Let's assume we have a carrier at $f_{c}$, and a message at $f_{m}$ ($f_{c} \gg f_{m}$). Amplitude modulation is used to modulate the message on the carrier. Let's call the modulated carrier $\text{rf}(t)$. At the receiver, we first have the antenna, followed by a polyphase filter, which creates an I and Q channel: $$ \begin{align} I_{pp}(t) &= \text{rf}(t) \\ Q_{pp}(t) &= \text{rf}\left( t-\frac{1}{4f_{c}} \right) \end{align} $$

After the polyphase filter there is a Zero-IF I/Q mixer circuit, which brings the carrier to DC and the message back into baseband. Note that the phase between the polyphase filter output and the local oscillator of the mixer is random.

$$ \begin{align} I_{bb}(t) &= \Re{\left((I_{pp}(t)+iQ_{pp}(t)) e^{-(i\omega_{lo}\,t+\theta_{rand}\,)}\right)} \\ Q_{bb}(t) &= \Im{\left((I_{pp}(t)+iQ_{pp}(t)) e^{-(i\omega_{lo}\,t+\theta_{rand}\,)}\right)} \end{align} $$

Now what I am wondering about is what is the phase between the baseband I and Q channel ($I_{bb}$ and $Q_{bb}$), after the mixer? First intuition for me was $90°$, however that cannot be the case, since the shift introduced by the polyphase filter is $90°$ with respect to $f_c$, and not $f_m$. So are I and Q going to be in-phase? Or might the phase also be 180°? Does it depend on the phase between polyphase filter output and the local oscillator?

  • $\begingroup$ Is there a LPF after the mixer? Also: I think things would be much clearer if you wrote down the equations for each signal, including their phase, and which phase differences you're interested in. $\endgroup$
    – MBaz
    Commented Feb 9, 2022 at 20:47
  • $\begingroup$ Thanks for the feedback, for this example there is no LPF $\endgroup$ Commented Feb 10, 2022 at 8:58
  • $\begingroup$ My suggestion: write $Q_{pp}(t)$ not as a time delay, but as a phase shift. Substitute into the baseband equations and simplify terms to find the phase. $\endgroup$
    – MBaz
    Commented Feb 10, 2022 at 14:42

1 Answer 1


The phase between the baseband I and Q channel will be 90°. The absolute phase between those nodes and other nodes in the system such as the mixer output can be anything, based on the delay in the system (and if a fixed delay is not typically a concern). Each component will add delay but to maintain quadrature in the I and Q paths (which would be the intention) the polyphase filters and everything else in the path would be designed to impose the same delay on the I and Q paths.

The representations given of the I and Q outputs at the polyphase filter are not accurate as they suggest time delay and phase are equivalent. To implement true quadrature we must maintain 90° phase relationship across the entire bandwidth of the channel and not just at $f_c$. Note that a true time delay is a linear phase with frequency, so if implemented as such to get quadrature at $f_c$, we will have necessary phase deviation on each side of $f_c$. Quadrature translation with 90° at all frequencies within our signal of interest is accomplished with a quadrature local oscillator, and the result of that is not equivalent to the time delay relationship the OP first introduced.

For signals that are sampled much higher than the bandwidth, then such approximations can be made, with tolerable quadrature error.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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