Electromagnetic Waves and Antennas

137 Pages·2010·1.03 MB·English

Electromagnetic Waves
and
Antennas

Exercise book

Sophocles J. Orfanidis1
Davide Ramaccia2
Alessandro Toscano2

1Department of Electrical & Computer Engineering
Rutgers University, Piscataway, NJ 08854
[email protected]
www.ece.rutgers.edu/~orfanidi/ewa

2Department of Applied Electronics, University "Roma Tre"
via della Vasca Navale, 84 00146, Rome, Italy
[email protected]
[email protected]
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
Table of Contents

Chapter1 Maxwell's Equations...................................................................................1
1.1 Exercise..........................................................................................................1
1.2 Exercise..........................................................................................................6
1.3 Exercise........................................................................................................12
1.4 Exercise........................................................................................................16
1.5 Exercise........................................................................................................18
1.6 Exercise........................................................................................................20
1.7 Exercise........................................................................................................25
1.8 Exercise........................................................................................................29
1.9 Exercise........................................................................................................30
1.10 Exercise........................................................................................................32
1.11 Exercise........................................................................................................42
1.12 Exercise........................................................................................................54
1.13 Exercise........................................................................................................56

D. Ramaccia and A. Toscano
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
Chapter1
Maxwell's Equations
1.1 Exercise
Prove the vector algebra identities:
a) A×(B×C) =B(A⋅C)-C(A⋅B)
It is possible to write the vectors in the form:
⎧A=A xˆ +A yˆ +A zˆ
x y z
⎪⎪
⎨B=Bxxˆ +Byyˆ +Bzzˆ (1.1.1)
⎪
C=C xˆ +C yˆ +C zˆ
⎪⎩ x y z
and to use the follow relationship:
xˆ yˆ zˆ
U×V = U U U =
x y z
(1.1.2)
V V V
x y z
= xˆ(U V −U V )−yˆ(U V −U V )+zˆ(U V −U V )
y z z y x z z x x y y x
Now we can prove the algebra identities with simply mathematical substitutions:
A×(B×C)=A×(xˆ(B C −B C )−yˆ(B C −B C )+zˆ(B C −B C ))=
y z z y x z z x x y y x
= xˆ(A (B C −B C )+A (B C −B C ))
y x y y x z x z z x
(1.1.3)
( ( ) ( ))
−yˆ A B C −B C −A B C −B C
x x y y x z y z z y
+zˆ(−A (B C −B C )−A (B C −B C ))
x x z z x y y z z y
Expanding the terms in (1.1.3), we have:
A×(B×C)=
( )
+xˆ A B C −A B C +A B C −A B C
y x y y y x z x z z z x
( )
+yˆ A B C −A B C +A B C −A B C (1.1.4)
x y x x x y z y z z z y
( )
+zˆ A B C −A B C −A B C +A B C
x z x x x z y y z y z y
Let us write eq. (1.1.4) in matrix form, separating the terms with the minus sign and the terms
with the plus sign:
D. Ramaccia and A. Toscano    Pag. 1
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
⎡ 0 B A C B A C ⎤ ⎡ 0 C A B C A B ⎤
y x x z x x y x x z x x
⎢ ⎥ ⎢ ⎥
A×(B×C)= ⎢B A C 0 B A C ⎥−⎢C A B 0 C A B ⎥(1.1.5)
x y y z y y x y y z y y
⎢ ⎥ ⎢ ⎥
B A C B A C 0 C A B C A B 0
⎢⎣ x z z y z z ⎥⎦ ⎢⎣ x z z y z z ⎥⎦
Note that the elements of the diagonal of each matrix are zero. Each term can be filled with the
product of the three component with the same subscript (a = A B C ):
ii i i i
⎡B A C B A C B A C ⎤ ⎡C A B C A B C A B ⎤
x x x y x x z x x x x x y x x z x x
⎢ ⎥ ⎢ ⎥
A×(B×C)=⎢B A C B A C B A C ⎥−⎢C A B C A B C A B ⎥=
x y y y y y z y y x y y y y y z y y
⎢ ⎥ ⎢ ⎥
B A C B A C B A C C A B C A B C A B
⎢⎣ x z z y z z z z z⎥⎦ ⎢⎣ x z z y z z z z z⎥⎦
( ) ( ) ( )
=+A C B xˆ+B yˆ+B zˆ +A C B xˆ+B yˆ+B zˆ +A C B xˆ+B yˆ+B zˆ −
x x x y z y y x y z z z x y z
−A B (C xˆ+C yˆ+C zˆ)−A B (C xˆ+C yˆ+C zˆ)−A B (C xˆ+C yˆ+C zˆ)= (1.1.6)
x x x y z y y x y z z z x y z
( ) ( )
=B A C +A C +A C −C A B +A B +A B =
x x y y z z x x y y z z
=B(A⋅C)−C(A⋅B)

