MoM - part 2

8/14/2019 MoM - part 2

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Method of Moments II: Linear Array of Dipoles

8 Array Formulation

The solution of the currents can also be solved for multiple elements in an array environment.

For an M-element array, figure 1, each element will have N segments of equal length. In order

to solve for the unknown current on each element in the array, all self coupling between

segments on the same dipole and all mutual coupling between segments on different elements

must be taken into account.

(1)

(2)

(3)

(4)

(5)

(6)

(M)

•

•

•

x

y

z

φ

θ

1M x −

0 x

1 x

Figure 1. Geometry of an M-element array of dipoles, the center of each element is

located on the x-axis.

The solution has the same form as in the isolated element case, i.e. there will be a set of linear

equations where the current on each segment (of each element) needs to be solved for in terms of

all other segments. The number of current coefficients to be solved for will be

equal to the product of the number of elements in the array and the number

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of segments on each element. The number of elements in the impedance

matrix will be equal to the square of the number of unknown currents; these

elements can then be grouped into MxM blocks where each block is an NxN

sub-matrix. Each block represents the coupling coefficients between two

array elements where the diagonal elements represent the self-coupling

coefficients. As before, the solution has the form of a set of linear equations,

where ( )m I is a Nx1 vector whose elements are the current distribution on the

m th element and ( )mV is a Nx1 vector whose elements represent the

integration using expansion functions for the m th element over the total

incident field on element m .

[ ][ ] [ ]=Z I V

11 12 1

21 22 2

1 2

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ]

[ ] [ ] [ ]

M

M

M M MM

S S S

S S S

S S S

=

Z

g g g

g g g

g g g g

g g g g

g g g g

g g g

(1)1

( )

[ ]

M

MN

I I

I I

= =

I

g gg gg g

(1)1

( )

[ ]

M

MN

V V

V V

= =

V

g gg gg g

The NxN blocks will be toeplitz symmetric because each antenna is identical

in its geometry; this is due to the fact that each antenna would act the same

if it were isolated and also that the thin wire kernel is dependent on the

distance between two segments. Each element is given by:

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( ) ( ) ( ), ,m n

m m n

z z

n F z z F z z dz dz Z K ρ ′∆∆

′ ′ ′= ′

∫ ∫

( ) ( ); ,m

itot m m

z al F z z E z z d V ρ

∆

′ ′= ∫

The expansion functions over each axial segment change every N rows, and

the basis functions over the cylindrical surface of each segment change

every N columns. For example, the generalized impedance matrix of a four-

element array will have 4 4 1 = sub-matrices as given below. The intersecting sub-matrix

corresponds to the block of coefficients which represent the coupling coefficients of array

element 3 with array expansion functions of element 2. The number of elements inside this sub-

matrix will depend on the number of basis/expansion functions taken. The intersecting sub-

matrix in figure 2 represents rows N+1 to 2N which correspond to columns 2N+1 to 3N; for N=7

these correspond to the row elements 8-14 and column elements 15-21 respectively.

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11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

S S S S

S S S S

S S S S

S S S S

Testing functionsfor 2 nd element

Basis functions for 3 rd element

(1)

(2 )

(3)

(4 )

[ ]

I

I

I

I

=

I

Current coefficientsfor 2 nd element

(Nx1 vector)

Current coefficientsfor all elements(4Nx1 vector)

z

1=n

2=n

3=n

7n =

4n =5n =6n =

( ) ( ) ( )1612

12 16 112 6, , , z z

F z z F z d a z dz K d Z z ′∆∆

′ ′ ′ = −

∫ ∫

1 2 N N m+ ≤ ≤

2 1 3 N N n+ ≤ ≤

12, 16

(1) (2) (3) (4)

7 N =

1514

21

8

d a−

Figure 2. Example of a four element array where N=7 basis and expansion functions have been taken.

In part I, the expression for the incident field on the surface of the dipole was

only applicable for an axial element where 0 ρ = . In general, the incident

field on any element due to another element in the array is given by thefollowing *.

