Circular Waveguide TM Mode
TM Mode in Circular Waveguide
A circular waveguide supports two fundamental types of electromagnetic modes: Transverse Electric (TE) modes and Transverse Magnetic (TM) modes. In a TM mode, the magnetic field has no longitudinal component, while the electric field has a longitudinal component.
Therefore, the defining condition for the TM mode is:
$ \boxed{H_z=0} $
and:
$ \boxed{E_z\neq0} $
The longitudinal electric field \(E_z\) is used as the starting point for deriving all the remaining field components, cutoff frequency, propagation constant, phase velocity, group velocity, guide wavelength, and TM wave impedance.
TM Mode Wave Equation in a Circular Waveguide
The electromagnetic fields inside a circular waveguide satisfy the wave equation. For the longitudinal electric field of the TM mode, the reduced wave equation is:
$ \left( \nabla_t^2+K_c^2 \right)E_z=0 $
where \(\nabla_t^2\) is the transverse Laplacian and \(K_c\) is the cutoff wave number.
In cylindrical coordinates, the transverse Laplacian is:
$ \nabla_t^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial\phi^2} $
Therefore, the TM wave equation becomes:
$ \frac{\partial^2E_z}{\partial r^2} + \frac{1}{r}\frac{\partial E_z}{\partial r} + \frac{1}{r^2} \frac{\partial^2E_z}{\partial\phi^2} + K_c^2E_z =0 $
Separating the Radial and Angular Dependence
The circular geometry suggests that the longitudinal electric field depends on both \(r\) and \(\phi\). We therefore assume a separable solution:
$ E_z(r,\phi,z) = R(r)\Phi(\phi)e^{-j\beta z} $
The factor \(e^{-j\beta z}\) represents propagation along the \(z\)-direction.
Since the \(z\)-dependence has already been separated, the transverse part can be written as:
$ E_z(r,\phi)=R(r)\Phi(\phi) $
Substituting this into the TM wave equation gives:
$ \frac{d^2R}{dr^2}\Phi + \frac{1}{r}\frac{dR}{dr}\Phi + \frac{1}{r^2}R\frac{d^2\Phi}{d\phi^2} + K_c^2R\Phi =0 $
Dividing the entire 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 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 $
The first, second, and fourth terms depend only on \(r\), while the third term depends only on \(\phi\). Therefore, each part must be equal to a separation constant.
Angular Equation
Let the separation constant be \(n^2\). The angular equation becomes:
$ \frac{d^2\Phi}{d\phi^2} + n^2\Phi =0 $
This is a standard second-order differential equation. Its general solution is:
$ \Phi(\phi) = C_1\cos(n\phi) + C_2\sin(n\phi) $
Because the physical field must have the same value after one complete revolution:
$ \Phi(\phi+2\pi)=\Phi(\phi) $
the order \(n\) must be an integer:
$ n=0,1,2,3,\ldots $
Thus, the angular dependence of the TM mode is described by:
$ \boxed{ \Phi(\phi) = C_1\cos(n\phi) + C_2\sin(n\phi) } $
Radial Equation
The remaining terms give the radial equation:
$ \frac{d^2R}{dr^2} + \frac{1}{r}\frac{dR}{dr} + \left( K_c^2-\frac{n^2}{r^2} \right)R =0 $
Multiplying by \(r^2\):
$ r^2\frac{d^2R}{dr^2} + r\frac{dR}{dr} + \left( K_c^2r^2-n^2 \right)R =0 $
This is Bessel's differential equation of order \(n\).
Therefore, its general 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.
Choosing the Physical Solution
The circular waveguide includes the center point \(r=0\). The Bessel function of the second kind \(Y_n(K_cr)\) becomes singular at \(r=0\).
