Related Concepts: 5.05 Dispersion, Phase & Group Velocity, Doppler Effect & Brewster’s Angle | 5.06 Radio Wave Propagation Modes & Ionospheric Effects | 5.07 Solved PYQ Numerical Bank - Waves & Propagation

5.08 Rectangular Waveguides & Wave Confinement

Study Checklist & Core Objectives

• Objective 1: Explain why single-conductor hollow waveguides cannot propagate Transverse Electromagnetic (TEM) waves.
• Objective 2: Solve the wave equation inside a hollow conductor under conducting boundary conditions to define TE and TM modes.
• Objective 3: Master the dominant $T{E}_{10}$ mode fields and calculate its cutoff frequency ($f_c$) and cutoff wavelength (${\lambda }_c$).
• Objective 4: Derive and calculate waveguide dispersion parameters (${\beta }_g$, ${\lambda }_g$, $u_p$, $u_g$, and $Z_{TE}$).
• Objective 5: Prove the high-yield relation $u_p{u}_g=u^2$.
• Objective 6: Solve typical exam numericals on rectangular waveguide dimensions and mode propagation boundaries.

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1. The Physics of Wave Confinement

Alright — up to this point, we have analyzed electromagnetic waves propagating in open, infinite media (like free space or lossy dielectrics). In those cases, the wave is transverse electromagnetic (TEM), meaning both the electric field $\vec{E}$ and magnetic field $\vec{H}$ are entirely perpendicular to the direction of propagation.

However, when we want to guide waves from one point to another at high frequencies (specifically microwave frequencies above $3\text{ GHz}$), open transmission lines suffer from massive radiation losses. To solve this, we use a waveguide {a hollow metallic tube of uniform cross-section used to guide electromagnetic energy}.

       ===================================  <- Metallic Wall (Conductor)
       ---->  ---->  ---->  ---->  ---->   <- Wave bounces between walls
       ===================================  <- Metallic Wall (Conductor)

The TEM Wave Impossibility Theorem

A single-conductor hollow waveguide cannot support TEM wave propagation.

• Mathematical Proof: For a TEM wave, the axial fields are zero: $E_z=0$ and $H_z=0$. Since $\nabla \cdot \vec{E}=0$ inside the hollow guide (charge-free region) and $\nabla \times \vec{E}=-j\omega \mu \vec{H}=0$ in the transverse plane, the transverse electric field behaves as a static field. This means $\vec{E}$ can be written as the gradient of a scalar potential, $\vec{E}=-{\nabla }_{t}V$, which satisfies Laplace’s equation: ${\nabla }_t^{2}V=0$ On the metallic boundary, the tangential electric field is zero, which forces the potential $V$ on the wall to be constant ($V=V_0$). According to the uniqueness theorem of electrostatics, a scalar potential that satisfies Laplace’s equation and is constant on the bounding surface must be constant everywhere inside that volume. Therefore, $V=V_0$ everywhere inside the waveguide, yielding: $\vec{E}=-{\nabla }_{t}V\equiv 0$ With no electric field, no wave can exist!

• Physical Intuition: Electric field lines must begin on positive charges and end on negative charges. In a single hollow tube, there is no second inner conductor (like in a coaxial cable) to hold the opposite charge. If a transverse electric field line tried to form, it would have to start on the wall and end on the wall. In a charge-free hollow space, this would violate Gauss’s Law because the field line would have to form a closed loop, which is physically impossible for a conservative field.

Therefore, waves inside a waveguide must propagate in modes that have an axial field component. This splits wave propagation into two families:

1. Transverse Electric (TE) Modes: $E_z=0,H_z\ne 0$ (electric field is entirely transverse to propagation direction).
2. Transverse Magnetic (TM) Modes: $H_z=0,E_z\ne 0$ (magnetic field is entirely transverse to propagation direction).

