Transverse electric(TE) wave in circular waveguides

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Circular waveguides offer implementation advantages over rectangular wave guide is that installation is much simpler when forming runs of turns and offsets. The uniform section in the gyrotron cavity is a circular waveguide. The working mode is usually a transverse electric mode in the cavity. Here is a calculator for the Transverse Electric mode in a circular waveguide. Please note that the radial wave number is an imaginal number when the working frequency is below the cutoff frequency.

There may be several modes in the cavity because the cavity is oversized. The frequencies of the modes can not be very difference because the excitation current is not wide band and the wave guide wavelength must be similar because it is the same cavity. So the wave guide wavelength is kept the same, and find the corresponding frequencies for different mode. If the difference of frequencies is less than 4GHz, the frequencies will be shown in the table.

Wave guide Theory

It is practical assume that the cross-sectional size and shape is constant along the wave-guide axis. With a sinusoidal time dependence e−jωte^{-j\omega t} for the fields inside the wave-guide, so ∂∂t=jω\frac{\partial}{\partial t}=j \omega and the Maxwell equations take the form

∇×E⃗=jωB⃗\nabla \times \vec E=j \omega \vec B

∇×H⃗=−jωD⃗\nabla \times \vec H=-j \omega\vec D

∇⋅B⃗=0\nabla \cdot \vec B=0

∇⋅D⃗=0\nabla \cdot \vec D=0

We assume the wave propagates along +z direction, so

E⃗(x,y,z,t)=E⃗(x,y)ej(ωt−kzz)\vec E(x,y,z,t)=\vec E(x,y) e^{j(\omega t-k_z z)} and H⃗(x,y,z,t)=H⃗(x,y)ej(ωt−kzz)\vec H(x,y,z,t)=\vec H(x,y)e^{j(\omega t-k_z z)}

There are 2 vector fields including 6 scalar quantity. For example the electrical field can be expanded in terms of the unit vector in the Cartesian coordinates

E⃗(x,y)=x^Ex(x,y)+y^Ey(x,y)+z^Ez(x,y)\vec E(x,y)=\hat x Ex(x,y)+\hat y Ey(x,y)+\hat z Ez(x,y)

where x^,y^andz^\hat x, \hat y and \hat z are the unit vector in the x,y and z direction respectively.

It is very useful to the express the transverse components of the field in terms of the parallel components by expand the two curl equations.

ikzEy(x,y)eitω−ikzz+∂yEz(x,y)eitω−ikzz=−iμωHx(x,y)eitω−ikzzi k_z Ey(x,y) e^{i t \omega -i k_z z}+\partial_y Ez(x,y) e^{i t \omega -i \text{kz} z}=-i \mu \omega Hx(x,y) e^{i t \omega -i k_z z}

−ikzEx(x,y)eitω−ikzz−∂xEz(x,y)eitω−ikzz=−iμωHy(x,y)eitω−ikzz-i k_z Ex(x,y) e^{i t \omega -i k_z z}-\partial_x Ez(x,y) e^{i t \omega -i \text{kz} z}=-i \mu \omega Hy(x,y) e^{i t \omega -i k_z z}

ikzHy(x,y)eitω−ikzz+∂yHz(x,y)eitω−ikzz=iϵωEx(x,y)eitω−ikzzi k_z Hy(x,y) e^{i t \omega -i k_z z}+\partial_y Hz(x,y) e^{i t \omega -i \text{kz} z}=i \epsilon \omega Ex(x,y) e^{i t \omega -i k_z z}

−ikzEx(x,y)eitω−ikzz−∂xHz(x,y)eitω−ikzz=iϵωEy(x,y)eitω−ikzz-i k_z Ex(x,y) e^{i t \omega -i k_z z}-\partial_x Hz(x,y) e^{i t \omega -i \text{kz} z}=i \epsilon \omega Ey(x,y) e^{i t \omega -i k_z z}

They are a linear equations for the transverse components Ex, Ey, Hx, Hy in terms of Ez and Hz. They are

Ex(x,y)=−ikz∂xEz(x,y)kc2−iμω∂yHz(x,y)kc2 Ex(x,y)=-\frac{i k_z \partial_x Ez(x,y)}{k_c^2}-\frac{i \mu \omega \partial_y Hz(x,y)}{k_c^2}

Ey(x,y)=−ikz∂yEz(x,y)kc2+iμω∂xHz(x,y)kc2 Ey(x,y)=-\frac{i k_z \partial_y Ez(x,y)}{k_c^2}+\frac{i \mu \omega \partial_x Hz(x,y)}{k_c^2}

