How does the use of a reflector affect the performance of a receive antenna?

[MarkWW]

Do receivers only receive signals or they emit emf?

 

[berkeman]

If they can also transmit, they are usually called "transceivers".

 

[jim hardy]

Do receivers only receive signals or they emit emf?

I'm not familiar with modern single IC radios so some of this might be obsolete.
but here goes anyway

Look up "Superheterodyne".
Receivers work on the principle of frequency conversion
where the incoming signal from the antenna is mixed with a second signal of a different(hetero) frequency(dyne as in dynamic).
Mixing is a mathematical multiplication
and if you multiply the two Fourier polynomials representing those two signals you get their sum, their difference, their product, and various other frequency terms.

Next part is clever
You can send the resulting smorgasbord of frequencies to a very selective amplifier that's tuned to a single frequency. It will amplify only that one frequency.
That amplifier's selectivity allows selective reception of one of the many frequencies present on the antenna, the one that produced its favorite frequency in the smorgasbord.. (That ought to be the difference frequency but things can go awry . I could offer a boring anecdote but won't.)

The 'second frequency' i mentioned is called "Local Oscillator" because it is produced inside the radio, ie locally .
In most radios it is of frequency higher than the radio station to be received, hence the term "Superheterodyne" meaning local oscillator is higher frequency than the station to be tuned.

In ordinary broadcast radios covering the 540 to 1650 khz AM broadcast band ,
that single frequency amplifier that is the key to selectivity is called the Intermediate Frequency Amplifier" "IF" for short and is tuned to 455 or 456 khz
meaning your local oscillator would be tunable over the range
540 + 455 = 995 khz
to
1650 + 455 = 2105 khz.

In FM band 88 to 108 mhz the IF is 10.7mhz or 10.8 mhz.

Standardizing IF frequencies let manufacturers make standard transformers for them that will interchange, a great conveniuence.

When you turn the dial to choose a station you are tuning the local oscillator's frequency not the IF amplifier's frequency as you'd think..

Now to your real question
The local oscillator makes some RF energy that escapes.
Take a pocket AM radio and tune it to a blank frequency between stations in the higher part of the band so you get just static. Turn up the volume.
Take a second radio and tune it to about 455 khz lower. Sweep its tuning up and down a little bit, you should hear a whistle on the first radio as you sweep the second's local oscillator past the frequency to which you tuned the first one.
At 890 KHZ, WLS in Chicago, your local oscillator should be about 1345 khz. If you live not close to Chicago give it a try.

In vacuum tube days that's how we checked to see if the local oscillator was working.

So yes, analog receivers emit some RF .
I suspect digital ones do too
ever hold a pocket transistor radio up to the back of a computer ?

old jim

 

[Baluncore]

There is a small amount of EM energy radiated from radio receivers. It usually originates from the first local oscillator, then escapes from the receiver along the the power supply lines, the audio output, or back out through the antenna cable.

During WWII, the 1st LO radiation from the Metox radar warning receivers used on U-boats could be detected and used to find the U-boat. https://en.wikipedia.org/wiki/Metox_radar_detector

In Britain the radiation from TV receivers was once used to detect the presence of television sets as part of the license enforcement.

Masthead amplifiers used on TV antennas sometimes oscillate and so block reception for others in the vicinity. I think they are banned in some countries because of that common nuisance radiation.

 

[MarkWW]

Thanks a lot!

 

[tech99]

Thanks a lot!

A receiving antenna re-radiates half the power by the way.

 

[davenn]

A receiving antenna re-radiates half the power by the way.

well, I have never heard that before !

you have a good link or 2 to support that ?

 

[tech99]

well, I have never heard that before !

you have a good link or 2 to support that ?

Radio Engineers Handbook, by Terman, page 786.
The induced current flows in the load resistance and the radiation resistance in series, so only half the power is delivered to the load. The I^2 R loss in the radiation resistance occurs because power is re-radiated.

 

[berkeman]

Radio Engineers Handbook, by Terman, page 786.
The induced current flows in the load resistance and the radiation resistance in series, so only half the power is delivered to the load. The I^2 R loss in the radiation resistance occurs because power is re-radiated.

Very interesting, and good to know. Does it apply to parabolic antennas too, or just open antennas?

 

[Baluncore]

A receiving antenna re-radiates half the power by the way.

If there is a matched transmission line between an antenna and a receiver, there should be no energy reflected by the receiver back up the transmission line towards the antenna.

