example 6.2

er=2.08                                                                % alumina

tand=0.0004

d=1.59e-3                                                            % substrate depth

Z0=50;f0=5e9;

 

A=Z0/60*((er+1)/2)^.5+(er-1)/(er+1)*(.23+.11/er)

B=120*pi*pi/(2*Z0*(er)^.5)

 

% guess W/d<2

Wd_below2=8*exp(A)/(exp(2*A)-2)    % W/d

 

% guess W/d>2

Wd_above2=2/pi*(B-1-log(2*B-1)+(er-1)/(2*er)*(log(B-1)+.39-.61/er))  % W/d

 

% Wd_below2>2 is contradictory therefore W/d>2

Wd=Wd_above2;W=Wd*d

ee=(er+1)/2+(er-1)/2*1/(1+12*d/W)^.5                

% effective relative permittivity ee within [1 er]

 

phi_target_deg=180;                                 

% in degree, phi=beta*L=(ee)^.5*k0*L

​

c0=2.998e8;

lambda0=c0/f0

lambda=lambda0/(ee)^.5

Length_resonator=lambda/2                                    % m

 

k0=2*pi*f0/c0                                                         % wave number

vp=c0/(ee)^.5                                                        % propagation velocity

beta=2*pi*f0*ee^.5/c0                                        % propagation constant

​

% 283pg print error

% 2*pi*5e9*1.8^.5/3e8 = 1.4049e+02, not 151.0 in book.

 

D=Length_resonator/lambda % D fraction effective lambda, not lambda0

 

Rs_Cu_10GHz=1.84e-2                                           % ohm

 

alpha_c_Np=Rs_Cu_10GHz/(Z0*W)                          % [Np/m]

alpha_d_Np=k0*er*(ee-1)*tand/(2*(ee)^.5*(er-1))     % [Np/m]

 

Np2dB=10*log10((exp(1))^2)

alpha_c=alpha_c_Np*Np2dB                     % dB/m conductor attenuation

alpha_d=alpha_d_Np*Np2dB                      % dB/m dielectric attenuation

 

alpha_Np=alpha_c_Np+alpha_d_Np

alpha=alpha_c+alpha_d                                           % dB/m

 

Q0=beta/(2*alpha_Np)