1. Aug 12, 2017

### MarkWW

2. Aug 12, 2017

### Staff: Mentor

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

3. Aug 12, 2017

### jim hardy

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.

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..

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

Last edited: Aug 12, 2017
4. Aug 12, 2017

### 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 licence 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.

5. Aug 12, 2017

### MarkWW

Thanks a lot!

6. Aug 13, 2017

### tech99

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

7. Aug 13, 2017

### davenn

well, I have never heard that before !

you have a good link or 2 to support that ?

8. Aug 13, 2017

### tech99

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.

9. Aug 13, 2017

### Staff: Mentor

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

10. Aug 13, 2017

### Baluncore

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.

11. Aug 13, 2017

### davenn

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

Dave

12. Aug 13, 2017

### Baluncore

Terman bases his analysis on the voltage induced in the antenna that causes a current to flow. The situation is now modelled 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.

13. Aug 14, 2017

### davenn

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 situation

Dave

14. Aug 14, 2017

### Baluncore

15. Aug 14, 2017

### 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.

16. Aug 14, 2017

### 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..

17. Aug 14, 2017

### davenn

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

without actual figures, that is meaningless

18. Aug 14, 2017

### Baluncore

19. Aug 14, 2017

### Staff: Mentor

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.

Last edited: Aug 14, 2017
20. Aug 14, 2017

### Staff: Mentor

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?