Consider the following super-heterodyne receiver enter image description here

The carrier frequency of the local oscillator is chosen to be

$$ f_{LO} = f_c + f_{IF} \tag{1} $$

where $f_c$ is the transmitted carrier frequency and $f_{IF}$ is the intermediate frequency. My goal is the derive the constraint of the RF filter bandwidth such that it rejects the image frequency, located at

$$ f_c' = f_c + 2 f_{IF} = f_{LO} + f_{IF} $$

To make it easier to grasp, it follows the picture of the spectrum of the transmitted signal (with bandwidth $B_c$) along with the unwanted image frequency and the RF filter bandwidth that we are trying to define enter image description here

For me, it is clear that

$$ f_c + \frac{B_{RF}}{2} < f_c + 2 f_{IF} - \frac{B_{c}}{2} $$ or $$ B_{RF} < 2 \left(2f_{IF} - \frac{B_{c}}{2} \right) $$

But the reference where I was reading⁽¹⁾ says that $B_{RF} < 2 f_{IF}$, which makes no sense to me. If anyone can clarify this point I would really appreciate that!

⁽¹⁾: Proakis, John G., and Masoud Salehi. Fundamentals of communication systems. Pearson Education India, 2007.


2 Answers 2


The OP is using $B_{RF}$ to refer to the passband of the front-end filter ahead of the mixer, and $B_c$ to refer to actual occupied bandwidth of the signal. With that definition, the OP is correct with his formula, as long as $B_{RF}$ is given by the rejection requirement for the alias region and not a passband for the signal of interest: for example a -50 dB bandwidth rather than the typical use of a -1 dB or -3 dB bandwidth. Further such filters can and often are asymmetrical favoring rejection where it is needed, so this approach is not the best way to define the RF filter requirement. Instead I recommend defining the passband to pass the channel of interest with a maximum ripple and group delay variation requirement, and a stop band defined with sufficient rejection as needed to reject the alias region.

I suspect that the text intended for $B_{RF}$ to represent the maximum possible theoretical bandwidth of the modulated RF signal itself (with use of impossible to realize brick-wall filters to reject the alias region). When $B_{RF}> 2f_{IF}$ the spectrums will overlap, meaning the spectrum of our RF signal of interest, with the spectrum of whatever is in the alias region, as both will land in the IF frequency region using a real down-converter. Even if there is no signal in the alias region, if the noise floor there is the same as the noise in our bandwidth of interest, the total noise will sum in power and a 3 dB penalty in noise figure will result.

limit condition

Filtering out the local oscillator (LO) feedthrough as Tim Wescott points out in his answer is certainly a benefit that we would also achieve with this condition, as we don't want the LO to reach the antenna and radiate. This is a feature but not a theoretical hard requirement as the text, based on the theoretical limits, not practical application, is presenting. A case in point is a Zero-IF implementation where such filtering of the LO is not possible (the LO is centered on the RF bandwidth), in which case other techniques are used such as the reverse isolation of the front-end low noise amplifier, in addition to carrier feedthrough nulling techniques at the mixer, and the additional use of isolators if further reverse isolation is needed. These same techniques can be applied here in an super-heterodyne architecture with an IF frequency, whereas the deleterious effects of aliasing cannot be avoided in a single mixer architecture if $B_{RF}>2f_{IF}$.

For the purposes of the anti-alias filter design, it is generally better to define a bandwidth region that you would pass, a bandwidth region that you would reject, and what is in between is the transition band. The complexity of the filter is driven by the transition band that we can allow between these two.

That said, the OP's interpretation of the frequency locations and bandwidths for rejection are correct as drawn.

