[资料共享]矩量法求解hallen's equation和Pocklington’s equation

(~fv;}}v  
clc o* e'D7  
clear |<%v`*  
lamuda=0.02; Lzx2An@R  
L=0.47*lamuda; 4jzjrG  
a=0.0001; B/Ltb^a  
k0=(2*pi)/lamuda; UYzNaw4/x  
N=81        ; V?~!Dp  
deltaz=L/N; HrUE?Sq  
z=(-L/2+deltaz/2):deltaz:(L/2-deltaz/2); { PS0.UZ  
Z1=zeros(N,N); c7nbHJi  
n1=120*pi; @Thrizh  
%hallen's equation; i/PL!'oq  
for m=1:N JD9
=gBN\?  
    for n=1:N z6|kEc"{  
        R=sqrt(a^2+(abs(m-n)*deltaz)^2); H;#C NB<e  
        Zmn=deltaz*(exp(-1j*k0*R))/(4*pi*R); J|([(  
        temp1=(sqrt(deltaz^2+4*a^2)+deltaz)/(sqrt(deltaz^2+4*a^2)-deltaz); AB<%GzW0(  
        temp2=(1j*k0*deltaz)/(4*pi); w"
L]?#  
        Zmm=(1/4*pi)*log(temp1)-temp2; 3CPSyF  
        if(m~=n) WwUHHm<v  
            Z1(m,n)=Zmn; !t?5U_on  
        else q9c-UQB(!  
            Z1(m,n)=Zmm; AuY*x;~  
        end *d31fBCk%  
    end .UvDew/Y  
end 7c8`D;A-K  
sm=cos(k0*z.'); N$u: !  
bm=(-1j/2*n1)*sin(k0*abs(z.')); z`,dEGfh^  
u=zeros(1,N); V5gr-^E  
u(1,1)=1; z G`|)  
u(1,N)=1; @?gRWH;Pq  
temp3=u*Z1^(-1)*bm; t_+owiF)M  
temp4=u*Z1^(-1)*sm; 3E}j*lo  
D1=-temp3/temp4; )_}xK={  
current1=D1*Z1^(-1)*sm+Z1^(-1)*bm; 9<o*aFgCa  
subplot(2,1,1); )5u#'5I>  
plot(abs(current1),'r'); Z6&bUZF$bE  
iu2O/l#r  
%Pocklington’s equation, O^_CqT%  
Z2=zeros(N,N); l nJ  
for p=1:N %
AA-G  
    for q=1:N YD0j&@.  
       R=sqrt(a^2+(abs(p-q)*deltaz)^2); +^%)QH>9   
       term1=(k0^2/(4*pi))*(exp(-1j*k0*R)/R)*deltaz; ,[3}t%
Da  
       temp1=abs(p-q)*deltaz-deltaz/2; qRq4PQ@  
       temp2=abs(p-q)*deltaz+deltaz/2; uH#X:Vne  
       R1=sqrt(temp1^2+a^2); -J0I2D  
       R2=sqrt(temp2^2+a^2); G/_xn5XDD  
       integral1=(temp1*(1+1j*k0*R1)*exp(-1j*k0*R1))/(4*pi*R1^3); #"C
!-kS'=  
       integral2=(temp2*(1+1j*k0*R2)*exp(-1j*k0*R2))/(4*pi*R2^3); m= %KaRI  
       term2=integral1-integral2; @.,'A[D!K  
       zmn=term1+term2; Hm+VGH'H?  
       Z2(p,q)=zmn; -fN5-AC  
&n .. )S}.QrG  
n*caP9B  

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