Power Normalization
The normalization constant is derived from the power flow condition:
\[P = \frac{1}{2} \text{Re} \int_0^a \int_0^b (\mathbf{E} \times \mathbf{H}^*) \cdot \mathbf{\hat{z}} dx dy = 1 \text{ Watt}\]
Power normalization in Rectangular Waveguides
For TE Modes
Using the field components (with the factors implemented in the package):
\[ E_x = F_0 \frac{\omega\mu}{k_c^2}\frac{mx}{a}\sin \left(\frac{mx}{a}\right) \cos\left(\frac{ny}{b}\right), \quad E_y = -F_0 \frac{\omega\mu}{k_c^2}\frac{ny}{b}{\epsilon} F_0 \cos\left(\frac{mx}{a}\right)\sin\left(\frac{ny}{b}\right)\]
\[ H_x = -F_0 \frac{\beta}{k_c^2}\frac{ny}{b}{\epsilon} F_0 \cos\left(\frac{mx}{a}\right)\sin\left(\frac{ny}{b}\right), \quad H_y = F_0 \frac{\beta}{k_c^2}\frac{mx}{a}\sin \left(\frac{mx}{a}\right) \cos\left(\frac{ny}{b}\right)\]
The z-component of the Poynting vector is:
\[S_z = \frac{1}{2} (E_x H_y^* - E_y H_x^*)\]
Substituting and simplifying:
\[S_z = \frac{1}{2} \frac{\beta F_0^2 \omega \mu}{k_c^4} \left[k_y^2 \sin^2(k_y y)\cos^2(k_x x) + k_x^2 \cos^2(k_y y)\sin^2(k_x x)\right]\]
The power is
\[P_z = \int_{0}^a\int_0^b \frac{1}{2} \frac{\beta F_0^2 \omega \mu}{k_c^4} \left[k_y^2 \sin^2(k_y y)\cos^2(k_x x) + k_x^2 \cos^2(k_y y)\sin^2(k_x x)\right]\]
Integrating over the cross-section and using the orthogonality of trigonometric functions:
\[\int_0^a \sin^2(k_x x) dx = \frac{a}{2}, \quad \int_0^a \cos^2(k_x x) dx = \frac{a}{2}\]
\[\int_0^b \cos^2(k_y y) dy = \frac{b}{2}, \quad \int_0^b \sin^2(k_y y) dy = \frac{b}{2}\]
Defining:
\[\delta_m = \begin{cases} 1 & \text{if } n \geq 1 \\ 2 & \text{if } n = 0 \end{cases}, \quad \delta_n = \begin{cases} 1 & \text{if } m \geq 1 \\ 2 & \text{if } m = 0 \end{cases}\]
We get:
\[P_z = \frac{1}{2} \frac{\beta F_0^2 \omega \mu}{k_c^4} \left( \int_{0}^a\int_0^b k_y^2 \sin^2(k_y y)\cos^2(k_x x) + \int_{0}^a\int_0^b k_x^2 \cos^2(k_y y)\sin^2(k_x x) \right)\]
\[P_z = \frac{1}{2} \frac{\beta F_0^2 \omega \mu}{k_c^4} \left( \delta_m k_y^2\frac{ab}{4} + \delta_n k_x^2 \frac{ab}{4}\right) = = \frac{1}{2} \frac{\beta F_0^2 \omega \mu}{k_c^4} \left( \delta_m \frac{n^2\pi^2}{b^2} \frac{ab}{4} + \delta_n \frac{m^2\pi^2}{a^2}\frac{ab}{4} \right)\]
\[P_z = \frac{1}{2} \frac{\beta F_0^2 \omega \mu \pi^2}{4k_c^4} \left( \delta_m n^2 \frac{a}{b} + \delta_n m^2\frac{b}{a} \right)\]
Setting $P = 1$ and solving for $F_0$:
\[F_0 = \sqrt{\frac{8k_c^4}{\beta \omega \mu \pi^2 \left( \delta_m n^2 \frac{a}{b} + \delta_n m^2\frac{b}{a} \right)}}\]
For TM modes:
\[F_0 = \sqrt{\frac{8k_c^4}{\beta \omega \varepsilon \pi^2 \left( \delta_m n^2 \frac{a}{b} + \delta_n m^2\frac{b}{a} \right)}}\]
Power normalization in Circular and Coaxial Waveguides
Field Representation
Let's define the potential function as
\[\psi(\rho) = A_m J_m(k_c\rho) + B_m Y_m(k_c\rho)\]
with derivative
\[\frac{\partial\psi}{\partial \rho} = \psi^{\prime}(\rho) = k_c \left( A_m J_m'(k_c\rho) + B_m Y_m'(k_c\rho) \right)\]
