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 \(e^{-j\omega t}\) for the fields inside the wave-guide, so \(\frac{\partial}{\partial t}=j \omega\) and the Maxwell equations take the form
\(\nabla \times \vec E=j \omega \vec B\)
\(\nabla \times \vec H=-j \omega\vec D\)
\(\nabla \cdot \vec B=0\)
\(\nabla \cdot \vec D=0\)
We assume the wave propagates along +z direction, so
\(\vec E(x,y,z,t)=\vec E(x,y) e^{j(\omega t-k_z z)}\) and \(\vec H(x,y,z,t)=\vec H(x,y)e^{j(\omega t-k_z z)}\)
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
\(\vec E(x,y)=\hat x Ex(x,y)+\hat y Ey(x,y)+\hat z Ez(x,y)\)
where \(\hat x, \hat y and \hat z\) 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.
\(i k_z Ey(x,y) e^{i t \omega -i k_z z}+\partial_y Ez(x,y) e^{i t \omega -i \text{kz} z}=-i \mu \omega Hx(x,y) e^{i t \omega -i k_z z}\)
\(-i k_z Ex(x,y) e^{i t \omega -i k_z z}-\partial_x Ez(x,y) e^{i t \omega -i \text{kz} z}=-i \mu \omega Hy(x,y) e^{i t \omega -i k_z z}\)
\(i k_z Hy(x,y) e^{i t \omega -i k_z z}+\partial_y Hz(x,y) e^{i t \omega -i \text{kz} z}=i \epsilon \omega Ex(x,y) e^{i t \omega -i k_z z}\)
\(-i k_z Ex(x,y) e^{i t \omega -i k_z z}-\partial_x Hz(x,y) e^{i t \omega -i \text{kz} z}=i \epsilon \omega Ey(x,y) e^{i t \omega -i k_z z}\)
They are a linear equations for the transverse components Ex, Ey, Hx, Hy in terms of Ez and Hz. They are
\( Ex(x,y)=-\frac{i k_z \partial_x Ez(x,y)}{k_c^2}-\frac{i \mu \omega \partial_y Hz(x,y)}{k_c^2}\)
\( Ey(x,y)=-\frac{i k_z \partial_y Ez(x,y)}{k_c^2}+\frac{i \mu \omega \partial_x Hz(x,y)}{k_c^2}\)
\( Hx(x,y)=\frac{i \epsilon \omega \partial_y Ez(x,y)}{k_c^2}-\frac{i k_z \partial_x Hz(x,y)}{k_c^2}\)
\( Hy(x,y)=-\frac{i \epsilon \omega \partial_x Ez(x,y)}{k_c^2}-\frac{i k_z \partial_y Hz(x,y)}{k_c^2}\)
Where \(k_c=k^2-k_z^2\) is the cutoff wave number.
Then we find the relationship between the transverse vector and the parallel scalar.
\(\begin{eqnarray}
\vec Et(x,y) &=&\hat x Ex(x,y)+\hat y Ey(x,y)\\
&=& \hat x \frac{1}{k_c^2}(-j k_z \partial_x Ez-j \omega \mu \partial_y Hz)+\hat y \frac{1}{k_c^2}(-j k_z \partial_y Ez+j\omega\mu\partial_x Hz) \\
&=& -\frac{j k_z}{k_c^2}(\hat x \partial_x Ez+\hat y \partial_y Ez)+\frac{j \omega \mu}{k_c^2}(-\hat x \partial_y Hz+\hat y \partial_x Hz)\\
&=& -\frac{j k_z}{k_c^2}\nabla_t Ez+\frac{j \omega \mu}{k_c^2}\hat z\times \nabla_t Hz \\
\end{eqnarray}\)
\(\begin{eqnarray}
\vec Ht(x,y) &=&\hat x Hx(x,y)+\hat y Hy(x,y)\\
&=& \hat x \frac{1}{k_c^2}(-j k_z \partial_x Hz+j \omega \epsilon \partial_y Ez)+\hat y -\frac{1}{k_c^2}(j k_z \partial_y Hz+j\omega\epsilon\partial_x Ez) \\
&=& -\frac{j k_z}{k_c^2}(\hat x \partial_x Hz+\hat y \partial_y Hz)+\frac{j \omega \epsilon}{k_c^2}(\hat x \partial_y Ez-\hat y \partial_x Ez)\\
&=& -\frac{j k_z}{k_c^2}\nabla_t Hz-\frac{j \omega \epsilon}{k_c^2}\hat z\times \nabla_t Ez \\
\end{eqnarray}\)
Where \(\nabla_t=\hat x\partial_x+\hat y \partial_y\) is the grad operator in the transverse plane.
