Q7Electromagnetics Waves

Question

Q.7 Derive an expression for the fields in a rectangular waveguide in the case of a Transverse Magnetic (TM) wave.

Answer 1

Governing equation: for TM modes (Hz = 0, Ez ≠ 0), the longitudinal electric field component Ez must satisfy the scalar Helmholtz equation in the waveguide's transverse cross-section:

(\frac{\partial ^{2}}{\partial x ^{2}} + \frac{\partial ^{2}}{\partial y ^{2}} + k_{c}^{2}) E_{z} = 0, k_{c}^{2} = k^{2} - β^{2}

Boundary conditions: since the tangential E-field must vanish on a perfectly conducting wall, and Ez is itself tangential to all four waveguide walls (x=0, x=a, y=0, y=b), the boundary condition for TM modes is Ez = 0 on all four walls — a homogeneous Dirichlet condition (unlike TE modes, where the corresponding condition on Hz is a Neumann condition, ∂Hz/∂n = 0).

Separation of variables: assume Ez(x,y,z) = X(x)Y(y)e^{−jβz}. Substituting into the Helmholtz equation and dividing through by XY separates the equation into two independent ordinary differential equations, X″/X = −kx² and Y″/Y = −ky², with kx²+ky² = kc². The general solutions are X(x) = A sin(kxx) + B cos(kxx) and Y(y) = C sin(kyy) + D cos(kyy). Applying Ez=0 at x=0 forces B=0; applying Ez=0 at y=0 forces D=0; applying Ez=0 at x=a forces kx = mπ/a (m=1,2,3,...; m=0 gives the trivial null solution, which is why TM modes require m≥1); applying Ez=0 at y=b forces ky = nπ/b (n=1,2,3,...; likewise n≥1 required) — confirming that TM modes require both indices nonzero (no TMm0 or TM0n modes exist, unlike TE modes).

E_{z} = E_{0} sin (\frac{mπ x}{a}) sin (\frac{nπ y}{b}) e^{- j β z}, k_{c}^{2} = (\frac{mπ}{a})^{2} + (\frac{nπ}{b})^{2}

Transverse field components: with Ez known and Hz=0, the transverse components follow from the standard TM-mode field relations (derived generally from Maxwell's equations decomposed into transverse/longitudinal parts, the same decomposition referenced in Part B, Q.1 of the companion 2024 Main/Back paper):

E_{x} = \frac{- j β}{k _{c}^{2}} \frac{\partial E _{z}}{\partial x}, E_{y} = \frac{- j β}{k _{c}^{2}} \frac{\partial E _{z}}{\partial y}, H_{x} = \frac{j ω ε}{k _{c}^{2}} \frac{\partial E _{z}}{\partial y}, H_{y} = \frac{- j ω ε}{k _{c}^{2}} \frac{\partial E _{z}}{\partial x}

Substituting the known Ez solution and differentiating gives the complete field pattern of any TMmn mode in closed form; the corresponding cutoff frequency is fc = (c/2)√[(m/a)²+(n/b)²] — the same formula as for TEmn, confirming the TE/TM degeneracy for m,n≥1 discussed in Part B, Q.6 of this same paper — and the lowest-order TM mode overall is TM11, since m=0 or n=0 is excluded for TM modes by the boundary conditions derived above.

Q7Electromagnetics Waves

Question

Answer