TE Mode in Rectangular Waveguide Derivation

TE Mode in Rectangular Waveguide Derivation 

In a rectangular waveguide, electromagnetic waves can propagate in different field configurations called modes. One of the most important modes is the Transverse Electric (TE) mode. In TE mode, the electric field has no component in the direction of propagation, while the magnetic field possesses a longitudinal component.

Therefore:

$ E_z = 0 $

$ H_z \neq 0 $

Since the longitudinal magnetic field component exists, the entire field structure can be derived from the magnetic field component \(H_z\).

Reduced Wave Equation for TE Mode

Starting from Maxwell's equations and applying the waveguide boundary conditions, the longitudinal magnetic field satisfies the two-dimensional Helmholtz equation:

$ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + K_c^2 \right) H_z = 0 \qquad ...(1) $

where:

• \(H_z\) is the longitudinal magnetic field component.
• \(K_c\) is the cutoff wave number.
• \(x\) and \(y\) are transverse coordinates of the rectangular waveguide.

The field varies along the direction of propagation (\(z\)-axis) as:

$ H_z(x,y,z) = H_z(x,y)e^{-j\beta z} $

where:

• \(\beta\) = phase propagation constant
• \(e^{-j\beta z}\) = travelling wave factor along the waveguide

The cutoff wave number is related to the free-space wave number and propagation constant by:

$ K_c^2 = k^2-\beta^2 $

or equivalently:

$ K_c^2 = \omega^2\mu\epsilon-\beta^2 $

Method of Separation of Variables

Equation (1) is a partial differential equation involving two independent variables, \(x\) and \(y\). To solve it, we use the method of separation of variables.

Assume:

$ H_z(x,y) = X(x)\,Y(y) $

This means the field distribution can be expressed as the product of two independent functions:

• \(X(x)\) depends only on \(x\)
• \(Y(y)\) depends only on \(y\)

Substituting into Equation (1):

$ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + K_c^2 \right) [X(x)Y(y)] = 0 $

Expanding:

$ Y \frac{d^2X}{dx^2} + X \frac{d^2Y}{dy^2} + K_c^2XY = 0 \qquad ...(2) $

Dividing throughout by \(XY\):

$ \frac{1}{X} \frac{d^2X}{dx^2} + \frac{1}{Y} \frac{d^2Y}{dy^2} + K_c^2 = 0 \qquad ...(3) $

Notice that:

• The first term depends only on \(x\).
• The second term depends only on \(y\).
• Their sum is a constant.

The only way this equation can hold for every value of \(x\) and \(y\) is if each term equals a constant.

Separation Constants

Introduce separation constants \(K_x\) and \(K_y\):

$ \frac{1}{X} \frac{d^2X}{dx^2} = -K_x^2 $

$ \frac{1}{Y} \frac{d^2Y}{dy^2} = -K_y^2 $

Substituting these into Equation (3):

$ -K_x^2 -K_y^2 + K_c^2 = 0 $

Therefore:

$ K_x^2+K_y^2 = K_c^2 $

This important relationship connects the cutoff wave number with the field variations along the width and height of the rectangular waveguide.

Ordinary Differential Equations

The separated equations become:

$ \frac{d^2X}{dx^2} + K_x^2X = 0 \qquad ...(4) $

$ \frac{d^2Y}{dy^2} + K_y^2Y = 0 \qquad ...(5) $

These are standard second-order differential equations whose solutions are sinusoidal functions.

Solving Equation (4):

$ X(x) = A_1\sin(K_xx) + A_2\cos(K_xx) $

Similarly, solving Equation (5):

$ Y(y) = B_1\sin(K_yy) + B_2\cos(K_yy) $

Combining both solutions gives the general expression for the longitudinal magnetic field:

$ H_z(x,y) = \left[ A_1\sin(K_xx) + A_2\cos(K_xx) \right] \left[ B_1\sin(K_yy) + B_2\cos(K_yy) \right] $

This is the general field solution before applying the waveguide boundary conditions.

