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Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in th...

posted 3y ago by Andy aka‭ · edited 3y ago by Andy aka‭

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#5: Post edited by

Andy aka‭ · 2020-10-10T15:52:35Z (over 3 years ago)

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Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
The ratio of E-field to H-field of a radio wave turns out to be volts per amp and that, as you probably know, is measured in ohms. In other words, the signal travelling towards your antenna experiences the "impedance of free space" and that is about 377 ohms. In effect your antenna is also an impedance transformer converting a signal with an impedance of 377 ohms to something around 50 ohms (antenna type dependent).
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
The ratio of E-field to H-field of a radio wave turns out to be volts per amp and that, as you probably know, is measured in ohms. In other words, the signal travelling towards your antenna experiences the "impedance of free space" and that is about 377 ohms. In effect your antenna is an impedance transformer converting a signal with an impedance of 377 ohms to something around 50 ohms (antenna type dependent). This effectively reduces the voltage by a ratio \$\sqrt{\frac{377}{50}}\$ if it were at all possible to measure voltage in free-space directly.
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

#4: Post edited by

Andy aka‭ · 2020-10-10T15:47:45Z (over 3 years ago)

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Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
The ratio of E-field to H-field of a radio wave turns out to be volts per amp and that, as you probably know, is measured in ohms. In other words, the signal travelling towards your antenna experiences the "impedance of free space" and that is about 377 ohms. In effect your antenna is also an impedance transformer.
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
The ratio of E-field to H-field of a radio wave turns out to be volts per amp and that, as you probably know, is measured in ohms. In other words, the signal travelling towards your antenna experiences the "impedance of free space" and that is about 377 ohms. In effect your antenna is also an impedance transformer converting a signal with an impedance of 377 ohms to something around 50 ohms (antenna type dependent).
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

#3: Post edited by

Andy aka‭ · 2020-10-10T15:46:43Z (over 3 years ago)

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Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
The ratio of E-field to H-field of a radio wave turns out to be volts per amp and that, as you probably know, is measured in ohms. In other words, the signal travelling towards your antenna experiences the "impedance of free space" and that is about 377 ohms. In effect your antenna is also an impedance transformer.
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

#2: Post edited by

Andy aka‭ · 2020-10-10T15:42:15Z (over 3 years ago)

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Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that net is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

Apart from the need to match impedances to prevent possibilities of signal reflections and the knock-on issue of signal nulls at your receiver LNA input, an antenna "expects" to be terminated in the "right" impedance in order to get best performance.
An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that "fishing net" is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).
To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.
Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.
But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

#1: Initial revision by

Andy aka‭ · 2020-10-10T15:40:13Z (over 3 years ago)

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An antenna is like a fishing net - it has an effective surface area (despite it possibly looking just like a single wire) and, that net is "capturing" both an electric field (volts per metre) and a magnetic field (amps per metre). Together, when multiplied you get watts per square metre hence, the antenna effective area is "capturing" watts (usually sub pico watts).

To convert that power into a signal-voltage, you need a resistor i.e. the load resistance required by the antenna to make it work most effectively.

Looking at things from another direction, 50 ohms (being a pretty normal standard for radio input and output impedance) **does not** generate a whole lot of thermal noise voltage for a given bandwidth and, this is important because, thermal noise can be a very significant factor in limiting how small a signal a particular radio design can effectively receive and demodulate. Keeping to a common impedance (i.e. 50 ohms or thereabouts) is a way of keeping signal-to-noise ratio as good as possible.

But, if you have a LNA that has 1 kohm input impedance (resistive) then there's absolutely no problem in using a filter that can convert impedance from around the 50 ohm mark up to 1 kohm. You wouldn't use resistors of course because that just wastes signal power but, you would use L-Pads, T-networks or \$\pi\$ networks to convert impedance without power loss.

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