I'm working in a project where I have to process ILS baseband signals. I have no problem obtaining DDM value from single frequency signals, nor the individual components of a dual frequency signal (DDM from clearance signal and DDM from course signal).

Now I need to merge those two individual DDM values (in dual frequency systems) to obtain an unified DDM value, similar to what Rohde equipment does (see figure below).

Rohde interface

I've been told that I can obtain this $\sum DDM$ value either:

  • As a function of individual DDM values: $\sum DDM = f(DDM_{clr}, DDM_{crs})$. But I'm not able to find the actual function $f(·)$.

  • Mixing clearance and course signals at baseband, so both components add up resulting in only one ILS-alike signal, and then calculate DDM from this new signal.

Are any of these methods valid? If not, how can I obtain $\sum DDM$ in a dual frequency system?

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    $\begingroup$ Hello Leroy, welcome to aviation.stackexchange.com! Your question is related to aviation (instrument landing system/ILS), but it is very much in the field of radio signal engineering (difference in depth of modulation/ DDM). It may take a while before someone comes along with an answer. Perhaps it is a better fit at another stackexchange site, although I am not entirely sure where (ham.stackexchange electronics.stackexchange or dsp.stackexchange ?) $\endgroup$ – DeltaLima Oct 30 '18 at 17:12
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    $\begingroup$ I'm voting to leave open. It's related to this question which was answered by @mins. He may be able to answer with authority. My interpretation of Annex 10 is that the SUM DDM is the is difference between the sums of the two 90 Hz tones and the sum of the two 150 Hz tones, but I can't say it with confidence. $\endgroup$ – Gerry Oct 30 '18 at 19:11
  • $\begingroup$ In the regions where the clearance signal dominates the course signal in power (receiver illuminated by one of the course sidelobes), the DDM is set to full scale deviation, no actual DDM is calculated, in the other region (receiver illuminated by the course main lobe) only the course signal is used to produce the DDM signal. When illuminated by a sidelobe, the receiver must still determine on which side of the main lobe this sidelobe is, to set the indicator on the correct extremity of the scale. The clearance DDM is used for this determination. I don't see why we would sum the two DDM. $\endgroup$ – mins Oct 31 '18 at 13:22
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    $\begingroup$ Thank you for the warm welcome. I may have stated the question from a slightier DSP perspective, but the core question is how that SUM DDM value is computed, which is more specific to aviation than DSP (I'm pretty sure DSP community would send this question back to this StackExchange). Then I can figure out the signal processing myself. @Gerry, could you reference me to which paragraph from Annex 10 you are talking about? Is it paragraph 2.7.x?. $\endgroup$ – Leroy Oct 31 '18 at 15:02
  • $\begingroup$ @mins, if I understood correctly you are saying to lock to receiver to the signal with higher power and basically ignore the other, right? I understand the logic behind it, and it's coherent with paragraph 2.7.1 from Annex 10. But our client asked for that SUM DDM, as done by Rohde (see figure from question), and I can't figure out how this value is obtained (they don't seem to be locking the receiver to the high power component either). Anyway, if this SUM DDM value is actually useless for aviation purposes, we could think about ignoring it. $\endgroup$ – Leroy Oct 31 '18 at 15:03

I don't think you're supposed to just average the two together somehow by themselves. The two signals are designed to be used in different zones and in practice have very different signal strengths at any point on approach.


The course pattern contains sidelobes that, without the clearance pattern, could cause an aircraft to select a wrong (sidelobe) course. Within the region of the true course (i.e., + 10 degrees about the designed procedural course), the signal strength due to the course array is greater than that from the clearance array. In the regions where the sidelobes exist, the clearance signal predominates. Airborne ILS receivers are designed to respond to the greater of the two signals.

enter image description here

This has to do with the background in processing these signals through analog means. Both the clearance and the course signals are within the passband of the receiver. Since this is an AM receiver and DDM is more or less a ratio of 90 Hz to 150 Hz, if the clearance or course is much stronger than the other, the stronger one prevails.

So in practice measuring the course signal DDM inside the clearance region doesn't reflect what's done in practice, and I'd guess it probably isn't even a guaranteed behavior of the transmitter. Do anything else instead in this region, like perhaps measuring just the field strength of the course signal.

But what about when slightly off the centerline? Then you'd get a mix of course and clearance signals. You could assume the ideal condition where a capture effect completely eliminates the weaker frequency. Another approximation presented in the admittedly confusing paper "Current Issues in Demanding ILS Ground and Flight Measurement Environments" by Gerhard Greving and L. Nelson Sponheimer is

\[ DDM_{total} = \frac {DDM_{clearance}* CR^x + DDM_{course} } {1+CR^x} \]

Although I can't quite make out what precisely CR and X is here.

Depending on the application, it may be more realistic to model the amount of distortion expected or permissible due to clearance band interference, especially due to multipath reflections of that signal. However, given your comments that probably won't help your specific application.

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Perhaps you should think about why two frequencies are used. One thought is to prevent shut down of the ILS caused by reflections. In doppler radar this effect is called a hole, and happens at unique slope elevations depending on the specific frequency. Phase cancellation happens when the reflected energy is exactly 180 degrees out of phase, similar to the functioning of noise cancellation headphones.

Localizers shut down after 10 seconds and Glide Slopes are set to 5. I've seen a parked aircraft at the far end of the run way shut down a localizer by remaining stationary for more than 10 seconds.

Using an average value may be a bad idea when one of the signals is affected by phase interference.

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  • $\begingroup$ Hi Javelin, thank you for your contribution. I didn't mention it in my first post but we are developing a measuring platform for ground inspection. So if it happens to be any phase interference, or other effects due to the environment, we actually need to measure how those are affecting the overall DDM. $\endgroup$ – Leroy Jan 24 '19 at 11:56

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