Hx(x,y)=iϵω∂yEz(x,y)kc2−ikz∂xHz(x,y)kc2 Hx(x,y)=\frac{i \epsilon \omega \partial_y Ez(x,y)}{k_c^2}-\frac{i k_z \partial_x Hz(x,y)}{k_c^2}

Hy(x,y)=−iϵω∂xEz(x,y)kc2−ikz∂yHz(x,y)kc2 Hy(x,y)=-\frac{i \epsilon \omega \partial_x Ez(x,y)}{k_c^2}-\frac{i k_z \partial_y Hz(x,y)}{k_c^2}

Where kc=k2−kz2k_c=k^2-k_z^2 is the cutoff wave number.

Then we find the relationship between the transverse vector and the parallel scalar.

\begin{eqnarray}<br> \vec Et(x,y) &amp;=&amp;\hat x Ex(x,y)+\hat y Ey(x,y)\\<br> &amp;=&amp; \hat x \frac{1}{k_c^2}(-j k_z \partial_x Ez-j \omega \mu \partial_y Hz)+\hat y \frac{1}{k_c^2}(-j k_z \partial_y Ez+j\omega\mu\partial_x Hz) \\<br>&amp;=&amp; -\frac{j k_z}{k_c^2}(\hat x \partial_x Ez+\hat y \partial_y Ez)+\frac{j \omega \mu}{k_c^2}(-\hat x \partial_y Hz+\hat y \partial_x Hz)\\<br> &amp;=&amp; -\frac{j k_z}{k_c^2}\nabla_t Ez+\frac{j \omega \mu}{k_c^2}\hat z\times \nabla_t Hz \\<br> \end{eqnarray}

\begin{eqnarray}<br>\vec Ht(x,y) &amp;=&amp;\hat x Hx(x,y)+\hat y Hy(x,y)\\<br>&amp;=&amp; \hat x \frac{1}{k_c^2}(-j k_z \partial_x Hz+j \omega \epsilon \partial_y Ez)+\hat y -\frac{1}{k_c^2}(j k_z \partial_y Hz+j\omega\epsilon\partial_x Ez) \\<br>&amp;=&amp; -\frac{j k_z}{k_c^2}(\hat x \partial_x Hz+\hat y \partial_y Hz)+\frac{j \omega \epsilon}{k_c^2}(\hat x \partial_y Ez-\hat y \partial_x Ez)\\<br>&amp;=&amp; -\frac{j k_z}{k_c^2}\nabla_t Hz-\frac{j \omega \epsilon}{k_c^2}\hat z\times \nabla_t Ez \\<br>\end{eqnarray}

Where ∇t=x^∂x+y^∂y\nabla_t=\hat x\partial_x+\hat y \partial_y is the grad operator in the transverse plane.

For the Transverse Electric (TE) waves, Ez=0, so

E⃗t(x,y)=jωμkc2z^×∇tHz\vec Et(x,y)=\frac{j \omega \mu}{k_c^2}\hat z\times \nabla_t Hz

H⃗t(x,y)=−jkzkc2∇tHz\vec Ht(x,y)=-\frac{j k_z}{k_c^2}\nabla_t Hz

So, H⃗t\vec Ht and ∇tHz\nabla_t Hz are in opposite direction. E⃗t\vec Et is perpendicular to them. Please note that z^×∇tHz\hat z\times \nabla_t Hz means rotate the vector 90 degree anticlockwise. ∇tHz×z^\nabla_t Hz \times \hat z means rotate the vector 90 degree clockwise.

It is similar to the Ohm’s law in the circuit, we define the wave impedance as the ratio of the magnitude of E⃗tandH⃗t\vec Et and \vec Ht, so

Z=∣E⃗t∣∣H⃗t∣=ωμkz=kkzμϵ=kkzZ0Z=\frac{|\vec Et|}{|\vec Ht|}=\frac{\omega \mu}{k_z}=\frac{k}{k_z}\sqrt{\frac{\mu}{\epsilon}}=\frac{k}{k_z}Z_0

where Z0=120πΩZ_0=120\pi\Omega is the impedance in the vacuum.