For the Transverse Electric (TE) waves, Ez=0, so
\(\vec Et(x,y)=\frac{j \omega \mu}{k_c^2}\hat z\times \nabla_t Hz\)
\(\vec Ht(x,y)=-\frac{j k_z}{k_c^2}\nabla_t Hz\)
So, \(\vec Ht\) and \(\nabla_t Hz\) are in opposite direction. \(\vec Et\) is perpendicular to them. Please note that \(\hat z\times \nabla_t Hz\) means rotate the vector 90 degree anticlockwise. \(\nabla_t Hz \times \hat z\) 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 \(\vec Et and \vec Ht\), so
\(Z=\frac{|\vec Et|}{|\vec Ht|}=\frac{\omega \mu}{k_z}=\frac{k}{k_z}\sqrt{\frac{\mu}{\epsilon}}=\frac{k}{k_z}Z_0\)
where \(Z_0=120\pi\Omega\) 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:
\(\begin{eqnarray}
\vec S &=& \frac{1}{2}(\vec E \times \vec H^*)\\
&=&\frac{1}{2}\vec Et \times(\vec Ht^*+\hat z H_z)\\
&=&\frac{1}{2}\frac{j \omega \mu}{k_c^2}\hat z\times \nabla_t Hz\times(-\frac{-j k_z}{k_c^2}\nabla_t Hz^*+\hat z Hz^*)\\
&=& \frac{ \omega\mu k_z}{2k_c^4}\hat z |\nabla_t Hz|^2+\frac{j \omega\mu}{2k_c^2} Hz^* \nabla_t Hz
\end{eqnarray}\)
To evaluate the total power flow P we integrate the axial component of S over the cross-sectional area A:
\(P=\iint_A\vec S \cdot \hat z da=\frac{ \omega\mu k_z}{2k_c^4}\iint_A(\nabla_t Hz)^* \cdot (\nabla_t Hz)da\)
Using the vector identity, \(\nabla\cdot(\phi \nabla \phi)=\phi \nabla^2 \phi+\nabla \phi \cdot \nabla \phi\) and the divergence theorem,
\(\iint_s \nabla\cdot \vec A ds=\oint_l \vec A\cdot \vec n dl\)
We obtained the Green’s first identity,
\(\iint_A \phi \nabla^2 \phi+\nabla \phi \cdot \nabla \phi da=\iint_A \nabla\cdot(\phi \nabla \phi) da=\oint_l \phi \frac{\partial \phi}{\partial n} dl\)
Let \(Hz(x,y,z)=\phi(x,y)e^{-j k_z z}\), then \(\nabla_t^2 \phi+k_c^2 \phi=0\) and \(\frac{\partial \phi}{\partial n}=0 \) at the boundary for TE mode, so it is
\(\iint_A \phi (-k_c^2 \phi)da+\iint_A\nabla \phi \cdot \nabla \phi da=0\)
so
\(\iint_A\nabla \phi \cdot \nabla \phi^* da=k_c^2\iint_A \phi \phi^* da \)
The total power iss
\(\begin{eqnarray}
P &=& \frac{ \omega\mu k_z}{2k_c^4}k_c^2\iint_A \phi ^2 da \\
&=& \frac{ \omega\mu k_z}{2k_c^2}\iint_A \phi ^2 da\\
&=& \frac{ \omega\mu \sqrt{k^2-k_c^2}}{2k_c^2}\iint_A \phi ^2 da\\
&=& \frac{ \omega\mu \sqrt{\omega^2\epsilon\mu-\omega_c^2\mu\epsilon}}{2\omega_c^2\mu\epsilon}\iint_A \phi ^2 da\\
&=& \frac{ \omega^2\mu \sqrt{1-\frac{\omega_c^2}{\omega^2}}}{2\omega_c^2\sqrt{\mu\epsilon}}\iint_A \phi ^2 da\\
&=& \frac{1}{2}\sqrt{\frac{\mu}{\epsilon}}(\frac{\omega}{\omega_c})^2\sqrt{1-\frac{\omega_c^2}{\omega^2}}\iint_A \phi ^2 da\\
&=&\frac{ Z_0}{2}(\frac{f}{f_c})^2\sqrt{1-\frac{f_c^2}{f^2}}\iint_A \phi ^2 da\\
\end{eqnarray}\)
Appendix Bessel Functions
Bessel functions are solutions to the Helmholtz equations in cylindrical coordinates,
\(\frac{1}{r}\frac{d}{dr}(r\frac{df}{dr})+(k^2-\frac{m^2}{r^2})f=0\)
where \(k^2\) 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 \(J_n(k r)\) and \(Y_n(kr)\).
