UNIVERSITÀ DI PISA

Electromagnetic RadiationsElectromagnetic Radiationsd Bi l i l I id Bi l i l I iand Biological Interactionsand Biological Interactions

“Laurea Magistrale” in “Laurea Magistrale” in BiomedicalBiomedical EngineeringEngineeringFirst First semestersemester (6 (6 creditscredits, 60 , 60 hourshours), ), academicacademic yearyear 2011/122011/12

Prof. Paolo Prof. Paolo NepaNepa

Reflection and transmission of Reflection and transmission of plane waves:plane waves:

p.nepa@iet.unipi.itp.nepa@iet.unipi.it

examples and numerical exercisesexamples and numerical exercises

1

EditedEdited byby Dr. Dr. AndaAnda GuraliucGuraliuc31/10/2011

Construction of a 60Hz highConstruction of a 60Hz high‐‐voltage power linevoltage power line

Design considerations:Design considerations:

the metal must have strength to support the long length of line.the metal must have good conductivity to reduce the ohmic losses

For the strengthen: steel is an option, but it has low conductivity ( ). The solutionconsists in using a core of steel (it provides strength) surrounded by a sheath of aluminum,which has a better conductivity ( ).

0 1steel cooper

.σ σ=

0 61l .σ σ=y ( )0 61alu minum cooper.σ σ

Steel core2δAluminum sheath

The 60Hz currents will flow primarily in the aluminum sheath as long as its thickness is a few skin depths.

@60Hz :1 1 1alu minum . cmf

δπ µσ

= ≈

For a sheath thickness of a few centimeters the majority of the current is confined in the aluminum sheath

73 63 10alu minum . S / mσ = ⋅

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For a sheath thickness of a few centimeters, the majority of the current is confined in the aluminum sheathand very little goes into the steel core.

Electromagnetic shieldingElectromagnetic shieldingShields are used to prevent unwanted EH‐fields and other electromagnetic disturbances fromt i t t d l t tientering a protected electromagnetic zone.

Shielded room

2δ

Ex: sensitive electronic equipment inside the shielded room is protected from theelectromagnetic fields generated by a radar transmitter outside the room.electromagnetic fields generated by a radar transmitter outside the room.

The transmitter fields will impinge on the shield wall and induce currents in the wall.If the conducting wall thickness is several skin depths, the induced current on the outsideill d i ifi tl b f hi th th id f th ll d i it l l dwill decay significantly before reaching the other side of the wall, reducing its level and

preventing its reception by the electronics inside the room.For a transmitter at 3GHz, and room walls of steel ( ) the steel skin

depth is:0 1 2000

r cooper r. ,σ σ µ= =

0 09 mδ µ=

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0 09. mδ µ=

Magnetically shielded room (with ferrites)Magnetically shielded room (with ferrites)Ex: magnetically shielded room can be used in medical studies like:storage of iron in the liver particles in lungs brain studies heart

To measure the iron concentration, the basic idea is to apply a fieldto the region of the liver and measure the field response produced by

storage of iron in the liver, particles in lungs, brain studies, heartfunctions.

g p p ymolecules in the body tissues. The response of the tissue incomparison with the strength of applied field is called magneticsusceptibility of the tissue. Hence to determine iron concentrationonly the susceptibility of the liver needs to be measuredonly the susceptibility of the liver needs to be measured.

The body produces magnetic fields in two main ways: by electric currents and byferromagnetic particles. The electric currents are the ion currents generated by the muscles,nerves and other organsnerves, and other organs.

The ion current generated by heart muscle, which provides the basis for theelectrocardiogram, also produces a magnetic field over the chest.

The ion current generated by the brain, which provides the basis for theelectroencephalogram, also produces a magnetic field over the head.

Ferromagnetic particles are insoluble contaminants of the body; the most important are theferromagnetic dust particles in the lungs (magnetite). Magnetic fields can give informationabout the internal organs not otherwise available

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about the internal organs not otherwise available.

These magnetic fields are very weak and can be measured using a magnetic shielded room.These magnetic fields are very weak and can be measured using a magnetic shielded room.

High frequencyHigh frequency resistanceresistance of a wireof a wire

Consider a circular cross section conductor of radius rw.

