The MESFET transistor
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 Î {2,3,4} :
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ai,j cos(xj+bi,j) = bi, 1 £ i £ n
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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:
ci = cos(xi), si= sin(xi), 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:
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Ai,j cj+ Bi,jsj = bi, 1 £ i £ n
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ci2 +si2 =1;
where the Aij's and the Bij's are polynomials with integer coefficients in
the cos(bi,j), sin(bi,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:
U4 s24+U3 s23+U2 s22+U1 s2+ U0=0
c2= |
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c1= |
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s1= |
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where the Uk's are
polynomials in Z[Aij,Aij,b1,b2] and the
Vi(s2)'s are polynomials in Z[Aij,Bij,b1,b2][s2]. For
example, V0(s2) is: 2(-B11A21+A11B21)
(A112A22B22s2-A112A22b2-A11B12A21A22s2
+A11b1A21A22
-A11A12A21B22s2+A11A12A21b2
-A12b1B212+A12B12B212s2+B11A12B21b2+B11b1B21A22
-B11A12B21B22s2-B11B12B21A22s2
-B112A22b2+B112A22B22s2+A12B12A212s2-A12b1
A212)
Two interesting problems remain to be solved with
respect to this problem:
- The simplification of the polynomials Ui or Vj(s2) would provide
a better solution: for example in the previous expression of
V0(s2) the first factor can easily be represented as
-B11A21+A11B21=a11a21sin(b21-b11)
- 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]
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D. Bini and B. Mourrain.
Polynomial test suite.
1996.