Likewise, a good antenna is a perfect matching network between the impedance of free space and the impedance of the transmission line, if matched it should not reflect energy.

The principle of reciprocity does not support the suggestion that half the power will be reflected by the antenna. A matched antenna does not reflect half the transmitted energy back down the line to the transmitter.

Radiation resistance is a mythical equivalent resistance that would radiate heat at the same rate that the real antenna radiates EM energy.

If half the energy was reflected, it would make stealth vehicles covered in 377 ohm “space cloth” (or radar absorbing paint) visible to radar.

 

[davenn]

Radio Engineers Handbook, by Terman, page 786.
The induced current flows in the load resistance and the radiation resistance in series, so only half the power is delivered to the load. The I^2 R loss in the radiation resistance occurs because power is re-radiated.

If there is a matched transmission line between an antenna and a receiver, there should be no energy reflected by the receiver back up the transmission line towards the antenna.

Likewise, a good antenna is a perfect matching network between the impedance of free space and the impedance of the transmission line, if matched it should not reflect energy.

@Baluncore , that's what I would have expected as well
I wonder if the quote tech99 has stated is taken out of context ?

Dave

 

[Baluncore]

I wonder if the quote tech99 has stated is taken out of context ?

Terman bases his analysis on the voltage induced in the antenna that causes a current to flow. The situation is now modeled as a current induced in the antenna elements and ignores voltage differences. That gets around the confusion of summing currents generated by voltages on elements with variable impedance.

Terman appears to have messed up the bottom of his equation (11) on page 786.
The proportion of the energy received then incorrectly becomes; RL / ( RL + Rr + Rl ).
But the radiation resistance Rr is not a real resistance and should not be added to the load resistance RL and the loss resistance Rl. That may be why it seems the proportion approaches 50% for low loss antennas. Removal of Rr from the equation fixes the problem and returns the proportion to 100%.

Terman then states that a reradiation takes place. I believe he is referring to a minor reradiation due only to the loss resistance Rl. The reradiation should not include Rr.
When this reradiation is conflated with the incorrect 50%, it sugests 50% reflected.

Terman does not write anything like “half the received energy is reflected”. That is a misrepresentation of his text.
I don't have time now for a proper analysis, that will have to wait.

 

[davenn]

Terman bases his analysis on the voltage induced in the antenna that causes a current to flow. The situation is now modeled as a current induced in the antenna elements and ignores voltage differences. That gets around the confusion of summing currents generated by voltages on elements with variable impedance.

Terman appears to have messed up the bottom of his equation (11) on page 786.
The proportion of the energy received then incorrectly becomes; RL / ( RL + Rr + Rl ).
But the radiation resistance Rr is not a real resistance and should not be added to the load resistance RL and the loss resistance Rl. That may be why it seems the proportion approaches 50% for low loss antennas. Removal of Rr from the equation fixes the problem and returns the proportion to 100%.

Terman then states that a reradiation takes place. I believe he is referring to a minor reradiation due only to the loss resistance Rl. The reradiation should not include Rr.
When this reradiation is conflated with the incorrect 50%, it sugests 50% reflected.

Terman does not write anything like “half the received energy is reflected”. That is a misrepresentation of his text.
I don't have time now for a proper analysis, that will have to wait.

Thanks
I don't have the publication, so wasn't able to see the full text ( possibly it is online?)
I was sure there was a problem there somewhere

You provided a good initial analysis of the situationDave

 

[Baluncore]

@davenn
A site worth visiting is; http://www.qsl.net/va3iul/Files/Old_Radio_Frequency_Books.htm

There is a link there to; Terman, Radio Engineers Handbook, 1943.
http://www.itermoionici.it/letteratura_files/Radio-Engineers-Handbook.pdf

 

[tech99]

May I reply to some of the points raised?

If you consider a parasitic element, it reflects all the power. If we add load resistance to it, we can extract part of the power, up to half, depending on the resistance added.

If we use a transmission line, I am not saying half the power is reflected from the load. The reflection occurs at the junction of the line and the antenna, and depends on the impedance presented by the line to the antenna, and it is at this point that up to 50% power transfer occurs.

If you consider the transmitting case, the maximum power is obtained from the transmitter when the source resistance of the transmitter equals that presented by the transmission line, and cannot be more than 50% for the case when adjustment is made for maximum power (better efficiency is possible under mismatched conditions when adjusted for lower output power).