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2. Rectangular Waveguide Geometry & TE/TM Modes

Let’s look at a standard rectangular waveguide. We align it along the $z$-axis (propagation direction), with the inner dimensions representing a width $a$ along the $x$-axis and a height $b$ along the $y$-axis. By convention, we design waveguides such that $a>b$.

       y ^
         |      +-----------------------+
         |      |                       |
       b |      |   Hollow Space        |
         |      |   (Dielectric or Air) |
         |      |                       |
         |      +-----------------------+
         +-----------------------------------> x
        0       a

Terminology & Concept Breakdown

• Boundary Conditions: Since the walls of the waveguide are made of a perfect conductor, the tangential component of the electric field (${\vec{E}}_{tan}$) and the normal component of the magnetic flux density (${\vec{B}}_{nor}$) must vanish on all four inner walls:
  • $E_y=0,E_z=0$ at $x=0,a$
  • $E_x=0,E_z=0$ at $y=0,b$
• Mode Indices ($m,n$): The integers $m$ and $n$ represent the number of half-wave variations (standing wave patterns) of the fields along the $x$-axis (width $a$) and $y$-axis (height $b$), respectively.
  • For example, a $T{E}_{10}$ mode has one half-sine wave variation along $x$ and zero variation along $y$.

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3. The Dominant $T{E}_{10}$ Mode Fields

To solve for the fields inside the guide, we solve the Helmholtz wave equation for the longitudinal field component: ${\nabla }^2{H}_z+k^2{H}_z=0$ Using the method of separation of variables and applying boundary conditions, we find the general expression for the cutoff wave number ($k_c$): $k_c^2={\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2$

A. Cutoff Frequency ($f_c$) & Cutoff Wavelength (${\lambda }_c$)

The cutoff frequency ($f_c$) {the critical threshold frequency below which a mode cannot propagate and instead decays exponentially} is given by: $f_{c,mn}=\frac{u}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}[\text{Hz}]$ where $u=\frac{1}{\sqrt{\mu ϵ}}$ is the speed of light in the dielectric medium filling the waveguide.

The corresponding cutoff wavelength (${\lambda }_c$) is: ${\lambda }_{c,mn}=\frac{u}{f_{c,mn}}=\frac{2}{\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}}[\text{m}]$

Why the $T{E}_{10}$ Mode is Dominant

The dominant mode is defined as the mode with the lowest cutoff frequency. Since we design waveguides with $a>b$ (typically $a=2b$), let’s compare the lowest order modes:

• For $T{E}_{10}$ ($m=1,n=0$): $f_{c,10}=\frac{u}{2a}$
• For $T{E}_{01}$ ($m=0,n=1$): $f_{c,01}=\frac{u}{2b}$
• For $T{E}_{20}$ ($m=2,n=0$): $f_{c,20}=\frac{u}{a}$

Since $a>b$, we have $\frac{u}{2a}<\frac{u}{2b}$. Thus, $T{E}_{10}$ has the absolute lowest cutoff frequency and acts as the dominant mode. Operating in the dominant mode ensures single-mode propagation, preventing signal distortion from multiple modes traveling at different velocities.

For the dominant $T{E}_{10}$ mode: $f_{c,10}=\frac{u}{2a},{\lambda }_{c,10}=2a$

B. $T{E}_{10}$ Field Equations

The non-zero electric and magnetic field phasor components for the $T{E}_{10}$ mode propagating in the $+z$ direction are given in terms of the peak electric field amplitude $E_{y0}$ by: $E_y=E_{y0}\sin \left(\frac{\pi x}{a}\right)e^{-j{\beta }_{g}z}$ $H_x=-\frac{E_{y0}}{Z_{TE}}\sin \left(\frac{\pi x}{a}\right)e^{-j{\beta }_{g}z}$ $H_z=j\frac{\pi E_{y0}}{\omega \mu a}\cos \left(\frac{\pi x}{a}\right)e^{-j{\beta }_{g}z}$ (All other components: $E_x=0,E_z=0,H_y=0$)

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4. Guide Parameters & Dispersion (The $u_p{u}_g=u^2$ Proof)

When a wave propagates inside a waveguide, it does not travel in a straight line; instead, it bounces obliquely off the conducting walls. This bouncing geometry alters the propagation constant along the guide axis ($z$-axis).

