# Binomial Current Distributions

Author: Edmund A. Laport

Example. A North American Regional Broadcasting Agreement class 2 station of 50 kilowatts is to be located on a class 1-B channel. The directive antenna must provide for the protection of the 500 microvolts per meter night sky-wave areas (50 percent of the time) of two class 1-B stations by suppression of radiation so that, at the protected sky-wave boundaries, the new station's signals will be less than 25 microvolts per meter all but 10 percent of the time. Referring to appropriate maps and propagation curves, it is determined that the field strengths at 1 mile cannot exceed the values in Table 2.10.

 TABLE 2.10 Azimuth (true) (degrees) Maximum permissible field at 1 mile along the ground (millivolts per meter) Station A 10 600 12 310 15 162 20 130 30 121 40 140 45 160 50 275 55 550 Station B 305 460 310 190 320 160 330 165 335 190 340 410

These data are represented graphically in the upper part of Fig. 2.44. Mid-angle of the open gap between zones of protection = 356 degrees. The axis for the array will be placed on this bearing. The angles of maximum suppression will be placed at ±40 degrees with respect to the array axis.

The directive pattern must be such as to fall well within these boundary values in operating at 50 kilowatts. The amount of signal suppression is rather extreme since 50 kilowatts into a nondirective one-fourth-wave radiator produces a field strength at 1 mile of about 1,200 millivolts per meter (175 millivolts per meter at 1 mile with 1 kilowatt).

We look at a chart of patterns for antenna pairs (Appendix IV-A) for a suggestion to start. It is noted that the angle between the two directions of greatest required signal suppression is about 80 degrees. We must then search for a possible pattern which will bring two symmetrical nulls 80 degrees apart, or 40 degrees from the axis of the array, also taking into account the client's desire for a maximum of radiation southwestward if possible. From the chart of patterns for radiator pairs we find one for S = 0.375λ, α = -90 degrees which looks promising. The null angles may not be correct, and we see, in proportion to the unit circle,that the field strength of the minor lobe is about one-half that of non-directive operation, which is too large.

 FIG. 2.44. Minor lobe reduction and null broadening using a squared pattern from a binomial array.

Also, the nulls are not wide enough for our needs. But we must examine the requirements for a null angle 40 degrees each side of the axis and find the spacing required to give this. This is done as follows:

As before, the equation for all these patterns is

We know β = 40 degrees and φ = 90 degrees and cos 40 = 0.77. Hence

For the cosine of an angle to be zero, the angle must be 90, 270, 450 degrees, etc. In this case we use 90 degrees. Then from which

Knowing S, we can then calculate the whole pattern as shown in Table 2.11. This pattern shows the minor lobe to be too large for use. However, if we square the values in the table, the amplitude of the minor lobe changes from 0.225 to 0.051 of maximum value for the pattern and the nulls are broadened. Let us then tabulate the squares of these values and get Table 2.12.

 TABLE 2.11 β (degrees) f( β) β (degrees) f( β) 0 0.225 100 0 819 10 0.208 110 0 906 20 0.156 .120 0.961 30 0.087 130 0.990 40 0.000 (minimum) 140 1.000 (maximum) 50 0.139 150 0.996 60 0.276 160 0.988 70 0.423 170 0 978 80 0.573 180 0.974 90 0.707

We must now determine the actual field strengths in the critical regions at 50 kilowatts to see whether or not we are within limits everywhere. To do this, we must change this pattern from a relative set of values to an absolute set of field strengths. This requires finding the root-mean-square value of the pattern. For simple arrays of this type this is obtained in two ways: (1) With a polar planimeter, measure the area of a polar plot of the above pattern, and construct a circle having exactly the same area.1

The radius of this circle is the relative field for the same power non-directional; and we know what this should be in millivolts per meter from the antenna efficiency and the power. From this we can assign values to all parts of the directive pattern. (2) All the squared pattern values of Table 2.12 can be squared again, added, and divided by the number of values added (19 in Table 2.12). Then the square root of this value is taken, and this is the radius of the root-mean-square circle, in proportion to the arbitrary dimensions of the pattern as calculated. This again is known to be so many millivolts per meter, which in our case is 1,200 millivolts per meter.2

In the above case, the radius of the root-mean-square circle is 0,645 = 1,200 millivolts per meter. The magnitude of the minor lobe at β = 0 is (0.051/0.645)·1,200 = 95 millivolts per meter. This is well under what is allowed.

 TABLE 2.12 β f2( β) β f2( β) 0 0.051 100 0.668 10 0.043 110 0.82 20 0.024 120 0.92 30 0 130 0.98 40 0 140 1.00 50 0.019 150 0.99 60 0.076 160 0.98 70 0.178 170 0.96 80 0.328 180 0.94 90 0.500 Root-mean-square value 0.645

The pattern is then calculated in field strengths by multiplying the above tabulated values by 1,860 (obtained by putting 1,200/0.645 = 1,860). The pattern is now plotted on another sheet in a convenient scale of millivolts per meter versus angle and oriented in azimuth so that the nulls fall at the correct geographical angles, as shown in Fig. 2.44. In this case the axis of the array and its pattern is 356 degrees (4 degrees west of true north). Upon further examination of protections over all the angles required, it is found to be satisfactory, with ample margins.

To square such a pattern, as required in the above problem, the radiating system is derived as shown in Fig. 2.45. The pattern would be squared if each radiator of the pair had a pattern the shape of that given in Table 2.11, instead of a circular pattern, as they actually have. To obtain this same effect, we must use a pair of pairs, each pair spaced and phased relatively in the same way, and on the same axis. When we do this, we find that radiator b of the first pair coincides with radiator c of the second pair. Again we get a pair of pairs by using only three radiators, since one does double duty, as shown by the current being twice as much as the current in the end radiators. The current magnitudes and phase angles are shown in the various steps and in the final array.

 FIG. 2.45. Synthesis of the binomial array having the pattern of Fig. 2.44.

The horizontal pattern for this array then, taking into account its geographical orientation at 356 degrees and its root-mean-square field value of 1,200 millivolts per meter for 50 kilowatts, is

 1) The method of obtaining the root-mean-square value of a pattern by integrating the horizontal pattern only, as was done here, is not sufficiently accurate for any but the very simple systems. The correct determination of root-mean-square pattern values requires integration of the pattern over the complete hemisphere enclosing the antenna system. For methods, consult refs. 44 and 45. 2) This disagrees with the value in Table 2.1 because in this case we are considering a typical value for 50 kilowatts input to the antenna instead of 50 kilowatts radiated power.

Last Update: 2011-03-19