While studying superheterodyne-based spectrum analyzers I ran into the concept of intermodulation products explained through a \$f_{RF}-f_{LO}\$ graph.
If it's not clear enough, the dashed lines are simply references \$f_{RF}=kf_{LO}\$ (\$k=1,2\$), while the others are \$f_{RF}=nf_{LO}\pm f_{IF}\$. \$f_{IF}\$ is the intermediate frequency of the system, at which the output of the mixer is filtered; \$f_{RF}\$ is the frequency of the input radio frequency signal; \$f_{LO}\$ is the variable frequency of the local oscillator.\$f_{LO}\$ is chosen in such a way that \$f_{LO} - f_{RF} = f_{IF}\$, so \$f_{RF,max} < f_{LO,min}\$ must hold: in the example, \$f_{RF} \in [0, 3]\, GHz\$ and \$f_{LO} \in [3, 6]\, GHz\$. No problem so far.
Then my lecturer said "if we set a fixed \$f_{LO}\$, then we have more than one curve at \$f_{IF}\$ output". That is true, of course, because \$f_{RF}=nf_{LO}\pm f_{IF}\$ holds and we get \$f_{IF}=\pm (f_{RF} - nf_{LO})\$. What I don't understand is the fact that to me the only segment that matters is the one that projects \$f_{LO}\$ in the range \$f_{RF} \in [0, 3]\, GHz\$ (so, the one marked as \$1-\$ in the picture), since we can suppose to provide a 3 GHz bandwidth spectrum analyzer with a signal that resides in that frequency range (I guess...). That would make more sense to me if we considered not only \$f_{RF}\$, but also the \$mf_{RF}\$ components of the mixer output, so that we could obtain \$f_{RF}=\frac{1}{m}(nf_{LO}\pm f_{IF})\$ in the spectrum analyzer range.
Of course, that doesn't make much sense, otherwise there would be no intermodulation products affecting the measurement. Where am I wrong?
