Circular Waveguide Derivation

Circular Waveguide Derivation Using Cylindrical Coordinates

A circular waveguide is a hollow conducting cylinder used for the transmission of microwave signals at very high frequencies. Unlike rectangular waveguides, circular waveguides are analyzed using cylindrical coordinates because of their circular symmetry. The electromagnetic field distribution inside the guide is obtained from Maxwell's equations and the wave equation expressed in cylindrical coordinates.

Wave Equation in Circular Waveguide

The wave equation for any field quantity \(\Psi\) is:

$ \nabla^2 \Psi + k^2\Psi = 0 $

where:

• \(k=\omega\sqrt{\mu\epsilon}\) is the propagation constant in the medium.
• \(\mu\) is the permeability of the medium.
• \(\epsilon\) is the permittivity of the medium.

For a circular waveguide, cylindrical coordinates \((r,\phi,z)\) are used. The Laplacian operator in cylindrical coordinates is:

$ \nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2} $

Substituting the cylindrical Laplacian into the wave equation gives:

$ \frac{\partial^2\Psi}{\partial r^2} + \frac{1}{r}\frac{\partial\Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2\Psi}{\partial \phi^2} + \frac{\partial^2\Psi}{\partial z^2} + k^2\Psi = 0 $

Propagation Along the z-Direction

Assume that the electromagnetic wave propagates along the z-axis. Therefore the field quantity may be written as:

$ \Psi(r,\phi,z) = \Psi(r,\phi)e^{-j\beta z} $

where:

• \(\beta\) is the phase propagation constant.
• \(e^{-j\beta z}\) represents wave propagation along the guide axis.

Substituting this expression into the wave equation converts the three-dimensional equation into a two-dimensional transverse equation:

$ \frac{\partial^2\Psi}{\partial r^2} + \frac{1}{r}\frac{\partial\Psi}{\partial r} + \frac{1}{r^2}\frac{\partial^2\Psi}{\partial \phi^2} + K_c^2\Psi = 0 $

where:

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

The quantity \(K_c\) is known as the cutoff wave number of the circular waveguide.

Separation of Variables

To solve this partial differential equation, assume that the field can be expressed as the product of two independent functions:

$ \Psi(r,\phi) = R(r)\Phi(\phi) $

Substituting into the wave equation:

$ \Phi \frac{d^2R}{dr^2} + \frac{\Phi}{r} \frac{dR}{dr} + \frac{R}{r^2} \frac{d^2\Phi}{d\phi^2} + K_c^2R\Phi = 0 $

Dividing throughout by \(R\Phi\):

$ \frac{1}{R} \frac{d^2R}{dr^2} + \frac{1}{rR} \frac{dR}{dr} + \frac{1}{r^2\Phi} \frac{d^2\Phi}{d\phi^2} + K_c^2 = 0 $

Multiplying by \(r^2\):

$ r^2 \frac{1}{R} \frac{d^2R}{dr^2} + r \frac{1}{R} \frac{dR}{dr} + \frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2} + K_c^2r^2 = 0 $

The first three terms depend on different independent variables. Therefore each term must be equal to a constant.

Let:

$ \frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2} = -n^2 $

where \(n\) is an integer called the mode number.

Angular Solution

The angular equation becomes:

$ \frac{d^2\Phi}{d\phi^2} + n^2\Phi = 0 $

Its solution is:

$ \Phi(\phi) = A\cos(n\phi) + B\sin(n\phi) $

where \(A\) and \(B\) are arbitrary constants.

Since the field must repeat after one complete revolution:

$ \Phi(\phi) = \Phi(\phi+2\pi) $

the value of \(n\) must be an integer:

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

Radial Differential Equation

Substituting the separation constant into the radial portion gives:

$ r^2 \frac{d^2R}{dr^2} + r \frac{dR}{dr} + (K_c^2r^2-n^2)R = 0 $

This equation is known as Bessel's Differential Equation.

The solutions of Bessel's equation are called Bessel functions and form the basis of circular waveguide analysis.

Solution of Bessel's Differential Equation

$ r^2 \frac{d^2R}{dr^2} + r \frac{dR}{dr} + (K_c^2r^2-n^2)R = 0 $

This equation is known as Bessel's differential equation. Its solution consists of two independent functions known as Bessel functions.

Therefore:

$ R(r) = C_1J_n(K_cr) + C_2N_n(K_cr) $

where:

• $ J_n(K_cr) $ is the Bessel function of the first kind.
• $ N_n(K_cr) $ is the Bessel function of the second kind (Neumann function).
• $ C_1,\;C_2 $ are arbitrary constants.