b) A⋅(B×C)= B⋅(C×A)= C⋅(A×B)
Using relationships (1.1.1) and (1.1.2), we can write:
A⋅(B×C)= A⋅(xˆ(B C −B C )−yˆ(B C −B C )+zˆ(B C −B C ))=
y z z y x z z x x y y x
(A B C −A B C )−(A B C −A B C )+(A B C −A B C )= (1.1.7)
x y z x z y y x z y z x z x y z y x
( ) ( )
A B C +A B C +A B C − A B C +A B C +A B C
x y z y z x z x y x z y y x z z y x
B⋅(C×A)= B⋅(xˆ(C A −C A )−yˆ(C A −C A )+zˆ(A C −A C ))=
y z z y x z z x x y y x
( ) ( ) ( )
B C A −B C A − B C A −B C A + B A C −B A C =
x y z x z y y x z y z x z x y z y x
(1.1.8)
( ) ( )
B C A +B C A +B C A − B C A +B C A +B C A =
x y z y z x z x y x z y y x z z y x
↑
order them
( ) ( )
A B C +A B C +A B C − A B C +A B C +A B C
x y z y z x z x y x z y y x z z y x
C⋅(A×B)=C⋅(xˆ(A B −A B )−yˆ(A B −A B )+zˆ(A B −A B ))=
y z z y x z z x x y y x
( ) ( ) ( )
C A B −C A B − C A B −C A B + C A B −C A B =
x y z x z y y x z y z x z x y z y x
(1.1.9)
( ) ( )
C A B +C A B +C A B − C A B +C A B +C A B =
x y z y z x z x y x z y y x z z y x
↑
order them
( ) ( )
A B C +A B C +A B C − A B C +A B C +A B C
x y z y z x z x y x z y y x z z y x
If we compare the last row of each expression, we note that they are identical so the algebra
identity is verified.
D. Ramaccia and A. Toscano    Pag. 2
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1

2 2 2 2
c) A×B + A⋅B = A B
Using relationships (1.1.1) and (1.1.2), we can write:

2
A×B2 + A⋅B2 = xˆ(A B −A B )−yˆ(A B −A B )+zˆ(A B −A B ) +
y z z y x z z x x y y x
( )2
+ A B +A B +A B =
x x y y z z
2
⎛⎜ (AyBz −AzBy)2 +(AxBz −AzBx)2 +(AxBy −AyBx)2 ⎞⎟ +(AxBx +AyBy +AzBz)2 =
⎝ ⎠
(A B −A B )2 +(A B −A B )2 +(A B −A B )2 +(A B +A B +A B )2 =
y z z y x z z x x y y x x x y y z z
2 2 2 2 2 2 2 2
A B +A B −2A B A B +A B +A B −2A B A B +
y z z y y z z y x z z x x z z x
2 2 2 2 ( )2
A B +A B −2A B A B + A B +A B +A B =
x y y x x y y x x x y y z z
2 2 2 2 2 2 2 2
A B +A B −2A B A B +A B +A B −2A B A B +
y z z y y z z y x z z x x z z x
2 2 2 2 2 2 2 2 2 2
A B +A B −2A B A B +A B +A B +A B +
x y y x x y y x x x y y z z
2A B A B +2A B A B +2A B A B =
x y y x x z z x x y y x
↑
cancel the opposites