DC: the current will be uniformly distributed over the wirecross section.the DC resistance per unit length is:

wrwr

δ

DC High frequency

wr δ> 2

1DC

w

rrσπ

= [Ω / m]wr δ<<

if the frequency of the current increases, the current will be concentrated close to the wireq ysurface

at higher frequency, where the wire radius is greater than skin depth, the cross‐sectional areaoccupied by the current is smaller, and the resistance per unit length is:

112HF

w

rrσ π δ

= [Ω / m]wr δ>>

Skin depth: 1 1f f

δπ µσ

= ∝

H Fr increases at a rate of f

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Radome designRadome design

RadomeRadome (Radar Dome) is a protective dielectric enclosure for amicrowave antenna. A ground‐based C‐band (NATO: 500‐1000MHz,IEEE 4 8GH ) i l di t d f i l l diIEEE:4‐8GHz) microwave landing system used for airplanes landingmust be protected from the weather by enclosing it in a radome.

Consider: the center frequency band 5GHz; thermoplastic PEI material (lossless, nonmagnetic, ε2r=3);Assume a flat planar radomeCalculate: (1) the minimum thickness of the radome that will give no reflections at 5GHz.

(2) for the frequency 4GHz, what percentage of the incident power is reflected?

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(3) for the frequency 6GHz, what percentage of the incident power is reflected?

Thickness of the Thickness of the radomeradome layer: design at 5 GHzlayer: design at 5 GHz

Incident wave

( )8

12 5 9

3 10 3 465 10 3@ GHz

c m / s . cmf Hz

λλε ε

⋅= = = =

Air

Radome layer

1ζ

2ζ2

4/λ@5GHz z=d

z=0

( )2 25 10 3

r rf Hzε ε ⋅

In order not to affect the operation of the microwave landing system:

λAir3 1

ζ ζ=z

2 1 2 32

d N ,N , , ....λ= =

2 1 73d λ2 1 73

2mind . cm= =

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Reflected power@4GHzReflected power@4GHz

4f GHz=Incident wave

4f GHz=Air

Ai

Radome layer

1ζ

2ζ

ζ ζ

24/λ

@4GHzz=d

z=0

7 5λ

8

1 9

3 10 7 54 10

m / s . cmHz

λ ⋅= =

⋅λAir

3 1ζ ζ=

z1

2

2

7 5 4 333

r

. . cmλλε

= = =

12

377 2183

ζζε

= = = Ω

21 732mind . cm λ

= ≠

23

rε

1

2 2

2

2 1 45k . cmπβλ

−= = =

( )( ) ( )( ) 22j k deζ ζ ζ ζ ζ ζ ζ ζ −+ + +( )( ) ( )( )( )( ) ( )( )

2

2

2 1 3 2 2 1 3 2

2

2 1 3 2 2 1 3 2

0.321 130j k d

ee

ζ ζ ζ ζ ζ ζ ζ ζΓ

ζ ζ ζ ζ ζ ζ ζ ζ −

− + + + −= = ∠ °

+ + + − −

Reflected power back to air:2100 100 10.3%

r

i

SS

Γ× = × =

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Reflected power@6GHzReflected power@6GHzIncident wave

6f GHz=Air

Ai

Radome layer

1ζ

2ζ

ζ ζ

24/λ

@6GHz

z=-d

z=0

6f GHz=

5λ

8

1 9

3 10 56 10

m / s cmHz

λ ⋅= =

⋅λAir

3 1ζ ζ=

z1

2

2

5 2 893

r

. cmλλε

= = =

12

377 2183

ζζε

= = = Ω

21 732mind . cm λ

= ≠

23

rε

1

2 2

2

2 2 17k . cmπβλ

−= = =

( )( ) ( )( ) 22j k deζ ζ ζ ζ ζ ζ ζ ζ −+ + +( )( ) ( )( )( )( ) ( )( )

2

2

2 1 3 2 2 1 3 2

2

2 1 3 2 2 1 3 2

0.321 130j k d

ee

ζ ζ ζ ζ ζ ζ ζ ζΓ

ζ ζ ζ ζ ζ ζ ζ ζ −

− + + + −= = ∠ °

+ + + − −

Th ff ti fl ti ffi i t i t i ll d th t f

Reflected power back to air:2100 100 10.3%

r

i

SS

Γ× = × =

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The effective reflection coefficient varies symmetrically around the center frequency , being down by the same amount in magnitude at 4GHz and 6GHz.

Reflectance of glassReflectance of glassConsider a beam of light is incident normally from air on a BK‐7 glass interface; BK‐7 l f ti i d 1 52glass refraction index n=1.52Calculate the reflection coefficient and the percent of incident energy reflectedAir BK‐7 glass

Incident wave 1 2 0 2

2 25r

; .µ µ µ ε= = = 0 0

2 2 2

11 52

r r

cn .ε µ

ε µ= = = =2 2 2

2 21r r v

µε µ

2 1 1 21 1 52 0 206n n . .ζ ζ− − −

Γ = = = = −

Reflected wave

2 1 1 21 1 52n n .ζ ζ+ + +

2100 100 4.26%r

i

SS

Γ× = × =S

4% of the incident power is reflected by the glass interface. In some applications, this loss may besignificant. A camera lens consists of three or more separate lenses, which represents six or more air‐glassinterfaces.

