TM Mode in Rectangular Waveguide
TM Mode in Rectangular Waveguide
In a Transverse Magnetic (TM) mode, the magnetic field component along the direction of propagation is zero while the electric field component along the direction of propagation exists.
Therefore, the defining conditions for TM mode are:
$ H_z = 0 $
$ E_z \neq 0 $
Since the longitudinal electric field component exists, the analysis of TM modes begins with the wave equation for \(E_z\).
The reduced wave equation for TM mode inside a rectangular waveguide is:
$ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + K_c^2 \right) E_z = 0 \qquad ...(1) $
where:
$ E_z(x,y,z) = E_z(x,y)e^{-j\beta z} $
and
$ K_c^2 = k^2-\beta^2 $
is the cutoff wave number.
Applying Separation of Variables
To solve Equation (1), assume that the longitudinal electric field can be expressed as the product of two independent functions:
$ E_z(x,y) = X(x)Y(y) $
Substituting into Equation (1):
$ 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) $
The first term depends only on \(x\), while the second term depends only on \(y\). Therefore, each term must be equal to a constant.
Let:
$ \frac{1}{X} \frac{d^2X}{dx^2} = -K_x^2 $
and
$ \frac{1}{Y} \frac{d^2Y}{dy^2} = -K_y^2 $
Substituting into Equation (3):
$ -K_x^2 - K_y^2 + K_c^2 = 0 $
Therefore:
$ K_x^2+K_y^2=K_c^2 \qquad ...(4) $
Separated Differential Equations
The wave equation has now been separated into two ordinary differential equations:
$ \frac{d^2X}{dx^2} + K_x^2X = 0 \qquad ...(5) $
and
$ \frac{d^2Y}{dy^2} + K_y^2Y = 0 \qquad ...(6) $
Solving the Wave Equation for TM Modes
$ \frac{d^2X}{dx^2}+K_x^2X=0 \qquad ...(5) $
$ \frac{d^2Y}{dy^2}+K_y^2Y=0 \qquad ...(6) $
Both equations are second-order linear differential equations with constant coefficients. Their solutions are sinusoidal in nature.
Solution of the X-Direction Equation
The solution of
$ \frac{d^2X}{dx^2}+K_x^2X=0 $
is:
$ X(x)=A_1\sin(K_xx)+A_2\cos(K_xx) $
where \(A_1\) and \(A_2\) are arbitrary constants.
Solution of the Y-Direction Equation
Similarly, the solution of
$ \frac{d^2Y}{dy^2}+K_y^2Y=0 $
is:
$ Y(y)=B_1\sin(K_yy)+B_2\cos(K_yy) $
where \(B_1\) and \(B_2\) are arbitrary constants.
General Solution for the Longitudinal Electric Field
Combining both solutions:
$ E_z(x,y) = \Big[ A_1\sin(K_xx)+A_2\cos(K_xx) \Big] \Big[ B_1\sin(K_yy)+B_2\cos(K_yy) \Big] $
Including the propagation factor along the z-direction:
$ E_z(x,y,z) = \Big[ A_1\sin(K_xx)+A_2\cos(K_xx) \Big] \Big[ B_1\sin(K_yy)+B_2\cos(K_yy) \Big] e^{-j\beta z} $
Applying Boundary Conditions
The walls of a rectangular waveguide are perfect conductors. Therefore, the tangential electric field must vanish at every conducting surface.
Since TM modes contain a longitudinal electric field component, the boundary condition becomes:
$ E_z=0 $
at all conducting walls.
Boundary Condition at x = 0
Substituting \(x=0\):
$ E_z(0,y) = \Big[ A_1\sin(0)+A_2\cos(0) \Big] \Big[ B_1\sin(K_yy)+B_2\cos(K_yy) \Big] $
Since:
$ \sin(0)=0 \qquad \cos(0)=1 $
we obtain:
$ A_2=0 $
Boundary Condition at x = a
Applying \(E_z=0\) at \(x=a\):
$ A_1\sin(K_xa)=0 $
For a non-trivial solution:
$ \sin(K_xa)=0 $
which gives:
$ K_xa=m\pi $
Therefore:
$ K_x=\frac{m\pi}{a} $
where:
$ m=1,2,3,\ldots $
Boundary Condition at y = 0
Applying \(E_z=0\) at \(y=0\):
$ B_2=0 $
Boundary Condition at y = b
Applying \(E_z=0\) at \(y=b\):
$ B_1\sin(K_yb)=0 $
For a non-trivial solution:
$ \sin(K_yb)=0 $
which gives:
$ K_yb=n\pi $
Therefore:
$ K_y=\frac{n\pi}{b} $
where:
$ n=1,2,3,\ldots $
TMmn Field Components in a Rectangular Waveguide
From the boundary conditions derived in the previous section, we obtained:
$ K_x=\frac{m\pi}{a} $
$ K_y=\frac{n\pi}{b} $
where:
$ m=1,2,3,\ldots $
$ n=1,2,3,\ldots $
Substituting these values into the general solution of the longitudinal electric field gives the final TM mode expression:
$ E_z(x,y,z) = E_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
where \(E_{mn}\) is an arbitrary amplitude constant.
