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Using approximate signal strength at a distance to estimate reception strength of a radio station
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I have some historical data on radio stations, but unfortunately, the dataset only has these variables:
the power of the transmitter, in watts
the coordinates of the radio tower
Unfortunately, I don't have any data on the characteristics of the antenna(s) or the radio towers themselves, e.g. their height. Anecdotally, the frequencies of these stations are from 500 to 1500 kilocycles, which I believe are AM frequencies.
I asked a similar question on the physics.SE and was told that, with an example of a 500 W transmitter, I can very roughly estimate the strength at a distance of, say, 10 km, as
At a distance of $10km$, that power is uniformly distributed over a sphere of radius $10km$. In other words, the power per $m^2$ will be
$frac5004 pi r^2$ in units of $fracwattm^2$.
In this example, this gives me approximately $4 times 10^-7 fractextwattsm^2$, or -33.98 dBm per square meter.
With that number in hand, is there anything I can say about a receiver's ability to pick up that signal? As in, a standard household radio receiver in the 1920s could, on average, audibly receive signals down to strength -25 dbm, so it probably wouldn't be able to pick up this signal?
Or is the more appropriate way to convert this power into $V/m$ and use an approximation of the audible area like in this article?
(Yes, I know this is an overly simplistic approximation; I'm trying to do the best I can with the extremely limited data that I have. It was recommended that I ask here, so any information is most welcome!)
history physics
asked 4 hours ago
Michael A
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up vote
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I have some historical data on radio stations, but unfortunately, the dataset only has these variables:
the power of the transmitter, in watts
the coordinates of the radio tower
Unfortunately, I don't have any data on the characteristics of the antenna(s) or the radio towers themselves, e.g. their height. Anecdotally, the frequencies of these stations are from 500 to 1500 kilocycles, which I believe are AM frequencies.
I asked a similar question on the physics.SE and was told that, with an example of a 500 W transmitter, I can very roughly estimate the strength at a distance of, say, 10 km, as
At a distance of $10km$, that power is uniformly distributed over a sphere of radius $10km$. In other words, the power per $m^2$ will be
$frac5004 pi r^2$ in units of $fracwattm^2$.
In this example, this gives me approximately $4 times 10^-7 fractextwattsm^2$, or -33.98 dBm per square meter.
With that number in hand, is there anything I can say about a receiver's ability to pick up that signal? As in, a standard household radio receiver in the 1920s could, on average, audibly receive signals down to strength -25 dbm, so it probably wouldn't be able to pick up this signal?
Or is the more appropriate way to convert this power into $V/m$ and use an approximation of the audible area like in this article?
(Yes, I know this is an overly simplistic approximation; I'm trying to do the best I can with the extremely limited data that I have. It was recommended that I ask here, so any information is most welcome!)
history physics
asked 4 hours ago
Michael A
1114
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It would be helpful to know the frequency range in question. Based on the ca. 1920 reference, are you talking about AM broadcast stations?
– Glenn W9IQ
4 hours ago
@GlennW9IQ According to some notes in the data, the frequencies run from about 500 kilocycles to about 1500 and are referred to as AM stations throughout.
– Michael A
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have some historical data on radio stations, but unfortunately, the dataset only has these variables:
the power of the transmitter, in watts
the coordinates of the radio tower
Unfortunately, I don't have any data on the characteristics of the antenna(s) or the radio towers themselves, e.g. their height. Anecdotally, the frequencies of these stations are from 500 to 1500 kilocycles, which I believe are AM frequencies.
I asked a similar question on the physics.SE and was told that, with an example of a 500 W transmitter, I can very roughly estimate the strength at a distance of, say, 10 km, as
At a distance of $10km$, that power is uniformly distributed over a sphere of radius $10km$. In other words, the power per $m^2$ will be
$frac5004 pi r^2$ in units of $fracwattm^2$.
In this example, this gives me approximately $4 times 10^-7 fractextwattsm^2$, or -33.98 dBm per square meter.
With that number in hand, is there anything I can say about a receiver's ability to pick up that signal? As in, a standard household radio receiver in the 1920s could, on average, audibly receive signals down to strength -25 dbm, so it probably wouldn't be able to pick up this signal?
Or is the more appropriate way to convert this power into $V/m$ and use an approximation of the audible area like in this article?
