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Whites, EE 481                         Lecture 10                      Page 1 of 10



    Lecture 10: TEM, TE, and TM Modes for
     Waveguides. Rectangular Waveguide.
We will now generalize our discussion of transmission lines by
considering EM waveguides. These are “pipes” that guide EM
waves. Coaxial cables, hollow metal pipes, and fiber optical
cables are all examples of waveguides.

We will assume that the waveguide is invariant in the z-
direction:
                         y
                                                         Metal walls

                     b

                                μ, ε
                                                     x
                 z                        a


and that the wave is propagating in z as e − j β z . (We could also
have assumed propagation in –z.)



                             Types of EM Waves

We will first develop an extremely interesting property of EM
waves that propagate in homogeneous waveguides. This will
lead to the concept of “modes” and their classification as
   • Transverse Electric and Magnetic (TEM),



                                © 2011 Keith W. Whites
Whites, EE 481                       Lecture 10                         Page 2 of 10



     •   Transverse Electric (TE), or
     •   Transverse Magnetic (TM).

Proceeding from the Maxwell curl equations:
                              ˆ
                              x    yˆ    ˆ
                                         z
                                         ∂        ∂    ∂
                 ∇ × E = − jωμ H ⇒                        = − jωμ H
                                         ∂x       ∂y   ∂z
                                         Ex       Ey   Ez
                             ∂Ez ∂E y
or                    ˆ
                      x:         −     = − jωμ H x
                              ∂y   ∂z
                            ⎛ ∂E ∂E ⎞
                      y : − ⎜ z − x ⎟ = − jωμ H y
                      ˆ
                            ⎝ ∂x    ∂z ⎠
                            ∂E y ∂Ex
                      ˆ
                      z:         −    = − jωμ H z
                             ∂x    ∂y

However, the spatial variation in z is known so that
                     ∂ ( e− jβ z )
                                   = − j β ( e− jβ z )
                         ∂z
Consequently, these curl equations simplify to
                     ∂Ez
                          + j β E y = − jωμ H x                       (3.3a),(1)
                      ∂y
                      ∂E
                   − z − j β Ex = − jωμ H y                       (3.3b),(2)
                       ∂x
                      ∂E y ∂Ex
                            −        = − jωμ H z                      (3.3c),(3)
                       ∂x        ∂y
Whites, EE 481                   Lecture 10                    Page 3 of 10



We can perform a similar expansion of Ampère’s equation
∇ × H = jωε E to obtain
                    ∂H z
                         + j β H y = jωε Ex   (3.4a),(4)
                     ∂y
                              ∂H z
                   − jβ H x −      = jωε E y  (3.4b),(5)
                               ∂x
                     ∂H y ∂H x
                          −       = jωε Ez    (3.5c),(6)
                      ∂x      ∂y

Now, (1)-(6) can be manipulated to produce simple algebraic
equations for the transverse (x and y) components of E and H .
For example, from (1):
                            j ⎛ ∂Ez         ⎞
                     Hx =           + jβ Ey ⎟
                          ωμ ⎜ ∂y
                              ⎝             ⎠
Substituting for Ey from (5) we find
                   j ⎡ ∂Ez        1 ⎛             ∂H z ⎞ ⎤
           Hx =       ⎢    + jβ      ⎜ − jβ H x −      ⎟⎥
                 ωμ ⎣ ∂y        jωε ⎝              ∂x ⎠ ⎦
                    j ∂Ez   β2     j β ∂H z
                 =        + 2 Hx − 2
                   ωμ ∂y ω με     ω με ∂x
                             j ⎛ ∂Ez       ∂H z ⎞
or,                  Hx =       ⎜ ωε    −β      ⎟            (3.5a),(7)
                            kc2 ⎝    ∂y     ∂x ⎠
where kc2 ≡ k 2 − β 2 and k 2 = ω 2 με .                          (3.6)

Similarly, we can show that
Whites, EE 481                  Lecture 10                Page 4 of 10



                              j ⎛ ∂Ez       ∂H z ⎞
                    Hy = −         ωε    +β
                             kc2 ⎜           ∂y ⎟
                                                        (3.5b),(8)
                                 ⎝    ∂x         ⎠
                           − j ⎛ ∂Ez      ∂H z ⎞
                    Ex =       ⎜ β   + ωμ      ⎟        (3.5c),(9)
                           kc2 ⎝ ∂x        ∂y ⎠
                            j ⎛     ∂E      ∂H z ⎞
                    Ey =         − β z + ωμ
                           kc2 ⎜             ∂x ⎟
                                                      (3.5d),(10)
                               ⎝     ∂y          ⎠

Most important point: From (7)-(10), we can see that all
transverse components of E and H can be determined from
only the axial components Ez and H z . It is this fact that allows
the mode designations TEM, TE, and TM.

