• We let COMSOL solve the problem and we check that we get the correct value of kz.
• We use Results>Derived values>Line integration and integrate |Js|2by choos-ing the square of the predefined surface current density emw.normJs2. We then calculate P (z) by Results>Derived values>Surface integration>emw.Poavz (power flow, time average, z-component). From these two integrals we obtain αp from (5.54).
5.9.2 Losses in waveguides with FEM: method 2
It is somewhat easier to use the impedance boundary condition to determine the attenu-ation. We then do the following steps in Comsol:
• We choose the, 2D>Electromagnetic waves>Mode analyzis.
• We draw the cross section of the waveguide.
• Define the materials. One should be air (or vacuum) and is for the interior of the waveguide. The other one is the material in the walls. In Geometric entity level: we mark Boundary for the metal material and add the boundaries of the waveguide.
• Right click on Electromagnetic waves and let all boundaries have Impedance Boundary Condition.
• In Study>Mode Analysis we define how many modes that are to be analyzed and the frequency that we are interested in. We also add the effective mode index where COMSOL starts to look for eigenfrequencies.
• We let COMSOL solve the eigenvalue problem. It then shows the electric field in the cross section of the waveguide for the different modes. It also gives the complex effective mode index n. From this we get α as α = ωIm{n}/c.
Example 5.9
Using impedance boundary conditions we get that a rectangular copper waveguide with dimension a = 0.3 m and b = 0.15 m has n = 0.7045 + i1.74· 10−5 for the TE10-mode at 704 MHz. This gives α = 0.0002565 Np/m. In ESS the distance between the klystrons and the cavities is approximately 20 meters. It means that the attenuation of the power is e−2α20=0.9898. Approximately one percent of the power is lost in the waveguides. The average power fed to the cavities is 5 MW which means that the loss is on the order of 50 kW. By increasing the size of the waveguide to 0.35× 0.175 m2 the effective mode index is changed to n = 0.7937 + i1.208· 10−5. Then e−2α20=0.9925 and the average loss is 38 kW.
Sources in waveguides 107
S z
V Ω 1 J
Ω 2
z1
z2
Figure 5.15: Waveguide with a source region V , confined by the surfaces Ω1, Ω2 and the lateral surface S.
conducting walls. The boundary to the cross section is assumed to be simply connected and hence there are no TEM-modes. It is easy to generalize to the case with non-perfectly conducting walls by using the theory in the previous section, and also to waveguides with TEM-modes. We assume a finite region in the waveguide confined between the cross-sections Ω1 and Ω2, see figure 5.15, where there is a known time-harmonic current density
J (r, t) = Re{J(r, ω)e−iωt}
Since the mode solutions that we developed and analyzed in section 5.4, constitute a complete set of functions in a source free region, the fields generated by the source is a superposition of propagating and non-propagating modes. The suitable vector valued expansion functions are given in (5.27), with the corresponding definitions in equations (5.24), (5.21) and (5.22).
As in the previous sections we use two indices for the modes, an index ν = TM,TE, and an n-index for the modes for each ν. As before the scalar functions vnand wn are the solutions to the eigenvalue problems in equations (5.4) and (5.5), normalized according to (5.26). The normalized vector functions satisfy the orthogonality relations in (5.37).
We let V be the volme between the surfaces Ω1 and Ω2 at z = z1 and z = z2, where z1 < z2, see figure 5.15. In the region z ≥ z2 the fields are propagating in the positive z-direction, since the sources of the fields are in the region z < z2. When z ≥ z2 the expansion of the fields reads
E+(r, ω) = X
ν=TM,TEn
a+nνE+nν(r, ω)
H+(r, ω) = X
ν=TM,TEn
a+nνH+nν(r, ω) z≥ z2 (5.55)
Analogously the fields are propagating in the negative z-direction when z ≤ z1 and then
E−(r, ω) = X
ν=TM,TEn
a−nνE−nν(r, ω)
H−(r, ω) = X
ν=TM,TEn
a−nνH−nν(r, ω) z≤ z1 (5.56)
To determine the coefficients a±nν we use the Lorentz’ reciprocity theorem, which we first derive. Let E and H be the fields generated by the current density J , i.e., E and
H satisfy (
∇ × E = iωµ0µH
∇ × H = J − iω0E
where the electric field E satisfies the boundary condition ˆn× E = 0 on the perfectly conducting surface of the waveguide. By definition the vector fields E±nν, H±nν for the modes satisfy the homogenous Maxwell equations (no current densities), i.e.,
(∇ × E±nν = iωµ0µH±nν
∇ × H±nν =−iω0E±nν and ˆn× E±nν = 0 on the perfectly conducting surfaces.
The following identity is obtained from the Maxwell field equations and vector formulas for the nabla-operator
∇ · E±nν× H − E × H±nν
= H · ∇ × E±nν
− E±nν· (∇ × H)
− H±nν· (∇ × E) + E · ∇ × H±nν
=−J · E±nν
We integrate this relation over the volume V and utilize the divergence theorem ZZ
S0
(E±nν× H − E × H±nν)· ˆn dS =− ZZZ
V
J· E±nνdv (5.57)
The surface S0 consists of the two cross-sections Ω1 and Ω2, and the part of the envi-ronmental surface S between the surfaces z = z1 and z = z2, see figure 5.15, and ˆn is the outward directed normal to S0. Since the environmental surface is the perfectly con-ducting wall, where ˆn× E±nν = ˆn× E = 0, the corresponding surface integral does not contribute. The integrals over Ω1 and Ω2 give contributions that can be determined from the expansions in equations (5.55) and (5.56) and the orthogonality relation (5.37).
