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矩量法计算带状线特征阻抗Matlab代码
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矩量法计算带状线特征阻抗Matlab代码
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发表于: 2008-01-15 01:39:48
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% RJA - 2/23/07
6-N?mSQU
%
#Z?A2r!1
% Translation of FORTRAN Code given in Fig. 5.9 to MATLAB. No attempt
{6|38$Rl
% whatsoever was made to make the code efficient. It is a pure
/?5 1D@
% translation.
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`2("gUCm
clear all; format compact;
Dp,L/1GQ8
<\rT%f}3^
%Compare to Table 5.2 page 313. These values were obtained with this
?+@n3]`0
%program. There is obviously a factor of 2 error between this and the
2=,lcWr
%table but the source of the error is unknown. Several modification to the
_S<3\%(0
%translation were made and noted in the code. Note that table 5.2 trends
V_:1EBzz
%down while this program trends up as n increases.
kCWV r
9-&Ttbb4)0
% n Zo Zo/2
md)c0Bg8~
%-------------------------------
x kx^%3dV
% 3 94.4473 47.2236
;DqWh0
% 7 96.8223 48.4112
:~&~y-14
% 11 97.4035 48.7017
JVvs-bK5
% 18 97.823 48.9115
{f3YsM;]C
% 39 98.23 49.115
DE}K~}sbd
% 59 98.3686 49.1843
BO~0ON0
% 100 98.49305 49.24652
?~BC#B\>o
% 1000 98.696565 49.348283
?uzRhC_)!
vyNxT* ,[K
36}?dRw#p
4KhV|#-;k
% *****************************************************************
LiN$ pwm
% USING MOMENTS METHOD,
HSjlD{R
% THIS PROGRAM DETERMINES THE CHARACTERISTIC IMPEDANCE
SYAyk
% OF A STRIP TRANSMISSION LINE
'N (:@]4N
% WITH CROSS-SECTION W X H
cLpYW7vZ[
% THE STRIPS MAITAINED AT 1 VOLT AND -1 VOLT.
PxWT1 !
%
#xsE3Wj-X
% ONE STRIP IS LOCATED ON THE Y = H/2 PLANE WHILE THE OTHER
KKja/p
% IS LOCATED ON THE Y = -H/2 PLANE.
eVjBGJ=2e
%
XKj|f`
% ALL DIMENSIONS ARE IS S.I. UNITS
% L$bf#
%
EH+"~-v)ae
% N IS THE NUMBER OF SUBSECTIONS INTO WHICH EACH STRIP IS DIVIDED
_<n~n]%
% *****************************************************************
I667Gz$j5
_C%3h5
% FIRST, SPECIFY THE PARAMETERS
$v4.sl:x
'\l"
NN = [3 7 11 18 39 59];
0gW"i&7c
S'Q@ScJ
disp(' ')
:3v}kLO7|
disp('n Zo')
`uv2H$
@Q)OGjaq
for N = NN
M8kPj8}{
CL=3.0e8; % SPEED OF LIGHT IN FREE SPACE
+ [iQLM?zo
ER=1.0;
M'7f O3&|
EO=8.8541878176E-12;
%a>&5V
H=2.0;
PS)4 I&;U
W=5.0;
2J7:\pR^
hO\<%0F
NT=2*N;
!`Fxa4i>
DELTA = W/(N);
q<xCb%#Jl
EU2$f
% SECOND, CALCULATE THE MATCH POINTS
%".HaI]
% AND THE COEFFICIENT MATRIX [A]
OcR$zlgs[v
n,vct<&z@
for K=1:N
p|X"@kuseO
X(K) = DELTA*(K - .5);
$O&b``
Y(K) = -H/2.0;
8g@<d^8@
X(K+N) = X(K);
lHN5Dr
Y(K+N) = H/2.0;
nsu@h
end
DrKP%BnS
B nFwlw
FACTOR = DELTA/(2.0*pi*EO);
LM:vsG
Y071Y:
for I=1:NT
L*Tj^q!t+
for J=1:NT
d~9A+m3b_
if(I==J) %eqn (5.98)
Ci@o|Y }tP
A(I,J) = -(log(DELTA) - 1.5)*FACTOR;
.0zY}`
else
"@e3EX7h
R = sqrt( (X(I) - X(J))^2 + (Y(I) - Y(J))^2 );
ePOG}k($/%
A(I,J) = -log(R)*FACTOR;
Zi *2nv'
end
<T` 7%$/E
end
y;35WtDVb
end
J{.{f
Nyku4r0
% NOW DETERMINE THE MATRIX OF CONSTANT VECTOR [B]
*>Ns_su7W
{%rA1g
for K=1:N
Q- cFtu-w
B(K)=1.0;
@6Y?\Wx$w
%The line below was commented!
Bd!bg|uO*
B(K+N)= -1.0; %rja -this line needed to be un-commented
wcrCEX=I>{
end
>8~.wXyoC
9P1OP Xv*p
% INVERT MATRIX A(I,J) AND CALCULATE MATRIX RO(N)
: |>h7v
% CONSISTING OF THE UNKNOWN ELEMENTS
tqz3zIQ
% ALSO CALCULATE THE TOTAL CHARGE Q,
wT6"U$cV
% THE CAPACITANCE C, AND THE CHARACTERISTIC IMPEDANCE Zo.
C.WX.Je
LdYB7T,
NIV=NT;
ep!Rf:
NMAX=100;
[g7L&`f9
A = inv(A);
]FJjgu<