A physical electromagnetic field cannot become infinite at the center of the waveguide. Therefore, the \(Y_n\) term must be discarded:
$ B=0 $
The radial field distribution is therefore:
$ \boxed{ R(r)=A J_n(K_cr) } $
Combining the radial and angular solutions, the longitudinal electric field can be written as:
$ E_z(r,\phi,z) = A J_n(K_cr) \left[ C_1\cos(n\phi) + C_2\sin(n\phi) \right] e^{-j\beta z} $
For one particular angular field orientation, we can choose the cosine solution:
$ \boxed{ E_z(r,\phi,z) = E_0J_n(K_cr)\cos(n\phi)e^{-j\beta z} } $
where \(E_0\) is an arbitrary amplitude constant.
Boundary Condition for a Perfectly Conducting Circular Wall
The radius of the circular waveguide is \(a\). Therefore, the conducting wall is located at:
$ r=a $
For a perfectly conducting wall, the tangential electric field must be zero:
$ E_t=0 $
At the circular wall, \(E_z\) is tangential to the conducting surface. Therefore:
$ E_z(a,\phi,z)=0 $
Substituting the expression for \(E_z\):
$ E_0J_n(K_ca)\cos(n\phi)e^{-j\beta z}=0 $
For a non-zero field, the amplitude and exponential terms cannot be responsible for making the field zero everywhere. Therefore, the required condition is:
$ J_n(K_ca)=0 $
The allowed values of \(K_ca\) are therefore the roots of the Bessel function:
$ J_n(X_{nm})=0 $
Hence:
$ K_ca=X_{nm} $
and therefore:
$ \boxed{ K_c=\frac{X_{nm}}{a} } $
Here, \(X_{nm}\) represents the \(m\)-th root of the Bessel function \(J_n(x)\).
Longitudinal Electric Field of TM Mode
The final longitudinal electric field expression for a TM mode in a circular waveguide is therefore:
$ \boxed{ E_z = E_0 J_n \left( \frac{X_{nm}}{a}r \right) \cos(n\phi) e^{-j\beta z} } $
The corresponding TM mode is identified as:
$ \boxed{ TM_{nm} } $
where \(n\) represents the angular variation and \(m\) identifies the radial root.
Unlike TE modes, where the boundary condition involves the derivative of the Bessel function, TM modes require the Bessel function itself to become zero at the conducting wall:
$ \boxed{ J_n(X_{nm})=0 } $
TM Mode Transverse Field Components in a Circular Waveguide
We have already obtained the longitudinal electric field for the TM mode:
$ E_z = E_0 J_n(K_cr) \cos(n\phi) e^{-j\beta z} $
where:
$ K_c=\frac{X_{nm}}{a} $
and \(X_{nm}\) is the \(m\)-th root of the Bessel function \(J_n(x)\).
For TM mode:
$ \boxed{H_z=0} $
and:
$ \boxed{E_z\neq0} $
Once \(E_z\) is known, the remaining transverse electric and magnetic field components can be obtained from Maxwell's equations.
Transverse Field Relations for TM Mode
For a wave propagating in the \(z\)-direction, the transverse electric field components for TM mode are:
$ E_t = -\frac{j\beta}{K_c^2} \nabla_t E_z $
In cylindrical coordinates:
$ \nabla_t E_z = \hat r\frac{\partial E_z}{\partial r} + \hat\phi \frac{1}{r} \frac{\partial E_z}{\partial\phi} $
Therefore:
$ E_r = -\frac{j\beta}{K_c^2} \frac{\partial E_z}{\partial r} $
and:
$ E_\phi = -\frac{j\beta}{K_c^2r} \frac{\partial E_z}{\partial\phi} $
For the transverse magnetic field:
$ H_t = \frac{j\omega\epsilon}{K_c^2} \left( \hat z\times\nabla_tE_z \right) $
This gives:
$ H_r = \frac{j\omega\epsilon}{K_c^2r} \frac{\partial E_z}{\partial\phi} $
and:
$ H_\phi = -\frac{j\omega\epsilon}{K_c^2} \frac{\partial E_z}{\partial r} $