Furthermore, we can use superposition to reduce the complexity
of the solution by considering each of these mode types
separately, then adding the fields together at the end.



             TE Modes and Rectangular Waveguides

A transverse electric (TE) wave has Ez = 0 and H z ≠ 0 .
Consequently, all E components are transverse to the direction
of propagation. Hence, in (7)-(10) with Ez = 0 , then all
transverse components of E and H are known once we find a
solution for only H z . Neat!
Whites, EE 481                         Lecture 10                      Page 5 of 10



For a rectangular waveguide, the solutions for Ex , E y , H x , H y ,
and H z are obtained in Section 3.3 of the text. The solution and
the solution process are interesting, but not needed in this
course.

What is found in that section is that
                            ⎛ mπ ⎞ ⎛ nπ ⎞
                                   2                2
                                                        m, n = 0,1,…
                 kc ,mn   = ⎜    ⎟ +⎜   ⎟                                  (11)
                            ⎝  a ⎠ ⎝ b ⎠                ( m = n ≠ 0)

Therefore,                    β = β mn = k 2 − kc2,mn                      (12)

These m and n indices indicate that only discrete solutions for
the transverse wavenumber (kc) are allowed. Physically, this
occurs because we’ve bounded the system in the x and y
directions. (A vaguely similar situation occurs in atoms, leading
to shell orbitals.)

Notice something important. From (11), we find that m = n = 0
means that kc ,00 = 0 . In (7)-(10), this implies infinite field
amplitudes, which is not a physical result. Consequently, the
m = n = 0 TE or TM modes are not allowed.

One exception might occur if Ez = H z = 0 since this leads to
indeterminate forms in (7)-(10). However, it can be shown that
inside hollow metallic waveguides when both m = n = 0 and
Whites, EE 481               Lecture 10                  Page 6 of 10



Ez = H z = 0 , then E = H = 0 . This means there is no TEM
mode.

Consequently, EM waves will propagate only when the
frequency is “large enough” since there is no TEM mode.
Otherwise β will be imaginary ( β → − jα ), leading to pure
attenuation and no propagation of the wave e − j β z → e −α z .

This turns out to be a general result. That is, for a hollow
conductor, EM waves will propagate only when the frequency is
large enough and exceeds some lower threshold. This minimum
propagation frequency is called the cutoff frequency f c ,mn .

It can be shown that guided EM waves require at least two
distinct conductors in order to support wave propagation all the
way down to 0+ Hz.

The cutoff frequencies for TE modes in a rectangular waveguide
are determined from (12) with β = 0 to be
                 1    ⎛ mπ ⎞ ⎛ nπ ⎞
                            2       2
                                      m, n = 0,1,…
      f c ,mn =       ⎜    ⎟ +⎜   ⎟                (3.84),(13)
                2π με ⎝ a ⎠ ⎝ b ⎠     (m = n ≠ 0)
In other words, these are the frequencies where β mn = 0 and
wave propagation begins when the frequency slightly exceeds
 f c ,mn .
Whites, EE 481               Lecture 10                  Page 7 of 10



For an X-band rectangular waveguide, the cross-sectional
dimensions are a = 2.286 cm and b = 1.016 cm. Using (13):

                 TEm,n Mode Cutoff Frequencies
                    m       n             fc,mn (GHz)
                    1       0                 6.562
                    2       0                13.123
                    0       1                14.764
                    1       1                16.156

In the X-band region ( ≈ 8.2-12.5 GHz), only the TE10 mode can
propagate in the waveguide regardless of how it is excited.
(We’ll also see shortly that no TM modes will propagate either.)
This is called single mode operation and is most often the
preferred application for hollow waveguides.

