exercise 1.8

c0=299792458        % [m/s] light speed

mu0=4*pi*1e-7     % [H/m] permeability
mur=1                   % [] relative permeability

etha0=120*pi         % [Ohm] air intrinsic impedance

E0=1                      % [V/m] incident electric field amplitude
f0=1e9                   % [Hz]
lambda0=c0/f0        % [m] wavelength
k0=2*pi/lambda0    % [m^-1] wave number


metal conductivity :  [S/m] = [1/Ohm] = [mho/m]

annealed (=deoxidized) copper conductivity 5.813e7 

sigma_Cu=59.6e6    % common copper conductivity             

metal skin depth

ds_Cu=(2/(2*pi*f0*mu0*mur*sigma_Cu))^.5

gamma : propagation constant

gamma_Cu=(1+j)*(2*pi*f0*mu0*sigma_Cu/2)^.5

approximation of gamma for good conductor

gamma_Cu=(1+1j)/ds_Cu

inside the metal

alpha_Cu=real(gamma_Cu)
beta_Cu=imag(gamma_Cu)

wave intrinsic impedance

etha_Cu=(1+1j)/(sigma_Cu*ds_Cu)

R : reflection coefficient metal surface
 

R=(etha_Cu-etha0)/(etha_Cu+etha0)


a) transmission loss [dB] air-Cu and Cu-air

T : transmission coefficient metal surface

T_air_Cu=2*etha_Cu/(etha_Cu+etha0)

assuming T_air_metal = T_metal_air

T_air_Cu_dB=10*log10(abs(T_air_Cu))


simple signal path budget

-150-10*log10(abs(T_air_Cu))-10*log10(abs(T_air_Cu))


b) dz to achieve 150dB

finding out required Cu thickness

syms dz
eq1=10*log10(exp(-2*alpha_Cu*dz))==-150-2*10*log10(abs(T_air_Cu))
z1=double(solve(eq1,dz))
 
z1_mm=z1*1e3


c.- Additional 1: How many sheets of common household tin foil would achieve same attenuation?

dz2=.2e-3                % [m] common tin foil thickness
sigma_tf=2.436e6     % [S/m]  tin foil electrical conductivity
 
ds2 << dz2 ok
ds_tf=(2/(2*pi*f0*mu0*mur*sigma_tf))^.5    % tin foil skin depth

gamma2 : tin foil propagation constant

gamma_tf=(1+j)*(2*pi*f0*mu0*sigma_tf/2)^.5

approximation of gamma for good conductor

gamma_tf=(1+1j)/ds_tf

inside the metal

alpha_tf=real(gamma_tf)
beta_tf=imag(gamma_tf)

wave intrinsic impedance

etha_tf=(1+1j)/(sigma_tf*ds_tf)

R : reflection coefficient metal surface 

R_air_tf=(etha_tf-etha0)/(etha_tf+etha0)


T : transmission coefficient air - metal surface

T_air_tf=2*etha_tf/(etha_tf+etha0)

2*10*log10(abs(T_air_tf))

how many layers

n=1;
while n*2*10*log10(abs(T_air_tf))>-150
   n=n+1;
end
n

So to achieve 150dB wrap it up at least thrice, not twice. Twice would be 140dB only.


d.- Additional 2: What happens if one (wrongly) assumes metal behaving as dielectric

It may be the case that one considers calculating transmission coefficient on the right hand side by just swapping intrinsic impedances as follows:
For copper

R_Cu_air=(etha0-etha_Cu)/(etha0+etha_Cu)
T_Cu_air=2*etha0/(etha0+etha_Cu)

20*log10(abs(T_air_Cu*T_Cu_air))

So apparently Cu, no matter how thick, it wouldn't reach -150dB Insertion Loss

for tin foil

R_tf_air=(etha0-etha_tf)/(etha0+etha_tf)
T_tf_air=2*etha0/(etha0+etha_tf)

20*log10(abs(T_air_tf*T_tf_air))

At this point one could reasonably ask the following 2 questions:

1.- So which one is the correct metal-air T coefficient?

2.- Although once a metal sheet is thicker than the skin depth then most of
the current remains in one side of the metal sheet, just using T12 T21 there's no influence of the actual thickness of the metal whatsoever in this Insertion Loss calculation.
metal thickness should somehow be taken into
account, because if the metal sheet is not thick enough, then there may be a
substantial amount of field reaching the other side. So why is is that
metal thickness is not considered with this dielectric-like way to
calculate insertion losses?


e.- Additional 3: Alternative calculation with exp(-alpha*z) and radiated field
So let's review the wave path:

1.- E0 V/m hits metal surface [X Y] plane

H0 is magnetic field associated incident E0. 