If 4% of the incoming energy reflects back every time light passes through one of these interfaces, up to22% of original energy is lost during each traverse of light through the lens.

These losses can be reduced by introducing antireflection coating on the glass surface.

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Coated glass surfaceCoated glass surfaceConsider a beam of light is incident normally from air on a BK‐7 glass interface; BK‐7 glass refraction indexn=1 52n=1.52Calculate the refractive index and the minimum thickness of a film (MgF2 n=1.38 ) to be deposited on theglass surface such that no normally incident visible light of free‐space wavelength 550nm (~545THz) isreflected.

1 550nmλ = (green light) ( )0r rn , /ε ζ ζ ε= =

Incident wave

Air1

1n =

1 3824/λ

z=0MgF2

2 1 3 1 1 52 1 23n n n . .= = × =9

2 1 550 10 0 1124 4 4 1 23mind d . mλ λ µ

−⋅= = = = =

( )2 1 3 2 1 3n n nζ ζ ζ= → =

( )0r rζ ζ

BK‐7 glass

2 1 38n .=

3 1 52n .=

2

@550nmz=d

MgF22

4 4 4 1 23min n .µ

⋅

In practice it is difficult to manufacture antireflection coating materials with desired refractive index, so a common material is MgF2 with n=1.38.

12

2

550 3991 38

nmn .λλ = = = 2 99 6

4mind d . nmλ= = =

377ζ = Ω1

377ζ = Ω

12

2

273 19.nζζ = = Ω

1 248 03ζζ Ω0 112.Γ = − 2100 100 1.26% 4%

r

i

SS

Γ× = × = <

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13

3

248 03.nζζ = = Ω S

Effectiveness over the visible spectrumEffectiveness over the visible spectrum

1 400nmλ = (violet light)

0 1485 163.Γ = ∠ °

2100 100 2.21% 4%r

i

SS

Γ× = × = <iS

1 750nmλ = (red light)

0 13 164 6. .Γ = ∠ °2100 100 1.7% 4%

r

i

SS

Γ× = × = <

( )2 1 41

( )1 2 1 09rn . .ε= = ≅

( )2 1 41rn .ε= = ≅

( )1 5 1 22rn . .ε= = ≅

31/10/2011 12

MultipleMultiple‐‐layered coatingslayered coatingsAntireflection coatings are required in most optical applications in order to reduce unwanted reflections

at the surface of optical elementsat the surface of optical elements.It is desirable to reduce the surface reflectivity (or reflection coefficient) to an extremely low value over

an extended spectral region to maintain proper color balance and provide optimum efficiency.A single layer is not enough since it reduce reflectivity only from 4% to 2.2%.To achieve the requirements a triple layer coating can be implementedTo achieve the requirements, a triple‐layer coating can be implemented.

Consider: a wideband nonmagnetic antireflection coating system for an air‐glass interface, including threelayers of coating materials.Calculate and plot: the reflection coefficient over the wavelength range 400‐750nm.

Incident wave

Air1

1n =

4/λ MgF2

1 550nmλ = (green light)

2 12

2

99 44 4

d . nmn

λ λ= = = 3 1

3

3

1152 2

d nmn

λ λ= = = 4 1

4

4

80 94 4

d . nmn

λ λ= = =

1377ζ = Ω

ζ ζ ζ2

1 38n .=2 4/λ MgF2

ZnS

C F

32 40n .=3 2/λ

4/λ

12

2

273 19.nζζ = = Ω 1

3

3

157nζζ = = Ω 1

4

4

222nζζ = = Ω 1

5

5

248nζζ = = Ω

BK‐7 glass5

1 52n .=

CeF34

1 70n .=4 4/λ

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http://www.kruschwitz.com/materials.htm

ReferencesReferences

1. G. Manara, A. Monorchio, P.Nepa, “Appunti di Campi Elettromagnetici”, , p , pp p g2. http://www.mathworks.com/matlabcentral/fileexchange/16724‐gui‐for‐tetm‐

electromagnetic‐plane‐waves‐propagation‐through‐multilayered‐structures3. U.S. Inan, A.S. Inan, “Electromagnetic waves”, Prentice Hall, 2000.

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