A brief description of the numerical technique which has been developed to solve the electromgnetic field problem which has given the user advantage of solving the complex problem more easily.

1. Choice of weighting function and
expansion function
MT15CMN008
28 October 2016 1Mohit Chimankar-MT15CMN08

2. INTRODUCTION
• In last decade various numerical techniques
have been developed to solve the
electromgnetic field problem which has given
the user advantage of solving the complex
problem more easily.
• A particular technique which has been applied
with great success is method of moments.
• It converts the general equation in the matrix
form which can be easily solved on computer.

3. Example:
• Given an operator equation
The MOM starts with expanding X in terms of set of known functions {Xi} with
unknown coefficients ‘αi’
Using exp function we are approx X with linear combination of
Substituting (2) in (1)

4. We want Rn to be minimal.
Hence,it is weighted to zero w.r.t. certain weighing
function {Wj} i.e.
By further simplifications we get
Now a set of weighting function{Wj} is desired

5. Since the objective is to solve for X which is approximated by a linear combination of
the expansion functions { x i } it is clear that for sufficiently large N , the
expansion functions must form a basis for D(A) (Only then theexpansion functions will
be able to represent any element X, in D(A)). Then, by definition Y, is in R(A) since it is
obtained by applying A to X,, and {Axi} must span R(A), as any Y, in R(A) can be written
as a linear combination of {Axi}. It is the duty of the weighting functions to make the
difference Y - Y, small.

6. Since the weighing function weights the residual to zero, we have
<Rn,Wj>=0
As error Yn-Y is orthogonal to the weighing functions we can say that weighting
function should be able to reproduce Yn and to some extent Y.
These requirements poses conditions on the weighing function, which are
a) The weighing function must be in the range of operator,more generally in the
domain of adjoint operator
b) As the weighing functions are orthogonal to error approximation, Wj should span
Yn.
c) As N->infinity, Yn->Y, therefore the weighing functions should be able to represent
the excitation Y in the limit.
About Adjoint operator:
The adjoint operator A * is defined by
(AJ, W) = (J,A*W) for all J in D(A) and W in D(A*). It is known from operator theory that
the domain of A* denoted by D(A*) covers R(A).
In particular from the orthogonal decomposition theorem
D(A*) = N(A *) +closure of R(A)
here N(A*) denotes the null space of A*. (i.e.. ifA* W = 0 then
W belongs to N(A*)).

7. A Numerical Example:
Let us try to solve with MOM with two different weighting functions. Choose the
expansion function as:

8. And the solution of αi gives the classical least squares, Fourier Series Solution
Above solution is a least square solution because Wj is
proportional to Axi
We obtain the approximate solution by assuming

9. Now if we were to apply Gallerkin method or weighting function would be
With this method some approximate solution may be obtained .
Then how close this solution is to the exact solution can be observed in the
graph..

10. 1)We have observed that the rate of convergence of the solution depends on the
choice of weighting functions.
2) As mentioned in the previous section, the weighting functions should be in the
range of the operator or, more generally, in the domain of the adjoint operator.
So we will check if the weighting function chosen in the gallerkin method
satisfies the condition
It is easy to show that the adjoint operator for the problem is
and the adjoint boundary condition is p(z = 1) = 0
Since for
odd j , sin (jπz)/2 is nonzero at z = 1
We can conclude that the
weighting function used here is not in the domain of the adjoint operator and hence
Galerkin’s method should not be applied.

12. Expansion Function
• Now we will see what mathematical
requirement should the expansion function
satisfy
• For this again we would use MOM
• First lets see which all conditions should
expansion functions satisfy

13. On the choice of Expansion Function
• The expansion functions should be in the domain of the operator in some
sense, i.e., they should satisfy the differentiability criterion and they must
satisfy the boundary conditions for an integro-differential operator. It is
also required that the total solution should also satisy the boundary
condition
• The expansion functions must be such that Axi form a complete set for the
range of the operator
It really does not matter whether the expansion functions are complete in
the domain of the operator.
What is important is that xi must be chosen in such a way that Axi is
complete
This will be demonstrated by an example.

14. A numerical example:
For the given boundary condition the obvious choice would be
Therefore, our approximation will be
Note that this expansion function satisfy both differentiability criteria and the
boundary condition
(1)

15. The above choice of expansion function leads to the solution
Looking at the solution it is quite clear that this solution doesn’t satisfy equation (1)
What could be the problem?
Perhaps the set {sin (iz)) does not form a complete set, even though they are
orthogonal in the interval [0, 2π]. Therefore, in addition to the sin terms,
we add the constant and the cos terms. This results in
where ao, ai, and bi are constants to be solved for. Now the total solution obtained has to
satisfy the boundary conditions of the problem
Now if we solve the problem again by Galerkin’s method or by the method of least
squares, we still obtain the solution as

16. But we already know that this equation doesn’t satisfy the equation (1)
The problem is that even though xi{ 1 , sin (iz), cos (iz)} form a complete set, Axi do not.
This is because Axi are merely {cos (iz), sin (iz)} .
The constant term is missing from Axi. Hence the representation in is not proper.
To have the constant term in Axi, the representation of ‘In’ must be of
the form
The final solution obtained using above equation gives us exact solution
So from this example we can say that its just not enough to choose the expansion functions
to be in a set of complete functions in domain of the operator but certain completeness
have to be satisfied for Axi.

17. CONCLUSION
• It is shown from a mathematical standpoint that there are certain rules
that should be followed in the choice of the weighting functions, and for a
given problem it is the operator that dictates which method (e.g.>
Galerkin‘s method or least squares method) to apply and it is not the
computational considerations.
• It is concluded that the weighting functions must be in the range of the
operator or, more generally, in the domain of the adjoint operator.
• Some of the mathematical restrictions on the expansion functions are
discussed. An example is given to illustrate that it is just not enough for
the expansion functions xi to be in the domain of the operator A. In
addition, it is required that Axj form a complete set.

18. REFERENCES
[1] “A Note on the Choice Weighting Functions in theMethod
of Moments” by T.K. Sarkar.
[2] “On the Choice of Expansion and Weighting Functions in the
Numerical Solution of Operator Equations” by T.K. Sarkar.
[3] “Electromagnetic Modelling and Measurements for Analysis
and Synthesis Problems - Google Books_files”

19. THANK YOU