There’s an old saying that we have one mouth and two ears so you can listen twice as much as you talk. However, talking and listening at the same time is fairly difficult and doing it with radio signals is especially hard. A company called Kumu Networks has an analog module that can use self-interference cancellation which allows transmitting and receiving on the same frequency with around 50 dB of the transmitted signal in the transceiver. You can see a video about Kumu’s claims its technology below.
You may think that cell phones and ham radio repeaters transmit and receive at the same time, which of course they do, but usually on different frequencies to avoid direct interference. A diplexer is a device that sorts out the two frequencies while a duplexer sorts them out by the direction of the signal, but they are tricky to use. A duplexer can operate on a single frequency in applications such as radar, and even then it is still very difficult to prevent leakage from the transmitter from overloading and desensitizing the receiver. Directional Coupler
While 50 dB might not sound like much, it is a factor of 100,000 which should open up new opportunities for radio transmitters and receivers to coexist even on the same frequency. The device is analog, so it uses circuitry to invert the transmitted signal and reincorporate it at the receiver.
IEEE Spectrum recently had a post claiming the company is releasing the K6 module that can be “easily installed in most any wireless system.” RF can seem like black magic, but we can envision how this should work in theory. You’d need to adjust the phase of the inversion network to match the phase delay between where you pick up the signal (presumably before power amplification) and where you mix the canceling signal with the receiver.
Sounds good, but then again, noise-canceling headphones sound straightforward too and we all know that even an expensive pair can leave something to be desired.
If your referring to power then dBW or dBm
If your referring to voltage the dBV
They’re referring to gain, the transmitted signal can be attenuated 50 dB by their signal cancellation filter
I think that he’s trying to say that the signal input to the receiver is 50dB down on that being transmitted. Hackaday would do well to pick up its game in properly describing these sort of technicalities IMO.
Yeah, I got it wrong with dBW and dBm as they’re absolute values.
Weather you talking about a power ratio or voltage ratio still needs to be stated.
W=V^2/R so they’re completely different.
“Weather you talking about a power ratio or voltage ratio still needs to be stated.”
No it doesn’t. One decibel means a ratio of powers equal to the tenth root of ten, full stop. In the special case that those different powers come from different voltages applied to the same impedance, then that implies a ratio of voltages equal to the twentieth root of ten.
A lot of engineering curriculum skips straight to quoting “20 log(V/Vref)” without mentioning the fact that bels and their decimal fractions are intrinsically a measure of power ratios, leading to precisely this sort of confusion about when a factor of 10 means 10 dB and when it means 20 dB.
The output of a transmitter is normally measured in power as it’s a power amplifier driving a simple and known impedance Signal in a receiver is normally measured as a voltage as its a voltage amplifier and often the impedances are complex and unknown.
So if 50dB is the correct way to explained this then –
Do you mean the voltage has dropped by 50dB in which case the power has dropped by 100dB
Or do you mean the power has dropped by 50dB in which case the voltage has dropped by 25dB
Right. Power. But it can be useful for voltage as 20 log (v ratio), or current ratio because power = E squared or I squared times a scalar. Also useful for 20 log distance or frequency ratios or constants for path attenuation.
“Do you mean the voltage has dropped by 50dB in which case the power has dropped by 100dB
Or do you mean the power has dropped by 50dB in which case the voltage has dropped by 25dB”
I mean that the power has dropped by 50 dB, in which case the voltage has dropped by 50 dB.
A drop of 50 dB means the ratio of powers is 10^-5. A drop of 50 dB also means, in the case of a constant impedance, that the ratio of voltages is 10^-2.5.
(Most engineers ignore the “in the case of a constant impedance” requirement for that voltage relationship to make any damn sense, and use a mangled concept of decibels that produces inconsistencies if you don’t always mangle the concept the right way in the right circumstances. Naturally, there are different communities that have different traditions for mangling it, and tend to generate nonsense when they try to cross-communicate. It may not be the worst sin against logic and mathematics that engineers are taught to commit, but it’s up there.)
Or to put it (hopefully) more simply, when you double the voltage, this quadruples the power, which is why a doubling of voltage is a 6dB increase. It’s not because “voltage” power is different from “watts” power; it’s because doubling the voltage IS THE SAME THING as quadrupling the power (+6dB). So no, it doesn’t matter whether you’re measuring volts or watts; -50dB is -50dB. And it IS -50, by the way; sign is important.
