Can someone explain me how quadrature signals (IQ) are received? This all makes sense when we are talking about transmitting, the to parts are summed up together and generate the actual signal, but for transmitting, I don’t see how we would get the two compunents if the signal. Thanks in advance.
I recommend you read this post from Elettra Venosa : https://www.eeweb.com/blog/elettra_venosa/comparison-of-iq-sampling-with-dual-core-adcs-vs-ping-pong-single-core-adc which does a good job of explaining it.
Yes, I’ve already seen that and I’m completely familiarized with this signals apart from one thing. What I don’t understand is how the I and the Q are calculated when receiving a signal. I have an image here that shows what is happening.
We have one signal and from that we calculate the I and Q components. They’re not equal (ignoring the 90° pase shift) and I haven’t found anything explaining how this calculation is done. I understand we have an local oscillator but I don’t understand it’s function in the scheme apart from shifting the signal, perhaps that’s it’s only function. But if the local oscillator’s only function is just shifting the signal, I don’t understand how come the I component is different from the Q component (again, ignoring the 90° pase shift).
Ignoring the 90 degree phase shift is the wrong thing to do, that’s the entire point of it!
A digital way of thinking about it is that each quadrature of the output signal is +I, +Q, -I, -Q. The analogue way of course is that the signals are multiplied by the cosine and sine of the LO at every point in time, so f(t)=I.cos(wt) + Q.sin(wt).
You can’t just recover I and Q from an instantaneous sampling of the signal. Instead, it’s recovered by performing the inverse function on the signal over an entire cycle of the LO. Conveniently, you can consider this as a -90 degree phase shift, and so it’s just inverting the Q signal.
As the LO frequency is much higher than the signal frequency, we can consider the I and Q inputs to be constant over a complete LO cycle, so if you re-apply the in I transform (cosine) then everywhere it’s in phase, you’ll see that signal, and when it’s not in phase you won’t. Similarly with the Q transform (sine). At any single point, you don’t have enough information to reconstruct either signal, but considering the entire cycle you do.
If you’re at all software minded, it’s worth looking at what happens with a single frequency in the digital equivalent, the DFT (which also happens to be the basis of the FFT) and there’s a great introduction here: http://blogs.zynaptiq.com/bernsee/dft-a-pied/ (I realise that might seem more complicated, but it’s a well written introduction to the FFT and if you can begin to understand that, the solution your problem should be apparent).
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Thanks for the explanation! I didn’t fully understand it but I’ll keep reading and trying to understand.
@ralferoo I’ve been reading about the subject but what I can’t really understand is how I can be different from Q (apart the 90° phase shift). For example, in this paper http://www.ni.com/white-paper/3896/en/ . The I and Q are different but I don’t understand how this is possible.
If anyone still has questions about this, check my question on DSP stackexchange.