To simplify our formulas, we define the dimensionless dispersion factor ($k_d$): $k_d=\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$

A. Summary of Guide Parameters

Parameter	Formula (LaTeX)	Physical Description
Guide Phase Constant	${\beta }_g=\beta \sqrt{1-{\left(\frac{f_c}{f}\right)}^2}=\beta k_d$	Phase variation rate along the guide ($z$-axis).
Guide Wavelength	${\lambda }_g=\frac{\lambda }{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}=\frac{\lambda }{k_d}$	Distance along the guide axis for a 2$\pi$ phase shift. Since $k_d<1$, ${\lambda }_g>\lambda$.
Phase Velocity	$u_p=\frac{\omega }{{\beta }_g}=\frac{u}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}=\frac{u}{k_d}$	Velocity of the constant phase fronts along the wall. Since $k_d<1$, $u_p>u$ (exceeds speed of light).
Group Velocity	$u_g=\frac{d\omega }{d{\beta }_g}=u\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}=u{k}_d$	Actual velocity of energy and information propagation. Since $k_d<1$, $u_g<u$.
TE Wave Impedance	$Z_{TE}=\frac{\eta }{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}=\frac{\eta }{k_d}$	Ratio of transverse fields ($E_y/-H_x$). Since $k_d<1$, $Z_{TE}>\eta$.
TM Wave Impedance	$Z_{TM}=\eta \sqrt{1-{\left(\frac{f_c}{f}\right)}^2}=\eta k_d$	Ratio of transverse fields ($E_x/H_y$). Since $k_d<1$, $Z_{TM}<\eta$.

Understanding $u_p>u$

The phase velocity $u_p$ measures the speed of the wave fronts intersecting the guide wall. It is a geometric velocity (like the speed of an intersection point of diagonal scissors closing) and carries no energy or information. Thus, having $u_p>c$ in free space does not violate Einstein’s theory of special relativity. The energy velocity is represented by the group velocity $u_g$, which is always less than the speed of light ($u_g<c$).

B. Master Derivation Proof: $u_p{u}_g=u^2$

[Highly High-Yield for Written Exams]

Step 1: Write the expressions for guide phase velocity ($u_p$) and group velocity ($u_g$): $u_p=\frac{u}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$ $u_g=u\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$

Step 2: Multiply both velocities together: $u_p\cdot u_g=\left(\frac{u}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\right)\cdot \left(u\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}\right)$

Step 3: The square root terms in the numerator and denominator cancel out perfectly: $u_p\cdot u_g=u\cdot u=u^2$

If the waveguide is filled with air/vacuum ($u=c$): $\boxed{u_p\cdot u_g=c^2}$ (Q.E.D.)

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5. Attenuation and Power Losses

A perfect waveguide would guide waves indefinitely. However, real waveguides experience losses due to two factors:

1. Dielectric Losses (${\alpha }_d$): Due to dissipation in the medium filling the guide.
2. Conductor Losses (${\alpha }_c$): Due to ohmic heat losses from currents flowing on the finite-conductivity walls.

A. Power Flow inside Guide ($P_{avg}$)

The average power transmitted along the guide is found by integrating the time-average Poynting vector over the guide cross-section: $P_{avg}={\int }_S{\vec{P}}_{avg}\cdot d\vec{s}=\frac{1}{2}{\int }_{x=0}^a{\int }_{y=0}^b\text{Re}\left(E_x{H}_y^{\ast }-E_y{H}_x^{\ast }\right)dydx$ For the dominant $T{E}_{10}$ mode, substituting the field equations yields: $P_{avg}=\frac{1}{2}{\int }_{x=0}^a{\int }_{y=0}^b\frac{|E_{y0}{|}^2}{Z_{TE}}{\sin }^2\left(\frac{\pi x}{a}\right)dydx=\frac{ab|E_{y0}{|}^2}{4Z_{TE}}[\text{W}]$