For the case of 377 Ohms-per-square resistive paint, it does not produce zero reflection, but only gives 3dB reduction. To obtain complete concealment we have to use a thicker coat, usually a quarter wavelength thick, to trap the radiation. The 377 Ohm surface is usually raised up a quarter of a wavelength above the metal surface.

Regarding a parabolic dish, I made one test over an actual 25 mile microwave path where it seemed to be reflecting a lot of power, maybe half as expected.

Regarding Terman's formula, I would suggest that Rr is a real resistance. If we take an open circuit receiving monopole, suppose the EMF between base and ground is 1 volt. If we add a resistor of, say 35 Ohms, the PD falls to 0.5 Volt. The resistance in the system is now 70 Ohms and the current in the system is now 1/70 Amps. The power in the load is I^2R = (1/70)^2 * 35 = 7mW. The current in the monopole radiates a power of I^2 * Rr = 7mW also.

 

[tech99]

Thank you for the edit.
I would also like to mention that the radiation from a receiving antenna is in phase opposition to the incoming wave. Both the re-radiated E and H fields oppose the incoming wave. For this reason, it opposes the incoming wave and reduces the intensity in its vicinity, whilst at the same time extracting up to half the incident energy..

 

[davenn]

The reflection occurs at the junction of the line and the antenna, and depends on the impedance presented by the line to the antenna, and it is at this point that up to 50% power transfer occurs.

ONLY if there is a mismatch ( as @Baluncore pointed out)

Regarding a parabolic dish, I made one test over an actual 25 mile microwave path where it seemed to be reflecting a lot of power, maybe half as expected.

without actual figures, that is meaningless

 

[Baluncore]

Regarding Terman's formula, I would suggest that Rr is a real resistance.

Radiation Resistance.—The radiation resistance referred to a certain point in an antenna system is the resistance which, if inserted at that point with the assumed current I₀ flowing, would dissipate the same energy as is actually radiated from the antenna system. Thus
Radiation resistance = radiated power / I₀²
Although this radiation resistance is a purely fictitious quantity, the antenna acts as though such a resistance were present, because the loss of energy by radiation is equivalent to a like amount of energy dissipated in a resistance.

 

[berkeman]

This is a very interesting discussion, and I'm glad we're keeping it based on the science. I may split this discussion out as a separate thread or re-name the thread. Please keep the good technical posts coming. Thank you to the participants.

 

[berkeman]

TBH, I've always wondered what the radiation pattern around a receiving antenna was like. The induced currents are real, and I'd assumed that those caused re-radiation of some kind. How could the moving charges not re-radiate something?

 

[tech99]

ONLY if there is a mismatch ( as @Baluncore pointed out)
without actual figures, that is meaningless

(a) If we connect some sort of load to a receiving antenna, it seems logical that the antenna must have a source resistance of some sort, as do all generators. If this were not the case, it would be possible to draw unlimited power from it.
(b) I am sorry my test involving reflection by a paraboloid antenna was not documented sufficiently to be regarded as evidence - it was just a comment for general interest.

 

[Baluncore]

(a) If we connect some sort of load to a receiving antenna, it seems logical that the antenna must have a source resistance of some sort, as do all generators. If this were not the case, it would be possible to draw unlimited power from it.

However logical it may seem to you, conservation of energy precludes that possibility. You are making an assumption that there is an infinite amount of incident energy available. You are also confusing a fixed voltage source having a fixed internal resistance, with a fixed impedance power source.

I agree that maximum power can be extracted from a fixed voltage source when the resistance of the load is equal to the internal resistance of the source. Then half the power is dissipated in the source internal resistance, the other half in the load.

But all that must be put aside when considering the propagation of EM energy, then the ratio of voltage to current is determined by impedance. The impedance is fixed by the electric and magnetic characteristics of the propagation medium. The intrinsic impedance of free space is Z_fs = Uo ⋅ c = 4×10^-7 ⋅ π ⋅ 299792458 = 376.73 ohm.
Z_fs sets the ratio of the electric to the magnetic field for EM waves in space. It does not set the product of the electric and magnetic fields, the product is the power, or the poynting vector.

An antenna can be seen as a transducer, a coupling network, or a transformer. It matches the intrinsic impedance of free space to the characteristic impedance of the feedline that couples the transmitter or load to the antenna.

