exercise 6.11
ZIP with MATLAB scripts and note:
exercise 6.11 notes:
c0=298792586 % m/s
mu0=1.256637061435917e-06
e0=8.854e-12
f0=11e10
lambda0=c0/f0;
k0=2*pi/lambda0
er=3
etha0=120*pi
A=1;B=1
a=.02;b=.01;d=.1
t=.025
dx=a/200;
x=[0:a/200:a];y=[0:b/200:b];z=[0:d/200:d];
[X Y Z]=meshgrid(x,y,z);
Ey=A*sin(pi*X/a).*sin(pi*Z/d);
Hx=-1j*A/(etha0)*sin(pi*X/a).*cos(pi*Z/d);
Hz=1j*pi*A/(k0*etha0*a)*cos(pi*X/a).*sin(pi*Z/d);
% since relative magnetic permeability is null, the integral is limited
% to |Ey| df0(t)/f0
df_rel=[]
for k=1:1:numel(y)
fun1=@(xsym,ysym,zsym)...
(er-1)*e0*(A*sin(pi*xsym/a)*sin(pi*zsym/d)).^2;
E1=triplequad(fun1,0,a,0,b,0,z(k));
fun2=@(xsym,ysym,zsym) (e0*(A*sin(pi*xsym/a)*sin(pi*zsym/d)).^2.+...
mu0*((-1j*A/(k0*etha0*a)*sin(pi*xsym/a).*cos(pi*zsym/d)).^2.+...
(1j*pi*A/(k0*etha0*a)*cos(pi*xsym/a).*sin(pi*zsym/d)).^2));
E2=triplequad(fun2,0,a,0,b,0,d);
df_rel=[df_rel -E1/E2];
end
df_rel(1)=[];df_rel(end)=[];
figure(1);plot(y([2:end-1]),df_rel);grid on;
xlabel('t');ylabel('df0')
title('frequency relative deviation with integral')
% df0(t)/f0 from direct expression
df_rel2=[]
for k=1:1:numel(z)
E1=(er-1)*e0*A^2*b*z(k)*a/4;
E2=a*b*d*e0/2*A^2;
df_rel2=[df_rel2 -E1/E2];
end
figure(2);plot(y,df_rel2);grid on;
xlabel('t');ylabel('df0')
title('frequency relative deviation with direct expression')
f0dev=f0+df_rel*f0
%% comparing with example 6.7
df_rel_example_6_7=[]
for k=1:1:numel(y)
fun1=@(xsym,ysym,zsym) …
(er-1)*e0*(A*sin(pi*xsym/a)*sin(pi*zsym/d)).^2;
E1=triplequad(fun1,0,a,0,b,0,z(k));
fun2=@(xsym,ysym,zsym) (e0*(A*sin(pi*xsym/a)*sin(pi*zsym/d)).^2.+...
mu0*((-1j*A/(k0*etha0*a)*sin(pi*xsym/a).*cos(pi*zsym/d)).^2.+...
(1j*pi*A/(k0*etha0*a)*cos(pi*xsym/a).*sin(pi*zsym/d)).^2));
E2=triplequad(fun2,0,a,0,b,0,d);
df_rel_example_6_7=[df_rel_example_6_7 -E1/E2];
end
df_rel_example_6_7(1)=[];df_rel_example_6_7(end)=[];
figure(3);plot(y([2:end-1]),df_rel_example_6_7);grid on;
xlabel('t');ylabel('df0')
title('frequency relative deviation with integral')
% df0(t)/f0 from direct expression (approximation)
df_rel2_example_6_7=[]
for k=1:1:numel(y)
E1=(er-1)*e0*A^2*a*y(k)*d/4;
E2=a*b*d*e0/2*A^2;
df_rel2_example_6_7=[df_rel2_example_6_7 -E1/E2];
end
figure(4);plot(y,df_rel2_example_6_7);grid on;
xlabel('t');ylabel('df0')
title('frequency relative deviation with direct expression')
f0dev=f0+df_rel_example_6_7*f0
In this exercise E1 and E2 are not electric fields but stored energy in the entire volume and in the dielectric material.
note that while in example 6.7 the direct expression (approximation) being integrated is
while for exercise 6.11 it's
this is the volume of dielectric material for both lay-outs.
(er-1)*e0*A^2*a*y(k)*d/4
(er-1)*e0*A^2*b*z(k)*a/4;
syms x z beta_a beta_d
Z0a=k0*etha0/beta_a
Z0d=k0*etha0/beta_d
Ey1=A*sin(pi*x/a)*sin(beta_a*z) % z within [0 d-t]
Ey2=B*sin(pi*x/a)*sin(beta_d*(d-z)) % z within [d-t d]
Hx1=-1j*A/Z0a*sin(pi*x/a)*cos(beta_a*z) % z within [0 d-t]
Hx2=-1j*B/Z0d*sin(pi*x/a)*cos(beta_d*z) % z within [d-t d]
% beta_a=(k0^2-(pi/a)^2)^.5
% beta_d=((er*k0)^2-(pi/a)^2)^.5
% on the surface of the dielectric slate: Ex1(z=t)=Ex2(z=t) and Hx1(z=t)=Hx2(z=t)
Ey1_t=A*sin(pi*x/a)*sin(beta_a*z) % z within [0 d-t]
Ey2_t=B*sin(pi*x/a)*sin(beta_d*(d-z)) % z within [d-t d]
Hx1_t=-1j*A/Z0a*sin(pi*x/a)*cos(beta_a*z) % z within [0 d-t]
Hx2_t=-1j*B/Z0d*sin(pi*x/a)*cos(beta_d*z) % z within [d-t d]
eq1=Ey1_t==Ey2_t
eq2=Hx1_t==Hx2_t
eqs=[eq1,eq2]
vars=[beta_a beta_d]
S=solve(eqs,vars,'ReturnConditions',true)
% when one reads 'transcendental' same as not that easy to solve
% as previous line shows, default approach returns void solution.
S.beta_a
S.beta_d
S.conditions
S.parameters
% all empty
z=d-t
%%
syms k0
beta_a=(k0^2-(pi/a)^2)^.5
beta_d=((er*k0)^2-(pi/a)^2)^.5
Z0a=k0*etha0/beta_a
Z0d=k0*etha0/beta_d
eq1=Z0a*tan(beta_a*(d-t))==Z0d*tan(beta_d*t)
solve(eq1,k0)
eq2=beta_d*tan(beta_a*(d-t))==beta_a*tan(beta_d*t)
solve(eq2,k0)
Warning: Unable to find explicit solution. For options, see help.
In solve (line 317)
S = struct with fields:
beta_a: [0×1 sym]
beta_d: [0×1 sym]
parameters: [1×0 sym]
conditions: [0×1 sym]
ans =Empty sym: 0-by-1
ans =Empty sym: 0-by-1
ans =Empty sym: 0-by-1
ans =Empty sym: 1-by-0
Warning: Cannot solve symbolically. Returning a numeric
approximation instead.
In solve (line 304)
=
-2.1949338757197986450804920190374