Derivation of \(E_r\)
Starting with:
$ E_r = -\frac{j\beta}{K_c^2} \frac{\partial E_z}{\partial r} $
Substitute:
$ E_z = E_0J_n(K_cr) \cos(n\phi)e^{-j\beta z} $
Taking the derivative with respect to \(r\):
$ \frac{\partial E_z}{\partial r} = E_0 \frac{d}{dr} J_n(K_cr) \cos(n\phi)e^{-j\beta z} $
Using the Bessel-function derivative:
$ \frac{d}{dr}J_n(K_cr) = K_cJ_n'(K_cr) $
we obtain:
$ \frac{\partial E_z}{\partial r} = E_0K_cJ_n'(K_cr) \cos(n\phi)e^{-j\beta z} $
Therefore:
$ E_r = -\frac{j\beta}{K_c^2} E_0K_cJ_n'(K_cr) \cos(n\phi)e^{-j\beta z} $
Canceling one \(K_c\):
$ \boxed{ E_r = -\frac{j\beta}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $
Derivation of \(E_\phi\)
Starting with:
$ E_\phi = -\frac{j\beta}{K_c^2r} \frac{\partial E_z}{\partial\phi} $
Differentiate \(E_z\) with respect to \(\phi\):
$ \frac{\partial E_z}{\partial\phi} = E_0J_n(K_cr) \frac{d}{d\phi} \cos(n\phi) e^{-j\beta z} $
Since:
$ \frac{d}{d\phi}\cos(n\phi) = -n\sin(n\phi) $
we obtain:
$ \frac{\partial E_z}{\partial\phi} = -nE_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} $
Substituting:
$ E_\phi = -\frac{j\beta}{K_c^2r} \left[ -nE_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} \right] $
Therefore:
$ \boxed{ E_\phi = \frac{j\beta n}{K_c^2r} E_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} } $
Derivation of \(H_r\)
For the radial magnetic field:
$ H_r = \frac{j\omega\epsilon}{K_c^2r} \frac{\partial E_z}{\partial\phi} $
We already obtained:
$ \frac{\partial E_z}{\partial\phi} = -nE_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} $
Therefore:
$ H_r = \frac{j\omega\epsilon}{K_c^2r} \left[ -nE_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} \right] $
Hence:
$ \boxed{ H_r = -\frac{j\omega\epsilon n}{K_c^2r} E_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} } $
Derivation of \(H_\phi\)
For the azimuthal magnetic field:
$ H_\phi = -\frac{j\omega\epsilon}{K_c^2} \frac{\partial E_z}{\partial r} $
We have already calculated:
$ \frac{\partial E_z}{\partial r} = E_0K_cJ_n'(K_cr) \cos(n\phi)e^{-j\beta z} $
Therefore:
$ H_\phi = -\frac{j\omega\epsilon}{K_c^2} E_0K_cJ_n'(K_cr) \cos(n\phi)e^{-j\beta z} $
Canceling one \(K_c\):
$ \boxed{ H_\phi = -\frac{j\omega\epsilon}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $
Complete TM Field Components
The longitudinal components are:
$ \boxed{ E_z = E_0J_n(K_cr) \cos(n\phi)e^{-j\beta z} } $
$ \boxed{ H_z=0 } $
The transverse electric-field components are:
$ \boxed{ E_r = -\frac{j\beta}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $
$ \boxed{ E_\phi = \frac{j\beta n}{K_c^2r} E_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} } $
The transverse magnetic-field components are:
$ \boxed{ H_r = -\frac{j\omega\epsilon n}{K_c^2r} E_0J_n(K_cr) \sin(n\phi)e^{-j\beta z} } $
$ \boxed{ H_\phi = -\frac{j\omega\epsilon}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $
Thus, the complete TM mode satisfies:
$ \boxed{ H_z=0,\qquad E_z\neq0 } $
Propagation Constant and Cutoff Frequency of TM Modes
The field distribution obtained for the TM mode contains the cutoff wave number \(K_c\). To determine whether the mode can actually propagate through the circular waveguide, we need to find the relationship between \(K_c\), the operating frequency, and the propagation constant \(\beta\).