On the other hand, at 15.5 GHz any combination of the first
three of these modes could exist and propagate inside a metal,
rectangular waveguide. Which combination actually exists will
depend on how the waveguide is excited.

Note that the TE11 mode (and all higher-ordered TE modes)
could not propagate. (We’ll also see next that no TM modes will
propagate at 15.5 GHz either.)
Whites, EE 481                Lecture 10                     Page 8 of 10



             TM Modes and Rectangular Waveguides

Conversely to TE modes, transverse magnetic (TM) modes have
H z = 0 and Ez ≠ 0 .

The expression for the cutoff frequencies of TM modes in a
rectangular waveguide
                            ⎛ mπ ⎞ ⎛ nπ ⎞
                                 2            2
                    1
          f c ,mn =         ⎜    ⎟ +⎜   ⎟     m, n = 1, 2…       (14)
                    2π με ⎝    a ⎠ ⎝ b ⎠
is very similar to that for TE modes given in (13).

It can be shown that if either m = 0 or n = 0 for TM modes,
then E = H = 0 . This means that no TM modes with m = 0 or
n = 0 are allowable in a rectangular waveguide.

For an X-band waveguide:

                  TMm,n Mode Cutoff Frequencies
                     m       n             fc,mn (GHz)
                     1       1                16.156
                     1       2                30.248
                     2       1                19.753

Therefore, no TM modes can propagate in an X-band
rectangular waveguide when f < 16.156 GHz.
Whites, EE 481                 Lecture 10                    Page 9 of 10




                        Dominant Mode

Note that from 6.56 GHz ≤ f ≤ 13.12 GHz in the X-band
rectangular waveguide, only the TE 10 mode can propagate. This
mode is called the dominant mode of the waveguide.

See Fig 3.9 in the text for plots of the electric and magnetic
fields associated with this mode.



                           TEM Mode

The transverse electric and magnetic (TEM) modes are
characterized by Ez = 0 and H z = 0 .

In order for this to occur, it can be shown from (3.4) and (3.5)
that it is necessary for f c = 0 . In other words, there is no cutoff
frequency for waveguides that support TEM waves.

Rectangular, circular, elliptical, and all hollow, metallic
waveguides cannot support TEM waves.

It can be shown that at least two separate conductors are
required for TEM waves. Examples of waveguides that will
Whites, EE 481               Lecture 10                 Page 10 of 10



allow TEM modes include coaxial cable, parallel plate
waveguide, stripline, and microstrip.

Microstrip is the type of microwave circuit interconnection that
we will use in this course. This “waveguide” will support the
“quasi-TEM” mode, which like regular TEM modes has no non-
zero cutoff frequency.

However, if the frequency is large enough, other modes will
begin to propagate on a microstrip. This is usually not a
desirable situation.

