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Electromagnetic Foundations
Gauss’ law ∇ i D = ρ
Conservation of magnetic flux ∇ i B = 0
Ampere’s law x = + t
H J^ D
Faraday’s law x = - t
E^ B
Traveling waves Taking the curl of Faraday
( ) =^ ( ) -^2 =^ -^ = - ( )
t t
∇ × ∇ × ∇ ∇ ⋅ ∇ ∇ × ⎛^ ∂^ ⎞ ∂ ∇ ×
E E E B B
In a charge-free (ρ=0) and current-free ( J =0) region.
2 2
2
t t = - t
μ
μ
∇ ∇ ∇ ∇ ∂^ ∇ × ∂ ∇ ×
∇ ∂ ∇ ×
E E E B H
E H
i
Substituting for ∇ × H using Ampere’s law ( J =0), we obtain the wave equation
2
2 2 2
t t
∇ ∂^ ⎛^ ∂ ⎞
∂ ⎜⎝^ ∂ ⎟⎠
E^ E
E E
This equation is called the wave equation. A wave equation for H can be found as well. Taking the curl of Ampere’s law and then substituting Faraday’s law [Again, in a charge- free (ρv =0) and current-free ( J =0) region.]
2 2
H^ H
Together, these equations predict that an electromagnetic wave can travel through space. These vector equations (one in E and one in H ) would each give three scalar equations (one for each component of the vectors) for a total of six scalar equations. Using Cartesian coordinates, one scalar equation for Ex would be
E 2 x (^) + E 2 x + E 2 x (^) = E 2 x x y z t
2 2 2 2
Waves traveling in the z direction would have the form
2 2 2 2 2 g = g t z^ g^ =^1 g v z v t
⎛ ± ⎞ ⇒^ ∂^ ∂
Two waves are possible: one traveling in the + a z direction with a speed of v = 1 με ,
the other traveling in the – a z direction with the same speed.
Plane Waves Assuming the electric vector to only have a component in the x-direction, Ex, which depends only on z and t.
E 2 x = E 2 x z t
2 2
Solutions are functions where the 2 nd^ derivative with respect to z are equal to their 2nd derivative with respect to t (aside from a constant multiplying factor). Sinusoids are an obvious choice (this particular choice travels in the + a z direction).
E = a x E cos ( ωt - β z + ϕ+)
Substituting this possible solution, 2 + 2 + x x 2 2
E cos ( t - z + ) = E cos ( t - z + ) = =
a a
Just as for sinusoidal steady state analysis in electrical circuits, harmonic analysis is provides a convenient notation.
E = a x E cos ( ωt - β z + ϕ ) = Re ⎡⎣^ a x Ee ϕ^ e β^ e ω⎤⎦^ = Re ⎡⎣ E^ eω⎤⎦
That is, for fields that vary harmonically with time, each Fourier component can be described as ( )
{ } {^ }
j t+ (^) j j t
j t
( )cos( t+ ) = Re ( )e = Re ( )e e
= Re e
ω ϕ ϕ ω
ω
E r E r E r
E E E E
Just as in phasor analysis, the derivative with respect to time corresponds to a jω multiplier in the frequency domain.
t =^ tRe^ {^ e^ j^ t^ }^ = Re j{^ ej^ t}
j t
E E E
E E
Example 2 An electric field, of amplitude 37.7 V/m points in the a z direction and travels along φ = 45° in the x-y plane. The wavelength is 700 nm and the medium is air.
i) Give the frequency domain description of E and H. Take the phase of E ^ to be zero degrees. ii) What is the frequency of the waves? Give their time-domain descriptions.
Solution
i) The wave number, or phase constant, β, is found from the wavelength.
= 2 = 2 9(10 ) m^6 - 700 nm
β π^ π
-j9(10^6 )(xcos45° + ysin45°) E = a z37.7 e V/m
The magnitude of the magnetic vector can be determined from the field analog of Ohm’s law and the direction of the vector can be determined from the fact that E × H is in the direction of travel (as shown in the diagram above). The magnetic vector travels with the electric vector (after all, we have a traveling EM wave of which the two vectors are part).