Relation Between \(K_c\), \(\beta\), and Frequency
For electromagnetic waves inside a waveguide, the wave number in the medium is:
$ k=\omega\sqrt{\mu\epsilon} $
The wave number can be separated into its transverse and longitudinal components:
$ k^2=K_c^2+\beta^2 $
Therefore:
$ \beta^2=k^2-K_c^2 $
Substituting:
$ k^2=\omega^2\mu\epsilon $
gives:
$ \beta^2 = \omega^2\mu\epsilon-K_c^2 $
Taking the square root:
$ \boxed{ \beta = \sqrt{ \omega^2\mu\epsilon-K_c^2 } } $
For the TM mode of a circular waveguide, the boundary condition gave:
$ K_c=\frac{X_{nm}}{a} $
Therefore:
$ K_c^2= \left( \frac{X_{nm}}{a} \right)^2 $
Substituting this into the propagation-constant equation:
$ \boxed{ \beta = \sqrt{ \omega^2\mu\epsilon - \left( \frac{X_{nm}}{a} \right)^2 } } $
Cutoff Condition for TM Mode
At cutoff, the wave is just at the boundary between propagation and attenuation. Therefore, the longitudinal phase constant becomes zero:
$ \beta=0 $
Starting from:
$ \beta^2 = \omega^2\mu\epsilon-K_c^2 $
putting \(\beta=0\):
$ 0 = \omega_c^2\mu\epsilon-K_c^2 $
Therefore:
$ \omega_c^2\mu\epsilon=K_c^2 $
Using:
$ K_c=\frac{X_{nm}}{a} $
we get:
$ \omega_c^2\mu\epsilon = \left( \frac{X_{nm}}{a} \right)^2 $
Taking the square root:
$ \omega_c\sqrt{\mu\epsilon} = \frac{X_{nm}}{a} $
Hence:
$ \boxed{ \omega_c = \frac{X_{nm}} {a\sqrt{\mu\epsilon}} } $
Since:
$ \omega_c=2\pi f_c $
we have:
$ 2\pi f_c = \frac{X_{nm}} {a\sqrt{\mu\epsilon}} $
Therefore, the cutoff frequency is:
$ \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} } $
Propagation Constant Above Cutoff
For propagation, the operating frequency must be greater than the cutoff frequency:
$ f>f_c $
Starting with:
$ \beta = \sqrt{ \omega^2\mu\epsilon-K_c^2 } $
At cutoff:
$ K_c^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} } $
Thus:
$ \boxed{ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } $
Since:
$ \frac{\omega_c}{\omega} = \frac{f_c}{f} $
the propagation constant can also be written as:
$ \boxed{ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } $
Three Operating Conditions for TM Modes
Case I: Below Cutoff
When the operating frequency is below the cutoff frequency:
$ f
or:
$ \omega^2\mu\epsilon
the quantity inside the square root is negative. Therefore, \(\beta\) becomes imaginary and the field does not propagate as a traveling wave.
The field instead decreases exponentially along the waveguide. This condition is called an evanescent mode.
Case II: At Cutoff
At the cutoff frequency:
$ f=f_c $
we have:
$ \omega_c^2\mu\epsilon=K_c^2 $
and therefore:
$ \boxed{\beta=0} $
The phase does not advance along the waveguide under the ideal cutoff condition.
Case III: Above Cutoff
When:
$ f>f_c $
or:
$ \omega^2\mu\epsilon>K_c^2 $
\(\beta\) becomes real and the TM mode propagates through the circular waveguide.
Therefore, the propagation condition is:
$ \boxed{ f>f_c } $
Cutoff Wavelength of TM Mode
The cutoff wavelength is related to the cutoff frequency by:
$ \lambda_c=\frac{c}{f_c} $
Using:
$ f_c=\frac{cX_{nm}}{2\pi a} $
we obtain:
$ \lambda_c = \frac{c} { \frac{cX_{nm}}{2\pi a} } $
Canceling \(c\):
$ \boxed{ \lambda_c = \frac{2\pi a}{X_{nm}} } $
Thus, the cutoff wavelength of a TM mode depends on the radius of the circular waveguide and the corresponding root of the Bessel function.