Also, let me point out that most engineers who deal with multiple impedances (i.e., all of them) are well aware of the effect of impedance on power. This is usually the reason for using interstage transformers – to step up the voltage when going from a low impedance output to a high impedance input. This may appear to produce gain if you’re measuring everything in volts, but it actually just decreases the loss that results from just dissipating power with a resistive impedance matching network. If they weren’t keenly aware of this, engineers wouldn’t use transformers at all in RF circuits.
Power is voltage squared divided by resistance.
Given that the resistance (impedance) doesn’t change this means halving the voltage reduces the power by a factor of 4, 2^2 = 4
Reducing the voltage by 10 (10dB) reduces the power by a factor of 100 (20dB).
Reducing the voltage by a factor of 100 (20dB) reduces the power by a factor of 10000 (40dB).
That’s just how the maths works, there is no escaping it.
You have 1000V into a 1 ohm load
Your voltage is 1 kV and your power is 1MW
You divide the voltage by 100
Now you have 10V and 100W
You went from 1000V to 10V which is 20dB
You went from 1MW to 100W which is 40dB
dB as a voltage is 10*Log(ratio)
dB as power is 20*Log(ratio)
The doubling is simply the square in the formula Power = Voltage squared divided by resistance.
ROB: “You went from 1000V to 10V which is 20dB”. Negative. 20 log(voltage ratio) is 40 dB. Apply 10 log to power ratio and 20 log to voltage, resistance, current, frequency, distance, apples and dollars. There is not a difference in dB between power and voltage. If you have a power that is 12.25 dB below another power, the voltages (at those 2 corresponding powers) have exactly the same dB difference, if the dB conversion factor is properly applied.
And that is exactly the point I raised.
From 1kV to 10V is s factor of 100
From 1MW to 100W is a factor of 10000
You say just use the right formula, well which is the right formula when it is not stated weather they are referring to a voltage or power reduction.
dB is strictly a ratio of 10*Log(ratio)
dBVo is also 10*Log(ratio) so dB is often substituted incorrectly as dBVo specifies it relates to a voltage and dB does not as it strictly a ratio
dBmo and dBWo is 20*Log(ratio) and it is specified that it relates to power.
Respectfully, you have it backwards. 10 log power (dBm and the like) 20 log voltage (dBuV, etc.) You can test it. Select an impedance for the power to work at. Select 2 power levels. 1. Determine the dB difference for the ratio of the powers (using 10 log) Calculate the voltage that would be at that impedance for the 2 power levels. 2. Calculate the dB difference for the ratio of the voltages (using 20 log) 1 and 2 match.
OK, I switched 10log and 20log in one post but that doesn’t matter because I demonstrated a difference
You say:2. Calculate the dB difference for the ratio of the voltages (using 20 log) 1 and 2 match.
Well that’s plane wrong dB can only be 10log by definition.
That’s precisely why we use dBm, dBW, dBmo, dBWo and even dBmop, to indicate this difference
ROB: Sometimes it’s hard to unlearn things that aren’t correct and I get that. Decibels as 10 log applies only to power ratios. https://en.m.wikipedia.org/wiki/Decibel Beyond that I’m not sure how to help.
When stretching the concept of a decibel to describe changes in non-power quantities, the implicit question is “What is the change in power due to changing this other thing, assuming everything else is constant?”
Want to represent differences in voltage in dB? Assume constant impedance, and power is proportional to V^2. Use 20 log(V/Vref).
Want to represent differences in bandwidth in dB? Assume constant spectral power density, and power is proportional to bandwidth. Use 10 log(BW/BWref).
Want to represent differences in distance in dB? Yeesh. You have to assume free-space propagation from a point source, where power goes as distance^(-2). Use -20 log(D/Dref).
From the very wikipedia page you linked –
The ISO Standard 80000-3:2006 defines the following quantities. The decibel (dB) is one-tenth of a bel: 1 dB = 0.1 B. The bel (B) is 1⁄2 ln(10) nepers: 1 B = 1⁄2 ln(10) Np. The neper is the change in the level of a field quantity when the field quantity changes by a factor of e, that is 1 Np = ln(e) = 1, thereby relating all of the units as nondimensional natural log of field-quantity ratios, 1 dB = 0.11513… Np = 0.11513…. Finally, the level of a quantity is the logarithm of the ratio of the value of that quantity to a reference value of the same kind of quantity.