( )
( )
6
6
y x -j9(10^ )(xcos45° + ysin45°)
y x -j9(10^ )(xcos45° + ysin45°)
= sin 45° - cos 45° 37.7 V/me 377 = 0.1 sin 45° - cos 45° e A/m
H a a
H a a
ii) The speed of propagation can be determined from the physical properties of the medium, in this case air. Knowing the speed of propagation and the wavelength permits the frequency to be determined.
p o 0 8 p (^) -12 - (^814)
v = 1 1 , for air
v = 1 3(10 ) m/s 8.854(10 ) F/m 4 (10 ) H/m
f = 3(10 ) m/s = 4.29(10 ) Hz 700(10 ) m
Reflection and Transmission How are EM waves reflected and transmitted when a plane dielectric boundary is encountered? Consider the situation below:
Define a transmission coefficient, T, and reflection coefficient, Γ.
t r i i
T = E^ = E
E E
Let E i (^) = a x E ei -j^ β^1 z
From EM theory, the boundary conditions that must be met are that the tangential components of the electric field and the magnetic field (assuming no surface currents) are equal on either side of the boundary.
Boundary conditions: E1tan (^) z=0 = E2tan (^) z=0 H1tan (^) z=0 = H2tanz=
The first equation is obtained from the boundary conditions on the electric vectors.
(E + E )i r a (^) x = E (^) t a x ⇒ E +i ΓE (^) i = ΤEi
The second is obtained from the boundary conditions on the magnetic vectors.
i r y t y i^ i^ i 1 1 2
(H - H ) = H E^ - E^ = E
a a ⇒ Γ^ Τ
So the result is two equations in Γ and Τ.
1 1 2
1 + = and 1 - =
Γ Τ Γ^ Τ
Solving for Γ and Τ.
2 1 2 2 1 2 1
= -^ =^2
Reflection and Transmission (σ 2 ≠0)
The conductivity in region 2 leads to complex values for the reflection and transmission coefficients as well as for the intrinsic impedance in region 2.
2 2 j^2 1 j^2 j 2 2 2 1 2 1
= = = e = -^ = e = 2 = e
η μ μ η ϕ^ η η^ η ϕ η ϕ
2 Γ^ Γ^ Γ^ Τ^ Τ Τ
2 2
^ ^
This results in a phase shift between the incident wave and waves that are reflected and transmitted. In addition, the fact that the intrinsic impedance in region 2 is complex leads to their being phase shifts between the electric vector and the magnetic vector in region 2.
( ) ( ) ( ) ( )
1 2
1 1
1
1
z + z - i x i r x i t x i i i z + i z +^ - i y r y t y 1 1 2
-j z j -j
-j z j -j
= E e = E e = E e
= E^ e = - E^ e = Ee^ η
β ϕ β ϕ
β ϕ β^ ϕ^ ϕ
β
β
Γ Τ
Γ Τ
E a E a E a
H a H a H a
The incident power is unchanged from the previous case,
2 i z i 1
= E
S a.
The reflected power density is
( ) (^ )^ (^ )
( ) ( )
1 1
1 1
z + (^) i z + r r r x i y 1 z + z +^2 r x i y i^ z i 1 1
j j
j -j
= 1 Re = 1 Re E e - Ee 2 2
= 1 Re E e - E^ e = - E 2 2
β ϕ β ϕ
β ϕ β ϕ
Γ Γ
Γ Γ
∗ × ∗^ ⎡⎢^ Γ × ⎜⎛^ Γ ⎞⎟⎤⎥ ⎢⎣ (^) ⎝ ⎠⎥⎦ ⎡ (^) Γ × Γ ⎤ Γ ⎢ ⎥ ⎣ ⎦
S E H a a
S a a a
The transmitted power density is
( ) (^ )^ (^ )
( ) ( )
1 1
1 1
z - (^) i z +^ - t t t x i y 2 z - (^) i z +^ -^2 2 i t x i y z 2 1
-j -j
-j j
= 1 Re = 1 Re E e Ee 2 2
= 1 Re E e E^ e = E cos 2 2
η
η
β ϕ β^ ϕ^ ϕ
β ϕ β^ ϕ^ ϕ η
Τ Τ
Τ Τ
∗ × ∗^ ⎡⎢^ Τ × ⎜⎛^ Τ ⎞⎟⎤⎥ ⎢⎣ (^) ⎝ ⎠⎥⎦ ⎡ (^) Τ ⎤ Τ ⎢ Τ^ × ⎥ ⎣ ⎦
S E H a a
S a a a
Standing Waves When reflection is present, the incident and reflected waves will interfere to produce standing waves in region 1.