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
A B +A B +A B +A B +A B +A B +A B +A B +A B =
y z z y x z z x x y y x x x y y z z
( 2 2 2)( 2 2 2) 2 2
A +A +A B +B +B = A B
x y z x y z

d) A=nˆ×A×nˆ +(nˆ⋅A)nˆ
Does it make a difference whether nˆ ×A×nˆ is taken to mean (nˆ ×A)×nˆ or nˆ ×(A×nˆ)?
The unit vector nˆcan be expressed as follow:
⎧nˆ = n xˆ +n yˆ +n zˆ
⎪ x y z
(1.1.10)
⎨
⎪nˆ = n2 +n2 +n2 =1
⎩ x y z
Let us begin considering the first case:
D. Ramaccia and A. Toscano    Pag. 3
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
(nˆ×A)×nˆ = ⎡xˆ(n A −n A )− yˆ(n A −n A )+zˆ(n A −n A )⎤×nˆ =
⎣ y z z y x z z x x y y x ⎦
+xˆ⎡(n A −n A )n −(n A −n A )n ⎤+
⎣ z x x z z x y y x y⎦
− yˆ ⎡(n A −n A )n −(n A −n A )n ⎤+
⎣ y z z y z x y y x x⎦
+zˆ⎡(n A −n A )n −(n A −n A )n ⎤ = (1.1.11)
⎣ y z z y y z x x z x⎦
+xˆ⎡n2A −n n A −n n A +n2A ⎤+
⎣ z x x z z x y y y x⎦
− yˆ ⎡n n A −n2A −n2A +n n A ⎤+
⎣ y z z z y x y y x x⎦
+zˆ⎡n2A −n n A −n n A +n2A ⎤
⎣ y z z y y z x x x z⎦
And now consider the second case:
nˆ×(A×nˆ)=nˆ×⎡xˆ(A n −A n )− yˆ(A n −A n )+zˆ(A n −A n )⎤ =
⎣ y z z y x z z x x y y x ⎦
+xˆ⎡n (A n −A n )−n (A n −A n )⎤+
⎣ y x y y x z z x x z ⎦
−yˆ ⎡n (A n −A n )−n (A n −A n )⎤+
⎣ x x y y x z y z z y ⎦
+zˆ⎡n (A n −A n )−n (A n −A n )⎤ = (1.1.12)
⎣ x z x x z y y z z y ⎦
+xˆ⎡A n2 −A n n −A n n +A n2⎤+
⎣ x y y y x z z x x z⎦
−yˆ ⎡A n n −A n2 −A n2 +A n n ⎤+
⎣ x x y y x y z z z y⎦
+zˆ⎡A n2 −A n n −A n n +A n2⎤
⎣ z x x x z y y z z y⎦
It is very easy to show that (nˆ × A)×nˆ = nˆ ×(A×nˆ).
The second term of the identity can be written as:
D. Ramaccia and A. Toscano    Pag. 4
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
( )( )
(nˆ ⋅A)nˆ = n A +n A +n A n xˆ +n yˆ +n zˆ =
x x y y z z x y z
+xˆ⎡n (n A +n A +n A )⎤+
⎣ x x x y y z z ⎦
+yˆ ⎡n (n A +n A +n A )⎤+
⎣ y x x y y z z ⎦
+zˆ⎡n (n A +n A +n A )⎤ =
⎣ z x x y y z z ⎦ (1.1.13)
+xˆ⎡n2A +n n A +n n A ⎤+
⎣ x x x y y x z z⎦
+yˆ ⎡n n A +n2A +n n A ⎤+
⎣ y x x y y y z z⎦
+zˆ⎡n n A +n n A +n2A ⎤
⎣ z x x z y y z z⎦
Adding the two results, we obtain:
nˆ×A×nˆ +(nˆ ⋅A)nˆ =
+xˆ⎡A n2 −A n n −A n n +A n2⎤+
⎣ x y y y x z z x x z⎦
−yˆ ⎡A n n −A n2 −A n2 +A n n ⎤+
⎣ x x y y x y z z z y⎦
+zˆ⎡A n2 −A n n −A n n +A n2⎤+
⎣ z x x x z y y z z y⎦
+xˆ⎡n2A +n n A +n n A ⎤+
⎣ x x x y y x z z⎦
+yˆ ⎡n n A +n2A +n n A ⎤+
⎣ y x x y y y z z⎦
+zˆ⎡n n A +n n A +n2A ⎤ =
⎣ z x x z y y z z⎦
↑
change signs in parentheses at first yˆ and add
+xˆ⎡A n2 −A n n −A n n +A n2 +n2A +n n A +n n A ⎤+
⎣ x y y y x z z x x z x x x y y x z z⎦
+yˆ ⎡A n2 +A n2 −A n n −A n n +n n A +n2A +n n A ⎤+
⎣ y x y z x x y z z y y x x y y y z z⎦
+zˆ⎡A n2 −A n n −A n n +A n2 +n n A +n n A +n2A ⎤ =
⎣ z x x x z y y z z y z x x z y y z z⎦
(1.1.14)
+xˆA ⎡n2 +n2 +n2⎤++yˆA ⎡n2 +n2 +n2⎤++zˆA ⎡n2 +n2 +n2⎤ = A
x ⎣ y z x⎦ y ⎣ x z y⎦ z ⎣ x y z⎦
D. Ramaccia and A. Toscano    Pag. 5
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1