ROB: Years ago I was picking dewberries in public easements and for the first 3 days I couldn’t find hardly any. After the 3rd day I found lots of them, even in areas I couldn’t find them before. Unlearning things is the same big picture thing. If you have 4 things in series with different dB gains and losses there is not a different answer in total loss based on whether you àre reading power levels or voltage levels. Your reference is about relationships of bels and dB and how logs work. Like I say not really next step on explanation other than that was covered.
If that’s what they meant (and I assume they did), then -50dB would have been the correct way of saying it.
I’m highly skeptical. But this is the almost the same techniques used by MIMO wifi routers. Turbo codes were met with the same skepticism by Viterbi himself when they were introduced at the IEEE conference. MIMO seemingly Violates Shannon’s theorem, however it does not when you consider that you’re putting out more power with more transmitters. Here, I can’t square it in my head, hopefully someone brighter than me ( not hard to find by the way )can explain it to all of us Beyond the slick sheet salesman / IPO explanation we got in the video.
Check out the IEEE article it’s way more clear. Basically, the analog component is doing this function: OutputToActualReceiver = InputFromAntenna – CurrentEmission. Estimating the current emission is all their black magic, but they assure that their “-CurrentEmission” term is very good so that 20 log(OutputToActualReceiver / InputFromAntenna) >= -50dB when the antenna is also emitting. This sounds very good (and indeed it is), but due to the current asymmetry of radio emission (there are cases with more than 120dB of difference between the emitted signal and the received signal from the other party), this is not enough to allow simultaneous emission and reception for the most case.
So, unless for very specific cases (like a radio jammer that would still want to listen to the jammed signal), there is no real use of this technology.
Also, please notice that even in the case that both Alice and Bob have this technology and are only using 50dB difference between emission and reception, then Eve can not receive anything anymore from Alice nor Bob (because she has no way to know about Alice and/or Bob emission profile, and can not thus cancel it). So this technology, if applicable, can only be done for 2 participant and not any other, so some kind of time or spectrum based multiplexing is still required.
So not so fast sounding the horns of triumph. Carrier cancellation has been commonplace for the last 10 years on VSAT communications where the transmitted signal is delayed and subtracted from the received signal to permit reuse of the same transponder frequency for both inroute and outroute carriers. They call it carrier in carrier. (or DoubleTalk) The only way to make it work is to use carefully selected FEC, modulation coding and power ratios which are less susceptible to carrier to noise interference. It reduces occupied bandwidth by about 40 not 50%. (because you have to spread to optimize s/n r) But because an uncontended megabit on VSAT is $ thousands per month, it is big news. In a real 2 way application, say a Motorola digital radio, you might have a transmit power of +47 dBm and a useful sensitivity in the range of -118 to -121 dBm. 50 dB carrier cancellation would make the world’s worst radio. (receive desensitivity at about -3 dBm). The most likely use for this is to augment other methods of cancellation, TDD or TDM or for links where C/N R isn’t critical.
I used to work at Comtech EF Data years ago. It was my first legit engineering job out of school. I was one of the firmware developers that worked on their modems and redundancy switches (w/ support for carrier-in-carrier). It was amusing looking at the non-volatile memory map and knowing that a particular bit was worth $30k+ (the one that enabled carrier-in-carrier.)
For those that are interested, here’s a whitepaper on carrier-in-carrier: http://www.comtechefdata.com/files/articles_papers/optimizing_satellite_communications_using_doubletalk_carrier-in-carrier___cdm-625_advanced_satellite_modem.pdf
The user manual for the CDM-625 modem also has a good section on carrier-in-carrier, including design considerations for engineering a link with it. One of the amusing things on that modem, is that with the packet processor, the fec cards, and the carrier-in-carrier card populated, it would have IIRC eight FPGAs covering all three of the major vendors. https://www.comtechefdata.com/files/mn-cdm-625_15_10-18.pdf
The carrier-in-carrier on the CDM-625 that I worked on had a pretty wide range of settings that it supported when it came to fec type, modulation type, code rate, etc.. It was more that there was a bunch of configurations that made no sense, and it was a convenient excuse when the responsible FPGA would fail that test at 125C =)
I researched in this topic years ago with software defined radio and the papers can achieve 120 dB with mixed HW/SW cancellation so I don’t get surprised by the 50 dB they claim. Also with 50 dB you can do much.
