How to calculate DDM in a ILS dual frequency system?

Question

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?

Answer 1

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.

From EVALUATION OF INSTRUMENT LANDING SYSTEM DDM CALIBRATION ACCURACIES [sic] by Dennis M. McCollum [PDF]:

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.

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.

Answer 2

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.