Phase Velocity and Group Velocity of TM Modes
Once the propagation constant \(\beta\) and cutoff frequency \(f_c\) are known, the phase velocity and group velocity of the TM mode can be derived directly.
Phase Velocity of TM Mode
The phase velocity is the velocity at which a constant phase of the electromagnetic wave travels along the waveguide.
It is defined as:
$ V_p=\frac{\omega}{\beta} $
For the TM mode, the propagation constant is:
$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } $
Substituting this into the phase velocity equation:
$ V_p = \frac{\omega} { \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } $
Canceling \(\omega\):
$ V_p = \frac{1} { \sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } $
Since:
$ \frac{1}{\sqrt{\mu\epsilon}}=c $
we obtain:
$ \boxed{ V_p = \frac{c} { \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } } $
Since:
$ \frac{\omega_c}{\omega} = \frac{f_c}{f} $
the phase velocity can also be written as:
$ \boxed{ V_p = \frac{c} { \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } } $
Therefore, the phase velocity of a TM mode is greater than the velocity of an electromagnetic wave in the medium.
As the operating frequency approaches cutoff:
$ f\rightarrow f_c^+ $
the denominator approaches zero and:
$ V_p\rightarrow\infty $
At frequencies much higher than cutoff:
$ f\gg f_c $
we have:
$ V_p\rightarrow c $
Group Velocity of TM Mode
The group velocity represents the velocity at which energy or information associated with the wave packet travels along the waveguide.
It is defined as:
$ V_g=\frac{d\omega}{d\beta} $
We start with the propagation relationship:
$ \beta^2 = \omega^2\mu\epsilon-K_c^2 $
Differentiating both sides with respect to \(\beta\):
$ 2\beta = 2\omega\mu\epsilon \frac{d\omega}{d\beta} $
Therefore:
$ \frac{d\omega}{d\beta} = \frac{\beta}{\omega\mu\epsilon} $
Since:
$ V_g=\frac{d\omega}{d\beta} $
we obtain:
$ V_g = \frac{\beta}{\omega\mu\epsilon} $
Using:
$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } $
we get:
$ V_g = \frac{ \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } { \omega\mu\epsilon } $
Canceling \(\omega\):
$ V_g = \frac{ \sqrt{\mu\epsilon} } { \mu\epsilon } \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } $
Since:
$ \frac{\sqrt{\mu\epsilon}}{\mu\epsilon} = \frac{1}{\sqrt{\mu\epsilon}} = c $
we obtain:
$ \boxed{ V_g = c \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } $
or, in terms of frequency:
$ \boxed{ V_g = c \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } $
Relationship Between Phase and Group Velocity
The phase velocity and group velocity have complementary 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 the two expressions:
$ V_pV_g = \frac{c} { \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } \cdot c \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } $
The square-root terms cancel:
$ \boxed{ V_pV_g=c^2 } $
Therefore:
$ \boxed{ V_g=\frac{c^2}{V_p} } $
This relationship is valid for an ideal, lossless waveguide.
Guide Wavelength of TM Mode
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 } } } $
Using the relationship:
$ \frac{f_c}{f} = \frac{\lambda_0}{\lambda_c} $
the guide wavelength can also be written as:
$ \boxed{ \lambda_g = \frac{\lambda_0} { \sqrt{ 1- \left( \frac{\lambda_0}{\lambda_c} \right)^2 } } } $
Rearranging gives the commonly used wavelength relationship:
$ \boxed{ \frac{1}{\lambda_g^2} = \frac{1}{\lambda_0^2} - \frac{1}{\lambda_c^2} } $
TM Wave Impedance in a Circular Waveguide
The wave impedance of a TM mode is defined as the ratio of the transverse electric field to the transverse magnetic field.