2.- T*E0 electric field gets in the metal and the wave in attenuation mode (heat through solids) moves towards the other side being attenuated along the path with exp(-alpha*z)
 
3.- the amplitude of the electric field reaching the other side of the copper sheet is E1=T*E0*exp(-alpha*dz) the amplitude of the magnetic field reaching the other side of the copper sheet is H1=1/etha*T*E0*exp(-alpha*dz)
 
example 1.3 solves radiated field from current sheet.
The magnetic field H1 that has reached the other side of the metal sheet
generates a surface current J0=cross(nz,H1) but this does not consider
the type of metal where the surface current is generated.
 
4.- on the right hand side surface of the metal sheet the surface current is
J1=sigma*E1
 
5.- EH fields that 'get out' of the metal sheet, radiated, back onto air:
H2=-J1/2*exp(-1j*k0*z)
E2=-etha0*J1/2*exp(-1j*k0*z)
 
|H1|=|H2|

E2=-etha0*sigma*T*E0*exp(-alpha*z)/2*exp(-1j*k0*z)
 
the target is to reach 150dB attenuation, therefore

|E0|^2/|E2|^2 [dB]= 150
 
comparing Cu and tf (tin foil) sheets same thickness z1 the apparently thick enough to reach 150dB

10*log10(abs(1/(-etha0*sigma_Cu*T_air_Cu*exp(-alpha_Cu*z1)/2 )^2))
 

10*log10(abs(1/(-etha0*sigma_tf*T_air_tf*exp(-alpha_tf*z1)/2 )^2))

copper is a better reflector than tin foil

10*log10(abs(T_air_Cu))
10*log10(abs(T_air_tf))

but for whatever field manages to cross the foil, for same thickness, copper is also a better radiator because higher conductivity causes higher surface current on the other side.
 
Using z1 thick Cu foil would not work to reach 150dB

So how thick should a copper foil be to reach 150dB 
 
syms dz3
eq2=abs(1/(-etha0*sigma_Cu*T_air_Cu*exp(-alpha_Cu*dz3)/2 )^2)==10^(150/10)
z3=double(solve(eq2,dz3))

comparing actual dz required to reach 150dB against the figure reached
assuming air -metal and metal-air 'symmetry'

z3/z1

about 4 times more than the theoretical skin depth calculation

and how thick should a tin foil be to reach 150dB 
 
syms dz4
eq3=abs(1/(-etha0*sigma_tf*T_air_tf*exp(-alpha_tf*dz4)/2 )^2)==10^(150/10)
z4=double(solve(eq3,dz4))

z4/z3

tf has to be 4.6 thicker than Cu


f.- Additional 4: Comparing thin films

alpha attenuates fields faster inside copper than inside tin foil

zm=[dz2/100:dz2/100:dz2];   % avoid null thickness
L1=exp(-alpha_Cu*zm);
L2=exp(-alpha_tf*zm);

figure;
plot(zm,L1,zm,L2);
title('\alpha_Cu vs \alpha_tinfoil')
grid on;
xlabel('z [m]')
legend({'Cu','tin foil'})

but

​% correction factor ; zero scale
k1=10*log10(abs(-etha0*sigma_Cu*T_air_Cu*exp(-alpha_Cu*zm(1))/2 ).^2);  

L1_total_dB=-10*log10(abs(-etha0*sigma_Cu*T_air_Cu*exp(-alpha_Cu*zm)/2 ).^2)+k1 ;

k2=10*log10(abs(-etha0*sigma_tf*T_air_tf*exp(-alpha_tf*zm(1))/2 ).^2)
L2_total_dB=-10*log10(abs(-etha0*sigma_tf*T_air_tf*exp(-alpha_tf*zm)/2 ).^2) +k2;

figure;
plot(zm,L1_total_dB,zm,L2_total_dB);
title('total attenuation comparison')
xlabel('z [m]')
ylabel('[dB]')
grid on;
legend({'Cu','tin foil'})

hitting a metal with a plane wave allows the calculation of transmitted
and reflected coefficients, but when transferring wave from metal back
to air, assuming 'symmetry' as suggested in the solutions manual, seems
precarious, because air-metal-air is not the same as air-dielectric-air.

Also simplifying all current to be contained in skin depth, may not be
correct if foil to thin.

So to achieve 150dB insertion loss, 1/4 mm copper would be a start value, not the ridiculously thin 17e-6 mm reached in the solutions manual, or 15.7e-6 mm initially calculated here following solutions manual assumptions for this exercise.