In the next part, the conducting wall boundary conditions will be applied to determine the allowable values of \(K_x\) and \(K_y\), leading to the TE_mn field expressions.

Rectangular Waveguide Dimensions

Consider a rectangular waveguide having:

• Width = \(a\)
• Height = \(b\)

The conducting walls are located at:

$ x=0,\quad x=a $

and

$ y=0,\quad y=b $

Since the walls are perfect conductors, the tangential electric field must vanish at the conducting surfaces.

Boundary Conditions for TE Mode

For TE mode:

$ E_z=0 $

The required boundary conditions become:

$ E_x=0 \qquad \text{at } y=0,b $

$ E_y=0 \qquad \text{at } x=0,a $

Using Maxwell's equations, the transverse electric fields can be expressed in terms of \(H_z\).

The field components are:

$ E_x = \frac{-j\omega\mu}{K_c^2} K_y \left[ A_1\sin(K_xx) + A_2\cos(K_xx) \right] \left[ -B_1\sin(K_yy) + B_2\cos(K_yy) \right] $

$ E_y = \frac{-j\omega\mu}{K_c^2} K_x \left[ -A_1\sin(K_xx) + A_2\cos(K_xx) \right] \left[ B_1\sin(K_yy) + B_2\cos(K_yy) \right] $

Applying Boundary Condition at y = 0

Since:

$ E_x=0 \qquad \text{at } y=0 $

Substituting \(y=0\):

$ -B_1\sin(K_y\cdot0) + B_2\cos(K_y\cdot0) = 0 $

Using:

$ \sin0=0 $

$ \cos0=1 $

Therefore:

$ B_2=0 $

Applying Boundary Condition at y = b

Again:

$ E_x=0 \qquad \text{at } y=b $

Substituting:

$ -B_1\sin(K_yb)=0 $

For a non-trivial field solution:

$ B_1\neq0 $

Hence:

$ \sin(K_yb)=0 $

The sine function becomes zero when:

$ K_yb=n\pi $

where:

$ n=0,1,2,3,\ldots $

Therefore:

$ K_y=\frac{n\pi}{b} $

Applying Boundary Condition at x = 0

Now:

$ E_y=0 \qquad \text{at } x=0 $

Substituting:

$ -A_1\sin(K_x\cdot0) + A_2\cos(K_x\cdot0) = 0 $

Therefore:

$ A_2=0 $

Applying Boundary Condition at x = a

Again:

$ E_y=0 \qquad \text{at } x=a $

Substituting:

$ -A_1\sin(K_xa)=0 $

For a non-zero field:

$ A_1\neq0 $

Hence:

$ \sin(K_xa)=0 $

Therefore:

$ K_xa=m\pi $

where:

$ m=0,1,2,3,\ldots $

Thus:

$ K_x=\frac{m\pi}{a} $

Allowed Wave Numbers

The boundary conditions restrict the field distribution to only certain discrete values:

$ K_x=\frac{m\pi}{a} $

$ K_y=\frac{n\pi}{b} $

where \(m\) and \(n\) are mode indices.

Final Expression for H_z

Substituting the boundary-condition constants into the general solution:

$ H_z(x,y,z) = A_{mn} \cos\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $

where:

• \(A_{mn}\) = arbitrary amplitude constant
• \(m\) = number of half-wave variations along the wider dimension \(a\)
• \(n\) = number of half-wave variations along the narrower dimension \(b\)

Electric Field Components

Substituting the allowed values of \(K_x\) and \(K_y\) into the field equations:

$ E_x = \frac{j\omega\mu n\pi} {K_c^2 b} A_{mn} \cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $

$ E_y = -\frac{j\omega\mu m\pi} {K_c^2 a} A_{mn} \sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $

In the next part, the magnetic field components \(H_x\) and \(H_y\) will be derived, followed by the cutoff wave number, propagation constant, and operating conditions of TE modes.

