Search for Communications Emitters

By Adamy, Dave

Communications EW – Part 10 A NARROWBAND SEARCH EXAMPLE

This month, we will consider a narrowband search example. We want to find a 25-kHz-wide communication signal that is between 30 and 88 MHz. We will assume that the signal is up for 0.5 sec. Note that a signal this short is probably a “key click,” which at one time was the shortest signal an intercept system had to worry about. In this example, times will be rounded to the nearest millisecond.

Our receiving antenna covers 360 degrees of azimuth, and the search receiver bandwidth is 25 kHz. The receiver must dwell at each tuning step for a time equal to the inverse of the bandwidth. To avoid band edge intercepts, we will overlap our tuning steps by 50 percent.

Dwell = 1/bandwidth = 1/ 25 kHz = 40 psec.

Figure 1 shows the search problem in the diagram format we discussed last month. Note the overlap of the receiver coverage that causes us to change frequency only 12.5 kHz with each tuning step.

For 100 percent probability of finding the signal of interest, the receiver must cover the whole 58 MHz in one half second. The number of bandwidths required to cover the signal range is:

58 MHz / 25 kHz = 2,320

With 50 percent overlap, the 58-MHz frequency range requires 4,640 tuning steps.

At 40 [mu]sec dwell per step, 4,640 steps require 186 msec.

This means that the receiver can find the signal of interest in less than one-half of the assumed minimum signal duration, so 100 percent probability of intercept is easily achieved.

However, this assumes that we have an optimum search and that the signal will be instantly recognized as our signal of interest. To make the problem more interesting, let’s assume that we have a processor that can recognize the modulation of the signal in 200 [mu]sec. This means that we must hold at each frequency for that dwell time, so it takes 928 msec to cover the 58 MHz search range.

200 [mu]sec x 4,640 = 928 msec

The search does not find the signal within the specified half second, as shown in Figure 2.

INCREASE THE RECEIVER BANDWIDTH

If the search receiver bandwidth is increased to 150 kHz (covering six target signal channels) and we assume that the 200 [mu]sec processing time also allows determination of the signal frequency within the search receiver bandwidth, the search is enhanced. See Figure 3. Now it takes only 773 steps to cover the frequency range of interest.

4,640 / 6 = 773

At 200 [mu]sec, per 150 kHz receiver tuning step, it takes only 155 msec to cover the 2,320 channels (with 50 percent overlap).

773 x 200 [mu]sec = 155 msec.

Note that this increase in bandwidth would reduce the receiver sensitivity by almost 8 dB. If the receiver gets much wider than this, the increasing probability of multiple signals in the bandwidth becomes problematic.

To make the problem even more interesting, let’s assume that our receiver is part of a direction-finding system and that we must determine the direction of arrival (DOA) of the signal of interest. Our direction finder requires 1 msec to determine the DOA. This adds only 1 msec to our search time – if there are no other signals present.

Last month we discussed the density of the tactical communications environment, and considered that a number often used in the evaluation of tactical systems is 10 percent channel occupancy. This means that our 58 MHz range of interest would be expected to contain:

2,320 x 0.1 = 232 signals

With 150 kHz search bandwidth and 50 percent overlap, the receiver can cover the 2,088 empty channels in 139 msec.

(2,088 /6) x 2 x 200 [mu]sec. = 139 msec.

However the 232 occupied channels will require an additional 232 msec. Thus, our search requires 371 msec (139 + 232), so our search strategy yields 100 percent probability of finding and determining the DOA for the target signal.

Note that we will return to this problem when we consider searching for frequency-hopping signals in a later column in this series.

SEARCH WITH A DIGITAL RECEIVER

You might want to review the discussion of digital receivers in the November and December 2006 “EW 101″ columns.

Here, we will consider only finding our signal of interest (between 30 and 88 MHz) using a fast Fourier transform (FFT) channelizer. Later, when discussing EW against frequency-hopping signals, we will return to this problem in a more complex receiving system. Let’s specify our digital receiver to use a standard VME bus format, which limits our data rate to 40 MBps. To provide the Nyquist sampling rate, we thus limit our input frequency band to 20 MHz. Figure 4 shows the block diagram of the digital receiver. Using the FFT, we will cover the full frequency search range of interest in three steps as shown in Figure 5.

FFT Timing Analysis

To generate a 1,000 channel channelizer, we must collect 2,000 data points for input to the FFT. If we use an I&Q digitizer, 1,000 I&Q points are 2,000 independent data points, and will allow a 1,000 channel spectral analysis. This means that 20 MHz of RF spectrum will be isolated to 20 kHz channels which are sufficient to show the operating frequency of the threat signal we are trying to find.

Since our digitizer operates at a 40 MBps rate, we sample once each 25 nsec. Thus, if we gather I&Q data in parallel channels, it will take us 25 [mu]sec to gather 1,000 points for input to the FFT.

Generating an FFT requires a digital signal processor (DSP) to perform:

n log2(n) complex adds and n/2 log2(n) complex multiplies

“n” is the number of points in the FFT. Because your scientific calculator does not typically do logs to the base 2, lets convert to logs to the base 10 (the plain “log” key on your calculator).

log2(n) = log10(n) / log10(2) – log10(n)/.30103

If a DSP performs one complex add or multiply in a single DSP cycle, the number of DSP cycles to complete the 1,000point FFT is:

1.5 n log10(n) / .30103 – 4.988 DSP log10(n) = approximately 15,000 DSP cycles

Let’s use a DSP that performs 600 million floating point operations per second (600MFLOPS/sec). This would produce the 1,000- point FFT in:

15,000 FLOPS / 600 MFLOPS/sec – 25 [mu]sec.

If our digital receiver has a buffer arrangement that allows us to collect data while we process it, we can produce a full 1,000- channel spectrum search of 20 MHz each 25 [mu]sec.

Since we must cover the 30- to 88-MHz band in three steps, the full spectrum search requires 75 [mu]sec.

WHAT’S NEXT

Next month, we will discuss the location of tactical communication threat emitters. For your comments and suggestions, Dave Adamy can be reached at dave@lynxpub.com.

Copyright Naylor, LLC Mar 2008

(c) 2008 Journal of Electronic Defense. Provided by ProQuest Information and Learning. All rights Reserved.