(Yes, I know this is an overly simplistic approximation; I'm trying to do the best I can with the extremely limited data that I have. It was recommended that I ask here, so any information is most welcome!)
history physics
asked 4 hours ago
Michael A
1114
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have some historical data on radio stations, but unfortunately, the dataset only has these variables:
the power of the transmitter, in watts
the coordinates of the radio tower
Unfortunately, I don't have any data on the characteristics of the antenna(s) or the radio towers themselves, e.g. their height. Anecdotally, the frequencies of these stations are from 500 to 1500 kilocycles, which I believe are AM frequencies.
I asked a similar question on the physics.SE and was told that, with an example of a 500 W transmitter, I can very roughly estimate the strength at a distance of, say, 10 km, as
At a distance of $10km$, that power is uniformly distributed over a sphere of radius $10km$. In other words, the power per $m^2$ will be
$frac5004 pi r^2$ in units of $fracwattm^2$.
In this example, this gives me approximately $4 times 10^-7 fractextwattsm^2$, or -33.98 dBm per square meter.
With that number in hand, is there anything I can say about a receiver's ability to pick up that signal? As in, a standard household radio receiver in the 1920s could, on average, audibly receive signals down to strength -25 dbm, so it probably wouldn't be able to pick up this signal?
Or is the more appropriate way to convert this power into $V/m$ and use an approximation of the audible area like in this article?
(Yes, I know this is an overly simplistic approximation; I'm trying to do the best I can with the extremely limited data that I have. It was recommended that I ask here, so any information is most welcome!)
history physics
history physics
asked 4 hours ago
Michael A
1114
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Michael A
1114
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
asked 4 hours ago
Michael A
1114
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Michael A
1114
asked 4 hours ago
Michael A
1114
1114
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Michael A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It would be helpful to know the frequency range in question. Based on the ca. 1920 reference, are you talking about AM broadcast stations?
– Glenn W9IQ
4 hours ago
@GlennW9IQ According to some notes in the data, the frequencies run from about 500 kilocycles to about 1500 and are referred to as AM stations throughout.
– Michael A
4 hours ago
add a comment |Â
It would be helpful to know the frequency range in question. Based on the ca. 1920 reference, are you talking about AM broadcast stations?
– Glenn W9IQ
4 hours ago
@GlennW9IQ According to some notes in the data, the frequencies run from about 500 kilocycles to about 1500 and are referred to as AM stations throughout.
– Michael A
4 hours ago
It would be helpful to know the frequency range in question. Based on the ca. 1920 reference, are you talking about AM broadcast stations?
– Glenn W9IQ
4 hours ago
It would be helpful to know the frequency range in question. Based on the ca. 1920 reference, are you talking about AM broadcast stations?
– Glenn W9IQ
4 hours ago
@GlennW9IQ According to some notes in the data, the frequencies run from about 500 kilocycles to about 1500 and are referred to as AM stations throughout.
– Michael A
4 hours ago
@GlennW9IQ According to some notes in the data, the frequencies run from about 500 kilocycles to about 1500 and are referred to as AM stations throughout.
– Michael A
4 hours ago
add a comment |Â
1 Answer
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up vote
3
down vote
We have some clues from a historical perspective. In the January 1917 edition of The Radio Experimentor, author H. Winfield Secor noted that approximately 50 kW (kilo-watts) of transmitter power was required to reach a 3,000 to 4,000 mile range with CW. He then went on to note that 10 μA (micro-amperes) is considered a weak receive signal while 20 μA is a strong signal. He also recorded an important fact that 10 μA is equivalent to 0.01 μW (micro-watts). We can then translate the 10 μA field signal level as equivalent to -50 dBm and the 20 uA signal as equivalent to -44 dBm. Take note that this is for CW receivers which will not be fully comparable to AM audio transmissions.
Just as a basis for comparison, modern radios easily achieve 0.25 μV (micro-volt) MDS (minimum discernible signal) sensitivity across a 50 ohm input impedance. This is equivalent to -119 dBm.
Secor does not make it clear if these references are the input power to the receiver or if this is the irradiance in the field of the receiver antenna. But the basis of the article was to assemble a compendium of competing methods of detection so I believe it would be safe to say that this was the required power at the input to the receiver terminals. Supporting this notion is the clarifying note from the author: 'Technically speaking, radio detectors are usually rated by the amount of electrical energy in ergs necessary to actuate them.' when he was referencing the earlier listed micro amperes.