Transverse Magnetic Field Components

Using the TE-mode field relations:

$ H_x = \frac{j\beta m\pi} {K_c^2 a} A_{mn} \sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $

$ H_y = \frac{j\beta n\pi} {K_c^2 b} A_{mn} \cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $

Since this is a TE mode:

$ E_z = 0 $

Thus all six field components are now completely defined.

Cutoff Wave Number

From the separation-of-variables procedure:

$ K_c^2 = K_x^2 + K_y^2 $

Substituting:

$ K_x = \frac{m\pi}{a} $

and

$ K_y = \frac{n\pi}{b} $

gives:

$ K_c^2 = \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 $

This is the cutoff wave number for the TE_mn mode.

Propagation Constant

For wave propagation inside a waveguide:

$ K_c^2 = \gamma^2 + \omega^2\mu\epsilon $

Therefore:

$ \gamma^2 = K_c^2 - \omega^2\mu\epsilon $

Substituting the expression for \(K_c\):

$ \gamma^2 = \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 - \omega^2\mu\epsilon $

Taking square root:

$ \gamma = \sqrt{ \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 - \omega^2\mu\epsilon } $

The propagation constant can also be written as:

$ \gamma = \alpha + j\beta $

where:

• \(\alpha\) = attenuation constant
• \(\beta\) = phase constant

Operating Conditions of a Rectangular Waveguide

The behavior of the wave inside the guide depends on the relationship between operating frequency and cutoff frequency.

Case I: Frequency Below Cutoff

When:

$ \omega^2\mu\epsilon < \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 $

the quantity inside the square root remains positive.

Therefore:

$ \gamma = \alpha $

and:

$ \beta = 0 $

Under this condition there is no travelling wave.

Instead, the field decays exponentially along the guide.

Hence the wave cannot propagate.

Case II: Critical Condition (Cutoff)

At cutoff:

$ \omega_c^2\mu\epsilon = \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 $

Substituting into the propagation constant:

$ \gamma=0 $

Thus:

$ \alpha=0 $

and

$ \beta=0 $

This condition defines the cutoff frequency of the mode.

Case III: Frequency Above Cutoff

When:

$ \omega^2\mu\epsilon > \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 $

the square-root quantity becomes negative.

Hence:

$ \gamma = j\beta $

For a perfectly conducting waveguide:

$ \alpha=0 $

and:

$ \beta = \sqrt{ \omega^2\mu\epsilon - \left[ \left( \frac{m\pi}{a} \right)^2 + \left( \frac{n\pi}{b} \right)^2 \right] } $

Under this condition, electromagnetic waves propagate successfully through the waveguide.

This is the normal operating region of a rectangular waveguide.

In the next part, the cutoff frequency, cutoff wavelength, phase velocity, and guide wavelength will be derived.

Derivation of Cutoff Frequency

At cutoff, the phase constant becomes zero:

$ \beta = 0 $

Substituting into the propagation equation:

$ \sqrt{ \omega_c^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } = 0 $

Squaring both sides:

$ \omega_c^2\mu\epsilon = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $

Therefore:

$ \omega_c = \frac{1}{\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

Since:

$ \omega_c = 2\pi f_c $

the cutoff frequency becomes:

$ f_c = \frac{1}{2\pi\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

Cutoff Wavelength

The cutoff wavelength is defined as:

$ \lambda_c = \frac{c}{f_c} $

Substituting the expression for \(f_c\):

$ \lambda_c = \frac{ c }{ \frac{1}{2\pi\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } } $

After simplification:

$ \lambda_c = \frac{2} { \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } } $

Phase Velocity

The phase velocity is defined as:

$ V_p = \frac{\omega}{\beta} $

Substituting the expression for \(\beta\):

$ V_p = \frac{\omega} { \sqrt{ \omega^2\mu\epsilon - \omega_c^2\mu\epsilon } } $

Factoring out \(\omega^2\mu\epsilon\):

$ V_p = \frac{\omega} { \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } } $

Therefore:

$ V_p = \frac{1}{\sqrt{\mu\epsilon}} \cdot \frac{1} { \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } } $

Since:

$ c=\frac{1}{\sqrt{\mu\epsilon}} $

the phase velocity becomes:

$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } } $

or:

$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Guide Wavelength

Guide wavelength is the distance travelled inside the waveguide corresponding to a phase shift of \(2\pi\) radians.

