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14.8.4 Toeplitz Matrices :-
Certain types of problems produce impedance matrices where there is a systematic repetition in the matrix elements. Often, this repetition can be used to decrease the impact of both the first and second terms in (14-66). Consider the straight wire in Fig. 14-3. If the segments are of equal length, all the values of the N2 matrix elements are contained in any one row of [Znud. say. the first one. All other rows are merely a rearranged version of the first. The remaining elements can be obtained by the rearrangement algorithm: = m > 2, n 1 (14-68)
Such a matrix is said to be a toeplitz matrix. Computer programs exist for solving toeplitz matrices that are considerably more efficient than those for solving a non- toeplitz matrix. For a toeplitz matrix, the first two terms in (14-66) become AN and BN93, respectively, and the execution time in (14-66) is reduced as in
t AN + B1N5I3 + CN2N; + DNN,N„ (14-69)
for which there is a significant improvement in the first term as well as the second. Toeplitz matrices can arise in the treatment of certain wire geometries. These are the straight wire (see Examples 14-1 and 14-2), the circular loop, and the helix. A toeplitz matrix can also arise in the treatment of geometries other than the wire, but these are outside the scope of this chapter.
14.5-1 In order to obtain some feeling for MoM, it is recommended that the student write a computer program to solve the following problem. Consider a straight dipole of length L (or monopole of length L/2) and radius a. Divide the dipole in N segments of equal length. each containing a pulse expansion function. (a) Use point-matching and the equation in Prob.
14.3-2 for the scattered field to compute the elements in the first row of the impedance matrix [Z.] as given in (14-26), noting that these are the only independence matrix elements since the matrix is toeplitz (see Sec. 14.8.4). Note that the integrand tends toward singularity when R = a, but even so one may numerically integrate through this region if reasonable care is taken. (b) Confirm the matrices in Example 14-1. Next, duplicate the curves in Fig. 14-9. (This exercise continues in Probs. 14.6-1 and 14.9-1.)
14.5-2 Starting with the electric vector potential and (14-30), derive (14-31).
14.6-1 (a) Extend the computer code of Prob. 14.5-1 to use pulse weighting functions in (14-40).
(b) Confirm the matrices in Example 14-2 and duplicate the curves in Fig. 14-13. (This exercise continues in Prob. 14.9-1.)
EXAMPLE 14-I Point-Matching on a Shod Dipole
“I lie purpose of this example is to illustrate the application of (14-26). An objective is to use MOM In determine the input impedance 4 of a short dipole with a length of 0.1i. and a radius of 0,0052 For yinvenience of illustration, choose N :.z 5, With reference to Fig. 148, the elements of V%,„„:, are calculated to he
679.5 L -89111 292,6 L89,97 33,1T3 L89,73″ 9.75 / 89 Oct 4 24 / ’47 ‘
IT INC 292,6 L89,97.- 679,5 / -89.99 292.6 / 89.97′ 33,03 L 89,73″ 9.75 L,89.09”
3103 L89,73″ 292.6 L89.97″ 679.5 L -89.991′ 292.6 L89.97″ 33.03 L89.73″
9.75 /_ 89.09° 33.03 L89.73″ 292.6 L89.97″ 679.5 L 292.6 L89.97′
4.24 L87.92′ 9.75 L89.09(- 33.03 L89,73t 292.6 L89.97′- 679.5 –89.W
For a 1-V excitation at the center of the short dipole (i.e., segment 3), the following voltage matrixtriV„,, is obtained using the frill source discussed in Sec. 14,5 with b/a = 2,3, and upon solving (14-28), the following current matrix is also obtained:
0.484 L–0.31 -3.128 L-0.04′ 67,938 L-01)02″ , = 3.128 L 0.484 L-0,31″
0.78 L89,54′ 1.48 L89.64t 2.35 L89.75″ 1.48 L89.64′ 0.78 L89.54″
On the other hand, if a 1-V delta gap excitation is used, V3 = 1 / A z = 1/0.02 = 50 LO”, and the resulting voltage and current matrices are
0 t 0′. 0 52 L89.54″ – LW 0.98 L89.64′ 50.0 trfa , tin] = 10 1.63 L89.76″ 0 Lfr 0.98 L89.64″ 0 LW 0.52 L89,54″