For TM modes, the longitudinal electric field is:
\[H_z = \psi(\rho) \cos(m\theta)\]
The transverse field components are:
\[E_\rho = -\frac{j\beta}{k_c^2} \psi^{\prime}(\rho) \cos(m\theta), \quad E_\theta = \frac{j\beta}{k_c^2} \frac{m}{\rho} \psi(\rho) \sin(m\theta)\]
\[H_\rho = -\frac{j\omega\varepsilon}{k_c^2} \frac{m}{\rho} \psi(\rho) \sin(m\theta), \quad H_\theta = -\frac{j\omega\varepsilon}{k_c^2} \psi^{\prime}(\rho) \cos(m\theta)\]
Power Flow
The time-average power density is:
\[S_z = \frac{1}{2} \left( E_\rho H_\theta^* - E_\theta H_\rho^*\right)\]
Substituting the field components:
\[S_z = \frac{1}{2}\frac{\beta\omega\varepsilon}{k_c^4} \left( {\psi^\prime}^2\text{cos}^2\left( m\theta\right) + \frac{m^2}{\rho^2}{\psi}^2\text{sin}^2\left( m\theta\right) \right)\]
The total power is
\[P = \frac{1}{2}\frac{\beta\omega\varepsilon}{k_c^4}\int_b^a \int_0^{2\pi} \left( {\psi^\prime}^2\text{cos}^2\left( m\theta\right) + \frac{m^2}{\rho^2}{\psi}^2\text{sin}^2\left( m\theta\right) \right) \,\rho\, d\rho d\theta\]
Angular integration
The angular integrals depend on the azimuthal mode number m:
\[\int_0^{2\pi} \cos^2(m\theta) \, d\theta = \begin{cases} \pi & \text{if } m \geq 1 \\ 2\pi & \text{if } m = 0 \end{cases}, \quad \int_0^{2\pi} \sin^2(m\theta) \, d\theta = \begin{cases} \pi & \text{if } m \geq 1 \\ 0 & \text{if } m = 0 \end{cases}\]
Thus, the power becomes:
\[P = \frac{1}{2} \frac{\beta\omega\varepsilon}{k_c^4} \cdot \pi \cdot I\]
where
\[I = \int_b^a \left[ (\psi^{\prime})^2 + \frac{m^2}{\rho^2} \psi^2 \right] \rho \, d\rho\]
Simplifying the Radial Integral
The integral I can be simplified using multiple identities and relations. Consider:
\[F = \psi\]
\[\frac{\partial}{\partial\rho}\left( \rho F F^\prime \right) = FF^\prime + \rho F^\prime F^\prime + \rho F F^{\prime\prime} = FF^\prime + \rho \left(F^\prime\right)^2+ \rho F F^{\prime\prime}\]
Rewriting the above expression:
\[\rho \left(F^\prime\right)^2 = \frac{\partial}{\partial\rho}\left( \rho F F^\prime \right) - \rho F F^{\prime\prime} - F F^\prime \]
Integrating on both sides:
\[\int_b^a \rho \left(F^\prime\right)^2 d\rho = \left[\rho F F^\prime\right]_b^a - \int_b^a F F^\prime d\rho - \int_b^a\rho F F^{\prime\prime} d\rho\]
We want to simplify the term with the second derivative. If we multiply Bessel's ODE we get:
\[\left( \rho^2 F^{\prime\prime} + \rho F^\prime + \left( k_c^2\rho^2 - m^2\right)F\right)\rho F = \left(0\right) \rho F\]
\[\rho F F^{\prime\prime} + F F^{\prime} + \left( k_c^2 \rho - \frac{m^2}{\rho}\right)F^2 = 0\]
\[\rho F F^{\prime\prime} = -F F^{\prime} - \left(k_c^2\rho-\frac{m^2}{\rho}\right)F^2\]
After algebraic manipulation with the previous results, we get:
\[\int_b^a \rho \left(F^\prime\right)^2 d\rho = \left[\rho F F^\prime\right]_b^a - \int_b^a F F^\prime d\rho - \int_b^a \left( -F F^{\prime} - \left(k_c^2\rho-\frac{m^2}{\rho}\right)F^2\right) d\rho\]