B. Conductor Wall Loss Attenuation (${\alpha }_c$)

The surface currents flowing on the guide walls satisfy ${\vec{J}}_s={\hat{a}}_n\times \vec{H}$. Since the walls have a finite surface resistance $R_s=\sqrt{\omega \mu /2{\sigma }_w}$, the power lost per unit length ($P_L$) due to ohmic wall heating is: $P_L=\frac{R_s}{2}{\oint }_{contour}|{\vec{J}}_s{|}^{2}dl$ The attenuation constant due to conductor losses (${\alpha }_c$) is then: ${\alpha }_c=\frac{P_L}{2P_{avg}}[\text{Np/m}]$

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6. Common Mistakes That Cost Marks

Avoid these exam pitfalls:

1. Using free-space speed ($c$) blindly: If the waveguide is filled with a dielectric material of relative permittivity ${ϵ}_r$ (e.g., polyethylene), the wave speed is $u=c/\sqrt{{ϵ}_r}$. You must use $u$ (not $c$) to calculate the cutoff frequency $f_c=u/2a$.
2. swapping $Z_{TE}$ and $Z_{TM}$ formulas: Remember that $Z_{TE}=\eta /k_d$ (which is $>\eta$) and $Z_{TM}=\eta k_d$ (which is $<\eta$). An easy memory trick is that TE waves have an electric field transverse to propagation, behaving like a series impedance expansion (TE is divided by $k_d$).
3. Forgetting to verify propagation ($f>f_c$): Always check if the operating frequency is above the cutoff frequency. If $f<f_c$, the wave cannot propagate. In this case, the guide parameter term $\sqrt{1-(f_c/f{)}^2}$ becomes imaginary, indicating an evanescent wave {a wave that attenuates exponentially without phase shift}.

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7. Worked PYQ Numerical Bank

Let’s walk through typical exam-style rectangular waveguide problems step-by-step.

Problem 1: X-Band Guide Parameters

A hollow rectangular waveguide with dimensions $a=2.286\text{ cm}$ and $b=1.016\text{ cm}$ is filled with air and operates in the dominant $T{E}_{10}$ mode at a frequency of $10\text{ GHz}$.

Calculate:

1. The cutoff frequency ($f_c$).
2. The guide wavelength (${\lambda }_g$).
3. The phase velocity ($u_p$) and group velocity ($u_g$).
4. The wave impedance ($Z_{TE}$).

Step-by-Step Solution:

1. Cutoff Frequency ($f_c$): The guide is air-filled, so $u=c\approx 3\times 10^8\text{ m/s}$. The width is $a=2.286\text{ cm}=0.02286\text{ m}$. $f_{c,10}=\frac{c}{2a}=\frac{3\times 10^8}{2\times 0.02286}=6.56\times 10^9\text{ Hz}=6.56\text{ GHz}$ Since the operating frequency $f=10\text{ GHz}>6.56\text{ GHz}$, the wave propagates.

2. Dispersion Factor ($k_d$): $k_d=\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}=\sqrt{1-{\left(\frac{6.56}{10}\right)}^2}=\sqrt{1-0.4303}\approx 0.755$

3. Guide Wavelength (${\lambda }_g$): First, find the free-space wavelength $\lambda$ at $10\text{ GHz}$: $\lambda =\frac{c}{f}=\frac{3\times 10^8}{10\times 10^9}=0.03\text{ m}=3\text{ cm}$ Now compute the guide wavelength: ${\lambda }_g=\frac{\lambda }{k_d}=\frac{3\text{ cm}}{0.755}\approx 3.97\text{ cm}$

4. Phase Velocity ($u_p$) and Group Velocity ($u_g$): $u_p=\frac{c}{k_d}=\frac{3\times 10^8}{0.755}\approx 3.97\times 10^8\text{ m/s}$ $u_g=c\cdot k_d=(3\times 10^8)\times 0.755\approx 2.27\times 10^8\text{ m/s}$ Sanity Check: $u_p\cdot u_g=(3.97\times 10^8)\cdot (2.27\times 10^8)\approx 9\times 10^{16}=c^2$. Correct!

