1
$\begingroup$

I'm working on a project to stream video between 2fpgas by using QAM. While mapping symbols with different amplitude pulses(PAM) a question popped up in my mind. In real life how does communicating wireless devices know which amplitude levels to agree on for digitally modulating-demodulating bitstreams? Is there a standard for that?

$\endgroup$
2
$\begingroup$

n real life how does communicating wireless devices know which amplitude levels to agree on for digitally modulating-demodulating bitstreams?

You can't "agree" on levels, since the received amplitude depends on the attenuation on the channel, and that depends on distance and a lot of other factors.

So, your receiver depends on detecting something that it knows the right amplitude (or power) of – for example, a preamble, or an average squared symbol, or something like that.

How that is done depends on the system you look at – generally, this is part of the synchronization and equalization efforts every system has to make. And it really differs! For example, some continuously transmitting system will not insert any specific symbols for that, but depend on the receiver doing statistics. Others have a "silence" period followed by known symbols; other systems are packetized and depend on preambles with specific autocorrelation properties (but not necessarily 100% defined content). Other systems depend on a pilot tone, and there's many more options.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Just a quick comment to add that the receiver may also do AGC. $\endgroup$ – MBaz Sep 17 at 22:59
  • $\begingroup$ there should be some sort of standard otherwise how all digital devices that communicates with QAM will understand each other. I'm really really confused $\endgroup$ – Bahattin Bademci Sep 18 at 2:04
  • $\begingroup$ @BahattinBademci there is more to it than device A uses QAM and device B uses QAM so they will be able to understand each other, that is not true. The above answer explains that parts of the rx chain are system dependent. For example, one system might be a wideband wireless system and the other might be a cabled system but both use QAM as the modulation. This does not mean that you can just interchange the receivers and expect good results $\endgroup$ – Engineer Sep 18 at 16:08
  • $\begingroup$ of course there is more to it. I know that but apart from this it is essentially how you are packing your bits to transmit which is QAM. this is how it starts right? You are basically deciding like for example 0010 will be 0.5V 1000 will be 1.45V and if you dont know how you packed there is no point what method you are using in the receiver cause you dont know what the incoming signal means. $\endgroup$ – Bahattin Bademci Sep 18 at 17:50
  • $\begingroup$ no. That's not how it starts. Nobody standardizes on a specific voltage set: You always have to decide based on a reference signal from the transmitter. $\endgroup$ – Marcus Müller Sep 18 at 18:16
1
$\begingroup$

The important part is that the symbol constellation is agreed upon between transmitter and receiver. The received amplitude will most likely randomly fluctuate over time, but eventually the receiver processing gets to a point where it needs to generate symbol estimates. For example, look at the transmission of a single symbol of 16 QAM. Lets say that the red symbol is transmitted.

enter image description here

The transmitter puts some gain on the signal and puts it out on the channel. At the other side, the receiver will do a bunch of processing that might be system dependent (equalization and synchronization for example) before it can get to a point where a symbol estimate can be made. At this point, you might have something that looks like this:

enter image description here

The scale of the symbols is different now. But this is just a scaling that can be undone to match the agreed upon constellation. So before the receiver does the symbol estimates it does the scaling to match.

enter image description here

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ and my question is that what are the amplitude values or range of values that decides which symbol it's at the end? $\endgroup$ – Bahattin Bademci Sep 18 at 17:54
  • $\begingroup$ @BahattinBademci the receiver is totally free to scale the receive signal. in the QAM16 example, it only matters that you make a 4×4 grid. There's no absolute level necessary anywhere. $\endgroup$ – Marcus Müller Sep 18 at 18:17
  • 1
    $\begingroup$ @MarcusMüller yes it's not absolute value but it's a range. so how receiver decides for example which range is 0101? it just can not be that the receiver does whatever it wants. $\endgroup$ – Bahattin Bademci Sep 18 at 18:22
  • 1
    $\begingroup$ it doesn't do what it wants. It gets a reference from its received signal, and adjusts the signal amplification (to fit its arbitrarily chosen decision boundaries) or adjusts the decision boundaries (to fit the unpredictably attenuated signal). No absolute levels or ranges can be involved anywhere. Hence, no standard needs to specify absolute ranges, it just needs to say "you use 16-QAM, and the transmitter sends a reference symbol, and then it's your responsibility as receiver to decide correctly" $\endgroup$ – Marcus Müller Sep 18 at 18:43
  • 1
    $\begingroup$ @BahattinBademci your reasoning about using voltages to decide on the bits is wrong, I'm not sure where you got that from but I've never heard of that (could you include a reference for that?). If you want to think in terms of received signal levels, the part that will matter is the ratio between the received signal power and the receiver noise power. A system will have a minimum SNR requirement (link budget) $\endgroup$ – Engineer Sep 19 at 11:54
-1
$\begingroup$

Digital communications are defined by standard. Here some of them:

  • 3GPP for mobile phones
  • ETSI with DVB for satellite broadcasting and terrestrial TV
  • IEEE 802.11 for Wifi

Inside each of those standards, there is a part which describes modulation schemes and constellation diagrams.

All wireless vendors implement them as it is written so they are compatible between themselves.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.