The flow of energy is described by the complex Poynting vector whose real part gives the time-averaged flux of energy:

\begin{eqnarray}<br>\vec S &amp;=&amp; \frac{1}{2}(\vec E \times \vec H^*)\\<br>&amp;=&amp;\frac{1}{2}\vec Et \times(\vec Ht^*+\hat z H_z)\\<br>&amp;=&amp;\frac{1}{2}\frac{j \omega \mu}{k_c^2}\hat z\times \nabla_t Hz\times(-\frac{-j k_z}{k_c^2}\nabla_t Hz^*+\hat z Hz^*)\\<br>&amp;=&amp; \frac{ \omega\mu k_z}{2k_c^4}\hat z |\nabla_t Hz|^2+\frac{j \omega\mu}{2k_c^2} Hz^* \nabla_t Hz<br>\end{eqnarray}

To evaluate the total power flow P we integrate the axial component of S over the cross-sectional area A:

P=∬AS⃗⋅z^da=ωμkz2kc4∬A(∇tHz)∗⋅(∇tHz)daP=\iint_A\vec S \cdot \hat z da=\frac{ \omega\mu k_z}{2k_c^4}\iint_A(\nabla_t Hz)^* \cdot (\nabla_t Hz)da

Using the vector identity, ∇⋅(ϕ∇ϕ)=ϕ∇2ϕ+∇ϕ⋅∇ϕ\nabla\cdot(\phi \nabla \phi)=\phi \nabla^2 \phi+\nabla \phi \cdot \nabla \phi and the divergence theorem,

∬s∇⋅A⃗ds=∮lA⃗⋅n⃗dl\iint_s \nabla\cdot \vec A ds=\oint_l \vec A\cdot \vec n dl

We obtained the Green’s first identity,

∬Aϕ∇2ϕ+∇ϕ⋅∇ϕda=∬A∇⋅(ϕ∇ϕ)da=∮lϕ∂ϕ∂ndl\iint_A \phi \nabla^2 \phi+\nabla \phi \cdot \nabla \phi da=\iint_A \nabla\cdot(\phi \nabla \phi) da=\oint_l \phi \frac{\partial \phi}{\partial n} dl

Let Hz(x,y,z)=ϕ(x,y)e−jkzzHz(x,y,z)=\phi(x,y)e^{-j k_z z}, then ∇t2ϕ+kc2ϕ=0\nabla_t^2 \phi+k_c^2 \phi=0 and ∂ϕ∂n=0\frac{\partial \phi}{\partial n}=0 at the boundary for TE mode, so it is

∬Aϕ(−kc2ϕ)da+∬A∇ϕ⋅∇ϕda=0\iint_A \phi (-k_c^2 \phi)da+\iint_A\nabla \phi \cdot \nabla \phi da=0

so

∬A∇ϕ⋅∇ϕ∗da=kc2∬Aϕϕ∗da\iint_A\nabla \phi \cdot \nabla \phi^* da=k_c^2\iint_A \phi \phi^* da

The total power iss

\begin{eqnarray}<br>P &amp;=&amp; \frac{ \omega\mu k_z}{2k_c^4}k_c^2\iint_A \phi ^2 da \\<br>&amp;=&amp; \frac{ \omega\mu k_z}{2k_c^2}\iint_A \phi ^2 da\\<br>&amp;=&amp; \frac{ \omega\mu \sqrt{k^2-k_c^2}}{2k_c^2}\iint_A \phi ^2 da\\<br>&amp;=&amp; \frac{ \omega\mu \sqrt{\omega^2\epsilon\mu-\omega_c^2\mu\epsilon}}{2\omega_c^2\mu\epsilon}\iint_A \phi ^2 da\\<br>&amp;=&amp; \frac{ \omega^2\mu \sqrt{1-\frac{\omega_c^2}{\omega^2}}}{2\omega_c^2\sqrt{\mu\epsilon}}\iint_A \phi ^2 da\\<br>&amp;=&amp; \frac{1}{2}\sqrt{\frac{\mu}{\epsilon}}(\frac{\omega}{\omega_c})^2\sqrt{1-\frac{\omega_c^2}{\omega^2}}\iint_A \phi ^2 da\\<br>&amp;=&amp;\frac{ Z_0}{2}(\frac{f}{f_c})^2\sqrt{1-\frac{f_c^2}{f^2}}\iint_A \phi ^2 da\\<br>\end{eqnarray}

Appendix Bessel Functions

Bessel functions are solutions to the Helmholtz equations in cylindrical coordinates,

1rddr(rdfdr)+(k2−m2r2)f=0\frac{1}{r}\frac{d}{dr}(r\frac{df}{dr})+(k^2-\frac{m^2}{r^2})f=0

where k2k^2 is real and m is an integer. The two independent solutions to this differential equation are called ordinary Bessel function of the first and second kind, written as Jn(kr)J_n(k r) and Yn(kr)Y_n(kr).