When operated with a transmitter, the antenna makes the intrinsic impedance of free space look to the transmitter feedline like the fictitious resistor, Rr, the radiation resistance. The small loss resistance Rl is in series with the hopefully bigger Rr, the same current flows through both. While Rl radiates I²⋅Rl heat, Rr remains "cold" because it is not a resistor at all, but represents a “port” through which EM energy is radiated. The efficiency of the transmit antenna is then about Rr / ( Rr + Rl ). For a matched antenna with a feedline, viewed from the transmitter, Rr will be close to the feedline impedance, Zo = Rr + Rl.

When operated as a receive antenna, the incident EM energy arrives with an E/H ratio of Z_fs. The antenna transforms that into a signal in the feedline with a V / I ratio of feedline impedance, Zo. The efficiency of the receive antenna is then about Zo / ( Zo + Rl ).

 

[tech99]

When operated as a receive antenna, the incident EM energy arrives with an E/H ratio of Z_fs. The antenna transforms that into a signal in the feedline with a V / I ratio of feedline impedance, Zo. The efficiency of the receive antenna is then about Zo / ( Zo + Rl ).

So to take an example, are you saying that if Zo is 35 Ohms and the input resistance of the receiver is also 35 Ohms, the efficiency of the antenna is 35/(35+35) = 0.5 or 50%? That seems to be the same number that I am suggesting. Except that I do not see a requirement to have the complication of a feeder in our discussion, as we can place the receiver right at the antenna terminals.

 

[tech99]

I can now see that by RL you mean loss resistance in the feeder rather than the input resistance of the receiver, so please ignore my last post.

It would be better for the purposes of analysing this problem to consider the receiver to be located at the antenna terminals and not to have a feeder involved.

Suppose I have an antenna that has a radiation resistance, referred to its terminals, of 50 Ohms. Are you supposing that it behaves differently to a signal generator with a 50 Ohm output when we connect it to a receiver? For the signal generator, the voltage falls by half when we connect it to the receiver. The efficiency of the system is 50%.

 

[berkeman]

It just occurred to me that we should be able to test this or model it. Does anybody have access to high-end antenna & EM modeling software? We could model what the field interactions are around a receiving antenna when presented with a plane wave at the antenna's resonant frequency (with the antenna properly terminated in 50 Ohms through a balun).

Alternately, we could set up a test with 2 terminated receiving antennas to see if there is multipath interaction between them. If there is no reflection, there can be no multipath nulls. If there are reflections off each antenna, we can show partial nulls in the RX signals as a function of separation...

 

[Baluncore]

It would be better for the purposes of analysing this problem to consider the receiver to be located at the antenna terminals and not to have a feeder involved.

No. Removal of the line confuses the forward and reflected signals on the line which makes anything possible. The feedline is there, matched at both ends, to move the load away from the antenna in space and time. It conceptually separates the two ends of the problem. It is failure to do that that confused Terman.

Suppose I have an antenna that has a radiation resistance, referred to its terminals, of 50 Ohms. Are you supposing that it behaves differently to a signal generator with a 50 Ohm output when we connect it to a receiver?

No. Conceptually, a transmit antenna has a radiation resistance, a receive antenna does not. A receive antenna hopefully transforms the intrinsic impedance of free space into the 50 ohm characteristic impedance of the line and load.

For the signal generator, the voltage falls by half when we connect it to the receiver. The efficiency of the system is 50%.

You are confusing internal series loss resistance with signal output impedance. A simple low power signal generator will probably have a low impedance voltage follower or buffer, with a 50 ohm series terminating resistor to the output. Then yes, the voltage will drop to 50% when connected to a 50 ohm load. To compensate for that, the buffer will have a voltage gain of two. Radio transmitters obviously do not have internal 50 ohm resistors wired in series with their outputs when driving 50 ohm lines and loads. Instead, transmitters produce voltage and current in the correct ratio to fit the line and load impedance.

When a generator is not connected, the system is not impedance matched and the efficiency of the system is zero. The generator output current is zero and the voltage is regulated, so the generator output sees an infinite impedance load. All generated energy is reflected back into the generator from the terminals.

When a 50 ohm generator is connected to the line and load, the ratio of voltage to current produced by the generator is 50 (ohms=volts/amps). That matches the line and receiver so no energy is reflected. Then the efficiency can approach 100%. If the generator had 1 ohm of real internal series loss resistance and a 50 ohm output impedance, it would have an efficiency closer to 50 / 51 = 98% when driving a 50 ohm load or line.