1.2 Exercise
Prove the vector analysis identities:
1. ∇×(∇φ)= 0
2. ∇⋅(φ∇ψ)= φ∇2ψ+∇φ⋅∇ψ (Green's first identity)
3. ∇⋅(φ∇ψ−ψ∇φ)=φ∇2ψ−ψ∇2φ (Green's second identity)
4. ∇⋅(φA)= (∇φ)⋅A +φ∇⋅A
5. ∇×(φA)= (∇φ)×A + φ∇×A
6. ∇⋅(∇×A)= 0
7. ∇ ⋅A ×B = B⋅(∇× A)− A ⋅(∇×B)
8. ∇×(∇×A)=∇(∇⋅A)−∇2A
First of all we have to express the operator ∇in general orthogonal coordinates in four common
applications. All vector components are presented with respect to the normalized base (eˆ ,eˆ ,eˆ ):
1 2 3
⎧ eˆ ∂φ eˆ ∂φ eˆ ∂φ
1 2 3
∇φ= + +
⎪
h ∂q h ∂q h ∂q
⎪ 1 1 2 2 3 3
⎪ 1 ⎡ ∂ ⎛h h ∂φ ⎞ ∂ ⎛h h ∂φ ⎞ ∂ ⎛h h ∂φ ⎞⎤
⎪∇2φ= ⎢ ⎜ 2 3 ⎟+ ⎜ 1 3 ⎟+ ⎜ 1 2 ⎟⎥
⎪ h1h2h3 ⎢⎣∂q1⎝ h1 ∂q1⎠ ∂q2 ⎝ h2 ∂q2 ⎠ ∂q3 ⎝ h3 ∂q3 ⎠⎥⎦
⎪
⎪ 1 ⎡ ∂ ∂ ∂ ⎤
∇⋅F = (Fh h )+ (F h h )+ (F h h )
⎪ ⎢ 1 2 3 2 1 3 3 1 2 ⎥
h h h ∂q ∂q ∂q
⎪ 1 2 3 ⎣ 1 2 3 ⎦
⎪⎪ h eˆ h eˆ h eˆ
1 1 2 2 3 3
⎨
⎪ 1 ∂ ∂ ∂
∇×F = =
⎪
h h h ∂q ∂q ∂q
1 2 3 1 2 3
⎪
⎪ h F h F h F
1 1 2 2 3 3
⎪
⎪ + eˆ1 ⎡ ∂ (h F )− ∂ (h F )⎤+ eˆ2 ⎡ ∂ (h F )− ∂ (h F )⎤+
⎪ ⎢ 3 3 2 2 ⎥ ⎢ 1 1 3 3 ⎥
h h ∂q ∂q h h ∂q ∂q
⎪ 2 3 ⎣ 2 3 ⎦ 1 3 ⎣ 3 1 ⎦
⎪ eˆ ⎡ ∂ ∂ ⎤
⎪ + 3 ⎢ (h2F2)− (h1F1)⎥
⎪⎩ h1h2 ⎣∂q1 ∂q2 ⎦ (1.2.1)
where (h ,h ,h )are the metric coefficients. For common geometries they are defined as follow:
1 2 3
D. Ramaccia and A. Toscano    Pag. 6
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
⎧h =1, h =1, h =1 (rectangular coordinates)
1 2 3
⎪
⎨h1 =1, h2 = r, h3 =1 (cylindrical coordinates) (1.2.2)
⎪
h =1, h = r, h = rsinϑ (spherical coordinates)
⎩ 1 2 3
For simplicity, the proves are done using rectangular coordinates (h =1, h =1, h =1):
1 2 3
• Identity n° 1
eˆ eˆ eˆ
1 2 3
∂ ∂ ∂
∇×(∇φ)= =
∂q ∂q ∂q
1 2 3
⎛ ∂φ ⎞ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
∂q ∂q ∂q
⎝ 1⎠ ⎝ 2 ⎠ ⎝ 3 ⎠
⎡ ⎛ ∂ ∂φ ∂ ∂φ ⎞ ⎛ ∂ ∂φ ∂ ∂φ ⎞ ⎤
⎢eˆ1⎜ − ⎟−eˆ2⎜ − ⎟+⎥
∂q ∂q ∂q ∂q ∂q ∂q ∂q ∂q
⎢ ⎝ 2 3 3 2 ⎠ ⎝ 1 3 3 1⎠ ⎥
= =0
⎢ ⎥
⎛ ∂ ∂φ ∂ ∂φ ⎞
⎢+eˆ3⎜ − ⎟ ⎥
⎢ ∂q ∂q ∂q ∂q ⎥
⎣ ⎝ 1 2 2 1⎠ ⎦
∂ ∂ ∂ ∂
For the property of linearity of the derivate operator φ = φ, so each term in the
∂q ∂q ∂q ∂q
i j j i
parentheses vanishes and also the result.

