5.3 Rectangular waveguide
5.3.3 Irises and waveguide discontinuities
Modes in a perfect, uniform waveguide are independent: each of them can exist alone in the waveguide since it satisfies Maxwell’s equations and all boundary conditions. If an obstacle is introduced, or a change in the cross section, new boundary conditions arise, which were not taken into account in the definition of the modes. In general then, all waveguide modes are excited
−0.040 −0.02 0 0.02 0.04 0.06 0.5
1 1.5 2
z (m) Z∞ d|I(z)|
A B
Figure 5.23. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10modal current vs. z at the frequency f = 5 GHz. The current is multiplied by the modal impedance for viewing convenience
−0.020 −0.01 0 0.01 0.02 0.03 0.04
0.5 1 1.5 2 2.5
z (m)
|V(z)|
A B
Figure 5.24. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal voltage vs. z at the frequency f = 9 GHz
by the discontinuity with suitable coefficients, such that the total electric and magnetic field, consisting of incident and scattered waves for all the modes, satisfy the new boundary conditions.
As a example, consider an iris in a rectangular waveguide, as shown in Fig. 5.28. Irises are metal windows, perpendicular to the guide axis. The thickness of irises is often small and can be neglected
−0.020 −0.01 0 0.01 0.02 0.03 0.04 0.5
1 1.5 2 2.5
z (m) Z∞ d|I(z)|
A B
Figure 5.25. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10modal current vs. z at the frequency f = 9 GHz. The current is multiplied by the modal impedance for viewing convenience
−0.020 −0.01 0 0.01 0.02 0.03 0.04
0.5 1 1.5 2 2.5 3
z (m)
|V(z)|
A B
Figure 5.26. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal voltage vs. z at the frequency for which S11= 0.
as a first approximation. Suppose that the fundamental mode T E10 is incident from the left. The boundary conditions to be satisfied on the iris are that the tangential electric field is zero on the metal parts of the iris and different from zero on the aperture. In this case the tangential field is the transverse field. If we recall the electric field distribution of T E10, shown in Fig. 5.14, we
−0.020 −0.01 0 0.01 0.02 0.03 0.04 0.5
1 1.5 2 2.5
z (m) Z∞ d|I(z)|
A B
Figure 5.27. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal current vs. z at the frequency for which S11 = 0. The current is multiplied by the modal impedance for viewing convenience
x y
A z
A
0, 0
kz Z∞ jB
TE
10ŵŽĚĂůůŝŶĞFigure 5.28. Example of an iris in a rectangular waveguide. Upper left: longitu-dinal view. Upper right: cross section view. Lower left: equivalent circuit for the fundamental mode (B > 0, capacitive iris).
realize that it is impossible to satisfy these boundary conditions with just the incident reflected and transmitted waves of this mode. Indeed, if the total Ey field must be zero on the the metal, certainly Eyr= −Eyi, that is the reflection coefficient must be Γ = −1. But the modal fields do not depend on y (the second subscript is n = 0), hence the total Ey field is zero also on the aperture and the transmitted field should be zero. This would imply that the iris behaves as a short circuit plate and this is contrary to experimental evidence: indeed, although the reflection coefficient can be high when the aperture is small, power is always transmitted through the hole. In conclusion, the behavior is more complicated and an infinite number of higher order modes are excited in order to satisfy the boundary conditions. In this way the reflected wave of T E10 need not be opposite to the incident wave and the reflection coefficient is smaller than one.
If the iris thickness is negligible, its fundamental mode equivalent circuit is an admittance con-nected in parallel on the T E10modal line. Indeed, the transverse electric field must be continuous on the complete cross section: equal to zero on the metal and different from zero (but continuous) on the aperture. Hence the modal voltage is continuous, V (A−) = V (A+), and the equivalent cir-cuit is a load in parallel. Note that this load takes into account only the effect the iris has on higher order modes. Indeed, this load is connected on the T E10 modal line and the iris effects on the propagation of this mode are taken into account directly by transmission line theory. Generally the waveguide is operated in single mode conditions, so that all higher order modes are below cut-off.
As well known, they do not contribute active power propagation but store electromagnetic energy.
The admittance that represents the iris on the T E10modal line must describe this behavior, hence it is a pure susceptance jB.
As an example of this procedure, we compute the scattering matrix of the iris for the fundamental mode, assuming that the susceptance B is known. Assume that the reference planes are in A− and in A+ and the reference impedances at both ports coincide with the modal impedance. Let b = BZ∞0 be the normalized susceptance (not to be confused with the small side of the waveg-uide!!). To compute S11 and S21, close port 2 on the reference impedance and compute reflection and transmission coefficients. They are
YA− = jB + Y∞0 yA− = jb + 1 S11= ΓA− = −yA−− 1
yA−+ 1 = −jb 2 + jb S21= 1 + ΓA− = 1 + −jb
2 + jb = 2 2 + jb
Then S12= S21 because of reciprocity and S22= S11because of the symmetry of the structure.
