# How to calculate DDM in a ILS dual frequency system?

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). 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?

• 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 ?) – DeltaLima Oct 30 '18 at 17:12
• 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. – Gerry Oct 30 '18 at 19:11
• 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. – mins Oct 31 '18 at 13:22
• 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?. – Leroy Oct 31 '18 at 15:02
• @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. – Leroy Oct 31 '18 at 15:03