• Identity n° 2
⎛ ∂ψ ∂ψ ∂ψ ⎞
∇⋅(φ∇ψ)=∇⋅⎜eˆ1φ +eˆ2φ +eˆ3φ ⎟ =
∂q ∂q ∂q
⎝ 1 2 3 ⎠
∂ ⎛ ∂ψ ⎞ ∂ ⎛ ∂ψ ⎞ ∂ ⎛ ∂ψ ⎞
= ⎜φ ⎟+ ⎜φ ⎟+ ⎜φ ⎟ =
∂q ∂q ∂q ∂q ∂q ∂q
1⎝ 1 ⎠ 2 ⎝ 2 ⎠ 3 ⎝ 3 ⎠

⎛ ∂φ ∂ψ ∂2ψ⎞ ⎛ ∂φ ∂ψ ∂2ψ⎞ ⎛ ∂φ ∂ψ ∂2ψ⎞
=⎜ +φ ⎟+⎜ +φ ⎟+⎜ +φ ⎟ =
⎜∂q ∂q ∂q ⎟ ⎜∂q ∂q ∂q ⎟ ⎜∂q ∂q ∂q ⎟
⎝ 1 1 1 ⎠ ⎝ 2 2 2 ⎠ ⎝ 3 3 3 ⎠
⎛∂2ψ ∂2ψ ∂2ψ⎞ ⎛ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ⎞
2
=φ⎜ + + ⎟+⎜ + + ⎟ = φ∇ ψ+∇φ⋅∇ψ
⎜⎝ ∂q1 ∂q2 ∂q3 ⎟⎠ ⎝∂q1 ∂q1 ∂q2 ∂q2 ∂q3 ∂q3 ⎠
• Identity n° 3
First of all we expand the sum inside parentheses:
⎧ ∂ψ ∂ψ ∂ψ
φ∇ψ =φeˆ +φeˆ +φeˆ
⎪ 1 2 3
⎪ ∂q1 ∂q2 ∂q3
⎨
∂φ ∂φ ∂φ
⎪ψ∇φ=ψeˆ +ψeˆ +ψeˆ
1 2 3
⎪ ∂q ∂q ∂q
⎩ 1 2 3