( ) (^) ( ) (^) ( ( ) )
1 1
1 1
1 1 1
1
1
z + z + 1 i r x i x i x x i j z +^ z + 1 x x i
j 1 x x i
-j z j -j z j -j z
-j z j^ -j
-j z
= + = E e + E e = 1- e + E e + e
= 1- e + E e e + e
= 1- e + 2 E e
β ϕ β ϕ
ϕ (^) β ϕ β ϕ
β β β
β
β
Γ Γ
Γ ⎛⎜^ Γ^ ⎞⎟^ ⎛⎜^ Γ⎞⎟ ⎝ ⎠ ⎝ ⎠
E E E a a a a
E a a
E a a
2 2 2
(^) cos 1 z + , where the 2nd term in the standing wave
In the time domain, we have
1 = Re^ (^1 e^ j^ ωt)^ =^ x (^ 1-^ )cos(^ t -^1 z) +^ x 2 E cosi 1 z +^ cos^ t + Γ ω β Γ ⎛^ β ϕΓ^ ⎞^ ⎛^ ω ϕΓ⎞ E E^ a a (^) ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
The standing wave ratio (SWR) is commonly used to characterize reflections from a surface. The SWR is defined as the ratio between the maximum amplitude electric field in region 1 to the minimum amplitude electric field.
max min
SWR = E^ = 1 -^ +2^ = 1 +^ = SWR - 1
E 1 - 1 - SWR + 1
Where, Γ is the magnitude of Γ^.
Example 3 i) Find the transmission and reflection coefficients for the situation shown below if medium 1 is air and medium 2 is PCB material (FR-4, σ=0, εr=4.4, μr=1). ii) Find the portion of the incident energy that is transported into the FR-4 material. iii) Find the standing wave ratio in medium 1.
Transmission Line Fundamentals
Transmission lines are a special case of solutions to Maxwell’s equations when the properties of the physical system do not change in one direction (typically taken to be the z-direction).
Employing frequency domain analysis is convenient when working with transmission lines. In frequency domain analysis, one assumes that all fields vary sinusoidally in time. Suppose a source a xEs cos ωt produces, in response, an electric vector which travels in the az direction.
E = a x E cos(o ωt - βz)
If the system is linear, the complex source a x Es (cos ωt + j sin ωt) would produce a complex response.
E = a x E o [ cos( ωt - βz) + j sin(ω t - βz)]
Using Euler’s formula ( e j^ α = cos α + j sinα ), this can be written as
j ( t - z ) -j z j t j t = (^) x E eo = (^) x E eo e = e E a^ ω^ β a β^ ω E ω
Example 4 An example will illustrate the form that Maxwell’s equations have in the frequency
domain. Using Faraday’s law with E = E ^ e j^ ωt^ and B = B ^ ej^ ωt.
j t (^ ej^ t) x = - x e = - t t
ω ∇ ∂ ⇒ ∇ ω^ ∂ ∂ ∂
B^ B
E E
Note that E ^ and B ^ have no time dependence and that ejωt^ has no spatial dependence.
This allows the curl to pass through ejωt^ and B ^ to pass through the time derivative.
( ) j t ej^ t jωt e x = - = -j e x = -j t
ω^ ω
E ^ B ^ B ^ E ^ B
In the frequency-domain version of Faraday shown above, E ^ and B ^ are the frequency domain field vectors. They are not the time domain vectors. The time domain vectors can be recovered by putting the time dependence back and taking the real part.
Taking the electric vector used in this example to demonstrate,
( ) ( ) (^) ( (^ )) { [^ ]}
j t -j z j t^ j^ t-^ z x o x o x o x o
= Re e = Re E e e = Re E e = Re E cos( t - z) + j sin( t - z) = E cos( t - z)
ω β ω^ ω^ β
E E a a E a a
This result can be generalized to express Maxwell’s equations in the frequency domain (current and charge-free)
= -j = j = 0 = 0
∇ ×
∇ ×
E B
H D
D
B
Combining the equations (as was done above in the time-domain), one obtains the Helmholtz wave equation
( 2 +^ ω με^2 ) = 0
E
H
Plane waves are solutions.
j z j z
(x,y,z, ) (^) = (x,y, )e (x,y,z, ) (x,y, )e
β β
± ±
E E
H H
It is important to note that the Laplacian operator can always be split into two pieces— one piece which depends only on a variable which varies in the direction of travel (in this case, z) and the other piece which depends only on variables which vary in directions transverse to that of travel (in this case, x and y).