Transverse Electric Field Components
Using Maxwell's curl equations, the transverse electric field components become:
$ E_x = -\frac{j\beta}{K_c^2} \frac{m\pi}{a} E_{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\beta}{K_c^2} \frac{n\pi}{b} E_{mn} \sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
Transverse Magnetic Field Components
Similarly, the transverse magnetic field components are:
$ H_x = \frac{j\omega\epsilon}{K_c^2} \frac{n\pi}{b} E_{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\omega\epsilon}{K_c^2} \frac{m\pi}{a} E_{mn} \cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
Longitudinal Magnetic Field
For TM mode, the longitudinal magnetic field component is always zero:
$ H_z=0 $
Cutoff Wave Number Relation
From the separation of variables procedure:
$ K_x^2+K_y^2=K_c^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 fundamental cutoff-wave-number equation for TM modes in a rectangular waveguide.
Relation Between Cutoff Wave Number and Propagation Constant
For guided electromagnetic waves:
$ K_c^2 = \gamma^2+\omega^2\mu\epsilon $
Substituting the cutoff-wave-number expression:
$ \gamma^2 = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 - \omega^2\mu\epsilon $
Therefore:
$ \gamma = \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 - \omega^2\mu\epsilon } $
Since:
$ \gamma=\alpha+j\beta $
the propagation behavior of the TM mode depends on the operating frequency relative to the cutoff frequency.
Propagation Characteristics of TM Modes in a Rectangular Waveguide
Case I: Below Cutoff Frequency
When the operating frequency is lower than the cutoff frequency:
$ \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 $
which means:
$ \alpha = \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 - \omega^2\mu\epsilon } $
Under this condition, no travelling wave exists inside the waveguide.
The electromagnetic field decays exponentially along the direction of propagation:
$ e^{-\alpha z} $
Hence the waveguide behaves as a blocking structure and does not transfer useful power.
Case II: Cutoff Condition
At the cutoff frequency:
$ \omega_c^2\mu\epsilon = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $
the propagation constant becomes:
$ \gamma=0 $
Therefore:
$ \alpha=0 $
and
$ \beta=0 $
This represents the transition point between attenuation and propagation.
Rearranging:
$ \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 } $
This equation determines the minimum frequency required for TMmn wave propagation.
Case III: Above Cutoff Frequency
When the operating frequency exceeds the cutoff frequency:
$ \omega^2\mu\epsilon > \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $
the quantity under the square root becomes negative.
Therefore:
$ \gamma = j\beta $
and
$ \alpha=0 $
For an ideal lossless waveguide there is no attenuation.
The phase constant becomes:
$ \beta = \sqrt{ \omega^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } $
Thus electromagnetic waves propagate through the guide with no attenuation.
The rectangular waveguide therefore behaves as a high-pass filter because only frequencies above the cutoff frequency can propagate.
TM Mode Propagation Conditions
Below Cutoff:
$ f
$ \gamma=\alpha $
$ \beta=0 $
No wave propagation occurs.
At Cutoff:
$ f=f_c $
$ \alpha=0 $
$ \beta=0 $
Transition point between attenuation and propagation.
Above Cutoff:
$ f>f_c $
$ \gamma=j\beta $
$ \alpha=0 $
Cutoff Wavelength (\(\lambda_c\))
The cutoff wavelength is related to 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}} $
we obtain:
$ \lambda_c = \frac{2} { \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } } $
This is the cutoff wavelength for the TMmn mode.
Phase Velocity (\(V_p\))
The phase velocity is defined as:
$ V_p=\frac{\omega}{\beta} $
For frequencies above cutoff:
$ \beta = \sqrt{ \omega^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } $
Substituting into the phase velocity equation:
$ V_p = \frac{\omega} { \sqrt{ \omega^2\mu\epsilon - \left[ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 \right] } } $
Using the cutoff condition:
$ \omega_c^2\mu\epsilon = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $
gives:
$ V_p = \frac{\omega} { \sqrt{ \omega^2\mu\epsilon - \omega_c^2\mu\epsilon } } $
Factoring \(\omega^2\mu\epsilon\):
$ 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}} $
therefore:
$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $
This shows that the phase velocity inside a waveguide is always greater than the velocity of light.
Guide Wavelength (\(\lambda_g\))
Guide wavelength is defined as:
$ \lambda_g=\frac{V_p}{f} $
Substituting the 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 } } $
Using:
$ \frac{f_c}{f} = \frac{\lambda_0}{\lambda_c} $
the guide wavelength becomes:
$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{\lambda_0}{\lambda_c}\right)^2 } } $
On simplification:
$ \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 is one of the most important waveguide relationships used in numerical problems.
TM Wave Impedance
For TM modes, the wave impedance is defined as:
$ Z_{TM} = \frac{E_x}{H_y} = -\frac{E_y}{H_x} = Z_g $
Substituting the TM field expressions:
$ Z_{TM} = \frac{\beta}{\omega\epsilon} $
Using:
$ \beta = \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } $
gives:
$ Z_{TM} = \frac{ \omega\sqrt{\mu\epsilon} \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } } {\omega\epsilon} $
Therefore:
$ Z_{TM} = \sqrt{\frac{\mu}{\epsilon}} \sqrt{ 1-\left(\frac{\omega_c}{\omega}\right)^2 } $
Since:
$ \eta=\sqrt{\frac{\mu}{\epsilon}} $
is the intrinsic impedance of the medium,
the final TM wave impedance becomes:
$ Z_{TM} = \eta \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $
Final Results for TM Modes
$ f_c = \frac{1}{2\pi\sqrt{\mu\epsilon}} \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } $
$ \lambda_c = \frac{2} { \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } } $
$ V_p = \frac{c} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $
$ \lambda_g = \frac{\lambda_0} { \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } } $
$ Z_{TM} = \eta \sqrt{ 1-\left(\frac{f_c}{f}\right)^2 } $