# Wideband receiver harmonic spur suppression

Click here to learn about percent bandwidth

Click here to learn about multiplexers

**New for August 2023**. When designing a receiver system over a wide bandwidth, you need to be aware that LNAs can generate sizable harmonic spurs, especially when they approach compression. On this page we present ways to analyze the problem and filter out the spurs.

First, let's do some math and compute the maximum number of harmonics in a given bandwidth. Recall that percent bandwidth can only be between 0 and 200%. A 200% BW receiver works down to DC and is a special case we will ignore for now. A DC signal has an infinite number of harmonics!

How many harmonics are possible in a given bandwidth? We will save you the math. Here is an equation that determines how many harmonics are possible for a signal at the lowest frequency in a wideband receiver. For an integer number of harmonics "N", bandwidth ("BW") is:

BW=2*(N-1)/(N+1)

Or, solved for N:

N=(2+BW)/(2-BW)

Note that in the latter case, N should be rounded up to the nearest integer.

Let's look at a plot of maximum harmonics versus bandwidth. You can see the trend toward infinity at 200% bandwidth.

Let's look at an example receiver. Suppose you needed to process 3 GHz to 18 GHz. The required bandwidth is (as described on this page) is:

BW=2*(FH-FL)/(FH+FL)

BW=2*(18-3)/(18+3)=142.9%

Let's calculate N, the maximum number of harmonics, while we are at it (you could cheat by looking at the above graph)... or, you could have just divided FH by FL and got the same result but where would the fun be in that?

N=(2+1.429)/(2-1.429)=6

For 142.9 bandwidth, N works out to be an exact integer, so no rounding is required. There should be up to six harmonics (including the fundamental signal) in the spectrum for a 3 GHz signal, as shown here:

We confess that we totally made up those amplitudes, don't put any stock in them. And we practically had to hand draw that plot as Excel does not have a "spectrum analyzer" plot type built in.

One thought you might have is this: if I have N possible harmonics, do I need a multiplexer with N filters to prevent spurs? Absolutely not! Before we get to that topic, let's look at two approaches for filtering a broadband signal.

## Switched filter approach

Here, four separate filters are used to break up the bandwidth. One nice thing about this approach is that the bands can overlay, and they can have steep skirts. But you will have to know which band your incoming signal is in and switch to the appropriate filter. That's probably OK in communication applications, but in electronic warfare, you will have to spend some time searching for signals. Note that you might want to put filters in front of your LNAs to keep them from saturating due to strong out-of-band signals.

## Multiplexer approach

Here the receiver band is divided up by filters in a multiplexer configuration, without any switches. An advantage of this approach is that the receiver will perform over the full instantaneous bandwidth (you don't have to search for signals using switches). The loss in front of the LNAs is typically lower, so the noise figure is lower than the switch filter approach.

A down-side of the multiplexer approach is you are likely to have dips in the frequency response at the transitions. But in a hold-my-beer moment, you could eliminate the dips if you phase match the adjacent paths at the transitions. Good luck!

## How many filters do you need to prevent harmonic spurs?

Here's how to proceed... so long as your filter paths have less than 67% bandwidth each, your receiver output will be harmonic free. 67% is the bandwidth where the lowest signal frequency and higher frequency are one octave apart. Your filters need to be sub-octave. So as not to confuse everyone, we will say that "M" filters are required to filter out the "N" harmonics.

Let's apply that to the 4-18 GHz example. To determine how many filters you need, start by assuming that they will all have 67% bandwidth. Thus, the first filter should have a maximum bandwidth of 3 to 6 GHz.

Then look at the second filter as a full octave: it will pass between 6 and 12 GHz.

You will need just one more filter, from 12 to 18 GHz, but it will only have 40% bandwidth... it would be silly to have the first two filters with more bandwidth than the final one. Your next job is to equalize the bandwidths of all three. How do you do that? The answer is best found using the magic of logarithms...

Take log10 of the total bandwidth ratio:

log10(18/3)=0.778

Now divide that by three (there will be three filters) and anti-log it:

10^(0.778/3)=1.817

That is the ratio of FH/FL for each of the three filters. The ratio has to be less than 2, as we want sub-octave filters to do the job of removing spurs, so this checks out.

Thus, the transition frequencies for the multiplexer are:

F1=3 GHz

F2=3x1.817=5.45 GHz

F3=5.45*1.817=9.91 GHz

F4=18 GHz.

Let's look at these frequencies on a linear scale. The low can filter serves 3 to 5.45 GHz, midband filter serves 5.45 to 9.91 GHz and high-hand filter serves 9.91 to 18 GHz.

Plotted on a log frequency scale you can easily see three filters all have the same bandwidth.

Questions or comments? send them in!

**Author : **Unknown Editor