Let’s see it work when the transmitter puts out 100W PEP (that’s +50dBm)… or even 30W constant (+44.7dBm).
My bet is the transmitter would literally fry the receiver if the two were allowed to connect. Normally in transceivers that powerful there’s a RF relay switching the antenna between receiver and transmitter.
Looking at teardowns of cell tower radios, they manage to do full duplex with very close channels using very precise filters.
Let’s see how well it works in a radar transmitter. In a megawatt level transmitter, an ATR (automatic transmit-receive relay) is made from a series of gas tubes across a waveguide. When the transmitter runs, the gas tubes fire (usually no active triggering needed, a megawatt usually suffices) and prevents power from going down that section of waveguide, which leads to the receiver. A succession of gas tubes and limiters is used to cut down the leakage to where the receiver LNA can handle it. It recovers in microseconds (you’re blind for that time, which could be miles).
A lower power transmitter of only a hundred kilowatts peak might use a magnetic circulator, which gets the power going in the right direction and the receive signal back to the receiver; but it might have only 25 dB isolation so a receiver protector is still needed. Some have radioactive gas tubes (e.g., Prometheum) to get the gas excited. It fires in nanoseconds so the receiver has to be tested to survive that leakage. Prometheum 145 has a half-life of maybe 20 years, so you need to replace the tubes periodically.
Magnetic circulators are 3 port devices. The transmit power goes in port 1, out port 2 to the antenna. Signal coming in from the antenna on port 2 goes to port 3, where your receiver sits. All through the magic of magnets and ferrite.
Circulators don’t help much with poor antenna VSWR though.
Circulators and isolators (which have the third port loaded) come in handy when you have an amplifier that doesn’t tolerate a funky load. They don’t push more power out of a bad VSWR antenna but dissipate it safely in a load.
The major difference is that with RADAR, you transmit for a few microseconds and then receive for some number of milliseconds. The TR switches are there, as you say, to help the receiver recover faster from the overload. But there’s more involved than the TR switch and whatever circulators or other hybrids are used: there are near reflections, which to these devices are indistinguishable from other received signals, so it is not uncommon in RADAR to have a sensitivity time constant so the receiver sensitivity is set lower, by a few tens of dB, increasing with time (which equals range). Again, this is totally different from full duplex communications. In the systems I worked on in the military, we had receiver sensitivities of around -110 dBm and transmit pulses of 2-5 MW (+93 to +97 dBm) for a path loss tolerance of about 205 dB. Which is pretty amazing, to me.
I wish their was a mention of the term ‘RF Circulator’ which as been feature here several times — not directly but in passing related to RFID and such. I was able to find SMD components with 50 dB attenuation (just check the first few manufacturers for rf circulators and sort by isolation spec)
As 12AU76L6GC points out, carrier cancellation is not a new technology. Old analog desk telephones had hybrid transformers in them that actually had to be deliberately unbalanced so that you’d hear SOME of your own voice. This was called “sidetone.” Otherwise you’d feel like you were talking in to a dead phone line.
Hybrid technologies have been extended well in to the short wave spectrum I used them extensively in the 1980s on analog microwave basebands, which were basically chunks of the radio spectrum from 60 kHz to 2540 kHz. They were also capable of around 50 dB of isolation.
Unfortunately most radio paths require a lot more dynamic range than that. Local signal cancellation technology has been around for many decades. Even with good antennas and transmission line systems, a path loss 100 dB is quite optimistic. Path losses of up to 150 dB are commonplace for 4G cellular systems.
The bottom line is that while 50 dB is good, it is still insufficient if you expect to hear anything but your own signal while you are transmitting.
BTW: in a conventional repeater (full duplex communication) configuration it isn’t just the transmit frequency which is filtered out of the receiver by the duplexer. Transmitters also generate considerable wideband noise actually on the receive frequencies, so a bandpass filter is used to pass the receive signal from the antenna *and* a band reject notch filter is used to block the transmitter from creating receive channel spurious noise from reaching the antenna port. Also, circulators are generally used in conjunction with passive resonant circuits (generally cavities) as an input protection device to protect the RF amplifier from reflected power due to unpredictable impedances and power reflections from the duplexer. Because of the really poor performance associated with second harmonic spurii, circulators are generally never used without additional LC or cavity filters except on very low power applications.
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