Recurrence formulas relate Bessel functions of different orderss:

Jm+1(x)+Jm−1=2mxJm(x)J_{m+1}(x)+J_{m-1}=\frac{2m}{x}J_m(x)

\begin{eqnarray}<br>J_m'(x) &amp;= &amp; \frac{1}{2}[J_{m-1}(x)-J_{m+1}(x)]\\<br>&amp;=&amp;\frac{1}{2}[J_{m-1}(x)-(\frac{2m}{x}J_m(x)-J_{m-1}(x))]\\<br>&amp;=&amp;J_{m-1}(x)-\frac{m}{x}J_m(x)\\<br>&amp;=&amp;\frac{1}{2}[(\frac{2m}{x}J_m(x)-J_{m+1}(x))-J_{m+1}(x)]\\<br>&amp;=&amp;\frac{m}{x}J_m(x)-J_{m+1}(x)\\<br>\end{eqnarray}

Jm+12+Jm−12+2Jm+1Jm−1=4m2x2Jm2J_{m+1}^2+J_{m-1}^2+2J_{m+1}J_{m-1}=\frac{4m^2}{x^2}J_m^2

Jm+12+Jm−12−2Jm+1Jm−1=4Jm’2J_{m+1}^2+J_{m-1}^2-2J_{m+1}J_{m-1}=4J_m’^2

so Jm+1Jm−1=m2x2Jm2−Jm’2J_{m+1}J_{m-1}=\frac{m^2}{x^2}J_m^2-J_m’^2

The following integral relations involving bessel function s are useful for getting the power at the waveguide:

∫0xJm2(kx)xdx=x22[Jm’2(kx)+(1−m2k2x2)Jm2(kx)]=x22[Jm2(kx)−Jm−1(kx)Jm+1(kx)]\int_0^x J_m^2(kx)x dx=\frac{x^2}{2}[J_m’^2(kx)+(1-\frac{m^2}{k^2x^2})J_m^2(kx)]=\frac{x^2}{2}[J_m^2(kx)-J_{m-1}(k x) J_{m+1}(k x)]

Prove:

\displaylinesddxx22[Jm2(kx)−Jm−1(kx)Jm+1(kx)]=<br>12x2(−12kJm−1(kx)(Jm(kx)−Jm+2(kx))−<br>12k(Jm−2(kx)−Jm(kx))Jm+1(kx)+k(Jm−1(kx)−Jm+1(kx))Jm(kx))+<br>x(Jm(kx)2−Jm−1(kx)Jm+1(kx))\displaylines{\frac{d}{dx}\frac{x^2}{2}[J_m^2(kx)-J_{m-1}(k x) J_{m+1}(k x)]=\\<br>\frac{1}{2} x^2 \left(-\frac{1}{2} k J_{m-1}(k x) (J_m(k x)-J_{m+2}(k x))-<br>\frac{1}{2} k (J_{m-2}(k x)-J_m(k x)) J_{m+1}(k x)+k (J_{m-1}(k x)-J_{m+1}(k x)) J_m(k x)\right)+\\<br>x \left(J_m(k x){}^2-J_{m-1}(k x) J_{m+1}(k x)\right)}

Expand the expression in the first term except x22\frac{x^2}{2} is

12k(−Jm−2(kx)Jm+1(kx)−Jm(kx)Jm+1(kx)+Jm−1(kx)Jm+2(kx)+Jm−1(kx)Jm(kx))\frac{1}{2} k (-J_{m-2}(k x) J_{m+1}(k x)-J_m(k x) J_{m+1}(k x)+J_{m-1}(k x) J_{m+2}(k x)+J_{m-1}(k x) J_m(k x))

=12k(Jm−1(kx)(Jm+2(kx)+Jm(kx))−Jm+1(kx)(Jm−2(kx)+Jm(kx)))=\frac{1}{2} k (J_{m-1}(k x) (J_{m+2}(k x)+J_m(k x))-J_{m+1}(k x) (J_{m-2}(k x)+J_m(k x)))

=12k((2(m+1))Jm−1(kx)Jm+1(kx)kx−(2(m−1))Jm+1(kx)Jm−1(kx)kx)=\frac{1}{2} k \left(\frac{(2 (m+1)) J_{m-1}(k x) J_{m+1}(k x)}{\text{kx}}-\frac{(2 (m-1)) J_{m+1}(k x) J_{m-1}(k x)}{\text{kx}}\right)

