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Normally shared frequency analogue1 radios (CB, aviation, etc.) can't receive when transmitting, because the signal being transmitted is much louder and will jam any signal from different transmitter.

Is it possible to somehow subtract the transmitted signal to be able to pick out signals from other transmitters even when transmitting? Is there any published design that allows receiving while transmitting or at least detecting that there is another transmission in progress?

1 Or digital, but with short packets collisions are less disruptive and there are other ways around the problem like slot allocation.

Jan Hudec
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    Stop and think about the enormous ratio between the voltage the antenna is being driven with while transmitting relative the tiny signal it picks up while receiving. – Olin Lathrop Mar 18 '14 at 18:13
  • Actually I think the first point to consider is what modulation you are using. For AM you would need a large dynamic range. For FM, its pretty much impossible by design. But in the digital domain, lots of methods exists. My favourite of which is CDMA, which uses orthogonal signal vectors (similar to how different frequencies are orthogonal under FT). – Aron Mar 19 '14 at 02:22
  • @Aron: That's why I mentioned analogue and gave frequency-modulated examples. – Jan Hudec Mar 19 '14 at 05:50
  • @drewbenn I believe that next gen will begin to use SDMA, which is really cool. Its already starting to show up in consumer Wifi. – Aron Mar 19 '14 at 07:10
  • @Aron: All that falls under the "other ways around the problem" for digital systems. I asked about analogue systems specifically because I wanted to know whether it is solvable in a way compatible with existing voice systems. – Jan Hudec Mar 19 '14 at 07:21
  • @drewbenn: Well, it's changing direction, so it still is just one frequency, but it won't behave like normal voice radio where everybody hears everybody. Which is a feature for many use-cases. – Jan Hudec Mar 19 '14 at 08:34
  • SDMA would in theory be able to resolve the local signal from the different remote signals. The typical way phased arrays work is by summing corralating signals across receivers (after a phase lag added) to recover the directional signal. Since the outward signal would have different phase characteristics, the signals would cancel out. – Aron Mar 21 '14 at 07:47

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This isn't trivial for a number of reasons, but it is possible. Part of the problem is that the receiver has to have a huge dynamic range to be able to receive the 1W signal and the tiny signal from afar just above the noise floor, and understand both without losing data.

One system that can help, though, is to remove the transmitted signal from the antenna before the receiver:

enter image description here

Some research points out that this can be made to work.

However, the point is that what you're asking is possible, but it's very difficult to do well.

Adam Davis
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  • +1 that's a good find dude and 73dB of cancellation is pretty good going although it's only a claim on their part. – Andy aka Mar 18 '14 at 20:45
  • @Andyaka Yes, I'll be interested to see others duplicate or improve on this work. – Adam Davis Mar 18 '14 at 21:01
  • It's going to get harder as you go higher in frequency because the link-loss between two radios is proportionally higher with increased frequency. Also, local reflections being variable are going be curved balls that might be a spoiler. – Andy aka Mar 18 '14 at 21:07
  • @Andyaka how do you figure link loss increases with frequency? – Phil Frost Mar 19 '14 at 11:26
  • S@Phil it's the Friis equations not me that figured this out. – Andy aka Mar 19 '14 at 11:37
  • What factors set the minimum practical turnaround time for transmit/receive? If the receiver can't blank its AGC circuit, it would have to reacquire the far-end signal level each tx/rx switch, but blanking the AGC shouldn't be hard. I would think that in a number of communications applications, being able to confirm that a packet header was received before sending the balance of the packet (but having the header recipient use the header to know precisely when and how to expect the packet, thus not needing a full second header) could improve efficiency. – supercat Aug 18 '15 at 16:33
  • @supercat I'm not sure what you mean. This system is full duplex - it can transmit and receive on the same frequency simultaneously. There is no turn around time. – Adam Davis Aug 18 '15 at 16:36
  • @AdamDavis: For true full-duplex operation, it's necessary that the amount of one's own signal that bleeds through be less than the amount of the far-end signal one receives. Most time-multiplex arrangements I've heard of use rather slow rx/tx switching. I was wondering if it would be possible to have an approach which used a much faster rx/tx switching time so that even within a packet the two parties would transmit alternately? – supercat Aug 18 '15 at 16:41
  • @Supercat That sounds like an interesting question. – Adam Davis Aug 18 '15 at 16:42
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Likely received power level at (say) 27MHz, 1km from a 1 watt transmitter can be calculated from the link loss equations of Friis. In free space and with isotropic antennas the link loss is: -

Link Loss (dB) = 32.5 + 20log\$_{10}\$(MHz) + 20log\$_{10}\$(kilo metres) = 32.5dB + 28.6dB + 0 = 61dB.

If the power output is 1 watt from that distance antenna (30dBm), the received power is -31dBm or 794 nW.

Trying to disentangle this tiny bit of power from the 1 watt you maybe transmitting would be a nightmare and this is the free-space scenario - over land and with objects in the way and trying to accommodate fades, the likely received power will be another 30dB lower.

Andy aka
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Typically full duplex radios utilize different RX/TX frequencies to avoid the mentioned problems of picking out the tiny received signal among the blasted transmission.

If you use something like time-slots, then it's not technically full duplex at that point. You can do that to get the illusion of full-duplex however. That depends on your application. For something like voice, which is low bandwidth, you can definitely time share the connection to provide the appearance of full-duplex.

Ryan Johnson
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  • How is time-slots any less full duplex than frequency slots? Surely any cell phone qualifies as "full duplex". Are TDMA phones actually not? – Phil Frost Mar 18 '14 at 19:17
  • @PhilFrost: Time-slots are not full duplex. And separate frequencies are not shared frequency. The question was explicitly not about either. – Jan Hudec Mar 18 '14 at 19:46
  • In an RF context, full duplex means transmitting while receiving at the same time instant. @jan-hudec sorry, somehow I missed the 'shared frequency' bit of your question. – Ryan Johnson Mar 18 '14 at 21:08
  • So, TDMA phones are not full-duplex communication devices, while CDMA or FDMA phones are? That does not make much since to me, since I can talk and listen simultaneously on any of these devices. – Phil Frost Mar 18 '14 at 22:30