For the TM mode:
$ Z_{TM} = \frac{E_r}{H_\phi} = -\frac{E_\phi}{H_r} $
We can derive the TM wave impedance using the transverse field expressions obtained earlier.
Using \(E_r\) and \(H_\phi\)
The radial electric field is:
$ E_r = -\frac{j\beta}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} $
The azimuthal magnetic field is:
$ H_\phi = -\frac{j\omega\epsilon}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} $
Taking their ratio:
$ Z_{TM} = \frac{E_r}{H_\phi} $
Substituting the two field expressions:
$ Z_{TM} = \frac{ -\frac{j\beta}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} }{ -\frac{j\omega\epsilon}{K_c} E_0J_n'(K_cr) \cos(n\phi)e^{-j\beta z} } $
The common terms cancel:
$ Z_{TM} = \frac{\beta}{\omega\epsilon} $
Therefore, the TM wave impedance is:
$ \boxed{ Z_{TM} = \frac{\beta}{\omega\epsilon} } $
Substituting the Propagation Constant
For a propagating TM mode:
$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } $
Substituting this into the wave impedance expression:
$ Z_{TM} = \frac{ \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } }{ \omega\epsilon } $
Canceling \(\omega\):
$ Z_{TM} = \frac{ \sqrt{\mu\epsilon} }{ \epsilon } \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } $
Since:
$ \frac{\sqrt{\mu\epsilon}}{\epsilon} = \sqrt{\frac{\mu}{\epsilon}} $
we obtain:
$ Z_{TM} = \sqrt{\frac{\mu}{\epsilon}} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } $
The intrinsic wave impedance of the medium is:
$ \eta = \sqrt{\frac{\mu}{\epsilon}} $
Therefore:
$ \boxed{ Z_{TM} = \eta \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } } $
Since:
$ \frac{\omega_c}{\omega} = \frac{f_c}{f} $
the TM wave impedance can also be written as:
$ \boxed{ Z_{TM} = \eta \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } $
Behavior of TM Wave Impedance
The TM wave impedance changes with the operating frequency relative to the cutoff frequency.
As the operating frequency approaches cutoff from above:
$ f\rightarrow f_c^+ $
we have:
$ \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } \rightarrow0 $
Therefore:
$ \boxed{ Z_{TM}\rightarrow0 } $
At frequencies much higher than the cutoff frequency:
$ f\gg f_c $
we have:
$ \left(\frac{f_c}{f}\right)^2\rightarrow0 $
and therefore:
$ \boxed{ Z_{TM}\rightarrow\eta } $
Thus, the TM wave impedance approaches the intrinsic impedance of the medium as the operating frequency becomes much greater than the cutoff frequency.
Final TM Circular Waveguide Results
The important results obtained from the complete TM-mode derivation are:
$ \boxed{ H_z=0,\qquad E_z\neq0 } $
$ \boxed{ K_c=\frac{X_{nm}}{a} } $
where \(X_{nm}\) is a root of:
$ \boxed{ J_n(X_{nm})=0 } $
The propagation constant is:
$ \boxed{ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } $
The cutoff frequency is:
$ \boxed{ f_c = \frac{cX_{nm}}{2\pi a} } $
The cutoff wavelength is:
$ \boxed{ \lambda_c = \frac{2\pi a}{X_{nm}} } $
The phase velocity is:
$ \boxed{ V_p = \frac{c} { \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } } $
The group velocity is:
$ \boxed{ V_g = c \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } $
The guide wavelength is:
$ \boxed{ \lambda_g = \frac{\lambda_0} { \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } } $
The TM wave impedance is:
$ \boxed{ Z_{TM} = \sqrt{\frac{\mu}{\epsilon}} \sqrt{ 1- \left( \frac{\omega_c}{\omega} \right)^2 } = \eta \sqrt{ 1- \left( \frac{f_c}{f} \right)^2 } } $
where:
$ \boxed{ \eta=\sqrt{\frac{\mu}{\epsilon}} } $
is the intrinsic wave impedance of the medium.