sistemi2exerciseschapter2

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PRELIMINARY VERSION 1

Exercises on plane waves

2.6 Exercise 1

Consider a plane wave in a dielectric with εr = 4, µr = 1, at the frequency f = 1GHz. The electricfield in the origin is

E(0) = 2x − jy + z V/m

1. Compute the phase velocity v ph, the wavelength λ, the wave impedance Z , the wavenumberk in deg/cm, the power density dP/dΣ.

2. Find the polarization of the electric field E0. Find the direction of propagation s knowingthat the phase of the wave decreases in the z direction.

3. Compute the magnetic field in the point P : (2,2,2)T √5 at the time t = T /4

Use the approximate values c = 3 · 108m/s and Z 0 = 377Ω.

Solution

Phase velocity:

v ph =1√εµ

=c√

εrµr=

c

2= 1.5 · 108 m/s

Wavelength

λ =v ph

f =

1.5 · 108

1.0 · 1010= 1.5 · 10−2 m

Wave impedance:Z =

µ

ε= Z 0

µr

εr=

Z 02

= 188.5Ω

Wavenumber:

k =2π

λ=

6.2832

1.5 · 10−2= 4.1888 rad/cm

=360

λ= 240.0 deg/cm

Power density

dP

dΣ=

1

2

|E0|2

Z =

1

2

|E 0x|2 + |E 0y|2 + |E 0z|2

188.5=

1

2

(4 + 1 + 1)

188.5= 0.01591 W/m2

Polarization:The real and imaginary part of E(0) are

E

0 = 2x + z E

0 = −y

They are not parallel, hence the polarization is not linear. Moreover

E

0 · E

0 = 0 but |E

0| = |E

0 |




hence the polarization is not circular. In conclusion, it is elliptical.Direction of propagation:Since the electric field is perpendicular to s, we have

s = ± E

0 × E

0

|E

0 × E

0 |= ± (2x + z) × (−y)

|(2x + z) × (−y)| = ±−2z + x√5

The phase factor is

exp[− j(kxx + kyy + kzz)]

If the phase is decreasing in the z direction, then kz = ksz > 0, hence

s =2z − x√

5

Phasor of magnetic field in the origin

H0 = Y s × E0 = Y

2z − x√

5

× (2x − jy + z) = Y

j

2√5

x +5√

5y + j

1√5

z

Phasor of magnetic field in P :

H(P ) = H0 exp

− j

2π

λs · rP

= H0 exp

− j

2π

λ

2z − x√

5

· (2x + 2y + 2z)

√5

=

= H0 exp

− j

2π

λ

1√5

(2 · 2 − 2)√

5

= H0 exp

− j

4π

λ

= H0 exp[− j 120]

Magnetic field in P at t = T /4

H(rP ,t) = R

H(rP )ejω0tt=T/4

= R

H0ej(90

−120)

= H

0 cos(−30) + H

0 sin(−30) = Y

5√

3

2√

5y − 2

2√

5x − 1

2√

5z

=

=1

188.5

− 1√

5x +

√15

2y − 1

2√

5z

2.7 Exercise 2

Consider a plane wave propagating in the z direction in a dielectric with εr = 4 and γ = 0.01 S/mat the frequency f = 1.0 GHz and E0 = x.

1. Compute the wavenumber k, the phase velocity v ph, the wavelength λ

2. Compute the wave impedance Z , the active power density in the origin dP/dΣ, the attenu-ation αdB in dB/m

Use c = 2.99792458 · 108 m/s, ε0 = 8.854 · 10−12 F/m.




Solution

Wavenumber:

k = ω

ε0εr − j γ ω

µ0 = ω√ε0µ0

εr − j γ

ωε0= k0

εr − j γ

ωε0

The free space wavelength and wavenumber are

λ0 =c

f = 0.29979 m k0 =

2π

λ= 20.9585 rad/m

Then

k = 20.9585

4 − j 0.1798 = 20.9585(2.0005 − j 0.0449) = 41.9275 − j 0.9416 m−1

The real and imaginary part are

β = 41.9275 rad/m α = 0.9416 Np/m

Phase velocity:

v ph =ω

β = 1.4986 · 108 m/s

Wavelength

λ =v ph

f = 0.1499 m

Wave impedance:

Z = µ

ε0εr − jγ/ω = µ

ε0 µ

rεr − jγ /(ωε0) = Z 0(0.4996 + j 0.0112) = 188.2227 + j 4.2270 Ω

Wave admittance

Y =1

Z =

1

188.2227 + j 4.2270= (5.3102 − j 0.1192) · 10−3 S

Notice that

RY = 1

RZ =1

188.2227= 5.3129 · 10−3 S

In this case the difference is small because the phase of Z is small but becomes enormous whenthis phase approaches π/2.

Active power density in the origin

dP

dΣ=

1

2RY |E0|2 =

1

25.3102 · 10−3(|E 0x|2 + |E 0y|2 + |E 0z|2) = 2.6551 · 10−3 W/m2

Attenuation:

α = Imk = 0.9416 Np/m

αdB = α 20log10 e = α 8.68589 = 8.1785 dB/m




2.8 Exercise 3

Consider a plane wave propagating in free space at the frequency f = 5 GHz. The electric field inthe points of the plane z = 0 has the value

E(x,y,z)|z=0 = E0 exp[− j(ξx + ηy)] ∀x, y

with ξ = k0/5 and η = k0/2 and it is known that this wave is a TE field, i.e. E0 has no z-component and that it carries the power density dP/dΣ > 0.Compute the direction of propagation s and the spherical angles that define this direction. Computealso the fields in the origin E0, H0.

Solution

The propagation factor of a plane wave is

exp[− jk · r] = exp[− j(kxx + kyy + kzz)]

hence, by inspection, we find kx = ξ and ky = η. From the dispersion relation

k2x + k2y + k2z = ω2ε0µ0 = k20

it follows that

kz =

k20 − k2x − k2y =

k20 − ξ2 − η2 = k0

1 − 1

25− 1

4= k0

√71

10

The sign of the square root is taken to be positive because power is flowing toward the regionz > 0. From this

s =k

k0=

1

5x +

1

2y +

√71

10z

The spherical angles that identify the direction s are found by recalling (A.2) and noting that|s| = 1

sx = sin ϑ cos ϕ

sy = sin ϑ sin ϕ

sz = cos ϑ

From this we find

ϑ = arccos sz = arccos√71

10= 32.5827 deg

ϕ = arccos sx

sin ϑ

= arcsin

sysin ϑ

= 68.1986 deg

The electric field in the origin, with magnitude

E 0 =

2Z 0

dP

dΣ




must be perpendicular to s because it is a plane wave and perpendicular to z because required inthe text. Then

E0 = E 0 s × z|s × z| = E 0 k × z|k × z| = E 0 (ξx + ηy + kzz) × z|k × z| = E 0 (ηx − ξy) ξ2 + η2

Thanks to the denominator, the vector multiplying E 0 has unit magnitude.

The magnetic field is computed by the impedance relation

H0 = Y 0s × E0 =Y 0k0

k × E0 =Y 0E 0

k0(ξx + ηy + kz z) × (ηx − ξy)

ξ2 + η2

=Y 0E 0

k0

ξ2 + η2

kz(ξx + ηy) − (ξ2 + η2)z

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