( ) 2 2 2 2 2 2
t +^ ω με^ -^ β^ = 0^ where^ t =^ -^ z 2
E
H
Special types of solutions exist in transmission lines (coaxial lines, microstrip, twin-lead) for which the a z component of E and H are zero. These solutions are transverse electromagnetic waves.
The equation above can be split into two pieces, both of which must be satisfied.
(^2) t = 0 and (^) ( ω με (^2) - β (^2) ) = 0
E E
H H
Gauss’ law, in integral form, reads
∫∫^ ε d^ = Qinside gaussiansurface
E i s
Applying Gauss’ law for the Gaussian surface shown, for a < ρ < b,
o o
o o
z + L 2 z + L 2
z = z = 0 z = z = 0
E d dz = a d dz
π π ρ ρ φ φ
∫ ∫ ε^ a^ iρ^ φ a^ ∫ ∫ ρ^ s φ
s 2 L E = 2 aL = a ρ
⇒ E a s
To find the magnetic field , first note that, by symmetry, the field must point in the a φ direction (still using cylindrical) and its magnitude can only depend on ρ. This allows the integral form of Ampere’s law to be readily used.
Ampere’s law, in integral form, reads
inside Amperianpath
v∫ H^ id = I l
Applying Ampere’s law for the Amperian path shown, for a < ρ < b, 2
= 0
H d = i = i 2
π ρ ρ φ φ
∫ a^ i a^ ⇒ H^ a
The transverse fields, found using statics, can are then used to find the traveling wave. In the frequency domain these fields are
j z t
j z t
= a^ = a e
= i^ = i e 2 2
β ρ ρ
β φ φ
±
±
E a s^ E a s
H a H a
With transmission lines, one is often not concerned with the details of the fields but rather is often more interested in the terminal current voltage equations.
In his work with transmission lines, Oliver Heaviside developed the telegrapher equations which involve traveling voltage and current waves rather than traveling electric and magnetic fields.
Voltage & Current Waves: Telegrapher’s equations
Using Faraday’s law, path (^) boundedarea by path
d = - d d ∫ (^) dt ∫∫ v E^ i^ l^ B^ i s , where d l^ and d s^ are related via
the right hand rule. Using frequency-domain field vectors, path (^) boundedarea by path
v∫ E ^^ i^ d^ l^ = -j^ ω ∫∫ B ^ id s.
Carrying out the path integral, Faraday’s law shows that
area area
d V(z+ z) - V(z) = -j d = -j I I
∫∫ ∫∫
B s B s
i (^) i
Where the quantity, area
d
I
∫∫ B ^ i s , being a ratio of flux to current is clearly the inductance
of the loop.
V(z+ z) - V(z) = -j^ Δ ^ ωI L
By dividing the the interval Δz, and taking the limit as Δz → 0, one obtains,
z 0 z 0 lim V(z+^ z) - V(z)^ = -j I lim L = -j I z z
L
Lossless Transmission Lines circuit model governing equations
dV = -j I dz
L
d I = -j V dz
C
Taking a derivative with respect to z of the first Telegrapher equation and substituting the second equation, one obtains:
2 2 2 2 (^22) 2
d V (^) = -j d I d V = -j -j V dz dz dz d V + LC V = 0 dz
L ⇒ L C
Similarly,
2 2 2 2 (^22) 2
d I (^) = -j dV d I = -j -j I dz dz dz d I (^) + LC I = 0 dz
C ⇒ C L
Solutions are traveling waves
V = V e^ ^ ^ +^ -j βz^ + V e^ -^ j β^ z^ where V^ +^ = V e+^ j^ ϕ v +^^ and V- = V e- j^ ϕ v − I = I e + -j βz^ + I e- j^ β^ z^ where I + (^) = I e+^ j^ ϕ i +^ and I -^ = I e- j^ ϕ i −
where β =ω LC is the phase constant.
Consider a voltage wave traveling in the +z direction and determine the relation between voltage and current. The ratio between the voltage and current waves is the transmission line’s characteristic impedance , Zc.
