Simulation 4 Log Periodic Antennae and Linear Uniform Arrays

Transcription

1 Simulation 4 Log Periodic Antennae and Linear Uniform Arrays In the first simulation one saw that a half-wavelength dipole has the expected performance, when operated at the nominal frequency. However, if the operating frequency deviates considerably from the nominal value, the antenna performance deteriorates. To compensate for this degradation, certain antennae are designed with a performance that is relatively insensitive to changes in the operating frequency, as long as the deviation is within design bounds. The Log Periodic Antennae (LPA) is one example of this type of antenna. In this work we will look at an LPA performance and also investigate certain details of linear uniform arrays which were overlooked during the previous simulations. 1. Log-Periodic Antenna As a first example, we will consider an LPA with 14 elements, designed to operate in the MHz range. There are many references that cover the design of such antennae, one example being Section of the Balanis text book. Another good reference is the American Radio Relay League (ARLL), very popular within the amateur radio community. On the internet one finds many examples of design specifications for these antennae. The address paginas.fe.up.pt/~amoura/aproweb/logperiodicantennadesign.pdf contains one article found this way. Make a copy of the file trab4_inicial.nec and give it the name trab4.nec. Open the file with the Notepad editor and it should have the following appearance. CM CM MHz 14 el 6.35 mm elements 2.5m boom 20 mm square CM This was erected on 23/9/2001, SWR as predicted up to 60 MHz CM FR CM RP CE GW GW GW GW GW AAM/JJC Pg.1 de 7

2 GW GW GW GW GW GW GW GW GW GE EX FR PQ -1 TL TL TL TL TL TL TL TL TL TL TL TL TL RP EN Use this file as input for the 4NEC2 program. Using the NEC editor, as in the other simulations, makes editing easier. The Symbols tab should have no entries. The Geometry tab will have 14 lines, all very similar, which are the lines above that have a 'GW' at the beginning. The first and last are here shown: Nr. Type Tag Seg X1 Y1 Z1 X2 Y2 Z2 Radius 1 Wire Wire AAM/JJC Pg.2 de 7

3 The Source/Load tab will have sub-windows. The top sub-window describes the Source and contains one entry: Nr. Type Tag Seg (opt) Real Imag Magn Phase (norm) 1 Voltage-src The bottom sub-window is labeled Trans-lines, with 13 lines, filled with the information from the lines that start (in the Notepad editor) with 'TL': Nr. Type Tag-1 Seg-1 Tag-2 Seg-2 Z0 Len End-1(G) (B) End-2(G) (B) 1 Trans-line Trans-line The Freq/Ground tab will indicate the Freq-start of 40 MHz, 60 for the number of steps, each of size 1 MHz. Finally, the Comments tab has the information that appears in the input file at the very top, in the lines that start with 'CM'. As will be seen, this antenna has a rather uniform performance, deviating from the expected only at the highest operating frequencies. The following spreadsheet table relates some of the parameters of the original file, as one can easily deduce. The parameter tau is supposed to be constant, and sets the size of the elements according to Li+1= tau * Li, as well as the separation between the same according to di, i+1= tau * Li-1, i X Y tau delta_x freq delta_f incf_f delta_x ratio AAM/JJC Pg.3 de 7

4 We will start by calculating the SWR of the antenna, from 40 to 110 MHz. Run the program, selecting the option Use original file. The window Imp./SWR/Gain should appear, after the run. P1 Using the criteria "the SWR does not exceed 2", measure the frequency band that meets this requirement. Observe the antenna gain plot by selecting 'Show > Forward Gain' on the window. Next, the antenna behavior at the various frequencies will be looked at. Instead of a frequency sweep, the frequency will be fixed at 48 MHz. One can either go to the Freq/Ground tab set the values: Frequency = 48 MHz and uncheck the 'Sweep' box or, if using the Notepad editor, change the line that starts with 'FR' to read: FR Run the program, but this time select Far Field Pattern. Note: depending on the version of the program being used, a message window may open with a warning about 'unequal segmentation'. Close that window, thus disregarding the message, and the program should continue. In the Geometry window select 'Currents > Current Magnitude', in order to view the current distribution in the various elements. Note: For a better view, with the window selected, hit Page Up once or more. Use the arrows to rotate the image, for a more detailed view. P2 For this frequency, which of the elements are more active? How do you justify such behavior? Make a sketch of the radiation diagram. Change the frequency to 60 MHz. P3 Repeat the previous question, for the new situation. Next, we will set the frequency at 77 MHz, where the antenna gain has an anomaly. Run the program and observe the radiation diagram. P4 What is the main difference between this diagram (along with its current distribution) and what was observed for the 60 MHz case? AAM/JJC Pg.4 de 7