The problem is the determination of the susceptance, which is a difficult one and requires the solu-tion of an integral equasolu-tion. Fig. 5.29 shows plots of the normalized susceptance of a symmetrical iris. It can be shown that when the edges are perpendicular to the electric field lines, B > 0 and the iris is said to be capacitive. Two sets of curves are shown for better reading accuracy: the right set uses the vertical axis on the left and is to be used for large apertures (d/b ' 1), the left one uses the axis on the right that measures essentially the inverse of the normalized susceptance.
Note that when the aperture is small, it resembles a short circuit, so the susceptance is large. If the edges of the iris are parallel to the electric field lines, the iris is said to be inductive, because its susceptance is negative, B < 0. Fig. 5.30 shows the normalized reactance of a symmetrical inductive iris. Again, if the aperture is small, the reactance is small, i.e. the susceptance is large as that of a short circuit.
It is interesting to compare the susceptance values of a capacitive and an inductive iris with the same aperture: the former is much smaller than the latter. As a numerical example, consider a WR90 (a = 2.286 cm, b = 1.016 cm) waveguide at f = 12 GHz. An inductive iris with an aperture of d = 0.3 cm has a normalized susceptance B/Y∞' −32, whereas a capacitive iris with the same aperture has a susceptance B/Y∞' 1.16.
Alternatively, consider a capacitive iris with a very small aperture, d = 0.5 mm, which has a suscep-tance B/Y∞= 4.3. An inductive iris with B/Y∞= −4.3 has a much wider aperture, d = 0.68 cm.
Such a susceptance produces
|S11| = 0.9067 |S21| = 0.4217
ď Ě
Figure 5.29. Normalized susceptance of a symmetrical capacitive iris. λg is the guided wavelength and Y0is the modal admittance.
A possible explanation of this phenomenon is related to the induced currents on the iris plates, which are parallel to the electric field since they are perpendicular to H (see (1.19)), which in turn is perpendicular to E. These induced currents radiate the scattered field but are disturbed
Ě Ă
Figure 5.30. Normalized reactance of a symmetrical inductive iris. λg is the guided wave-length and Z0is the modal impedance. The curves are parametrized by a/λ where λ = c/f is the free space wavelength.
(interrupted) by the aperture in the case of capacitive iris so that their effect is small. In the case of inductive iris they are not interrupted and are then very efficient.
Irises are reactive circuit elements and can be used to build matching devices: in practice, they substitute stubs in waveguide technology. Moreover they can be used to build filters. A simple example consists of two irises separated by a distance of about λg/2. They form a cavity, coupled to the waveguide by means of the apertures: For the center frequency the structure is transparent, otherwise it is highly reflecting, so that it behaves as a bandpass filter. Cavities are the equivalent of LC resonators in waveguide technology.
Mathematical Basics
A.1 Coordinate systems and algebra of vector fields
The position of a point in space is specified by means of a coordinate system, that is a one to one correspondence between points in space and triples of real numbers. The most common are cartesian, cylindrical and spherical ones. It is to be remarked that all coordinate systems are equivalent: a specific one is chosen according to the geometry of the problem at hand. If the domain has a symmetry, it is convenient that the coordinate system has the same symmetry. For example, to describe the position of points on the surface of the Earth, a spherical system is more convenient than a cartesian one. To describe the magnetic field created by a straight wire carrying an electric current, a cylindrical system is better suited than a cartesian one. To describe the position of objects in a room, a cartesian system is more appropriate than a spherical one. Fig.
A.1 shows the main characteristics of the cylindrical reference system. Fig. A.2 describes the spherical reference system.
The relationship between cylindrical (ρ, ϕ, z) and cartesian coordinates is x = ρ cos ϕ
y = ρ sin ϕ (A.1)
and that between spherical (r, ϑ, ϕ) and cartesian coordinates is x = r sin ϑ cos ϕ
y = r sin ϑ sin ϕ z = r cos ϑ
(A.2)
Every coordinate system has a set of three fundamental unit vectors, that will be denoted by carets. The expressions of the cylindrical unit vectors in the cartesian basis are
ˆ
ρ = cos ϕˆx + sin ϕˆy ˆ
ϕ = − sin ϕˆx + cos ϕˆy (A.3)
Clearly the unit vectors are different in each point. The origin is a singular point of the system of coordinates, since the unit vectors are not defined there. The corresponding expressions for the
1
Figure A.1. Cylindrical coordinate reference system. (a) The three mutually perpendicular sur-faces defining the position of a point. (b) The three unit vectors in the point P ; in the text ˆρ, ˆϕ, ˆz are used in place of aρ, aφ, az. (c) The elementary volume dV at P : the sides are dρ, ρdφ, dz.