=2(Jm−1(kx)Jm+1(kx))x=\frac{2 (J_{m-1}(k x) J_{m+1}(k x))}{x}

So, the original differential is

12x2⋅2(Jm−1(kx)Jm+1(kx))x+x(Jm(kx)2−Jm−1(kx)Jm+1(kx))\frac{1}{2} x^2\cdot\frac{2 (J_{m-1}(k x) J_{m+1}(k x))}{x}+x \left(J_m(k x){}^2-J_{m-1}(k x) J_{m+1}(k x)\right)

=x(Jm−1(kx)Jm+1(kx))+x(Jm(kx)2−Jm−1(kx)Jm+1(kx))=x (J_{m-1}(k x) J_{m+1}(k x))+x \left(J_m(k x){}^2-J_{m-1}(k x) J_{m+1}(k x)\right)

=xJm(kx)2=xJ_m(k x){}^2

If we integrate the time averaged Poynting vector directory to obtain the power, we will meet a more complex integration about the Bessel functions,

∫0pmn’[Jm’2(x)+m2x2Jm2(x)]xdx=12(pmn’2−m2)Jm2(pmn)\int_0^{p_{mn}’}[J_m’^2(x)+\frac{m^2}{x^2}J_m^2(x)]xdx=\frac{1}{2}(p_{mn}’^2-m^2)J_m^2(p_{mn})

Prove: According to the Green’s first identity in the waveguide,

∬A∇ϕ⋅∇ϕ∗da=kc2∬Aϕϕ∗da\iint_A\nabla \phi \cdot \nabla \phi^* da=k_c^2\iint_A \phi \phi^* da

Let the scalar function ϕ=CJm(kcx)e−jmθ\phi=C J_m(k_c x)e^{-j m\theta}, where Hz=ϕe−jkzzHz=\phi e^{-j k_z z} for the TE mode in the waveguide.

∇ϕ=CkcJm′(kcr)e−jmθe^r−jmrCJm(kcr)e−jmθe^θ\nabla \phi=C k_c J_m'(k_c r)e^{-j m \theta}\hat e_r-\frac{j m}{r}C J_m(k_c r) e^{-j m \theta}\hat e_{\theta}

∇ϕ∗=CkcJm′(kcr)ejmθe^r+jmrCJm(kcr)ejmθe^θ\nabla \phi^*=C k_c J_m'(k_c r)e^{j m \theta}\hat e_r+\frac{j m}{r}C J_m(k_c r) e^{j m \theta}\hat e_{\theta}

I=∬A∇ϕ⋅∇ϕ∗da=∫02πdθ∫0R(z)rC2kc2Jm’2(kcr)+m2rC2Jm2(kcr)drI=\iint_A\nabla \phi \cdot \nabla \phi^* da=\int_0^{2\pi}d\theta\int_0^{R(z)} r C^2 k_c^2 J_m’^2(k_c r)+\frac{m^2}{r}C^2 J_m^2(k_c r) dr

I=2πC2∫0R(z)rkc2Jm’2(kcr)+m2rJm2(kcr)drI=2\pi C^2 \int_0^{R(z)} r k_c^2 J_m’^2(k_c r)+\frac{m^2}{r} J_m^2(k_c r) dr

Let x=kcrx=k_c r then dx=kcdrdx=k_c dr

I=2πC2∫0P’mnxJm’2(x)+m2xJm2(x)dxI=2\pi C^2 \int_0^{P’_{mn}} x J_m’^2 (x)+\frac{m^2}{x} J_m^2 (x) dx

The right side

I=kc2∬Aϕϕ∗da=kc2∫02πdθ∫0R(z)C2Jm2(kcr)rdrI=k_c^2\iint_A \phi \phi^* da=k_c^2 \int_0^{2 \pi}d\theta \int_0^{R(z)}C^2 J_m^2(k_c r) r dr

=2πC2kc2∫0P’mnJm2(x)1kc2xdx=2\pi C^2 k_c^2\int_0^{P’_{mn}} J_m^2(x) \frac{1}{k_c^2}x dx

=2πC2∫0P’mnJm2(x)xdx=πC2(P’mn2−m2)Jm2(P’mn)=2\pi C^2 \int_0^{P’_{mn}} J_m^2(x) x dx=\pi C^2 (P’^2_{mn}-m^2)J_m^2(P’_{mn})

Compare the two side, so

∫0pmn’[Jm’2(x)+m2x2Jm2(x)]xdx=12(pmn’2−m2)Jm2(pmn)\int_0^{p_{mn}’}[J_m’^2(x)+\frac{m^2}{x^2}J_m^2(x)]xdx=\frac{1}{2}(p_{mn}’^2-m^2)J_m^2(p_{mn})

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