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Assume we transmit a bandpass signal over an AWGN channel adding the Gaussian noise contribution $n$

\begin{equation} Acos(\omega_ct) + n(t) \end{equation}

Further, the bandpass signal is generated by IQ-modulation. If we perform homodyne IQ demodulation + lowpass filtering at the receiver should the resulting in-phase & quadrature component not be affected by the same lowpass noise contribution? Instead the equivalent complex lowpass noise is modeled as

\begin{equation} N_{LP}=X+jY \end{equation} where the imaginary & real part are both Gaussian distributed and independent processes. Hence the real & imaginary part of the transmitted symbol are affected by different noise contributions.

Why is that?

fl0ta''
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It is simply because each sample of $n(t)$ has a random magnitude and phase by definition given as $n(t) = |n(t)|e^{j\phi(t)}$. With real and imaginary components as follows:

$$|n(t)|e^{j\phi(t)} =|n(t)|\cos(phi(t))+j|n(t)|\sin(phi(t)) $$

Real: $I(t) = |n(t)|\cos(\phi(t))$
Imag: $Q(t) = |n(t)|\sin(\phi(t))$

Since the phase and magnitude are independent for each sample, the I and Q components will also be completely independent. Given any I value, there is no constraint or dependence on what the Q value can be for that sample, and vice versa. (A dependence would exist if given the phase, or given the magnitude. And given both then there would be a one-one mapping between I and Q)

Consider the opposite case if the I and Q components of a Gaussian Noise process were dependent such as I = kQ, the resulting noise would stay on a fixed angle passing through the origin (such as staying on the 45° line if I = Q) rather than adding a random magnitude and phase to each sample.

Dan Boschen
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  • This is not zero correlation. The zero inner product shows that the real and imaginary parts have zero inner product over time, not over the set of all possible outcomes. In fact, this form shows that the real and imaginary parts are not stochastically independent; rather, they are simply π/2 radians out of phase, so knowledge of one part completely determines the other part. – Joe Mack May 17 '20 at 23:49
  • @JoeMac yes over time (and to be precise over an integer number of 2 pi cycles) the inner product is zero so over time there is zero correlation; the OP’s process is over time so this is consistent with that. I don’t see how if you had knowledge of one you can possibly determine the other; they are completely independent of each other. A(t) is complex with real and imaginary components. – Dan Boschen May 18 '20 at 00:23
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    Apologies; I didn't realize your A(t) (which must be the random detail of the signal) was complex-valued. In that case, the formulas for the real and imaginary parts are incorrect. Let A(t) = α(t) + jβ(t). For each t, α(t) and β(t) are RVs, and the Re part of the signal/process at time t is α(t)cos(ωt) - β(t)sin(ωt), while the Im part is β(t)cos(ωt) + α(t)sin(ωt). For these to be independent as RVs, then E[Re part × Im part] =E[Re part]×E[Im part] must be true for each t, where E is expectation, not time average. (1/2) – Joe Mack May 18 '20 at 01:06
  • • E[Re part × Im part] = (E[α(t)^2] - E[β(t)^2])cos(ωt)sin(ωt) + E[α(t)β(t)](cos^2(ωt) - sin^2(ωt)) • E[Re part] = E[α(t)]cos(ωt) - E[β(t)]sin(ωt) • E[Im part] = E[α(t)]sin(ωt) + E[β(t)]cos(ωt) • More algebra reveals relationships that must hold among the expected values of α(t), β(t), and products thereof, in order for E[Re part × Im part] =E[Re part]×E[Im part] to be true. But these algebraic requirements don't seem to be explained by the physical model. I suspect, rather, that independence is an assumption that simplifies models just enough to keep them tractable. (2/2) – Joe Mack May 18 '20 at 01:23
  • @Dan Thanks for the answer and the discussion! I get your point about the orthogonality of the sine and cosine. Also actual measurements obviously show that real and imaginary part are disturbed by independent gaussian noise processes. However, I don't completely get it in term of the physical process. If I receive my bandpass signal as specified above and I mix it down to baseband on two paths, i.e. once with a cosine (real part) and once with a sine (imaginary part) would I not mathematically see that both real & imaginary part are affected by the same noise contribution? – fl0ta'' May 18 '20 at 11:28
  • @fl0ta'' each noise sample that is added to a signal sample will be at a random amplitude and phase. Given the magnitude and phase is random, you could have ANY Q value for any given I value- there is no constraint nor any correlation / dependence. – Dan Boschen May 18 '20 at 12:51
  • (It we constrain either magnitude or phase then we would get a dependence, but then that would not be a complex AWGN process. – Dan Boschen May 18 '20 at 12:52
  • Ok, that makes it very clear if you put it like that. Many thanks! – fl0ta'' May 18 '20 at 12:55
  • @JoeMac yes! Thank you, will update to show the correction – Dan Boschen May 18 '20 at 13:01
  • @JoeMac I simplified it to be simply the magnitude |n(t)| to show the I and Q components as representative of sine and cosine components, but removed everything showing temporal correlation as I got your point that about the time averaging not being a factor. Does this look right to you now? – Dan Boschen May 18 '20 at 19:37
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    @Dan: I think I get the model: real-valued over-the-air noise made complex by basebanding. I hate to admit it, but I think that as long as one component has just sine and the other has just cosine of the same angle, they can't be independent. My non-rigorous intuition is the following. Fix time. Then "roll a die" and let the random angle ϕ assume a particular value. Now |n(t)| is another random variable (value still undetermined, because it is independent of ϕ), as are |n(t)|cos(ϕ) and |n(t)|sin(ϕ). Those last two are just scalar multiples of each other, so they can't be independent. – Joe Mack May 18 '20 at 20:19
  • @Dan I do have a quick follow-up comment: say I want to simulate a bandpass channel with AWGN in GNUradio with IQ-modulation & demodulation. How would I model the noise in passband? Because adding a Gaussian noise source in passband leads to precisely the odd situation you described where " the resulting noise would stay on a fixed angle passing through the origin (such as staying on the 45° line if I = Q) rather than adding a random magnitude and phase to each sample." – fl0ta'' May 19 '20 at 11:20
  • @Joe Yes, that is exactly my intuition as well. So where exactly are we thinking wrongly? – fl0ta'' May 19 '20 at 11:24
  • We need to avoid extended discussion as it will flag (and then ultimately annoy) the moderator. I will create a chat room so we can continue the interesting discussion. Please join the room here to continue: https://chat.stackexchange.com/rooms/108221/independence-of-i-and-q – Dan Boschen May 19 '20 at 11:41