For the field E+nν on the surface Ω1 we use the expansions in equations (5.56), (5.21) and (5.22). The orthogonality relation (5.39) gives
ZZ
Ω1
(E+nν× H − E × H+nν)· ˆn dS
=− X
n0 ν0=TM,TE
a−n0ν0
ZZ
Ω1
E+nν × H−n0ν0 − E−n0ν0× H+nν
· ˆz dS
= X
n0 ν0=TM,TE
a−n0ν0ei(kz n−kz n0)z1 ZZ
Ω1
(ET nν× HT n0ν0+ ET n0ν0 × HT nν)· ˆz dS
=−4a−nνUnν
Note that on the surface Ω1, the outward directed normal is −ˆz. On the surface Ω2 we
Sources in waveguides 109
use the expansion (5.55) and get ZZ
Ω2
(E+nν× H − E × H+nν)· ˆn dS
= X
n0 ν0=TM,TE
a+n0ν0
ZZ
Ω2
E+nν× H+n0ν0 − E+n0ν0 × H+nν
· ˆz dS
= X
n0 ν0=TM,TE
a+n0ν0ei(kz n+kz n0)z2 ZZ
Ω2
(ET nν× HT n0ν0 − ET n0ν0× HT nν)· ˆz dS = 0
By using (5.57) we obtain the expression a−nν = 1
4Unν
ZZZ
V
J · E+nνdv
where Unν is given by (5.40). In the same manner we get the coefficients for E−nν ZZ
Ω1
(E−nν× H − E × H−nν)· ˆn dS
=− X
n0 ν0=TM,TE
a−n0ν0
ZZ
Ω1
E−nν× H−n0ν0− E−n0ν0× H−nν
· ˆz dS
= X
n0 ν0=TM,TE
a−n0ν0e−i(kz n+kz n0)z1 ZZ
Ω1
(ET nν× HT n0ν0 − ET n0ν0× HT nν)· ˆz dS = 0
and ZZ
Ω2
(E−nν× H − E × H−nν)· ˆn dS
= X
n0 ν0=TM,TE
a+n0ν0
ZZ
Ω2
E−nν× H+n0ν0 − E+n0ν0 × H−nν
· ˆz dS
= X
n0 ν0=TM,TE
a+n0ν0e−i(kz n−kz n0)z2 ZZ
Ω2
(ET nν× HT n0ν0 + ET n0ν0× HT nν)· ˆz dS
=−4a+nνUnν This leads to
a+nν = 1 4Unν
ZZZ
V
J · E−nνdv The final result is:
a±nν = 1 4Unν
ZZZ
V
J · E∓nνdv (5.58)
Example 5.10
Consider a current density J that is confined to a thin wire along the curve C with tangential unit vector ˆτ . The volume integrals in (5.58) are reduced to line integrals.
a±nν = 1 4Unν
Z
C
I(ρ)E∓nν· dr (5.59)
where dr = ˆτ dl.
Example 5.11
If the curve C is in a transverse plane z = z0 we see from equations (5.59), (5.21), (5.22) and (5.23) that a+nν = a−nνexp(−2ikzz0).
Example 5.12
A small, planar, closed wire with r-independent time-harmonic current I has its center in r0. The wire is represented by a magnetic elementary dipole m = IA ˆn, where A is the area of the planar surface spanned by the wire and ˆn is the unit normal to A directed according to the right hand rule. Since the loop is small H±nν(r)' H±nν(r0) in A. From (5.59), Stoke’s theorem, and the induction law we get
a±nν = I 4Unν
I
C
E∓nν· dr = I 4Unν
ZZ
Ω
(∇ × E∓nν)· ˆn dS
' I
4Unνiωµ0µH∓nν(r0)· ZZ
Ω
ˆ
n dS = iωµ0µ
4Unν H∓nν(r0)· m
Example 5.13
Consider an electric elementary dipole, p = iI dr/ω where the vector dr is much smaller than the wavelength and the dimensions of the waveguide. When this dipole is placed in a waveguide we see from (5.59) that
a±nν =− iω 4Unν
p· E∓nν
Assume that the dipole is placed in the point r0. The field from the dipole is given by
E(r) =
−iω 4
X
ν=TM,TEn
1
Unν(p· E−nν(r0))E+nν(r), z > z0
−iω 4
X
ν=TM,TEn
1
Unν(p· E+nν(r0))E−nν(r), z < z0 This is rewritten as
E(r) = iω
4 p· G(r, r0) The functionG is given by
G(r, r0) =
− P
ν=TM,TEn
1
UnνE−nν(r0)E+nν(r), z > z0
− P
ν=TM,TEn
1 Unν
E+nν(r0)E−nν(r), z < z0
Mode matching method 111
V H
Ω Ω
A (z)+ B (z)+
B (z)
-
-A (z) z
z
0
Figure 5.16: The geometry for the mode matching method.
and is called the Green dyadic, which is a vectorial analogue of the Green function for scalar fields.