Figure A.2. Spherical coordinate reference system. (a) Definition of the spherical coordinates.
(b) The three mutually perpendicular surfaces defining the position of a point. (c) The three unit vectors in the point P ; in the text ˆr, ˆϑ, ˆϕ are used in place of ar, aθ, aφ. (d) The elementary volume dV at P : the sides are dr, rdϑ, r sin ϑdϕ.
spherical system are
and again in each point the basis of the unit vectors is different. As in the previous case, the origin is a singular point. The cartesian system is different in this respect: indeed, the basis of unit vectors is the same in each point, ˆx,ˆy,ˆz. Moreover, no point is singular.
Suppose that a certain vector field E(r) is given in a certain region of space. This means that in every point r the vector E(r) is defined: we can imagine it as an arrow with the tail in the point r. This is the abstract picture (coordinate-free). According to the system of coordinates, in the point r a basis of unit vectors is defined and three numbers are associated to the vector, i.e. its components:
E(r) = Exx + Eˆ yˆy + Ezˆz = Eρρ + Eˆ ϕϕ + Eˆ zz = Eˆ rr + Eˆ ϑϑ + Eˆ ϕϕˆ (A.5) Components can also be arranged in a 3 × 1 matrix (column vector) so that the previous equations can also be written as
E(r) ↔
Eq.(A.5) can be conveniently written as
E(r) =¡
Notice that the row vector contains the unit vectors instead of numbers.
Also eq.(A.3) can be written in matrix form
¡ ρˆ ϕ ˆˆ z ¢
As for the spherical system
¡ ˆr ϑˆ ϕˆ ¢
In this way it is a simple matter to find the relationship between cylindrical and cartesian compo-nents of a vector. From (A.7) and (A.8)
E(r) =¡
from which we get, on comparing the second and the fourth term of this chain of equalities
where the change of basis matrix is
PCaCy =
The inverse transformation is
Eρ
Since both the cartesian and the cylindrical basis are orthonormal, the change of basis matrix is unitary, hence the inverse of it coincides with its transpose.
The transformation of the components of a vector between the spherical and cartesian bases is obtained analogously. From (A.7) and (A.9)
E(r) =¡
from which we get, on comparing the second and the fourth term of this chain of equalities
where the change of basis matrix is
PCaS =
The inverse transformation is
Er
Vectors can be multiplied by a scalar, i.e. a number: the result is a vector with the same direction.
Two vectors can be multiplied in two ways. The scalar product of two arbitrary vectors is a scalar, the vector product is a vector. In the former case
A · B = |A||B| cos α
= AxBx+ AyBy+ AzBz= AρBρ+ AϕBϕ+ AzBz= ArBr+ AϑBϑ+ AϕBϕ
(A.13)
where ϑ is the angle between the two vectors. The scalar product of orthogonal vectors is obviously zero. The first definition is coordinate-free. The other expressions are so simple because the bases are orthonormal. In matrix form
A · B =¡
and the central 3 × 3 matrix is the identity matrix since the cartesian basis is orthonormal.
Two vectors can also be multiplied so that the result is a vector A × B (or, more appropriately, A ∧ B) :
|A × B| = |A||B| sin α (A.15)
and the direction is orthogonal to the plane on which the two vectors lie and points in the direction of the “right-hand-rule”. The vector product of parallel vectors is obviously zero. The magnitude of the vector product has the geometrical meaning of surface of the parallelogram with sides A and B. The easiest way to introduce it is by specifying the vector products of unit basis vectors:
from which we see very clearly that the vector product is anticommutative. Recalling (A.7) again
A × B =¡
The last expression is the traditional one (although only mnemonic!), where the determinant must be expanded using the elements of the first line.
Similar expressions hold for the cylindrical and spherical coordinates. In particular, for the cylindrical system
and for the spherical one
These definitions can be easily remembered by the method of Fig. A.3 A useful identity to know
Figure A.3. Correct order of unit vectors is the one that allows to expand a double vector product
A × (B × C) = (A · C) B − (A · B) C
Vector and scalar product can be combined in the mixed product A · (B × C). This number is the volume of the (skewed) parallelepiped with edges defined by A, B and C; it is clearly zero if the three vectors lie in the same plane.
An important concept associated to vector fields is that of field line. A field line is a line such that its tangent at any point r is parallel to the vector field E(r) in that point. They are useful tools for visualizing vector fields. If the vector field is regular, for each point only a field line passes.