( )2

0

1, ,

4 ln( / )

b j Ri

z

a

e E z d

b a R

π β

φ ρ

ρ φ φ π

−

′= ′=

− ′=

∫

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* C. M. Butler, L. L. Tsai, “An Alternate Frill Field Formulation”, IEEE Transaction on Antennasand Propagation , Jan 1973.

The separation distance from each differential element of the magnetic frill

(described by the primed coordinates) to a field point (given by the unprimed

coordinates) is given by R. Then the distance between two points in

cylindrical coordinates is:

2 2 22 cos( ) ( ) R z z ρ ρ ρρ φ φ ′ ′ ′ ′= + − − + −

For the location of the array we have that the generator (magnetic frill) of

each element will be located in the xy -plane and centered on the negative x -

axis, thus 0 z ′ = . The location of each element in the array is also parallel to

the z -axis and so φ π = which means that cos( ) cos( )φ φ φ ′ ′− → − .

2 2 22 cos R z ρ ρ ρρ φ ′ ′ ′= + + +

For the first element the total incident field would be given by the following sum where the value

of ρ depends on the distance between the two elements.

( ) ( ) ( ) ( ),1 1 2 0 20, 0, , ,i i i itotal M M E z E z E x z E x z −= + + +g g g

Likewise, for the second element:

( ) ( ) ( ) ( ) ( ),2 1 0 2 3 0 30, , 0, , ,i i i i itotal M M E z E x z E z E x z E x z −= + + + +g g g

When the inner product of each side is taken, via integration by an

expansion function over each side, the voltage vector corresponding to the

m th axial segment for the first element will have the form:

( ) ( ) ( ) ( ),1 1 2 0 2( ) 0, ( ) 0, , ,m m

i i i im total m M M

z z

F z E z dz F z E z E x z E x z dz −∆ ∆

= + + + ∫ ∫ g g g

9 Two Element Array

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The integral over the m th axial segment taken over both the expansion functions for the first and

second elements can be given as:

( ) ( ) ( ) ( )1 1 1 11

, ,m m

N i

n m n m z nn z z

I F z f z z dz F z E z z dz = ∆ ∆

′ ′=∑∫ ∫

( ) ( ) ( ) ( )2 2 2 21

, ,m m

N i

n m n m z nn z z

I F z f z z dz F z E z z dz = ∆ ∆

′ ′=∑ ∫ ∫

Their sum is:

( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 2 2 1 1 2 21 1

, , , ,m m m

N N i i

n m n n m n m z n z nn n z z z

I F z f z z dz I F z f z z dz F z E z z E z z = =∆ ∆ ∆

′ ′ ′ ′ + = + ∑ ∑∫ ∫ ∫

There are now 2N unknown current values and so you now need to solve for the 4N 2 elements in

the generalized impedance matrix and 2N elements in the generalized voltage matrix. The

impedance matrix will consist of four NxN blocks which will have a block

toeplitz symmetry which means that 11 22[ ] [ ]S S = and 12 21[ ] [ ]S S = .

11 12

21 22

[ ] [ ][ ]

[ ] [ ]

S S

S S

=

Z

Some examples of the impedance matrix elements are given below for a two

element array.

z

1=n

2=n

3=n

7n =

4n =5n =

6n =

( ) ( ) ( )55

2,5 ,,m z z

n Z K F z a z dz F z z dz ′∆∆

= ′ ′ ′∫ ∫

( ) ( ) ( )55

2,14 , ,m z z

n F z z F z d a z dz Z K dz ∆ ′∆

= ′ ′ ′−

∫ ∫

( ) ( ) ( )55

13,5 , ,m z z

n F z z F z d a z dz Z K dz ∆ ′∆

= ′ ′ ′−

∫ ∫

( ) ( ) ( )55

13,14 ,,m z z

n Z F z a z F z z z z K d d ′∆∆

′ ′ ′ =

∫ ∫

MoM - part 2

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