The MESFET transistor

                                                         mesfet [1]
A Mediavilla (DICOM) and L. Gonzalez-Vega

The design of the MESFET transistor requires the solution of the following nonlinear system of equations for n ∈ {2, 3, 4} :

∑_{j=1}^{n} a_{i,j} cos(x_{j} + β_{i,j}) = b_{i}, 1 ≤ i ≤ n

This system arises when computing the coefficients linked to the derivatives in the Taylor development of the nonlinearities of the capacity appearing in the circuit model for the microwave transistor MESFET. These coefficients are very important when determining the transistor behaviour in the intermodulation distortion and, currently, almost no model takes care of this nonlinearity.

This system is converted to an algebraic one by expanding the cosine function and by introducing the new variables:

c_{i} = cos(x_{i}), s_{i} = sin(x_{i}), 1 ≤ i ≤ n

In this way the algebraic problem is reduced to the resolution of the following polynomial system of 2n equations with 2n unknowns:

∑_{j=1}^{n} A_{i,j}c_{j} + B_{i,j}s_{j} = b_{i}, 1 ≤ i ≤ n

c_{i}^{2} + s_{i}^{2} = 1;

where the Aij’s and the Bij’s are polynomials with integer coefficients in the cos(β_{i,j}), sin(β_{i,j}) and aij. In this case, only real solutions are required.

Example 1:

mesfet := proc(n)
local eq,C,S,i;
  eq := seq(c[i]^2 + s[i]^2-1,i=1..n);
  C := vector(n,i->c[i]);
  S := vector(n,i->s[i]);
  [eq,op(convert(
     evalm(matrix(n,n,(i,j)->A[i,j]) &* C +matrix(n,n,(i,j)->B[i,j]) &* S)
        ,list))];
end:

Problem: Given A and B, compute the real solutions.

• Solution by L. Gonzalez-Vega

The case n = 2 is very easily solved and the solution is given by the following:

U_{4}s_{2}^{4} + U_{3}s_{2}^{3} + U_{2}s_{2}^{2} + U_{1}s_{2} + U_{0} = 0

c_{2} = \frac{V_{1}(s_{2})}{V_{0}(s_{2})}

c_{1} = \frac{V_{2}(s_{2})}{V_{0}(s_{2})}

s_{1} = \frac{V_{3}(s_{2})}{V_{0}(s_{2})}

where the U_{k}’s are polynomials in ℤ[A_{ij}, A_{ij}, b_{1}, b_{2}] and the V_{i}(s_{2})’s are polynomials in ℤ[A_{ij}, B_{ij}, b_{1}, b_{2}][s_{2}]. For example, V_{0}(s_{2}) is:

2(−B_{11}A_{21} + A_{11}B_{21})(A_{11}^{2}A_{22}B_{22}s_{2} − A_{11}^{2}A_{22}b_{2} − A_{11}B_{12}A_{21}A_{22}s_{2} + A_{11}b_{1}A_{21}A_{22}

−A_{11}A_{12}A_{21}B_{22}s_{2} + A_{11}A_{12}A_{21}b_{2} − A_{12}b_{1}B_{21}^{2} + A_{12}B_{12}B_{21}^{2}s_{2} + B_{11}A_{12}B_{21}b_{2} + B_{11}b_{1}B_{21}A_{22}

−B_{11}A_{12}B_{21}B_{22}s_{2} − B_{11}B_{12}B_{21}A_{22}s_{2} − B_{11}^{2}A_{22}b_{2} + B_{11}^{2}A_{22}B_{22}s_{2} + A_{12}B_{12}A_{21}^{2}s_{2} − A_{12}b_{1}A_{21}^{2})

Two interesting problems remain to be solved with respect to this problem:

1. The simplification of the polynomials U_{i} or V_{j}(s_{2}) would provide a better solution: for example in the previous expression of V_{0}(s_{2}) the first factor can easily be represented as

−B_{11}A_{21} + A_{11}B_{21} = a_{11}a_{21} sin(β_{21} − β_{11})

2. The solution of the cases n = 3 and n = 4 requires the manipulation of very big parametric expressions and one possible way of solving this problem could be the simplification question addressed before.

References

[1] D. Bini and B. Mourrain. Polynomial test suite. 1996.