To compare your tabular data with this historical reference, you would need to presume that the equivalent receive and transmit antenna gains for the path in question are used in the comparison. You would also need to presume that the path loss is equal. Then, as you noted in your question, you can compare potential distances simply as a function of power and the inverse of distance squared.
Do note that during the 1915 to 1930 period, there were significant advances in detector technology and receiver architectures. As a result, there could be a significant error range in your estimates depending upon the exact period of comparison.
answered 3 hours ago
Glenn W9IQ
12k1738
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
We have some clues from a historical perspective. In the January 1917 edition of The Radio Experimentor, author H. Winfield Secor noted that approximately 50 kW (kilo-watts) of transmitter power was required to reach a 3,000 to 4,000 mile range with CW. He then went on to note that 10 μA (micro-amperes) is considered a weak receive signal while 20 μA is a strong signal. He also recorded an important fact that 10 μA is equivalent to 0.01 μW (micro-watts). We can then translate the 10 μA field signal level as equivalent to -50 dBm and the 20 uA signal as equivalent to -44 dBm. Take note that this is for CW receivers which will not be fully comparable to AM audio transmissions.
Just as a basis for comparison, modern radios easily achieve 0.25 μV (micro-volt) MDS (minimum discernible signal) sensitivity across a 50 ohm input impedance. This is equivalent to -119 dBm.
Secor does not make it clear if these references are the input power to the receiver or if this is the irradiance in the field of the receiver antenna. But the basis of the article was to assemble a compendium of competing methods of detection so I believe it would be safe to say that this was the required power at the input to the receiver terminals. Supporting this notion is the clarifying note from the author: 'Technically speaking, radio detectors are usually rated by the amount of electrical energy in ergs necessary to actuate them.' when he was referencing the earlier listed micro amperes.
To compare your tabular data with this historical reference, you would need to presume that the equivalent receive and transmit antenna gains for the path in question are used in the comparison. You would also need to presume that the path loss is equal. Then, as you noted in your question, you can compare potential distances simply as a function of power and the inverse of distance squared.
Do note that during the 1915 to 1930 period, there were significant advances in detector technology and receiver architectures. As a result, there could be a significant error range in your estimates depending upon the exact period of comparison.
answered 3 hours ago
Glenn W9IQ
12k1738
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
add a comment |Â
up vote
3
down vote
We have some clues from a historical perspective. In the January 1917 edition of The Radio Experimentor, author H. Winfield Secor noted that approximately 50 kW (kilo-watts) of transmitter power was required to reach a 3,000 to 4,000 mile range with CW. He then went on to note that 10 μA (micro-amperes) is considered a weak receive signal while 20 μA is a strong signal. He also recorded an important fact that 10 μA is equivalent to 0.01 μW (micro-watts). We can then translate the 10 μA field signal level as equivalent to -50 dBm and the 20 uA signal as equivalent to -44 dBm. Take note that this is for CW receivers which will not be fully comparable to AM audio transmissions.
Just as a basis for comparison, modern radios easily achieve 0.25 μV (micro-volt) MDS (minimum discernible signal) sensitivity across a 50 ohm input impedance. This is equivalent to -119 dBm.
Secor does not make it clear if these references are the input power to the receiver or if this is the irradiance in the field of the receiver antenna. But the basis of the article was to assemble a compendium of competing methods of detection so I believe it would be safe to say that this was the required power at the input to the receiver terminals. Supporting this notion is the clarifying note from the author: 'Technically speaking, radio detectors are usually rated by the amount of electrical energy in ergs necessary to actuate them.' when he was referencing the earlier listed micro amperes.
To compare your tabular data with this historical reference, you would need to presume that the equivalent receive and transmit antenna gains for the path in question are used in the comparison. You would also need to presume that the path loss is equal. Then, as you noted in your question, you can compare potential distances simply as a function of power and the inverse of distance squared.
Do note that during the 1915 to 1930 period, there were significant advances in detector technology and receiver architectures. As a result, there could be a significant error range in your estimates depending upon the exact period of comparison.
answered 3 hours ago
Glenn W9IQ
12k1738
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have some clues from a historical perspective. In the January 1917 edition of The Radio Experimentor, author H. Winfield Secor noted that approximately 50 kW (kilo-watts) of transmitter power was required to reach a 3,000 to 4,000 mile range with CW. He then went on to note that 10 μA (micro-amperes) is considered a weak receive signal while 20 μA is a strong signal. He also recorded an important fact that 10 μA is equivalent to 0.01 μW (micro-watts). We can then translate the 10 μA field signal level as equivalent to -50 dBm and the 20 uA signal as equivalent to -44 dBm. Take note that this is for CW receivers which will not be fully comparable to AM audio transmissions.