\[\int_b^a \rho \left(F^\prime\right)^2 d\rho = \left[\rho F F^\prime\right]_b^a - \int_b^a F F^\prime d\rho + \int_b^a F F^{\prime}d\rho +\int_b^a \left(k_c^2\rho-\frac{m^2}{\rho}\right)F^2 d\rho\]
\[\int_b^a \rho \left(F^\prime\right)^2 d\rho = \left[\rho F F^\prime\right]_b^a +\int_b^a k_c^2\rho F^2 d\rho - \int_b^a \frac{m^2}{\rho}F^2 d\rho\]
If we reagroup the terms we can get exactly $I$ on the right hand side
\[I = \int_b^a \rho \left(F^\prime\right)^2 d\rho + \int_b^a \frac{m^2}{\rho}F^2 d\rho = \left[\rho F F^\prime\right]_b^a + k_c^2\int_b^a \rho F^2 d\rho\]
Now we just need to solve the last part of the rhs:
\[k_c^2\int_b^a \rho F^2 d\rho\]
Bessel Function Integrals
The function $F$ depends of $\rho$. With the change of variable $x = k_c \rho$ we get
\[\frac{k_c^2}{k_c^2}\int_{x_b}^{x_a} x F(x)^2 dx\]
The remaining integral can be evaluated using the identity:
\[\int z C_\mu(az)D_\mu(az) = \frac{1}{4} z^2 \left( 2 C_\mu(az)D_\mu(az) - C_{\mu-1}(az)D_{\mu+1}(az) - C_{\mu+1}(az)D_{\mu-1}(az)\right)\]
where $C_\mu$ and $D_\mu$ are cylindrical functions($J_\mu$, $Y_\mu$ or $H_\mu$).
For $m = 0$ the result is almost identical. The derivation follows the same steps, yielding:
\[I = \int_b^a \rho \left(F^\prime\right)^2 d\rho = \left[\rho F F^\prime\right]_b^a + k_c^2\int_b^a \rho F^2 d\rho\]
which is exactly the same radial integral result. The only difference appears in the angular integration: instead of multiplying by $\pi$, we now multiply by $2\pi$.
Circular Waveguide
For circular waveguides ($b = 0$, $B_m = 0$), the boundary term vanishes at $\rho = 0$ and
\[\psi = Am J_m(k_c\rho)\]
\[I = k_c a J_m(k_c a) J_m'(k_c a) + \frac{(k_c a)^2}{2} \left[ J_m^2(k_c a) - J_{m+1}(k_c a) J_{m-1}(k_c a) \right]\]
TM Modes
\[F_0 = \sqrt{ \frac{2}{\displaystyle \frac{\omega\mu\beta}{k_c^4} \pi I } }\]
When $m=0$
\[F_0 = \sqrt{ \frac{2}{\displaystyle \frac{\omega\mu\beta}{k_c^4} 2\pi I } }\]
TE Modes
\[F_0 = \sqrt{ \frac{2}{\displaystyle \frac{\omega\varepsilon\beta}{k_c^4} \pi I } }\]
When $m=0$
\[F_0 = \sqrt{ \frac{2}{\displaystyle \frac{\omega\varepsilon\beta}{k_c^4} 2\pi I } }\]
Coaxial Waveguide
For coaxial waveguides, both Bessel functions are required since the domain excludes the origin:
TM Modes
\[B_m = -\frac{J_m(k_c b)}{Y_m(k_c b)}\]
TE Modes
\[B_m = -\frac{J_m'(k_c b)}{Y_m'(k_c b)}\]
The radial function becomes
\[F = A_m J_m(k_c\rho) + B_m Y_m(k_c\rho)\]
\[F^{\prime} = k_c \left[ A_m J_m'(k_c\rho) + B_m Y_m'(k_c\rho) \right]\]
The integral I has two parts:
\[I_1 = \left[ xF(x)F^{\prime}(x)\right]_b^a = \left( a F(a) F^{\prime}(a)\right) - \left( b F(b) F^{\prime}(b)\right) \]
and
\[I_2 =\int_b^a x F^2(x) dx = \int_b^a A_m^2J_m^2(x) + B_m^2 Y_m^2(x) + 2A_mB_mJ_m(x)Y_m(x) dx\]
\[I_2 =\int_b^a x F^2(x) dx = A_m^2\int_b^a J_m^2(x)dx + B_m^2 \int_b^aY_m^2(x)dx + 2A_mB_m \int_b^a J_m(x)Y_m(x) dx\]
All this integrals can be solved with closed form solutions.
