TE Mode in Circular Waveguide

TE Mode in Circular Waveguide

The TE mode in a circular waveguide is an important mode of electromagnetic wave propagation in microwave engineering. In Transverse Electric (TE) mode, the electric field has no component in the direction of propagation, while the magnetic field has a longitudinal component.

For a circular waveguide, the analysis is carried out using cylindrical coordinates \((r,\phi,z)\), where \(r\) is the radial distance from the center, \(\phi\) is the angular coordinate, and \(z\) is the direction of wave propagation.

The main objective of this derivation is to determine the longitudinal magnetic field \(H_z\), apply the conducting-wall boundary condition, and obtain the allowed TE modes, cutoff frequency, propagation constant, wave impedance, and transverse field components.

TE Mode Condition in Circular Waveguide

For a Transverse Electric mode, the electric field has no longitudinal component. Therefore:

$ E_z=0 $

while the magnetic field has a longitudinal component:

$ H_z\neq0 $

Thus, the TE mode is characterized by:

$ E_z=0,\qquad H_z\neq0 $

Since \(H_z\) is non-zero, it is used as the starting point for determining the electromagnetic field distribution inside the circular waveguide.

Reduced Wave Equation for \(H_z\)

For a uniform waveguide, the longitudinal magnetic field satisfies the wave equation. In cylindrical coordinates, the Laplacian operator 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} $

For TE mode, the longitudinal field can be represented as:

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

where \(\beta\) is the propagation constant along the \(z\)-direction.

Because:

$ \frac{\partial^2}{\partial z^2}e^{-j\beta z} = -\beta^2e^{-j\beta z} $

the dependence on the \(z\)-coordinate can be separated from the transverse field distribution.

The reduced wave equation for \(H_z(r,\phi)\) becomes:

$ \left[ \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial\phi^2} + K_c^2 \right]H_z = 0 $

where \(K_c\) is the cutoff wave number.

The cutoff wave number is related to the propagation constant by:

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

where:

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

Therefore:

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

This reduced equation is the starting point for finding the allowed TE modes of a circular waveguide.

Separation of Variables

The reduced wave equation for the longitudinal magnetic field is:

$ \left[ \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial\phi^2} + K_c^2 \right]H_z=0 $

Since the circular waveguide has two transverse coordinates, \(r\) and \(\phi\), we separate the longitudinal magnetic field into a radial part and an angular part:

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

Substituting this into the reduced wave equation gives:

$ \frac{\partial^2}{\partial r^2} \left[R(r)\Phi(\phi)\right] + \frac{1}{r} \frac{\partial}{\partial r} \left[R(r)\Phi(\phi)\right] + \frac{1}{r^2} \frac{\partial^2}{\partial\phi^2} \left[R(r)\Phi(\phi)\right] + K_c^2R(r)\Phi(\phi) = 0 $

Because \(\Phi(\phi)\) does not depend on \(r\), the first radial derivative becomes:

$ \frac{\partial}{\partial r} \left[R(r)\Phi(\phi)\right] = \Phi(\phi)\frac{dR}{dr} $

Similarly, the second radial derivative is:

$ \frac{\partial^2}{\partial r^2} \left[R(r)\Phi(\phi)\right] = \Phi(\phi)\frac{d^2R}{dr^2} $

For the angular derivative, \(R(r)\) is independent of \(\phi\), so:

$ \frac{\partial^2}{\partial\phi^2} \left[R(r)\Phi(\phi)\right] = R(r)\frac{d^2\Phi}{d\phi^2} $

Therefore:

$ \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 the complete equation 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 the equation by \(r^2\):

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

Rearranging the radial and angular terms:

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

The left side depends only on \(r\), while the right side depends only on \(\phi\). Since the two sides must be equal for every value of \(r\) and \(\phi\), both sides must be equal to the same separation constant.

Let the separation constant be \(n^2\). Thus:

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

Therefore, the angular differential equation becomes:

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

Angular Solution

The differential equation:

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

has the general solution:

$ \Phi(\phi) = C_1\cos(n\phi) + C_2\sin(n\phi) $

Because the physical field must be single-valued after one complete revolution around the circular waveguide, the field at \(\phi\) and \(\phi+2\pi\) must be identical:

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

This condition requires:

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

Thus, \(n\) is an integer representing the angular variation of the field around the circular waveguide.

Radial Differential Equation

The radial part obtained from the separation process is:

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

Multiplying by \(R\):

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

Dividing by \(r^2\):

$ \frac{d^2R}{dr^2} + \frac{1}{r}\frac{dR}{dr} + \left( K_c^2-\frac{n^2}{r^2} \right)R = 0 $

This is Bessel's differential equation in cylindrical coordinates.

