Dominant and Determinant Modes
Dominant and Determinant Modes in Rectangular and Circular Waveguides
After deriving the TE and TM modes for rectangular and circular waveguides, the next step is to determine which modes are allowed and which mode has the lowest cutoff frequency. These two ideas are closely related, but they are not the same.
Mode determination tells us which TE and TM modes can exist in a particular waveguide and what their cutoff frequencies are. The dominant mode is the mode with the lowest cutoff frequency among all the allowed modes.
Mode Determination in a Waveguide
A waveguide does not allow every possible field distribution to propagate. The conducting walls impose boundary conditions on the electric and magnetic fields. These boundary conditions restrict the possible values of the mode indices and therefore determine the allowed modes.
For both rectangular and circular waveguides, the two basic types of modes are:
- TE mode: \(E_z=0\) and \(H_z\neq0\)
- TM mode: \(E_z\neq0\) and \(H_z=0\)
The allowed modes are obtained by applying the appropriate boundary conditions to the longitudinal field component.
1. TE and TM Modes in a Rectangular Waveguide
Consider a rectangular waveguide having dimensions \(a\) and \(b\), where \(a\) is normally the wider dimension and \(b\) is the narrower dimension.
The cutoff wave number for a rectangular waveguide is:
$ K_c^2 = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $
Therefore:
$ K_c = \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $
The corresponding cutoff frequency is:
$ f_c = \frac{c}{2} \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } $
The values of \(m\) and \(n\) determine the field variation across the width and height of the rectangular waveguide.
TE Modes in a Rectangular Waveguide
For a TE mode:
$ E_z=0 $
and:
$ H_z\neq0 $
The longitudinal magnetic field obtained from the boundary conditions has the form:
$ H_z = A_{mn} \cos\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
The corresponding cutoff frequency is:
$ \boxed{ f_c = \frac{c}{2} \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } } $
Allowed Values of \(m\) and \(n\) for TE Modes
For TE modes, one of the indices is allowed to be zero. However, both indices cannot be zero simultaneously.
Therefore:
$ m=0,\quad n=1,2,3,\ldots $
or:
$ m=1,2,3,\ldots,\quad n=0 $
and modes such as:
$ TE_{10},\quad TE_{01},\quad TE_{20},\quad TE_{11},\quad TE_{21} $
can exist.
However:
Why $ TE_{00} $ doesnt exist?
Why Does \(TE_{00}\) Not Exist?
For a TE mode, the longitudinal electric field is zero:
$ E_z=0 $
while the longitudinal magnetic field is non-zero:
$ H_z\neq0 $
For a rectangular waveguide, the longitudinal magnetic field for a TE mode is:
$ H_z = A_{mn} \cos\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
If we take \(m=0\) and \(n=0\), the field becomes:
$ H_z = A_{00}e^{-j\beta z} $
This means that \(H_z\) has no variation in either the \(x\)-direction or the \(y\)-direction. Therefore, the transverse derivatives of \(H_z\) are zero:
$ \frac{\partial H_z}{\partial x}=0 $
$ \frac{\partial H_z}{\partial y}=0 $
The transverse electric and magnetic fields of a TE mode are obtained from these derivatives. Therefore:
$ E_x=E_y=H_x=H_y=0 $
Since TE mode already requires:
$ E_z=0 $
all electric and magnetic field components would reduce to zero except for a longitudinal \(H_z\) component that has no transverse field variation. Such a field cannot satisfy the requirements for a propagating electromagnetic mode in the rectangular waveguide.
The same conclusion can also be seen from the cutoff wave number:
$ K_c^2 = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $
For \(m=0\) and \(n=0\):
$ K_c^2=0 $
$ K_c=0 $
This would give a cutoff frequency of:
$ f_c=0 $
However, \(TE_{00}\) does not represent a valid transverse electromagnetic field distribution in a hollow rectangular waveguide. Therefore, the \(TE_{00}\) designation is not an allowed propagating waveguide mode.
Hence:
$ \boxed{TE_{00}\text{ does not exist}} $
For TE modes, one index may be zero, but both indices cannot be zero simultaneously. Thus, modes such as \(TE_{10}\), \(TE_{01}\), and \(TE_{20}\) can exist, whereas \(TE_{00}\) cannot.
$ TE_{00} $
is not considered a valid propagating mode because it does not produce the required transverse field distribution.
Determining the Dominant TE Mode of a Rectangular Waveguide
The dominant mode is found by comparing the cutoff frequencies of all allowed modes.
For the \(TE_{10}\) mode:
$ m=1,\qquad n=0 $
Substituting into the cutoff-frequency equation:
$ f_{c10} = \frac{c}{2} \sqrt{ \left(\frac{1}{a}\right)^2+0 } $
Therefore:
$ \boxed{ f_{c10}=\frac{c}{2a} } $
For the \(TE_{01}\) mode:
$ m=0,\qquad n=1 $
Therefore:
$ \boxed{ f_{c01}=\frac{c}{2b} } $
Since the wider dimension is normally:
$ a>b $
we have:
$ \frac{1}{a}<\frac{1}{b} $
and therefore:
$ f_{c10}
The \(TE_{10}\) mode has the lowest cutoff frequency.