The normalization constant $F_0$ has exactly the same expression as in circular waveguides, only changing the computation of I.
Radial and Wedge Waveguide
In radial and wedge waveguides, power propagates radially along the $\rho$ direction.
Power Flow Components
The radial component of the Poynting vector is:
\[S_\rho = \frac{1}{2} \left( E_\phi H_z^* - E_z H_\phi^*\right)\]
TE modes $E_z = 0$ so $S_\rho = \frac{1}{2} E_\phi H_z^*$
TM modes $H_z = 0$ so $S_\rho = \frac{1}{2} E_z H_\phi^*$.
Field Representation
Let's define the field functions:
\[\psi(\rho) = A_{mn}H_m^{(2)}(k_c\rho)\]
\[\phi_m(\varphi) = \begin{cases} \cos(m\varphi) \\ \sin(m\varphi) \end{cases}\]
\[Z_n(z) = \begin{cases} \sin(\frac{n\pi}{h}z) \\ \cos(\frac{n\pi}{h}z) \end{cases}\]
where the longitudinal field $E_z$ or $H_z$ (TM or TE modes) is defined as $\psi(\rho)\phi_m(\varphi)Z_n(z)$.
Power Integration
For TE modes, integrating the power flow:
\[P = \frac{1}{2} \int_0^{2\pi} \int_0^h \frac{j\mu\omega}{k_c^2}\psi^{\prime}\text{cos}(m\varphi)\text{sin}(\frac{n\pi}{h}z) \psi^*\text{sin}(\frac{n\pi}{h}z)\text{cos}(m\varphi) \rho d\varphi dz\]
The longitudinal and angular integrals yield:
\[\int_0^h \text{sin}(\frac{n\pi}{h}z) dz = \frac{h}{2}\]
\[\int_0^{2\pi}\text{cos}^2(m\varphi) d\varphi = \alpha_m =\begin{cases} \pi & \text{if } m \geq 1 \\ 2\pi & \text{if } m = 0 \end{cases}\]
Thus, the total power simplifies to:
\[P = \frac{1}{2} \text{Re} \left[ \frac{j\mu\omega}{k_c^2} \cdot \frac{h}{2} \cdot \alpha_m \cdot \rho \cdot \psi(\rho) \psi^{\prime}(\rho) \right]\]
\[F_0 = \sqrt\frac{1}{P}\]
The same procedure and result applies to TM modes, replacing $\mu$ by $\varepsilon$.
Spherical Modes
The Poynting vector in spherical coordinates has the radial component:
\[S_r = (\mathbf{E} \times \mathbf{H}^*) \cdot \hat{\mathbf{r}} = E_\theta H_\phi^* - E_\phi H_\theta^*\]
The angular integrals are:
\[\oint |Y_{lm}|^2 d\Omega = 1,\quad \oint |\nabla_\perp Y_{lm}|^2 d\Omega = l(l+1)\]
The radiated power can be expressed as:
\[P \propto \frac{r^2}{2\omega\mu\varepsilon^2} \Re\left[i R_l(kr) R_l'(kr)^*\right] l(l+1)\]
with $R_l(kr) = h_l^{(1)}(kr)$ or $h_l^{(2)}(kr)$ for incoming or outgoing waves, respectively.
\[P = \left| \frac{k r^2}{2\omega\mu\varepsilon^2} \Re\left[i R R'\right] l(l+1) \right|\]
The normalization factor for 1W power is:
\[F_0 = \sqrt\frac{1}{P}\]