5 2. Linear uniform array with four elements We will use now an array with four elements, similar to what was used in the previous simulations. Make a copy of the file trab4_4el_inicial.nec giving it the name trab4_4el_anew.nec This file will have the following tabs: The Symbols tab should have two entries and look like this: Nr Symbols and equations comment 1 len=0.5 length of the antenna 2 d=0.25 Spacing between the elements The Geometry tab entries should be: Nr. Type Tag Seg X1 Y1 Z1 X2 Y2 Z2 Radius 1 Wire len/2 -d/2-d 0 len/2 -d/2-d Wire len/2 -d/2 0 len/2 -d/ Wire len/2 d/2 0 len/2 d/ Wire len/2 d/2+d 0 len/2 d/2+d.0001 The Source/Load tab is: Nr. Type Tag Seg (opt) Real Imag Magn Phase (norm) Comment 1 Curr-src Current src 2 Curr-src Current src 3 Curr-src Current src 4 Curr-src Current src For practice, run the simulation and observe the radiation diagram. The array analysis made here will make use of the mapping z = e jψ = e j(ß+kdcosδ). To follow the developments, let's consider the following table : δ ψ = ß + kd cos δ = 0+(2π/λ)(λ/2) cos δ Direction around the unit circle 0 π cos δ=π Starting point. π/2 0 Moving clockwise (=ψ diminishes) π -π Moving clockwise (=ψ diminishes) 3π/2 0 Moving counterclockwise (=ψ increases) 2π π : repeat value; starting point Moving counterclockwise (=ψ increases) In order to use this table, one should remember the unit circle showing the location of the polynomial zeroes associated with any uniform linear array that has N AAM/JJC Pg.5 de 7

6 elements. If N=4, which is the present case, the unit circle contains N-1=3 zeroes, at π/2, π and 3π/2, which is the same as -π/2. When δ varies from 0 to 360 around the physical array, the correspondent displacement around the circle is dictated by the parameters ß (phase) and d (separation), as the table shows: the phase (ß) controls the starting point and the separation (d) controls how much of the unit circle is covered. The calculations become now trivial. Note: The angle δ indicates the radial line around the array, beginning with the line that contains the various elements. In the present configuration, given the orientation of the antennae, the angle δ coincides with the angle θ of the radiation diagram. Having familiarized ourselves with this technique, let's change the separation between elements to d = 1,25 and calculate the diagram. The corresponding table is necessary and it looks like: δ ψ = ß + kd cos δ = 0+(2π/λ)(5λ/4) cos δ Direction around the unit circle 0 (5/2)π Starting point. π/2 0 Moving clockwise (=ψ diminishes) π -(5/2)π Moving clockwise (=ψ diminishes) 3π/2 0 Moving counterclockwise (=ψ increases) 2π (5/2)π : repeat value; starting point Moving counterclockwise (=ψ increases) P5 Why is it that in this case there is a null at θ = 0? Let's say that the exact location of the nulls and maxima of the diagram are needed. As an example, the null located on the interval 120 < θ < 130. To the angle θ = 90 corresponds the value ψ = 0, which yields a maximum. As θ increases to 180, ψ decreases to (5/2)π, passing through the values π/2, π, 3π/2, 2π. One sees that the diagram starts from a maximum, passes by three consecutive nulls, passes again by a maximum and finishes at another null, at (5/2)π. The sought null is the third in this sequence, that is, when ψ = 3π/2. To obtain the angle δ: 3π/2 = (5/2)πcosδ <=> cosδ = 3/5 <=> δ = P6 Determine the angle at which occurs the maximum in the third quadrant. AAM/JJC Pg.6 de 7

7 In order to obtain the diagram value for any angle δ, it suffices to determine the corresponding angle ψ in the unit circle, calculate the coordinates of the point in the circle, calculate the distances from the point to the three zeroes in the circle and multiply the distances. Example: what is the diagram value for δ = 130? To this value of δ corresponds the angle ψ = , with coordinates (0,329; 0,944). Calculating the distances to the zeroes located at (0;1), (-1;0) and (0;-1) and multiplying said distances, the desired value is obtained, namely, 11 db below the maximum. Note: The obtained calculated has to be normalized by the maximum array value, which is 4 in the present case. This unit-circle technique calculates only the "array factor" of the antennae ensemble, i.e. assumes that on the plane investigated each antenna radiates uniformly in each radial direction. If that is not the case (= the elements all have the same individual diagram, which favors certain directions), the array factor has to be multiplied by the individual diagram. P7 Using the above described technique, calculate the diagram value for δ = 45 and compare it with the simulation value, which shows 3 db below the maximum. AAM/JJC Pg.7 de 7