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481 lecture10

  • 1. Whites, EE 481 Lecture 10 Page 1 of 10 Lecture 10: TEM, TE, and TM Modes for Waveguides. Rectangular Waveguide. We will now generalize our discussion of transmission lines by considering EM waveguides. These are “pipes” that guide EM waves. Coaxial cables, hollow metal pipes, and fiber optical cables are all examples of waveguides. We will assume that the waveguide is invariant in the z- direction: y Metal walls b μ, ε x z a and that the wave is propagating in z as e − j β z . (We could also have assumed propagation in –z.) Types of EM Waves We will first develop an extremely interesting property of EM waves that propagate in homogeneous waveguides. This will lead to the concept of “modes” and their classification as • Transverse Electric and Magnetic (TEM), © 2011 Keith W. Whites
  • 2. Whites, EE 481 Lecture 10 Page 2 of 10 • Transverse Electric (TE), or • Transverse Magnetic (TM). Proceeding from the Maxwell curl equations: ˆ x yˆ ˆ z ∂ ∂ ∂ ∇ × E = − jωμ H ⇒ = − jωμ H ∂x ∂y ∂z Ex Ey Ez ∂Ez ∂E y or ˆ x: − = − jωμ H x ∂y ∂z ⎛ ∂E ∂E ⎞ y : − ⎜ z − x ⎟ = − jωμ H y ˆ ⎝ ∂x ∂z ⎠ ∂E y ∂Ex ˆ z: − = − jωμ H z ∂x ∂y However, the spatial variation in z is known so that ∂ ( e− jβ z ) = − j β ( e− jβ z ) ∂z Consequently, these curl equations simplify to ∂Ez + j β E y = − jωμ H x (3.3a),(1) ∂y ∂E − z − j β Ex = − jωμ H y (3.3b),(2) ∂x ∂E y ∂Ex − = − jωμ H z (3.3c),(3) ∂x ∂y
  • 3. Whites, EE 481 Lecture 10 Page 3 of 10 We can perform a similar expansion of Ampère’s equation ∇ × H = jωε E to obtain ∂H z + j β H y = jωε Ex (3.4a),(4) ∂y ∂H z − jβ H x − = jωε E y (3.4b),(5) ∂x ∂H y ∂H x − = jωε Ez (3.5c),(6) ∂x ∂y Now, (1)-(6) can be manipulated to produce simple algebraic equations for the transverse (x and y) components of E and H . For example, from (1): j ⎛ ∂Ez ⎞ Hx = + jβ Ey ⎟ ωμ ⎜ ∂y ⎝ ⎠ Substituting for Ey from (5) we find j ⎡ ∂Ez 1 ⎛ ∂H z ⎞ ⎤ Hx = ⎢ + jβ ⎜ − jβ H x − ⎟⎥ ωμ ⎣ ∂y jωε ⎝ ∂x ⎠ ⎦ j ∂Ez β2 j β ∂H z = + 2 Hx − 2 ωμ ∂y ω με ω με ∂x j ⎛ ∂Ez ∂H z ⎞ or, Hx = ⎜ ωε −β ⎟ (3.5a),(7) kc2 ⎝ ∂y ∂x ⎠ where kc2 ≡ k 2 − β 2 and k 2 = ω 2 με . (3.6) Similarly, we can show that
  • 4. Whites, EE 481 Lecture 10 Page 4 of 10 j ⎛ ∂Ez ∂H z ⎞ Hy = − ωε +β kc2 ⎜ ∂y ⎟ (3.5b),(8) ⎝ ∂x ⎠ − j ⎛ ∂Ez ∂H z ⎞ Ex = ⎜ β + ωμ ⎟ (3.5c),(9) kc2 ⎝ ∂x ∂y ⎠ j ⎛ ∂E ∂H z ⎞ Ey = − β z + ωμ kc2 ⎜ ∂x ⎟ (3.5d),(10) ⎝ ∂y ⎠ Most important point: From (7)-(10), we can see that all transverse components of E and H can be determined from only the axial components Ez and H z . It is this fact that allows the mode designations TEM, TE, and TM. Furthermore, we can use superposition to reduce the complexity of the solution by considering each of these mode types separately, then adding the fields together at the end. TE Modes and Rectangular Waveguides A transverse electric (TE) wave has Ez = 0 and H z ≠ 0 . Consequently, all E components are transverse to the direction of propagation. Hence, in (7)-(10) with Ez = 0 , then all transverse components of E and H are known once we find a solution for only H z . Neat!
  • 5. Whites, EE 481 Lecture 10 Page 5 of 10 For a rectangular waveguide, the solutions for Ex , E y , H x , H y , and H z are obtained in Section 3.3 of the text. The solution and the solution process are interesting, but not needed in this course. What is found in that section is that ⎛ mπ ⎞ ⎛ nπ ⎞ 2 2 m, n = 0,1,… kc ,mn = ⎜ ⎟ +⎜ ⎟ (11) ⎝ a ⎠ ⎝ b ⎠ ( m = n ≠ 0) Therefore, β = β mn = k 2 − kc2,mn (12) These m and n indices indicate that only discrete solutions for the transverse wavenumber (kc) are allowed. Physically, this occurs because we’ve bounded the system in the x and y directions. (A vaguely similar situation occurs in atoms, leading to shell orbitals.) Notice something important. From (11), we find that m = n = 0 means that kc ,00 = 0 . In (7)-(10), this implies infinite field amplitudes, which is not a physical result. Consequently, the m = n = 0 TE or TM modes are not allowed. One exception might occur if Ez = H z = 0 since this leads to indeterminate forms in (7)-(10). However, it can be shown that inside hollow metallic waveguides when both m = n = 0 and
