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exercise 6.11

ZIP with MATLAB scripts and note:

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pozar_06_exercise_11_question.jpg

 exercise 6.11 notes:

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pozar_06_exercise_11_circuit.jpg

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

001 fig1.jpg
002 fig2 approx.jpg

%% 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

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004.jpg

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

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