If you terminate a 50 ohm line in a 50 ohm resistor, the ratio of V / I on the line is the same as the ratio of V / I across the resistor. There is no mismatched V or I so there is no reflection.

On a transmission line, impedance is the ratio of V to I of the traveling wave, while power is the product of V and I.

A 50 volt traveling wave signal on a 50 ohm line will have a traveling wave current of 1 amp. The product of V and I is power = 50 watt propagated.

If the voltage on a matched line was 1000 V, the current would be 1000V / 50R = 20A. The power would then be 1kV * 20 A = 20 kW. If the line had a zero series loss resistance the line would not get hot and the energy going in at one end would all come out at the other some time later. With just one ohm of loss resistance, the line would dissipate I²R = 20² = 400 watts of the 20 kW provided. That is an energy transmission efficiency of about 19k6 / 20k0 = 98%.

For RF we use matched lines with fixed impedance and low loss resistance for several reasons. RF energy is valuable and we want efficient transfer. When matched at both ends, the length of the line is no longer important and so the system is frequency independent. Each time energy is reflected by a mismatch, that energy must be circulated through the line loss resistance twice before it gets another chance to be radiated.

We live in a universe of mismatched impedances. Impedance is decided by the electric and magnetic characteristics of the various things around us. If you can't see it, then there is no impedance mismatch so there is no energy reflected or scattered in your direction. An ideal antenna that transforms an EM wave in the intrinsic impedance of free space to the impedance of the connected line and load will not reflect energy.

 

[tech99]

I have just noticed another good description given in Antennas (second edition) by John D Kraus, page 31 section 2.14. He explains at length the absorption of energy by dipoles and apertures, including the 377 Ohm absorbing paint, and seems to confirm that a paraboloid will reflect half the energy.

 

[Baluncore]

I have just noticed another good description given in Antennas (second edition) by John D Kraus, page 31 section 2.14. He explains at length the absorption of energy by dipoles and apertures, including the 377 Ohm absorbing paint, ...

Krause ends section 2:14 with the statement; “Although the above discussion of scattering aperture is applicable to a single dipole (λ/2 or shorter), it does not apply in general.”

Kraus does not support your original assertion that a receive antenna reflects half the incident energy.
Kraus does point out that half the energy can be reflected when a receive antenna is operated into a load having the same impedance as the antenna loss resistance. But that deliberate analogy with a generator having an internal resistance, makes the same assumption that the voltage is fixed, independent of load, and that unlimited power is available. Receive antennas have a fixed power available and maximise energy efficiency by minimising loss resistance relative to output impedance.

… and seems to confirm that a paraboloid will reflect half the energy.

Kraus does not confirm that a paraboloid will reflect half the incident energy.
A “paraboloid” is used as a mirror to direct axial energy onto a transducer at the focus. A parabolic reflector is made of conductive metal, so it reflects close to 100% of the incident energy. You must separate the analysis of the reflective surface from the analysis of the transducer at the focus.

The re-radiated or scattered field of an absorbing antenna may be considered as interfering with the incident field so that a shadow may be cast behind the antenna …

While that is certainly an interesting idea, a conductive screen reflector could make the same shadow. When a signal is canceled by having the reverse phase signal added, that simple cancellation would appear to annihilate two sources of energy, so it must be impossible. There must be other directional effects that conserve energy.

 

[berkeman]

Are you guys really going to make me go test this? I'm kinda busy right now at work...

 

[Baluncore]

Are you guys really going to make me go test this? I'm kinda busy right now at work...

What is the precise hypothesis you want to test ?
Can the principle of reciprocity be trumped by one member's belief ?
When you transmit, does half the power come back down the matched feedline ?

 

[anorlunda]

There is a small amount of EM energy radiated from radio receivers. It usually originates from the first local oscillator, then escapes from the receiver along the the power supply lines, the audio output, or back out through the antenna cable.

During WWII, the 1st LO radiation from the Metox radar warning receivers used on U-boats could be detected and used to find the U-boat. https://en.wikipedia.org/wiki/Metox_radar_detector

In Britain the radiation from TV receivers was once used to detect the presence of television sets as part of the license enforcement.

Masthead amplifiers used on TV antennas sometimes oscillate and so block reception for others in the vicinity. I think they are banned in some countries because of that common nuisance radiation.

Thanks for the wonderful reminder.