Introducing the variable:

$ x=K_cr $

the radial equation takes the standard Bessel form:

$ x^2\frac{d^2R}{dx^2} + x\frac{dR}{dx} + (x^2-n^2)R = 0 $

Therefore, the general radial solution is:

$ R(r) = A J_n(K_cr) + B Y_n(K_cr) $

where \(J_n\) is the Bessel function of the first kind and \(Y_n\) is the Bessel function of the second kind.

At the center of the circular waveguide, \(r=0\). The Bessel function \(Y_n(K_cr)\) becomes singular at \(r=0\), while \(J_n(K_cr)\) remains finite.

Since the electromagnetic field must remain finite inside the waveguide, the \(Y_n\) term must be discarded. Therefore:

$ B=0 $

and the radial solution becomes:

$ R(r)=A J_n(K_cr) $

Combining the radial and angular solutions, the longitudinal magnetic field is:

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

Including propagation in the \(z\)-direction:

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

Applying the Boundary Condition at the Conducting Wall

The circular waveguide has a perfectly conducting cylindrical wall at a radius \(r=a\). For a perfect electric conductor, the tangential component of the electric field must be zero at the conducting surface.

Therefore, at the wall:

$ E_\phi=0 \qquad\text{at}\qquad r=a $

For TE mode:

$ E_z=0 $

The transverse electric field components can be obtained from Maxwell's equations. For TE mode, the azimuthal electric field is related to the longitudinal magnetic field by:

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

The boundary condition therefore requires:

$ \frac{\partial H_z}{\partial r}=0 \qquad\text{at}\qquad r=a $

From the longitudinal field expression:

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

Differentiating with respect to \(r\):

$ \frac{\partial H_z}{\partial r} = \left[ A\cos(n\phi)+B\sin(n\phi) \right] K_cJ_n'(K_cr)e^{-j\beta z} $

At the conducting wall \(r=a\):

$ \frac{\partial H_z}{\partial r}\bigg|_{r=a}=0 $

Therefore:

$ K_cJ_n'(K_ca)=0 $

For a propagating mode, \(K_c\neq0\), so:

$ J_n'(K_ca)=0 $

Bessel Function Root Condition for TE Modes

The condition:

$ J_n'(K_ca)=0 $

means that \(K_ca\) must be one of the roots of the derivative of the Bessel function \(J_n\).

These roots are commonly denoted by:

$ X'_{nm} $

where \(n\) represents the angular mode number and \(m\) represents the radial mode number.

Therefore:

$ K_ca=X'_{nm} $

and hence:

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

This equation determines the allowed cutoff wave numbers of the TE modes in a circular waveguide.

Longitudinal Magnetic Field for TE Modes

Using:

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

the longitudinal magnetic field can be written as:

$ H_z(r,\phi,z) = H_0 J_n \left( \frac{X'_{nm}}{a}r \right) \cos(n\phi)e^{-j\beta z} $

where \(H_0\) is an arbitrary amplitude constant.

The sine form can also be used:

$ H_z(r,\phi,z) = H_0 J_n \left( \frac{X'_{nm}}{a}r \right) \sin(n\phi)e^{-j\beta z} $

The sine and cosine forms represent the possible angular field orientations. They have the same cutoff frequency for a given \(n\) and \(m\).

Cutoff Wave Number

The cutoff wave number is therefore:

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

The general relation between the wave number, propagation constant, and cutoff wave number is:

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

where:

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

Therefore:

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

Substituting the circular-waveguide cutoff wave number:

$ \beta^2 = \omega^2\mu\epsilon - \left( \frac{X'_{nm}}{a} \right)^2 $

Hence, the propagation constant is:

$ \boxed{ \beta = \sqrt{ \omega^2\mu\epsilon - \left( \frac{X'_{nm}}{a} \right)^2 } } $

Cutoff Frequency of TE Modes in a Circular Waveguide

At the cutoff frequency, the wave is just at the boundary between propagation and attenuation. Therefore, the propagation constant becomes zero:

$ \beta=0 $

We already obtained the propagation constant as:

$ \beta = \sqrt{ \omega^2\mu\epsilon - \left( \frac{X'_{nm}}{a} \right)^2 } $

At cutoff, putting \(\beta=0\):

$ 0= \sqrt{ \omega_c^2\mu\epsilon - \left( \frac{X'_{nm}}{a} \right)^2 } $

Squaring both sides gives:

$ \omega_c^2\mu\epsilon = \left( \frac{X'_{nm}}{a} \right)^2 $

Therefore:

$ \omega_c = \frac{X'_{nm}} {a\sqrt{\mu\epsilon}} $

Since:

$ \omega_c=2\pi f_c $

the cutoff frequency is:

$ 2\pi f_c = \frac{X'_{nm}} {a\sqrt{\mu\epsilon}} $

Hence:

$ \boxed{ f_c= \frac{X'_{nm}} {2\pi a\sqrt{\mu\epsilon}} } $

For a waveguide filled with a medium where:

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

the cutoff frequency becomes:

$ \boxed{ f_c= \frac{cX'_{nm}}{2\pi a} } $

Thus, the cutoff frequency of each TE mode depends on the radius \(a\) of the circular waveguide and the corresponding root \(X'_{nm}\) of the derivative of the Bessel function.