-j z
-j z + -j z + -j z
-j z c
using V = V e in the Telegrapher eq. dV = -j I dz
-j V e = -j I I = -j^ V e = -j V e -j -j
I = V e Z =
β
β β β
β
L,
L LC
L L
L
L C C
Lossy Transmission Lines Transmission lines serve to guide electromagnetic waves. As the waves travel along the transmission lines, there are three main sources of loss: 1) Joule heating or copper losses, 2) dielectric losses, and 3) radiation.
Joule Heating – as the EM wave travels along the conductors of the transmission line, the electromagnetic wave enters the conductors—typically a metal—where it decays exponentially. Even though the wave decays once it enters the TL conductors, the EM wave does enter them to a degree. Inside the conductor, the EM field accelerates free electrons, which then collide with metal atoms to make them vibrate. The result is that the EM wave heats the metal. This process is called Joule heating and is usually the dominant loss mechanism in transmission lines.
Dielectric Losses – a dielectric is formed by fixed positive nuclei to which are bound negatively charged electrons. In an ideal dielectric, an applied electric field results in charge separation as shown below. Notice that, for an ideal dielectric, the force and
motion are in the same direction for ¼ of a period (or equivalently the voltage and current have the same sign for ¼ of a period) and then are in opposite directions for the next ¼ period.
The result is that work is done on the dielectric (energy is stored) for ¼ of a period which is then given back in the next ¼ period. This is the physical basis of capacitance.
For a non-ideal dielectric, the charges have non-zero mass and their motion lags that shown. The result is that the current is no longer 90° out of phase with the voltage which results in energy loss from the EM wave.
Low Loss Transmission Lines
When transmission lines are lossy, nearly always they are low-loss ( R ω L and
G ω C ). The general case is seldom and will not be treated here. For the more
common low loss case, the expression for the propagation constant, γ, can be
simplified.
= = ( + j )( + j )= j 1 - j 1 - j
j 1 - j 1 - j for and 2 2
≅ ⎛^ ⎞⎛^ ⎞
ZY R L G C LC R^ G
L C
LC R^ G R L G C
L C
^
This can be further simplified.
γ j ω 1 - 4 ω 2 - j 2 ω - j 2 ω
≅ ⎛^ ⎞
LC GR^ R^ G
LC L C
The second-order term can be neglected.
j 1 - j - j = + + j 2 2 2 2
= + j + + j 2 2
≅ ⎛^ ⎞
LC R^ G^ R^ C^ G^ L LC
L C L C
R C G L LC
L C
Most often copper loss dominates the attenuation constant. In this case
= + j + j (low-loss where is determined by copper loss) 2
γ α β ≅ R^ C^ ω LC α
L
What is the nature of this propagation constant? First, the phase constant is the same as that found for the lossless case. This gives the important result that the phase velocity of the wave is the same as for the lossless case.
p
v = ω =^1
β LC
Second, α is usually small, which might lead one to question whether it is really necessary or whether one might assume that α = 0. Since the attenuation factor is the product of αz, the answer to this question depends on 1) how long the transmission line is and 2) what magnitude changes are considered important in a particular application.
Example 6 Consider a transmission line with
L = 5 μH/m, C = 0.02 μF/m, R = 0.01 Ω/m, G = 0 and f = 1 GHz
i) How long would the transmission line need to be for a voltage wave to suffer 5% attenuation?
ii) What is the speed of propagation?
iii) What is the wavelength?
iv) What is the characteristic impedance?
Solutions Using these values, the propagation constant can be determined.
= + j + j = ( 0.000316 + j1986.9) m-
γ α β ≅ R^ C^ ω LC
L
i) Let the length of line = d.
e-0.000316d = 0.95 ⇒ d = 162.3 m
In this case, even though α is quite small, if d = 162.3 m, there will be 5% attenuation. α, even if small, cannot be ignored for long lines.
ii) v p= ω^ = 2 πf = = 3.16(10 ) m/s^6
LC
iii) p 2 v = = = 3.16 mm f
iv) Strictly speaking the characteristic equation for a lossy line is complex.
-6 - c Z = = + j = 15.8 - j2.5(10 ) = 15.8 -9.1(10 )° 15.
Z R L ⎡⎣ ⎤ Ω⎦ ∠ Ω ≅ Ω
Y G C
This example makes it clear that, even though Z^ c is actually complex, it is so nearly real
that it hardly matters in nearly all practical cases. Therefore, to determine the characteristic impedance, one would use
Zc = = 15.8 Ω
L
C