D. Ramaccia and A. Toscano    Pag. 7
S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1
so
⎛ ∂ψ ∂φ ⎞ ⎛ ∂ψ ∂φ ⎞ ⎛ ∂ψ ∂φ ⎞
(φ∇ψ−ψ∇φ)=eˆ1⎜φ −ψ ⎟+eˆ2⎜φ −ψ ⎟+eˆ3⎜φ −ψ ⎟
∂q ∂q ∂q ∂q ∂q ∂q
⎝ 1 1⎠ ⎝ 2 2 ⎠ ⎝ 3 3 ⎠
Now we can apply the dot product:
⎡ ∂ ⎛ ∂ψ ∂φ ⎞ ∂ ⎛ ∂ψ ∂φ ⎞ ∂ ⎛ ∂ψ ∂φ ⎞⎤
∇⋅(φ∇ψ−ψ∇φ)= ⎢ ⎜φ −ψ ⎟+ ⎜φ −ψ ⎟+ ⎜φ −ψ ⎟⎥ =
⎣⎢∂q1 ⎝ ∂q1 ∂q1 ⎠ ∂q2 ⎝ ∂q2 ∂q2 ⎠ ∂q3 ⎝ ∂q3 ∂q3 ⎠⎦⎥
⎛ ∂φ ∂ψ ∂2ψ ∂ψ ∂φ ∂2φ⎞ ⎛ ∂φ ∂ψ ∂2ψ ∂ψ ∂φ ∂2φ⎞
= +⎜ +φ − −ψ ⎟+⎜ +φ − −ψ ⎟+
⎜∂q ∂q ∂q ∂q ∂q ∂q ⎟ ⎜∂q ∂q ∂q ∂q ∂q ∂q ⎟
⎝ 1 1 1 1 1 1 ⎠ ⎝ 2 2 2 2 2 2 ⎠

⎛ ∂φ ∂ψ ∂2ψ ∂ψ ∂φ ∂2φ⎞
+⎜ +φ − −ψ ⎟ =
⎜⎝∂q3 ∂q3 ∂q3 ∂q3 ∂q3 ∂q3 ⎟⎠ ↑
cancel opposite terms in parentheses
⎛∂2ψ ∂2ψ ∂2ψ⎞ ⎛∂2φ ∂2φ ∂2φ⎞
2 2
=φ⎜ + + ⎟−ψ⎜ + + ⎟ = φ∇ ψ−ψ∇ φ
⎜ ∂q ∂q ∂q ⎟ ⎜ ∂q ∂q ∂q ⎟
⎝ 1 2 3 ⎠ ⎝ 1 2 3 ⎠

• Identity n°4
⎡ ∂ ∂ ∂ ⎤
∇⋅(φA)=∇⋅(φA1eˆ1 +φA2eˆ2 +φA3eˆ3)= ⎢ (φA1)+ (φA2)+ (φA3)⎥ =
∂q ∂q ∂q
⎣ 1 2 3 ⎦
⎡⎛ ∂A ∂φ ⎞ ⎛ ∂A ∂φ ⎞ ⎛ ∂A ∂φ ⎞⎤
=⎢⎜φ 1 +A1 ⎟+⎜φ 2 +A2 ⎟+⎜φ 3 +A3 ⎟⎥ =
⎢⎣⎝ ∂q1 ∂q1⎠ ⎝ ∂q2 ∂q2 ⎠ ⎝ ∂q3 ∂q3 ⎠⎥⎦
⎛ ∂φ ∂φ ∂φ ⎞ ⎛∂A ∂A ∂A ⎞
=⎜A1 +A2 +A3 ⎟+φ⎜ 1 + 2 + 3 ⎟ =(∇φ)⋅A+φ∇⋅A
∂q ∂q ∂q ∂q ∂q ∂q
⎝ 1 2 3 ⎠ ⎝ 1 2 3 ⎠
• Identity n° 5
eˆ eˆ eˆ
1 2 3
∂ ∂ ∂
∇×(φA)= =
∂q ∂q ∂q
1 2 3
φA φA φA
1 2 3
D. Ramaccia and A. Toscano    Pag. 8