Hence, the dominant mode of a conventional rectangular waveguide is:
$ \boxed{TE_{10}} $
TM Modes in a Rectangular Waveguide
For a TM mode:
$ E_z\neq0 $
and:
$ H_z=0 $
The longitudinal electric field must satisfy the conducting-wall boundary conditions.
For a rectangular waveguide, this leads to a field distribution of the form:
$ E_z = E_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
The sine functions are important because the longitudinal electric field must vanish at the conducting walls.
At \(x=0\):
$ \sin\left(\frac{m\pi(0)}{a}\right)=0 $
and at \(x=a\):
$ \sin(m\pi)=0 $
Similarly, at \(y=0\):
$ \sin\left(\frac{n\pi(0)}{b}\right)=0 $
and at \(y=b\):
$ \sin(n\pi)=0 $
Therefore:
$ m=1,2,3,\ldots $
and:
$ n=1,2,3,\ldots $
Both indices must therefore be non-zero for a TM mode.
Allowed TM Modes in a Rectangular Waveguide
Because both \(m\) and \(n\) must be non-zero, the following modes can exist:
- \(TM_{11}\)
- \(TM_{12}\)
- \(TM_{21}\)
- \(TM_{22}\)
- and higher-order TM modes.
Modes such as:
$ TM_{10},\qquad TM_{01},\qquad TM_{20} $
cannot exist because one of the indices is zero.
Why $ TM_{10},\qquad TM_{01},\qquad TM_{20} $ cannot exist?
Modes such as:
$ TM_{10},\qquad TM_{01},\qquad TM_{20} $
cannot exist in a rectangular waveguide because, for TM modes, both mode indices must be non-zero.
For a rectangular waveguide, the longitudinal electric field of a TM mode is given by:
$ E_z = E_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
The sine terms arise from the boundary conditions at the conducting walls. The longitudinal electric field must be zero at the walls.
Case 1: \(m=0\)
If \(m=0\), the first sine term becomes:
$ \sin\left(\frac{0\pi x}{a}\right) = \sin(0) = 0 $
Therefore, the complete longitudinal electric field becomes:
$ E_z = E_{0n} \cdot 0 \cdot \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
$ \therefore\quad E_z=0 $
But a TM mode requires:
$ E_z\neq0 $
Therefore, a TM mode with \(m=0\) cannot exist.
$ \boxed{ TM_{01},\ TM_{02},\ TM_{03},\ldots \text{ do not exist} } $
Case 2: \(n=0\)
Similarly, if \(n=0\), the second sine term becomes:
$ \sin\left(\frac{0\pi y}{b}\right) = \sin(0) = 0 $
Therefore:
$ E_z=0 $
Again, this contradicts the basic requirement for a TM mode:
$ E_z\neq0 $
Therefore, a TM mode with \(n=0\) cannot exist.
$ \boxed{ TM_{10},\ TM_{20},\ TM_{30},\ldots \text{ do not exist} } $
Case 3: \(m=0,\ n=0\)
If both indices are zero:
$ m=0,\qquad n=0 $
both sine terms become zero:
$ \sin(0)=0 $
and therefore:
$ E_z=0 $
Thus:
$ \boxed{ TM_{00}\text{ does not exist} } $
Condition for TM Modes
Therefore, both mode indices must be non-zero:
$ \boxed{ m\neq0,\qquad n\neq0 } $
Since the mode indices are positive integers, the allowed values are:
$ m=1,2,3,\ldots $
$ n=1,2,3,\ldots $
Hence, the lowest possible TM mode is:
$ \boxed{ TM_{11} } $
Determining the Dominant TM Mode of a Rectangular Waveguide
The lowest possible values of the two indices are:
$ m=1,\qquad n=1 $
Therefore, the lowest-order TM mode is:
$ TM_{11} $
Its cutoff frequency is:
$ f_{c11} = \frac{c}{2} \sqrt{ \left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 } $
Since every allowed TM mode has \(m\geq1\) and \(n\geq1\), increasing either index increases the cutoff frequency.
Therefore:
$ \boxed{ TM_{11} } $
is the dominant TM mode of a rectangular waveguide.
Rectangular Waveguide: Final Mode Results
| Mode Type | Condition | Lowest Mode |
|---|---|---|
| TE | One index may be zero, but not both | \(TE_{10}\) |
| TM | \(m\neq0,\ n\neq0\) | \(TM_{11}\) |
For a conventional rectangular waveguide where \(a>b\):
$ \boxed{ f_c(TE_{10}) < f_c(TM_{11}) } $
Therefore, \(TE_{10}\) is not only the dominant TE mode but also the dominant mode of the rectangular waveguide overall.