  • 6. Whites, EE 481 Lecture 10 Page 6 of 10 Ez = H z = 0 , then E = H = 0 . This means there is no TEM mode. Consequently, EM waves will propagate only when the frequency is “large enough” since there is no TEM mode. Otherwise β will be imaginary ( β → − jα ), leading to pure attenuation and no propagation of the wave e − j β z → e −α z . This turns out to be a general result. That is, for a hollow conductor, EM waves will propagate only when the frequency is large enough and exceeds some lower threshold. This minimum propagation frequency is called the cutoff frequency f c ,mn . It can be shown that guided EM waves require at least two distinct conductors in order to support wave propagation all the way down to 0+ Hz. The cutoff frequencies for TE modes in a rectangular waveguide are determined from (12) with β = 0 to be 1 ⎛ mπ ⎞ ⎛ nπ ⎞ 2 2 m, n = 0,1,… f c ,mn = ⎜ ⎟ +⎜ ⎟ (3.84),(13) 2π με ⎝ a ⎠ ⎝ b ⎠ (m = n ≠ 0) In other words, these are the frequencies where β mn = 0 and wave propagation begins when the frequency slightly exceeds f c ,mn .
  • 7. Whites, EE 481 Lecture 10 Page 7 of 10 For an X-band rectangular waveguide, the cross-sectional dimensions are a = 2.286 cm and b = 1.016 cm. Using (13): TEm,n Mode Cutoff Frequencies m n fc,mn (GHz) 1 0 6.562 2 0 13.123 0 1 14.764 1 1 16.156 In the X-band region ( ≈ 8.2-12.5 GHz), only the TE10 mode can propagate in the waveguide regardless of how it is excited. (We’ll also see shortly that no TM modes will propagate either.) This is called single mode operation and is most often the preferred application for hollow waveguides. On the other hand, at 15.5 GHz any combination of the first three of these modes could exist and propagate inside a metal, rectangular waveguide. Which combination actually exists will depend on how the waveguide is excited. Note that the TE11 mode (and all higher-ordered TE modes) could not propagate. (We’ll also see next that no TM modes will propagate at 15.5 GHz either.)
  • 8. Whites, EE 481 Lecture 10 Page 8 of 10 TM Modes and Rectangular Waveguides Conversely to TE modes, transverse magnetic (TM) modes have H z = 0 and Ez ≠ 0 . The expression for the cutoff frequencies of TM modes in a rectangular waveguide ⎛ mπ ⎞ ⎛ nπ ⎞ 2 2 1 f c ,mn = ⎜ ⎟ +⎜ ⎟ m, n = 1, 2… (14) 2π με ⎝ a ⎠ ⎝ b ⎠ is very similar to that for TE modes given in (13). It can be shown that if either m = 0 or n = 0 for TM modes, then E = H = 0 . This means that no TM modes with m = 0 or n = 0 are allowable in a rectangular waveguide. For an X-band waveguide: TMm,n Mode Cutoff Frequencies m n fc,mn (GHz) 1 1 16.156 1 2 30.248 2 1 19.753 Therefore, no TM modes can propagate in an X-band rectangular waveguide when f < 16.156 GHz.
  • 9. Whites, EE 481 Lecture 10 Page 9 of 10 Dominant Mode Note that from 6.56 GHz ≤ f ≤ 13.12 GHz in the X-band rectangular waveguide, only the TE 10 mode can propagate. This mode is called the dominant mode of the waveguide. See Fig 3.9 in the text for plots of the electric and magnetic fields associated with this mode. TEM Mode The transverse electric and magnetic (TEM) modes are characterized by Ez = 0 and H z = 0 . In order for this to occur, it can be shown from (3.4) and (3.5) that it is necessary for f c = 0 . In other words, there is no cutoff frequency for waveguides that support TEM waves. Rectangular, circular, elliptical, and all hollow, metallic waveguides cannot support TEM waves. It can be shown that at least two separate conductors are required for TEM waves. Examples of waveguides that will
  • 10. Whites, EE 481 Lecture 10 Page 10 of 10 allow TEM modes include coaxial cable, parallel plate waveguide, stripline, and microstrip. Microstrip is the type of microwave circuit interconnection that we will use in this course. This “waveguide” will support the “quasi-TEM” mode, which like regular TEM modes has no non- zero cutoff frequency. However, if the frequency is large enough, other modes will begin to propagate on a microstrip. This is usually not a desirable situation.