Physical Requirement at the Centre of the Waveguide

The center of a circular waveguide corresponds to:

$ r=0 $

The electromagnetic field inside the guide must remain finite everywhere.

The Bessel function of the second kind becomes infinite at:

$ r=0 $

Therefore:

$ N_n(K_cr)\rightarrow\infty \quad\text{at}\quad r=0 $

Since an infinite field is physically impossible, the Neumann function must be discarded.

Thus:

$ C_2=0 $

and the radial solution reduces to:

$ R(r) = C_1J_n(K_cr) $

Complete Field Solution

Combining the radial and angular solutions:

$ \Psi(r,\phi) = J_n(K_cr) \left[ A\cos(n\phi) + B\sin(n\phi) \right] $

Including propagation along the z-direction:

$ \Psi(r,\phi,z) = J_n(K_cr) \left[ A\cos(n\phi) + B\sin(n\phi) \right] e^{-j\beta z} $

This represents the general field distribution inside a circular waveguide.

Boundary Condition at the Conducting Wall

Let the radius of the circular waveguide be:

$ r=a $

Since the wall is a perfect conductor, the tangential electric field at the conducting surface must be zero.

This boundary condition determines the allowed values of:

$ K_c $

and therefore determines the cutoff frequencies of the various modes.

The resulting equations involve the zeros of Bessel functions or the zeros of their derivatives.

These roots are usually represented by:

$ X_{nm} $

or

$ X'_{nm} $

depending on whether TM or TE modes are being considered.

Cutoff Wave Number

The cutoff wave number is therefore:

$ K_c = \frac{X_{nm}}{a} $

or

$ K_c = \frac{X'_{nm}}{a} $

where:

• $ a $ is the radius of the circular waveguide.
• $ X_{nm} $ is the m-th zero of the Bessel function.
• $ X'_{nm} $ is the m-th zero of the derivative of the Bessel function.

Propagation Constant in Circular Waveguide

The propagation characteristics of a circular waveguide are determined by the relationship between the propagation constant, the cutoff wave number, and the wave number of the medium.

From electromagnetic wave theory:

$ k=\omega\sqrt{\mu\epsilon} $

where:

• \(\omega\) = angular frequency
• \(\mu\) = permeability of the medium
• \(\epsilon\) = permittivity of the medium

The cutoff wave number obtained from the boundary conditions is:

$ K_c=\frac{X_{nm}}{a} $

or

$ K_c=\frac{X'_{nm}}{a} $

depending on the mode under consideration.

The propagation constant of the waveguide is:

$ \gamma=\alpha+j\beta $

where:

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

The relationship between the propagation constant and cutoff wave number is:

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

Substituting:

$ k=\omega\sqrt{\mu\epsilon} $

gives:

$ \gamma = \sqrt{ K_c^2 - \omega^2\mu\epsilon } $

Operating Conditions of a Circular Waveguide

The nature of wave propagation depends on the relative values of \(K_c\) and \(\omega\sqrt{\mu\epsilon}\).

Case I: Below Cutoff Frequency

If:

$ \omega^2\mu\epsilon < K_c^2 $

then:

$ \gamma = \alpha = \sqrt{ K_c^2 - \omega^2\mu\epsilon } $

and:

$ \beta=0 $

The wave does not propagate through the guide. Instead, the field decays exponentially along the guide axis.

The field variation becomes:

$ e^{-\alpha z} $

This condition is called the evanescent mode region.

Case II: Cutoff Condition

At cutoff:

$ \omega_c^2\mu\epsilon = K_c^2 $

Therefore:

$ \gamma=0 $

and:

$ \alpha=0 $

$ \beta=0 $

No propagation occurs at the cutoff frequency.

Case III: Above Cutoff Frequency

If:

$ \omega^2\mu\epsilon > K_c^2 $

then:

$ \gamma = j\beta $

where:

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

For a perfectly conducting waveguide:

$ \alpha=0 $

and the wave propagates without attenuation.

The field variation becomes:

$ e^{-j\beta z} $

which represents a travelling electromagnetic wave.

Cutoff Frequency of Circular Waveguide

At cutoff:

$ \omega_c^2\mu\epsilon = K_c^2 $

Substituting:

$ K_c=\frac{X_{nm}}{a} $

gives:

$ \omega_c = \frac{X_{nm}} {a\sqrt{\mu\epsilon}} $

Since:

$ \omega_c=2\pi f_c $

the cutoff frequency becomes:

$ f_c = \frac{X_{nm}} {2\pi a\sqrt{\mu\epsilon}} $

or:

$ f_c = \frac{c\,X_{nm}} {2\pi a} $

where:

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

is the velocity of electromagnetic waves in the medium.