Propagation Constant Above Cutoff

The propagation constant is:

$ \beta = \sqrt{ \omega^2\mu\epsilon - \left( \frac{X'_{nm}}{a} \right)^2 } $

From the cutoff-frequency relationship:

$ \omega_c = \frac{X'_{nm}} {a\sqrt{\mu\epsilon}} $

we can write:

$ \left( \frac{X'_{nm}}{a} \right)^2 = \omega_c^2\mu\epsilon $

Therefore:

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

Taking \(\omega^2\mu\epsilon\) outside the square root:

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

Therefore:

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

Since \(\omega=2\pi f\), the same expression can be written in terms of frequency:

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

For propagation, the operating frequency must be greater than the cutoff frequency:

$ f>f_c $

Phase Velocity of TE Mode

The phase velocity is the velocity at which a constant phase point of the wave travels along the waveguide. It is given by:

$ 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 } } $

Canceling \(\omega\):

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

Since:

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

we obtain:

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

Because the denominator is less than one for \(f>f_c\), the phase velocity in a waveguide can be greater than the velocity of electromagnetic waves in the filling medium. This does not represent the velocity of energy or information transmission.

Group Velocity of TE Mode

The group velocity represents the velocity associated with the propagation of energy and a modulated wave packet. For a waveguide, it is related to the phase velocity by:

$ V_gV_p=c^2 $

Therefore:

$ V_g=\frac{c^2}{V_p} $

Substituting the phase velocity:

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

Hence:

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

The relationship between phase velocity and group velocity is therefore:

$ \boxed{ V_pV_g=c^2 } $

Guide Wavelength

The guide wavelength is the distance along the waveguide corresponding to a phase change of \(2\pi\) radians. It is related to the propagation constant 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:

$ \omega=2\pi f $

and:

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

we obtain:

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

TE Wave Impedance in a Circular Waveguide

The wave impedance of a waveguide is the ratio of the transverse electric field to the transverse magnetic field. For the TE mode, it is defined as:

$ Z_{TE} = \frac{E_x}{H_y} = -\frac{E_y}{H_x} $

We can obtain the TE wave impedance directly from Maxwell's equations.

Starting from Maxwell's Equation

For a source-free, homogeneous medium, Maxwell's curl equation in phasor form is:

$ \nabla\times E=-j\omega\mu H $

For TE mode:

$ E_z=0 $

and:

$ H_z\neq0 $

The transverse field components are related to the longitudinal magnetic field by:

$ E_t = -\frac{j\omega\mu}{K_c^2} \left( \hat{z}\times\nabla_t H_z \right) $

and:

$ H_t = -\frac{j\beta}{K_c^2} \nabla_t H_z $

where \(\nabla_t\) represents the transverse gradient:

$ \nabla_t = \hat r\frac{\partial}{\partial r} + \hat\phi \frac{1}{r} \frac{\partial}{\partial\phi} $

The ratio between the transverse electric and magnetic fields therefore becomes:

$ Z_{TE} = \frac{E_t}{H_t} $

Substituting the transverse field relationships:

$ Z_{TE} = \frac{ \frac{\omega\mu}{K_c^2} }{ \frac{\beta}{K_c^2} } $

The \(K_c^2\) terms cancel:

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

Therefore, the TE wave impedance is:

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

Substituting the Propagation Constant

For a circular waveguide, the propagation constant above cutoff is:

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

Substituting this into the TE wave impedance equation:

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

Canceling \(\omega\):

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

Since:

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

we obtain:

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

The intrinsic wave impedance of the medium is:

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

Therefore, the TE wave impedance becomes:

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

Using angular frequency instead of ordinary frequency, the same result is:

$ \boxed{ Z_{TE} = \frac{\eta} { \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } } $

Behavior of TE Wave Impedance

The expression shows that the TE wave impedance depends strongly on the operating frequency relative to the cutoff frequency.

As the operating frequency approaches cutoff:

$ f\rightarrow f_c^+ $

the denominator approaches zero:

$ \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } \rightarrow0 $

Therefore:

$ Z_{TE}\rightarrow\infty $

At frequencies much higher than cutoff:

$ f\gg f_c $

we have:

$ \left(\frac{f_c}{f}\right)^2\rightarrow0 $

and therefore:

$ Z_{TE}\rightarrow\eta $

Thus, the TE wave impedance is very high near cutoff and approaches the intrinsic impedance of the medium as the operating frequency becomes much greater than the cutoff frequency.