Just as a basis for comparison, modern radios easily achieve 0.25 μV (micro-volt) MDS (minimum discernible signal) sensitivity across a 50 ohm input impedance. This is equivalent to -119 dBm.
Secor does not make it clear if these references are the input power to the receiver or if this is the irradiance in the field of the receiver antenna. But the basis of the article was to assemble a compendium of competing methods of detection so I believe it would be safe to say that this was the required power at the input to the receiver terminals. Supporting this notion is the clarifying note from the author: 'Technically speaking, radio detectors are usually rated by the amount of electrical energy in ergs necessary to actuate them.' when he was referencing the earlier listed micro amperes.
To compare your tabular data with this historical reference, you would need to presume that the equivalent receive and transmit antenna gains for the path in question are used in the comparison. You would also need to presume that the path loss is equal. Then, as you noted in your question, you can compare potential distances simply as a function of power and the inverse of distance squared.
Do note that during the 1915 to 1930 period, there were significant advances in detector technology and receiver architectures. As a result, there could be a significant error range in your estimates depending upon the exact period of comparison.
answered 3 hours ago
Glenn W9IQ
12k1738
We have some clues from a historical perspective. In the January 1917 edition of The Radio Experimentor, author H. Winfield Secor noted that approximately 50 kW (kilo-watts) of transmitter power was required to reach a 3,000 to 4,000 mile range with CW. He then went on to note that 10 μA (micro-amperes) is considered a weak receive signal while 20 μA is a strong signal. He also recorded an important fact that 10 μA is equivalent to 0.01 μW (micro-watts). We can then translate the 10 μA field signal level as equivalent to -50 dBm and the 20 uA signal as equivalent to -44 dBm. Take note that this is for CW receivers which will not be fully comparable to AM audio transmissions.
Just as a basis for comparison, modern radios easily achieve 0.25 μV (micro-volt) MDS (minimum discernible signal) sensitivity across a 50 ohm input impedance. This is equivalent to -119 dBm.
Secor does not make it clear if these references are the input power to the receiver or if this is the irradiance in the field of the receiver antenna. But the basis of the article was to assemble a compendium of competing methods of detection so I believe it would be safe to say that this was the required power at the input to the receiver terminals. Supporting this notion is the clarifying note from the author: 'Technically speaking, radio detectors are usually rated by the amount of electrical energy in ergs necessary to actuate them.' when he was referencing the earlier listed micro amperes.
To compare your tabular data with this historical reference, you would need to presume that the equivalent receive and transmit antenna gains for the path in question are used in the comparison. You would also need to presume that the path loss is equal. Then, as you noted in your question, you can compare potential distances simply as a function of power and the inverse of distance squared.
Do note that during the 1915 to 1930 period, there were significant advances in detector technology and receiver architectures. As a result, there could be a significant error range in your estimates depending upon the exact period of comparison.
answered 3 hours ago
Glenn W9IQ
12k1738
answered 3 hours ago
Glenn W9IQ
12k1738
answered 3 hours ago
Glenn W9IQ
12k1738
answered 3 hours ago
Glenn W9IQ
12k1738
12k1738
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
add a comment |Â
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
This is extremely helpful; thank you! Perhaps this is a separate question, but I'm a little confused about the units. How is 10 μA equivalent to 0.01 μW? It's necessary to have some impedance in ohms in the calculation as well, right? I didn't think amperes and watts were directly comparable. Also, by my math using the formulas here, 0.01 μW is equivalent to -80 dBm.
– Michael A
2 hours ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
@MichaelA I will add the formulas to support my answer later today.
– Glenn W9IQ
13 mins ago
add a comment |Â
Michael A is a new contributor. Be nice, and check out our Code of Conduct.
Michael A is a new contributor. Be nice, and check out our Code of Conduct.
Michael A is a new contributor. Be nice, and check out our Code of Conduct.
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It would be helpful to know the frequency range in question. Based on the ca. 1920 reference, are you talking about AM broadcast stations?
– Glenn W9IQ
4 hours ago
@GlennW9IQ According to some notes in the data, the frequencies run from about 500 kilocycles to about 1500 and are referred to as AM stations throughout.
– Michael A
4 hours ago