5. Wave Impedance ($Z_{TE}$): In air, ${\eta }_0\approx 120\pi \approx 377\Omega$. $Z_{TE}=\frac{{\eta }_0}{k_d}=\frac{377}{0.755}\approx 499.3\Omega$

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Problem 2: Higher Order Modes & Propagation Sizing

An air-filled rectangular waveguide has dimensions $a=4\text{ cm}$ and $b=2\text{ cm}$. A signal of frequency $5\text{ GHz}$ is launched into the guide. Determine which of the following modes can propagate in the guide: $T{E}_{10}$, $T{E}_{01}$, $T{E}_{20}$, and $T{M}_{11}$.

Step-by-Step Solution:

Write the general cutoff frequency formula for an air-filled guide ($u=c=3\times 10^{10}\text{ cm/s}$): $f_{c,mn}=\frac{c}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$ Substitute $a=4\text{ cm}$ and $b=2\text{ cm}$: $f_{c,mn}=\frac{3\times 10^{10}}{2}\sqrt{{\left(\frac{m}{4}\right)}^2+{\left(\frac{n}{2}\right)}^2}=15\sqrt{\frac{m^2}{16}+\frac{n^2}{4}}[\text{GHz}]$

Calculate cutoff frequencies for each mode:

1. $T{E}_{10}$ Mode ($m=1,n=0$): $f_{c,10}=15\sqrt{\frac{1}{16}+0}=\frac{15}{4}=3.75\text{ GHz}$
2. $T{E}_{01}$ Mode ($m=0,n=1$): $f_{c,01}=15\sqrt{0+\frac{1}{4}}=\frac{15}{2}=7.50\text{ GHz}$
3. $T{E}_{20}$ Mode ($m=2,n=0$): $f_{c,20}=15\sqrt{\frac{4}{16}+0}=15\sqrt{\frac{1}{4}}=7.50\text{ GHz}$
4. $T{M}_{11}$ Mode ($m=1,n=1$): $f_{c,11}=15\sqrt{\frac{1}{16}+\frac{1}{4}}=15\sqrt{\frac{5}{16}}=\frac{15\sqrt{5}}{4}\approx 8.385\text{ GHz}$

Mode Propagation Evaluation: A mode can propagate if and only if the operating frequency is greater than its cutoff frequency ($f>f_c$). The operating frequency is $f=5\text{ GHz}$.

• For $T{E}_{10}$: $f=5\text{ GHz}>f_c=3.75\text{ GHz}⟹$ Propagates
• For $T{E}_{01}$: $f=5\text{ GHz}<f_c=7.50\text{ GHz}⟹$ Evanescent (Blocked)
• For $T{E}_{20}$: $f=5\text{ GHz}<f_c=7.50\text{ GHz}⟹$ Evanescent (Blocked)
• For $T{M}_{11}$: $f=5\text{ GHz}<f_c=8.385\text{ GHz}⟹$ Evanescent (Blocked)

Conclusion: Only the dominant $T{E}_{10}$ mode can propagate through the waveguide at $5\text{ GHz}$.

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8. Self-Check Before Moving On

• Can you prove mathematically why TEM waves cannot propagate inside a single hollow conductor?
• Can you explain the physical meaning of mode indices $m$ and $n$ in a rectangular waveguide?
• Can you define the dominant mode and explain why $T{E}_{10}$ is dominant when $a>b$?
• Can you calculate the cutoff frequency and cutoff wavelength for the dominant mode of a given guide?
• Can you derive the high-yield relation $u_p\cdot u_g=u^2$?
• Can you explain why $u_p>c$ does not violate the theory of relativity?
• Can you write the expressions for guide parameters (${\beta }_g,{\lambda }_g,Z_{TE}$) and use them in numerical calculations?
• Can you evaluate which modes will propagate in a waveguide given its dimensions and operating frequency?

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Source: ECE 2105 Syllabus, Matthew Sadiku’s Elements of Electromagnetics (Ch. 11), David K. Cheng’s Field and Wave Electromagnetics (Ch. 10), and Lecture Slides L 14 & L 15.