Transverse Field Components

For the TE mode in a circular waveguide, the longitudinal electric field is zero:

$ \boxed{E_z=0} $

The longitudinal magnetic field is:

$ H_z = H_0 J_n(K_cr) \cos(n\phi) e^{-j\beta z} $

where:

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

Using the transverse field relations, the radial electric field is:

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

Since:

$ \frac{\partial}{\partial\phi}\cos(n\phi) = -n\sin(n\phi) $

we obtain:

$ \boxed{ E_r = -\frac{j\omega\mu n}{K_c^2r} H_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} } $

The azimuthal electric field is:

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

Using:

$ \frac{d}{dr}J_n(K_cr) = K_cJ_n'(K_cr) $

we obtain:

$ \boxed{ E_\phi = -\frac{j\omega\mu}{K_c} H_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $

The radial magnetic field is:

$ H_r = -\frac{j\beta}{K_c^2} \frac{\partial H_z}{\partial r} $

Therefore:

$ \boxed{ H_r = -\frac{j\beta}{K_c} H_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $

Finally, the azimuthal magnetic field is:

$ H_\phi = -\frac{j\beta}{K_c^2r} \frac{\partial H_z}{\partial\phi} $

and hence:

$ \boxed{ H_\phi = \frac{j\beta n}{K_c^2r} H_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} } $

Therefore, the complete TE field is characterized by:

$ \boxed{ E_z=0,\qquad H_z\neq0 } $

together with the transverse components \(E_r\), \(E_\phi\), \(H_r\), and \(H_\phi\) obtained above.

TE Mode Cutoff Condition in a Circular Waveguide

The cutoff condition determines whether a particular TE mode can propagate through the circular waveguide. For a circular waveguide, the cutoff condition is determined by the roots of the derivative of the Bessel function.

For TE modes, the cutoff wave number is:

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

where \(X'_{nm}\) represents the \(m\)-th root of the derivative of the Bessel function \(J_n'(x)\), and \(a\) is the radius of the circular waveguide.

The propagation constant is:

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

Substituting \(K_c=X'_{nm}/a\):

$ \beta = \sqrt{ \omega^2\mu\epsilon - \left( \frac{X'_{nm}}{a} \right)^2 } $

Case I: Below Cutoff

When the operating frequency is lower than the cutoff frequency:

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

the quantity inside the square root becomes negative. Therefore, the propagation constant becomes imaginary:

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

In this condition, a propagating wave cannot exist. Instead, the field decreases exponentially along the waveguide. This is called an evanescent field.

Case II: At Cutoff

At the cutoff frequency:

$ \omega=\omega_c $

and the propagation constant becomes zero:

$ \beta=0 $

Therefore:

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

Using:

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

we obtain:

$ \omega_c^2\mu\epsilon = \left( \frac{X'_{nm}}{a} \right)^2 $

Taking the square root:

$ \omega_c = \frac{X'_{nm}} {a\sqrt{\mu\epsilon}} $

Since:

$ \omega_c=2\pi f_c $

the cutoff frequency is:

$ \boxed{ f_c = \frac{X'_{nm}} {2\pi a\sqrt{\mu\epsilon}} } $

For a medium where \(c=1/\sqrt{\mu\epsilon}\):

$ \boxed{ f_c = \frac{cX'_{nm}}{2\pi a} } $

Case III: Above Cutoff

When the operating frequency is greater than the cutoff frequency:

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

the propagation constant is real and the wave propagates through the waveguide.

Starting with:

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

Using:

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

we get:

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

Taking \(\omega^2\mu\epsilon\) outside the square root:

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

Since \(\omega_c/\omega=f_c/f\):

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

Thus, a TE mode propagates only when:

$ \boxed{f>f_c} $

TE Mode Cutoff Wavelength

The cutoff wavelength is related to the cutoff frequency by:

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

Substituting:

$ f_c=\frac{cX'_{nm}}{2\pi a} $

gives:

$ \lambda_c = \frac{c} { \frac{cX'_{nm}}{2\pi a} } $

Canceling \(c\):

$ \boxed{ \lambda_c= \frac{2\pi a}{X'_{nm}} } $

Therefore, for each TE mode, the cutoff wavelength is directly determined by the waveguide radius and the corresponding Bessel-function derivative root.

TE Mode

• For \(f
• For \(f=f_c\), the propagation constant is zero.
• For \(f>f_c\), the mode propagates through the circular waveguide.
• The TE cutoff condition is determined by \(J_n'(K_ca)=0\).
• The cutoff wave number is \(K_c=X'_{nm}/a\).
• The cutoff frequency is \(f_c=cX'_{nm}/(2\pi a)\).
• The cutoff wavelength is \(\lambda_c=2\pi a/X'_{nm}\).