It is given by:

$ \lambda_g = \frac{V_p}{f} $

Substituting the expression for \(V_p\):

$ \lambda_g = \frac{1}{f} \cdot \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Since:

$ \lambda_0=\frac{c}{f} $

we obtain:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Using:

$ \frac{f_c}{f} = \frac{\lambda_0}{\lambda_c} $

the guide wavelength can also be written as:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{\lambda_0}{\lambda_c}\right)^2 } } $

Relationship Between λ₀, λ_c and λ_g

Starting from:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{\lambda_0}{\lambda_c}\right)^2 } } $

Squaring both sides:

$ \lambda_g^2 = \frac{\lambda_0^2} { 1-\frac{\lambda_0^2}{\lambda_c^2} } $

Rearranging:

$ \frac{1}{\lambda_g^2} = \frac{1}{\lambda_0^2} - \frac{1}{\lambda_c^2} $

or:

$ \frac{1}{\lambda_0^2} = \frac{1}{\lambda_c^2} + \frac{1}{\lambda_g^2} $

This important relation connects free-space wavelength, cutoff wavelength, and guide wavelength for any rectangular waveguide mode.

Propagation Constant Above Cutoff

For a perfectly conducting rectangular waveguide operating above cutoff frequency:

$ \alpha = 0 $

and the propagation constant becomes:

$ \gamma=j\beta $

where the phase constant is:

$ \beta = \sqrt{ \omega^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } $

This equation describes wave propagation inside the guide when the operating frequency is higher than the cutoff frequency.

Condition at Cutoff Frequency

At cutoff, the electromagnetic wave is just about to propagate. Therefore:

$ \beta = 0 $

Substituting this condition into the phase constant equation:

$ \sqrt{ \omega_c^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } = 0 $

Squaring both sides:

$ \omega_c^2\mu\epsilon = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $

Taking square root:

$ \omega_c = \frac{1}{\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

Derivation of Cutoff Frequency

Since:

$ \omega_c=2\pi f_c $

Substituting:

$ 2\pi f_c = \frac{1}{\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

Therefore:

$ f_c = \frac{1}{2\pi\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

This equation gives the cutoff frequency of the TE_mn mode in a rectangular waveguide.

Physical Meaning of Cutoff Frequency

The cutoff frequency represents the minimum frequency required for wave propagation inside the waveguide.

• If \(f > f_c\), the wave propagates.
• If \(f = f_c\), the wave is at the threshold of propagation.
• If \(f < f_c\), the wave becomes evanescent and decays exponentially.

This behavior makes a waveguide act as a natural high-pass filter.

Derivation of Cutoff Wavelength

The cutoff wavelength is related to the cutoff frequency by:

$ \lambda_c=\frac{c}{f_c} $

Substituting the expression for \(f_c\):

$ \lambda_c = \frac{ c }{ \frac{1}{2\pi\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } } $

Using:

$ c=\frac{1}{\sqrt{\mu\epsilon}} $

the expression simplifies to:

$ \lambda_c = \frac{2} { \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } } $

Importance of Cutoff Wavelength

The cutoff wavelength determines whether a particular mode can exist inside the waveguide.

For successful propagation:

$ \lambda_0 < \lambda_c $

where:

• \(\lambda_0\) = free-space wavelength
• \(\lambda_c\) = cutoff wavelength

If the operating wavelength exceeds the cutoff wavelength, the wave cannot propagate and becomes exponentially attenuated.