When I lived in Sweden, the state owned TV showed ads with Jack booted thugs with a radio direction finder going down the street. When they found a cheat who didn't pay the TV tax, they would break down the doors rape and kill (just kidding)

 

[berkeman]

What is the precise hypothesis you want to test ?
Can the principle of reciprocity be trumped by one member's belief ?
When you transmit, does half the power come back down the matched feedline ?

It seems there is a disagreement whether there is a reflection off of a receiving (dipole or monopole) antenna when an EM wave passes by. Both arguments seem plausible about whether there is a reflection or not, and it is pretty straightforward to test. If there is a reflection, then as an EM wave (at their resonant frequency) passes by two antennas, you will be able to adjust their spacing to get a moderate null in the magnitude of the Rx signal. If there is no reflection, there will be no multipath null that can be found.

 

[jim hardy]

Hmmm just thinking about a Yagi

The directors and reflectors surely reradiate but then they're unterminated.

Does the discussion need to distinguish single versus multi element antennas ?

 

[berkeman]

The directors and reflectors surely reradiate but then they're unterminated.

Yeah, I think we would all agree on what happens with unterminated or shorted antenna elements.

Does the discussion need to distinguish single versus multi element antennas ?

I think we can stay with simple dipoles and monopoles for now -- that makes it easier to test or simulate.

 

[Baluncore]

There will be some scattering or reflection, but it will not always be 50% of the incident energy.
The extreme 50% belief comes from the simplistic analysis of getting maximum power into a load from a fixed voltage source with a fixed source resistance.

The efficient transfer of the limited energy incident on a receive antenna requires the source or loss resistance be small compared with the load resistance.

There is a shadow immediately behind a receive antenna that is quickly filled with energy refracted in from the part of the wavefront that passed outside the antenna aperture. That makes a larger but shallower shadow in the region beyond the antenna.

I cannot see how a multipath test would work. If you place a second antenna within the near field of the first, the antennas will couple and confuse the analysis. Standing waves will make it a challenge.

Does the discussion need to distinguish single versus multi element antennas ?

Yes. A single dipole or a scattering aperture only get one chance to trap the passing energy. By adding a reflector and or directors the effective aperture increases along with the gain. That is why we do not use simple dipole antennas except for very long wavelengths.

 

[berkeman]

If you place a second antenna within the near field of the first, the antennas will couple and confuse the analysis. Standing waves will make it a challenge.

I wasn't proposing placing them in the near field. They can be far enough away not to distort each other's near field, say a few wavelengths away. The depth of the null would indicate what fraction of the incident energy was being reflected away (or scattered sideways I guess).

There will be some scattering or reflection, but it will not always be 50% of the incident energy.

That seems reasonable.

It occurs to me that such scattering must be accounted for in phased array receive antenna configurations. I"ll have a look around to see if the literature addresses inter-antenna scattering in phased array antenna systems...

 

[tech99]

Krause ends section 2:14 with the statement; “Although the above discussion of scattering aperture is applicable to a single dipole (λ/2 or shorter), it does not apply in general.”

Kraus does not support your original assertion that a receive antenna reflects half the incident energy.
Kraus does point out that half the energy can be reflected when a receive antenna is operated into a load having the same impedance as the antenna loss resistance. But that deliberate analogy with a generator having an internal resistance, makes the same assumption that the voltage is fixed, independent of load, and that unlimited power is available. Receive antennas have a fixed power available and maximise energy efficiency by minimising loss resistance relative to output impedance.Kraus does not confirm that a paraboloid will reflect half the incident energy.
A “paraboloid” is used as a mirror to direct axial energy onto a transducer at the focus. A parabolic reflector is made of conductive metal, so it reflects close to 100% of the incident energy. You must separate the analysis of the reflective surface from the analysis of the transducer at the focus.

While that is certainly an interesting idea, a conductive screen reflector could make the same shadow. When a signal is canceled by having the reverse phase signal added, that simple cancellation would appear to annihilate two sources of energy, so it must be impossible. There must be other directional effects that conserve energy.

It is unclear what Kraus means by saying it does not apply in general. He seems to be distinguishing between aperture antennas and dipoles, where the collecting area is greater than the physical area.

Antenna reflects half power. I have never said that unlimited power is available. I said unlimited power is not available. The max that can be extracted is half.

Paraboloid. If the energy arrives at the focus, then the feed unit will re-radiate half the power and it will create a beam going back towards the transmitter.

