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I am reading a chapter on VCO noise in "Design of CMOS phase-locked loops from circuit level to architecture level by Behzad Razavi";I am confused when the upconverted noise is writen as $N_I\cos\omega t - N_Q\sin\omega t$, which means the baseband noise(complex envelope of the upconverted noise) have a form of $N_I+jN_Q$; I have this confusion because I think the complex envelope has symmetric spectrum which implies it is real signal so Nq should be 0; please help me understand this, thanks! enter image description here

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The noise is actually never upconverted, it was already there in the bandpass channel. The textbook is simply modeling the noise as a quadrature signal.

To recap:

  • The transmitted passband signal is usually assumed to be noise-free.
  • The passband signal is real, and it occupies frequencies from $-W_2$ to $-W_1$ and from $W_1$ to $W_2$.
  • The receiver adds noise to the signal, usually in the first few stages of the analog front-end. This noise is passband: it occupies the same spectrum as the received signal.
  • The passband noise goes through the quadrature receiver and its complex envelope is calculated. In the process, the spectrum from $-W_2$ to $-W_1$ is discarded, and the spectrum from $W_1$ to $W_2$ is shifted to baseband.
  • As a result, the spectrum of the complex envelope of the noise is not symmetric.
  • In conclusion, we can think of the passband noise as a quadrature signal.

Here's an illustration:

enter image description here

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  • $\begingroup$ Thanks for answering, I still have question: why baseband one can be writen as a complex number: ni+nq*j,even it has a symmetric spectrum(implies it is real signal) $\endgroup$
    – Ziyuan Ning
    Commented Apr 26, 2021 at 23:15
  • $\begingroup$ The complex envelope of baseband noise is real; you are quite correct about that. What the text you're reading is doing, however, is bringing passband noise back to baseband, using a quadrature receiver. Then, each of the I and Q branches of the receiver will see different noise. The noise in the I branch is the real part of its complex envelope, and the noise in the Q branch is the complex part. $\endgroup$
    – MBaz
    Commented Apr 26, 2021 at 23:26
  • $\begingroup$ I don't understand the confict between "complex envelope has complex part" and "The complex envelope of baseband noise is real" $\endgroup$
    – Ziyuan Ning
    Commented Apr 26, 2021 at 23:35
  • $\begingroup$ Let me try a different approach. Consider a receiver that consists of a bandpass filter with center frequency $f_c$ and bandwidth $W$, followed by a quadrature receiver. The input is white Gaussian noise $n(t)$. Can you calculate the receiver's output? It is $n_I(t)$ in the I branch and $n_Q(t)$ in the Q branch; this can be written as a complex envelope $n_I(t)+jn_Q(t)$. $n_I(t)$ and $n_Q(t)$ are the result of filtering and downconverting the white noise at the input. $\endgroup$
    – MBaz
    Commented Apr 27, 2021 at 1:23
  • $\begingroup$ Yet another approach, this time from the transmitter. You generate two different baseband Gaussian random noise signals, $n_I(t)$ and $n_Q(t)$, each with bandwidth $B = W/2$. You upconvert them using a quadrature transmitter to a center frequency $f_c$; the bandpass noise will have bandwidht $W$. The important point is: This bandpass noise is statistically indistinguishable from the noise after the receiver's BPF (assuming equal variance, zero mean, etc). $\endgroup$
    – MBaz
    Commented Apr 27, 2021 at 1:28

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