Key Results Derived So Far

Cutoff angular frequency:

$ \omega_c = \frac{1}{\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

Cutoff frequency:

$ f_c = \frac{1}{2\pi\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $

Cutoff wavelength:

$ \lambda_c = \frac{2} { \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } } $

In the next part, we will derive the phase velocity \(V_p\), guide wavelength \(\lambda_g\), and the important relationship:

$ \frac{1}{\lambda_0^2} = \frac{1}{\lambda_c^2} + \frac{1}{\lambda_g^2} $

Phase Velocity in a Rectangular Waveguide

The phase velocity is defined as the velocity at which a constant phase point of the electromagnetic wave travels along the waveguide.

Mathematically:

$ V_p=\frac{\omega}{\beta} $

From the TE-mode propagation constant:

$ \beta = \sqrt{ \omega^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } $

Using the cutoff-frequency condition:

$ \omega_c^2\mu\epsilon = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $

Substituting into the expression for \(\beta\):

$ \beta = \sqrt{ \omega^2\mu\epsilon - \omega_c^2\mu\epsilon } $

Factoring:

$ \beta = \sqrt{\mu\epsilon} \sqrt{\omega^2-\omega_c^2} $

Substituting into the phase velocity equation:

$ V_p = \frac{\omega} { \sqrt{\mu\epsilon} \sqrt{\omega^2-\omega_c^2} } $

Dividing numerator and denominator by \(\omega\):

$ V_p = \frac{1} { \sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } } $

Since:

$ c=\frac{1}{\sqrt{\mu\epsilon}} $

Therefore:

$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } } $

Expressing in terms of frequency:

$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Interpretation of Phase Velocity

An interesting characteristic of waveguides is that the phase velocity is always greater than the speed of light.

Since:

$ 0< \left(\frac{f_c}{f}\right)^2 <1 $

the denominator is always less than one, which makes:

$ V_p>c $

This does not violate relativity because energy and information travel with group velocity, not phase velocity.

Guide Wavelength

Guide wavelength is the distance travelled by the wave inside the waveguide corresponding to a phase change of \(2\pi\) radians.

It is defined as:

$ \lambda_g = \frac{V_p}{f} $

Substituting the expression for phase velocity:

$ \lambda_g = \frac{1}{f} \cdot \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Since:

$ \lambda_0=\frac{c}{f} $

we obtain:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

This is the standard guide-wavelength equation for rectangular waveguides.

Alternative Form Using Cutoff Wavelength

Since:

$ \frac{f_c}{f} = \frac{\lambda_0}{\lambda_c} $

the guide wavelength can also be written as:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{\lambda_0}{\lambda_c}\right)^2 } } $

Relationship Between λ₀, λ_c, and λ_g

Starting from:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{\lambda_0}{\lambda_c}\right)^2 } } $

Squaring both sides:

$ \lambda_g^2 = \frac{\lambda_0^2} { 1-\frac{\lambda_0^2}{\lambda_c^2} } $

Taking reciprocal:

$ \frac{1}{\lambda_g^2} = \frac{ 1-\frac{\lambda_0^2}{\lambda_c^2} } {\lambda_0^2} $

Expanding:

$ \frac{1}{\lambda_g^2} = \frac{1}{\lambda_0^2} - \frac{1}{\lambda_c^2} $

Rearranging:

$ \frac{1}{\lambda_0^2} = \frac{1}{\lambda_c^2} + \frac{1}{\lambda_g^2} $

This equation is one of the most important wavelength relationships in rectangular waveguide theory.

In Short:

Phase velocity:

$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Guide wavelength:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Alternative guide wavelength form:

$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{\lambda_0}{\lambda_c}\right)^2 } } $

Waveguide wavelength relation:

$ \frac{1}{\lambda_0^2} = \frac{1}{\lambda_c^2} + \frac{1}{\lambda_g^2} $