Phase Constant Above Cutoff Frequency

When the operating frequency is greater than the cutoff frequency, electromagnetic waves propagate through the circular waveguide.

Under this condition:

$ f>f_c $

and the propagation constant becomes:

$ \gamma=j\beta $

where the phase constant is:

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

Using the cutoff condition:

$ K_c^2=\omega_c^2\mu\epsilon $

the phase constant can be written as:

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

or:

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

Since:

$ \frac{\omega_c}{\omega} = \frac{f_c}{f} $

therefore:

$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $

Phase Velocity

Phase velocity is defined as the velocity at which a constant phase point travels along the waveguide.

Mathematically:

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

Substituting the expression for \(\beta\):

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

Hence:

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

Since:

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

therefore:

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

Guide Wavelength

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

It is given by:

$ \lambda_g = \frac{2\pi}{\beta} $

Substituting the expression for \(\beta\):

$ \lambda_g = \frac{2\pi} { \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Since:

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

the guide wavelength becomes:

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

Cutoff Wavelength

The cutoff wavelength is related to the cutoff frequency by:

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

Substituting:

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

into the guide wavelength expression:

$ \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-\left( \frac{\lambda_0}{\lambda_c} \right)^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 relationship connects the free-space wavelength, cutoff wavelength, and guide wavelength of a circular waveguide.

Group Velocity

Group velocity is the velocity with which energy or information propagates through the circular waveguide.

The group velocity is defined as:

$ V_g = \frac{d\omega}{d\beta} $

Using the propagation relation:

$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $

the group velocity becomes:

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

where:

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

is the velocity of electromagnetic waves in the medium.

Relationship Between Phase Velocity and Group Velocity

From the expressions:

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

and

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

Multiplying:

$ V_pV_g = c^2 $

Thus, in a circular waveguide, phase velocity and group velocity satisfy:

$ V_pV_g=c^2 $

Wave Impedance in Circular Waveguide

Wave impedance is defined as the ratio of electric field intensity to magnetic field intensity.

The value depends on whether the wave is propagating in TE mode or TM mode.

TE Wave Impedance

For Transverse Electric (TE) waves:

$ E_z=0 $

and the wave impedance is:

$ Z_{TE} = \frac{\omega\mu}{\beta} $

Substituting:

$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $

gives:

$ Z_{TE} = \frac{\omega\mu} { \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Therefore:

$ Z_{TE} = \sqrt{\frac{\mu}{\epsilon}} \, \frac{1} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

Since:

$ \eta = \sqrt{\frac{\mu}{\epsilon}} $

is the intrinsic impedance of the medium,

$ Z_{TE} = \frac{\eta} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $

TM Wave Impedance

For Transverse Magnetic (TM) waves:

$ H_z=0 $

and the wave impedance is:

$ Z_{TM} = \frac{\beta}{\omega\epsilon} $

Substituting the value of \(\beta\):

$ Z_{TM} = \frac{ \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } {\omega\epsilon} $

Therefore:

$ Z_{TM} = \sqrt{\frac{\mu}{\epsilon}} \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $

or:

$ Z_{TM} = \eta \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $

Transverse Field Components in Circular Waveguide

Once the longitudinal field component is known, all remaining field components can be obtained from Maxwell's equations.

The transverse electric field components are:

$ E_r = -\frac{\gamma}{K_c^2} \frac{\partial E_z}{\partial r} - \frac{j\omega\mu}{K_c^2r} \frac{\partial H_z}{\partial \phi} $

$ E_\phi = -\frac{\gamma}{K_c^2r} \frac{\partial E_z}{\partial \phi} + \frac{j\omega\mu}{K_c^2} \frac{\partial H_z}{\partial r} $

The transverse magnetic field components are:

$ H_r = -\frac{\gamma}{K_c^2} \frac{\partial H_z}{\partial r} + \frac{j\omega\epsilon}{K_c^2r} \frac{\partial E_z}{\partial \phi} $

$ H_\phi = -\frac{\gamma}{K_c^2r} \frac{\partial H_z}{\partial \phi} - \frac{j\omega\epsilon}{K_c^2} \frac{\partial E_z}{\partial r} $

These equations represent the complete field relationships inside a circular waveguide and form the basis for the derivation of TE and TM modes.