As far as filter design requiements, The passband shoud be defined to pass $B_{RF}$ with whatever distortion requirements are necessary, and the stop band is to start at $2f_{IF} - B_{RF}/2$ offset from $f_c$, and should be at least $B_{RF}$ wide, and everything else is "don't care", as I show this in the graphic below.

filter requirement

With that understood, and bringing in practical considerations, we see that the real requirement is $B_{RF}>2KB_{IF}$ with $K>1$ as a complexity constant. The closer $K$ is to 1, the more impossible it is to design an actual filter since there will be no transition band. For example below shows the filter requirement with $K=1.1$. Cases like this with small $K$ is a motivation for using an image reject mixer.

tight filter constraint

  • $\begingroup$ "There is no distinction between $B_c$ and $B_{RF}$". Well, AFAIK the RF filter brandwidth ($B_{RF}$) is usually much wider than the signal bandwidth ($B_c$), that is, $B_{RF} \gg B_c$. Even in my pic, you can see that $B_{RF}$ is much wider than $B_c$. Can you explain more why you make no distinction between both? $\endgroup$ Apr 7 at 19:54
  • 1
    $\begingroup$ @RubemPacelli Yes agreed, but it would be the stop band not passband that I would want to define in that case. Of course if they are the same, it's a brickwall filter which isn't feasible. We care about REJECTING the image, not passing it, which means we need a rejection criteria at that frequency. So it isn't about the filter passband but it's stopband that I'm trying to clarify. $\endgroup$ Apr 7 at 20:01
  • 1
    $\begingroup$ RF bandwidth means the bandwidth of the RF filter. Indeed, my post make it a little bit ambiguous as it could be referring to the bandwidth of the transmitted signal. Let me make it clearer $\endgroup$ Apr 7 at 20:03
  • 1
    $\begingroup$ Note that of lower frequency of the alias region corresponds to $f_{LO}$. Hence, we would also be rejecting the carrier frequency and avoiding the local radiator effect... $\endgroup$ Apr 8 at 10:18
  • 1
    $\begingroup$ and yes that is correct. You get the benefit that it rejects the LO from radiating but my point on that is it is not a driving requirement. We can avoid that radiation using other methods--- but if the RF occupied BW overlaps with the alias, we can't recover from the damage. $\endgroup$ Apr 8 at 10:19

If $B_{RF} \ge 2f_{IF}$ then the RF filter will include the LO frequency. Regardless of any theory, in practice there will be a strong tendency for your receiver to be (hopefully very) local radiator of the LO frequency.

So the authors are probably giving that rule to dodge the LO signal, which automatically takes care of the images.

Note that what you really want to do is inventory every single possible source of spurious responses, including every reasonable harmonic and intermodulation product of your RF and IF (or IF's, if you have a multi-stage conversion scheme). Then make a giant table with estimates of what's going to leak where and how it's going to get beaten down by the various filters in the radio.

Then you can include a prediction of all the spurious responses in your proposed design.

  • $\begingroup$ Interesting, what do you mean by "local radiator"? $\endgroup$ Apr 7 at 20:15
  • $\begingroup$ Unless you go to great lengths to shield it, a little bit of your LO radiation leaks out, and can get back into the input. $\endgroup$
    – TimWescott
    Apr 8 at 0:40
  • $\begingroup$ I think we would be more concerned about it leaking out and getting into someone else's input! (The LO is at our input regardless and the leakage typically wouldn't be near enough to saturate our front end, and it won't create any other intermods that aren't at the mixer output already). Regardless this is all a really good and important additional consideration in radio design; managing all spurious signals and a big part of frequency planning and architecture, thanks for bringing it up Tim! $\endgroup$ Apr 8 at 10:41
  • $\begingroup$ "concerned about it leaking out and getting into someone else's input" -- I bet every consumer AM radio receiver manufactured back in the tube days, and quite a few today, leak a bit of their LO signal. Probably FM, too. I haven't actually set one up next to a spectrum analyzer, but if you look at the circuits and the layout, it's inevitable. Professional radios are going to be much better, of course. $\endgroup$
    – TimWescott
    Apr 8 at 16:33

Your Answer

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

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