Conductive screen. It would reflect the power back to the transmitter, whereas a 377 Ohm sheet backed by a reflector does not do so.

May I mention that an omni directional receiving antenna will reduce the incoming EM wave all around it, not just as a shadow. This is because it radiates a cancelling wave itself.

Of course, as I mentioned previously, parasitic elements reflect all the incident power, and power can be extracted as desired by altering the terminating resistor, up to a max of half.

 

[Baluncore]

I wasn't proposing placing them in the near field. They can be far enough away not to distort each other's near field, say a few wavelengths away.

The near field is usually specified as being out to 60 wavelengths. Now I am confused.

It is unclear what Kraus means by saying it does not apply in general. He seems to be distinguishing between aperture antennas and dipoles, where the collecting area is greater than the physical area.

I believe Kraus is considering only isolated short dipoles and apertures in section 14:2, Kraus is being careful to make sure the analysis in that section is not applied to antennas in general.

Conductive screen. It would reflect the power back to the transmitter, whereas a 377 Ohm sheet backed by a reflector does not do so.

Correct. Now consider replacing the space cloth with an array of dipoles in front of the screen. (Ideally spaced λ/4 from the screen).

Paraboloid. If the energy arrives at the focus, then the feed unit will re-radiate half the power and it will create a beam going back towards the transmitter.

As above, the antenna at the focus of a paraboloid is also operated against a small screen. You must analyse the antenna at the focus with the screen, independently to the paraboloid. Then multiply "the pattern of the antenna at the focus" by "the array factor of the parabolic reflector aperture".

May I mention that an omni directional receiving antenna will reduce the incoming EM wave all around it, not just as a shadow. This is because it radiates a cancelling wave itself.

But while the direct energy arrives from one direction, the scattered energy is re-radiated in the radial pattern of the antenna. That forms standing waves in all directions, except directly behind the antenna in the close shadow.
So it does not reduce the field with destructive interference everywhere near the antenna. In many places, the waves will sum to increase the field by constructive interference.

 

[berkeman]

The near field is usually specified as being out to 60 wavelengths. Now I am confused.

Me too. Can you provide a reference link? The more normal definition is a couple of wavelengths. I will also look for a link.

 

[berkeman]

https://en.wikipedia.org/wiki/Near_and_far_field

resulting in relative lack of near-field effects within a few wavelengths of the radiator.

 

[berkeman]

From my Google searching... I guess I need to subscribe to the IEEE to get to this whole paper...

http://ieeexplore.ieee.org/document/1296172/

Abstract:
This paper discusses the amount of power, which is scattered and absorbed by a receiving antenna and in particular, whether an antenna can absorb the entire power incident upon it. The absorbed and scattered power from dipole arrays in either free space, or over ground plane is considered. By defining a suitable "aperture efficiency" for the receiving case, a dipole array without a ground plane can best absorb half of the incident power (scattering the rest), while an array over a ground plane can absorb all of the incident power. It is shown how aperture efficiency varies with load impedance, which is of practical interest for array designers.

 

[Baluncore]

I guess I need to subscribe to the IEEE to get to this whole paper...

The conclusion is reproduced, outside the paywall, at the end of the abstract page.
D. Pozar. “Scattered and absorbed powers in receiving antennas”. 2004.
“In closing, I think that Allan Love's intuition that it is possible for a receiving antenna to “capture” all available incident power, without re-radiating or scattering any of that incident power, is correct, and is demonstrated by these results. In fact, of course, practical antenna performance would be seriously hampered if this were not the case.”

 

[Baluncore]

Following the 2004 paper by David Pozar, “Scattered and Absorbed Powers in Receiving Antennas”, the refinements and discussion continued. It reaches some maturity by 2009 in an interesting paper by Do-Hoon Kwon and David M. Pozar. "Optimal Characteristics of an Arbitrary Receive Antenna". The fundamental conclusions of the paper appear to be that;
1. An isolated dipole or an array of dipoles will have an aperture efficiency of 50%.
2. Placing the dipole or array over a ground plane will increase the aperture efficiency to 100%.

For those of you visiting Tohoku University see; http://www.sawaya.ecei.tohoku.ac.jp/common/item/pdf/doctor/100506.pdf

 

[tech99]

Me too. Can you provide a reference link? The more normal definition is a couple of wavelengths. I will also look for a link.

