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?

  • 1
    $\begingroup$ While not an exact duplicate as a question, the answers to How do I add AWGN to an I and Q representation of a signal? describe in detail why $X$ and $Y$ can be considered to be independent Gaussian processes so that the real and imaginary parts of the transmitted symbol are affected by different noise contributions (even though both noises derive from the same bandpass noise process). $\endgroup$ Commented May 19, 2020 at 19:58

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Joe Mack
    Commented May 17, 2020 at 23:49
  • $\begingroup$ @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. $\endgroup$ Commented May 18, 2020 at 0:23
  • 2
    $\begingroup$ 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) $\endgroup$
    – Joe Mack
    Commented May 18, 2020 at 1:06
  • $\begingroup$ 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) $\endgroup$
    – Joe Mack
    Commented May 18, 2020 at 1:23
  • 1
    $\begingroup$ @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. $\endgroup$
    – Joe Mack
    Commented May 18, 2020 at 20:19

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