Suppose I use an IQ mixer to downconvert an RF (i.e. 8 GHz) signal to IF (i.e. 50 MHz). For example, the RF signal might be \$V_\text{RF}(t) = A \cos((\Omega + \omega) t + \phi)\$ and the LO signal could be at frequency \$\Omega \$, resulting in IF signals

\begin{align} V_I(t) &= I \cos(\omega t + \phi) \\ V_Q(t) &= Q \sin(\omega t + \phi) \end{align}

where we've assumed that there are antialiasing filters on the I and Q ports rejecting the signals at \$2\Omega + \omega\$.

How is the mixer's "conversion loss" defined in terms of the RF amplitude \$A\$ and the IF amplitudes \$I\$ and \$Q\$? In other words, if I know my RF and I know the mixer's rated "conversion loss", what what is the equation giving the IF power in terms of the RF power?


An ideal mixer will mix an RF signal \$A\cos ((\Omega + \omega)t + \phi)\$ with a LO signal \$B\cos (\Omega t + \phi)\$ to give an In-phase output of \$\frac{AB}{2}\cos (\omega t + \phi)\$ + \$\frac{AB}{2}\cos ((2 \Omega + \omega) t + \phi)\$ making the conversion loss (after rejecting the \$2 \Omega + \omega\$ component) equal to \$-20 \log \frac{B}{2}\$ dB if the resistance at the ports is 1 Ohm (LOSS = RF Power - IF Power).

The quadrature phase would be zero because the original input RF signal has no quadrature component. If the input RF signal also had a quadrature component (i.e RF = \$A\cos ((\Omega + \omega)t + \phi) + A\sin ((\Omega + \omega)t + \phi)\$ then the loss suffered by the quadrature will be similar to the In-phase component's loss, nothing changes.

To answer the question you posed in your comments about defining IF power - The IF power is the power of the IF component i.e in our examples it is the power of the component at frequency \$\omega\$ ONLY, the other high frequency components do not factor in any of the power calculations. Also, the presence of both an I and Q component do not change the power equations in any way, the I and Q components have 2 different signal flows and are mixed by two different mixers.

An non-ideal (i.e real ) mixer on the other hand will have other losses which are brought about by a lot of reasons such as non-ideal port isolation (this leads to feed-through). These reasons make it difficult to have an analytical expression for loss as it will depend on a lot of electrical and environmental properties like temperature and the like. A mixer datasheet like the one you linked to gives you the conversion loss you can expect when using it under different environmental conditions and with different levels of LO power.

It is also important to note that you can't just reject the \$2\Omega +\omega\$ components and expect to get your converted signal, you have to reject all other components except the component at the IF frequency. This is because the mixer will produce many so-called spurious responses at \$m\Omega + n\omega\$, where m,n = 0, 1, 2, etc, you have to reject all these responses except the one you want (i.e the one at m = 0, n = 1 ) to ensure you recover the required signal.


The conversion loss will usually be specified at some specific LO power. If the LO is not at that power, then the conversion loss may be different.

The conversion loss simply relates the IF and RF powers, such that IF power = RF power - conversion loss. You can convert the power to amplitude if you like with \$P = V^2 / Z\$, considering the impedance \$Z\$ of the ports (where \$V\$ is the RMS voltage).

  • \$\begingroup\$ Defining conversion loss as IF - RF (in dB) would be fine if the IF power were defined unambiguously. In the case presented in the OP, there are not only two different IF signals (I and Q), but there are also higher frequency parts of each of those IF signals which are stated to be rejected by antialiasing filters. To answer the question unambiguously, could you specify exactly what you mean by the IF frequency? For example, perhaps conversion gain is \$P_I / P_\text{RF}\$ where \$P_I\$ is the total power of the I port including the \$\Omega + \omega\$ part... \$\endgroup\$ – DanielSank Dec 1 '16 at 1:02
  • \$\begingroup\$ typo: "If frequency" should have been "IF power". \$\endgroup\$ – DanielSank Dec 1 '16 at 1:10

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