $lambda$ antenna. Many of these antennas are used in a multiband configuration. Their popularity is due to the convenience of feeding it from one end.
answered 7 hours ago
Glenn W9IQ
11.6k1738
edited 1 hour ago
answered 7 hours ago
Glenn W9IQ
11.6k1738
answered 7 hours ago
Glenn W9IQ
11.6k1738
answered 7 hours ago
Glenn W9IQ
11.6k1738
11.6k1738
add a comment |Â
add a comment |Â
Nicolas Dufour is a new contributor. Be nice, and check out our Code of Conduct.
Nicolas Dufour is a new contributor. Be nice, and check out our Code of Conduct.
Nicolas Dufour is a new contributor. Be nice, and check out our Code of Conduct.
Nicolas Dufour is a new contributor. Be nice, and check out our Code of Conduct.
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3
Welcome to Amateur Radio stackexchange.
– mike65535
22 hours ago
Thank you! Lots to learn and experiment!
– Nicolas Dufour
22 hours ago