Sometimes in a region two vector fields are given and there is functional dependence of one on the other. The simplest relationship that can exist is a linear one, so that the principle of superposition holds. An example in electromagnetism is the relationship existing between the electric field and the electric induction in a dielectric. Or the one between an electric current and the electric and magnetic fields it produces. Referring to the former example, we can write in abstract form
D = L{E}
However just as in the case of vectors, it is generally useful to introduce a basis to carry out computations. Using a cartesian basis, the previous equation becomes, because of linearity
D = Dxˆx + Dyˆy + Dzˆz = ExL{ˆx} + EyL{ˆy} + EzL{ˆz}
In matrix form
Dx
Now L{ˆx} is a (abstract) vector with three components along the three axes that can be collected in a column vector. The same can be done for the other two elements so that a 3 × 3 matrix arises and the previous equation can be written, with a more physical notation
The row index denotes the component, the column index the unit vector. The matrix ε represents the abstract linear operator L in the cartesian basis:
L ↔
and eq.(A.20) can be rephrased as
D = εE
Following in the reverse direction the steps leading from (A.5) to (A.6), this equation can be written L = εxxˆxˆx + εxyxˆˆy + εxzˆxˆz + εyxˆyˆx + εyyˆyˆy + εyzˆyˆz + εzxˆzˆx + εzyˆzˆy + εzzˆzˆz (A.21) Symbols such as ˆxˆy are called (unit) dyads and play just the role of placeholders to denote a specific row and column. However, it is simple to give rules to operate with scalar products on dyads so as to reproduce the results that could be obtained by the matrix formalism. The rules are the following
E · ˆxˆydef= (E · ˆx) ˆy = Exyˆ ˆ
xˆy · Edef= ˆx (ˆy · E) = Eyˆx
(A.22)
Eq.(A.21) is called the dyadic form of the linear operator L.
A dyad must not necessarily be formed by two unit vectors. If A, B are arbitrary vectors, we form the dyad AB with the rule of operation
E · ABdef= (E · A) B AB · Edef= A (B · E)
(A.23)
If we want to obtain the matrix that represents the abstract operator L = AB in the cartesian basis, we can use a distributive law of the “side-by-side-placement” of two vectors
L = AB = (Axˆx + Ayˆy + Azˆz) (Bxˆx + Byˆy + Bzˆz)
= AxBxˆxˆx + AxByˆxˆy + AxBzxˆˆz + AyBxˆyˆx + . . . + AzBzˆzˆz (A.24) The matrix form can be obtained immediately
L ↔
It is possible to introduce the vector product of dyads and vectors: it is just enough to extend (A.23):
E × ABdef= (E × A) B AB × Edef= A (B × E)
(A.25)
Useful linear operators are
• the identity I, such that
I · A = A · I = A
for any vector A. Its dyadic and matrix form in cartesian coordinates are
I = ˆxˆx + ˆyˆy + ˆzˆz ↔
• the identity transverse to ˆz, i.e. the projection operator onto the x,y plane, such that Itz· A = A − Azˆz
• the identity transverse to ˆr, i.e. the projection operator onto the tangent plane to the sphere r =const., such that Itr· A = A − Arˆr
Consider again the vector product of two vectors A × B = C. Since the vector C depends linearly on B, we can say that it is the result of the application of a linear operator on B. This operator can be found by recalling the second line of (A.17)
A × B =¡
First we have carried out the multiplication of the row vector times the (formal) matrix, obtaining a row vector whose elements are abstract vectors. Their components can be interpreted as columns of the successive matrix. Alternatively,
A × B = A × I · B
so that the matrix we are looking for is the one representing A × I, i.e.
A × I = (A × ˆx) ˆx + (A × ˆy) ˆy + (A × ˆz) ˆz Then
(A × ˆx) = (Axˆx + Ayy + Aˆ zˆz) × ˆx = −Ayˆz + Azˆy (A × ˆy) = (Axˆx + Ayy + Aˆ zˆz) × ˆx = Axˆz − Azˆx
(A × ˆz) = (Axˆx + Ayy + Aˆ zˆz) × ˆz = −Axˆy + Ayˆx
(A.27)
Substituting in the previous equation we get
A × I = (−Ayˆz + Azy)ˆˆx + (Axˆz − Azˆx)ˆy + (−Axˆy + Ayˆx)ˆz
= −Ayˆzˆx + Azyˆˆx + Axˆzˆy − Azˆxˆy − Axyˆˆz + Ayˆxˆz In conclusion
A × I ↔
0 −Az Ay
Az 0 −Ax
−Ay Ax 0
As a direct application of this result, we consider another useful operator to be added to the previous list. It is ˆr × I = ˆr × Itr:
ˆr × I ↔
0 0 0
0 0 −1
0 1 0
↔ ˆϕ ˆϑ − ˆϑ ˆϕ
where we have used the spherical basis and clearly ˆr ↔ (1 0 0)T in this basis.