We need to distinguish two types of near field:-

(a) The Radiation Near Field, which is the region in front of a directional antenna where the pattern is not fully formed. It may be considered as extending to a distance equal to the Rayleigh Distance, approximately (Diameter^2) / 2*Lambda. It is also called the Fresnel region. In this region, an aperture antenna has an essentially parallel beam.

(b) The Reactive Near Field, or Induction Field, located very close to the antenna, where the fields due to the voltages and currents on the antenna are predominant. Usually extends to Lambda/2*pi. These fields contain stored energy rather than radiated energy.

I think Berkman was intending (b).

 

[tech99]

Following the 2004 paper by David Pozar, “Scattered and Absorbed Powers in Receiving Antennas”, the refinements and discussion continued. It reaches some maturity by 2009 in an interesting paper by Do-Hoon Kwon and David M. Pozar. "Optimal Characteristics of an Arbitrary Receive Antenna". The fundamental conclusions of the paper appear to be that;
1. An isolated dipole or an array of dipoles will have an aperture efficiency of 50%.
2. Placing the dipole or array over a ground plane will increase the aperture efficiency to 100%.

For those of you visiting Tohoku University see; http://www.sawaya.ecei.tohoku.ac.jp/common/item/pdf/doctor/100506.pdf

On this topic, Kraus says that a flat metal sheet has an aperture that collects the energy over four times its area. When configured as a receiving antenna, such as a paraboloid with a resistor at its feedpoint, the antenna can then have a maximum aperture equal to its physical area.

 

[tech99]

As above, the antenna at the focus of a paraboloid is also operated against a small screen. You must analyse the antenna at the focus with the screen, independently to the paraboloid. Then multiply "the pattern of the antenna at the focus" by "the array factor of the parabolic reflector aperture".

On a point of clarification, pattern multiplication applies to the case of a broadside array of unit antennas, where the overall pattern is the product of the patterns for the complete aperture and that of individual radiators. In the case of the paraboloid, if we increase the gain of the feed, it causes it to have a narrower beam, and this reduces the illuminated area of the dish, lowering the overall gain of the system and broadening its pattern. It is not correct to multiply the patterns (or gains) of the feed and reflector.

 

[Baluncore]

On this topic, Kraus says that a flat metal sheet has an aperture that collects the energy over four times its area.

Where does Krause write that? The devil is in the detail. Kraus has several different definitions of aperture; A_physical, A_geometric, A_collecting, A_scattering, A_effective, A_maximum-_effective, A_receive-effective, and A_transmit-effective. It is important to use consistent definitions which makes mixing definitions from different references very risky. You can hide almost anything behind an aperture definition.

It is not correct to multiply the patterns (or gains) of the feed and reflector.

Do you think pattern multiplication only applies to discrete point arrays and not to continuous apertures? Do you have a reference that shows the product of an illuminated continuous aperture and the driven element gives incorrect results?

 

[sophiecentaur]

It is not correct to multiply the patterns (or gains) of the feed and reflector.

Because it is not an element factor times and array factor; it's all one bit radiator and not a 'separable variable' problem. Each element on the surface of the dish is fed differently and 'pointing in a different direction'.

 

[tech99]

Do you think pattern multiplication only applies to discrete point arrays and not to continuous apertures? Do you have a reference that shows the product of an illuminated continuous aperture and the driven element gives incorrect results?

It would be very difficult for me to find a paper describing something that is incorrect.

Just for clarification, the aperture of a dish antenna can be considered as an array of Huygens Sources.
The feed is just one means of creating, so far as possible, a uniform amplitude and phase across the aperture. But significant radiation from the feed does not directly reach the receiver, and so it does not form an array in conjunction with the aperture. Therefore, we cannot apply pattern multiplication because the two sources do not constitute an array.

 

[sophiecentaur]

Do you think pattern multiplication only applies to discrete point arrays and not to continuous apertures? Do you have a reference that shows the product of an illuminated continuous aperture and the driven element gives incorrect results?

Pattern multiplication can only be used where the individual sources have identical radiation patterns (i.e. main beam direction). When you are using a paraboloid, this is not the case. Fourier Optics tells us that a paraboloid (or a lens) produces the inverse Fourier transform at infinity of an object at the focus. Altering the directivity of the feed will actually have the inverse effect on the overall beam width which is not a 'multiplicative' effect. (The very opposite, in fact.)
There are situations where a continuous set of radiators can be analysed by multiplying but it's not a general thing.