Recurrence formulas relate Bessel functions of different orderss:
\(J_{m+1}(x)+J_{m-1}=\frac{2m}{x}J_m(x)\)
\(\begin{eqnarray}
J_m'(x) &= & \frac{1}{2}[J_{m-1}(x)-J_{m+1}(x)]\\
&=&\frac{1}{2}[J_{m-1}(x)-(\frac{2m}{x}J_m(x)-J_{m-1}(x))]\\
&=&J_{m-1}(x)-\frac{m}{x}J_m(x)\\
&=&\frac{1}{2}[(\frac{2m}{x}J_m(x)-J_{m+1}(x))-J_{m+1}(x)]\\
&=&\frac{m}{x}J_m(x)-J_{m+1}(x)\\
\end{eqnarray}\)
\(J_{m+1}^2+J_{m-1}^2+2J_{m+1}J_{m-1}=\frac{4m^2}{x^2}J_m^2\)
\(J_{m+1}^2+J_{m-1}^2-2J_{m+1}J_{m-1}=4J_m’^2\)
so \(J_{m+1}J_{m-1}=\frac{m^2}{x^2}J_m^2-J_m’^2\)
The following integral relations involving bessel function s are useful for getting the power at the waveguide:
\(\int_0^x J_m^2(kx)x dx=\frac{x^2}{2}[J_m’^2(kx)+(1-\frac{m^2}{k^2x^2})J_m^2(kx)]=\frac{x^2}{2}[J_m^2(kx)-J_{m-1}(k x) J_{m+1}(k x)]\)
Prove:
\(\displaylines{\frac{d}{dx}\frac{x^2}{2}[J_m^2(kx)-J_{m-1}(k x) J_{m+1}(k x)]=\\
\frac{1}{2} x^2 \left(-\frac{1}{2} k J_{m-1}(k x) (J_m(k x)-J_{m+2}(k x))-
\frac{1}{2} k (J_{m-2}(k x)-J_m(k x)) J_{m+1}(k x)+k (J_{m-1}(k x)-J_{m+1}(k x)) J_m(k x)\right)+\\
x \left(J_m(k x){}^2-J_{m-1}(k x) J_{m+1}(k x)\right)}\)
Expand the expression in the first term except \(\frac{x^2}{2}\) is
\(\frac{1}{2} k (-J_{m-2}(k x) J_{m+1}(k x)-J_m(k x) J_{m+1}(k x)+J_{m-1}(k x) J_{m+2}(k x)+J_{m-1}(k x) J_m(k x))\)
\(=\frac{1}{2} k (J_{m-1}(k x) (J_{m+2}(k x)+J_m(k x))-J_{m+1}(k x) (J_{m-2}(k x)+J_m(k x)))\)
\(=\frac{1}{2} k \left(\frac{(2 (m+1)) J_{m-1}(k x) J_{m+1}(k x)}{\text{kx}}-\frac{(2 (m-1)) J_{m+1}(k x) J_{m-1}(k x)}{\text{kx}}\right)\)
\(=\frac{2 (J_{m-1}(k x) J_{m+1}(k x))}{x}\)
So, the original differential is
\(\frac{1}{2} x^2\cdot\frac{2 (J_{m-1}(k x) J_{m+1}(k x))}{x}+x \left(J_m(k x){}^2-J_{m-1}(k x) J_{m+1}(k x)\right)\)
\(=x (J_{m-1}(k x) J_{m+1}(k x))+x \left(J_m(k x){}^2-J_{m-1}(k x) J_{m+1}(k x)\right)\)
\(=xJ_m(k x){}^2\)
If we integrate the time averaged Poynting vector directory to obtain the power, we will meet a more complex integration about the Bessel functions,
\(\int_0^{p_{mn}’}[J_m’^2(x)+\frac{m^2}{x^2}J_m^2(x)]xdx=\frac{1}{2}(p_{mn}’^2-m^2)J_m^2(p_{mn})\)
Prove: According to the Green’s first identity in the waveguide,
\(\iint_A\nabla \phi \cdot \nabla \phi^* da=k_c^2\iint_A \phi \phi^* da\)
Let the scalar function \(\phi=C J_m(k_c x)e^{-j m\theta}\), where \(Hz=\phi e^{-j k_z z}\) for the TE mode in the waveguide.
\(\nabla \phi=C k_c J_m'(k_c r)e^{-j m \theta}\hat e_r-\frac{j m}{r}C J_m(k_c r) e^{-j m \theta}\hat e_{\theta}\)
\(\nabla \phi^*=C k_c J_m'(k_c r)e^{j m \theta}\hat e_r+\frac{j m}{r}C J_m(k_c r) e^{j m \theta}\hat e_{\theta}\)
\(I=\iint_A\nabla \phi \cdot \nabla \phi^* da=\int_0^{2\pi}d\theta\int_0^{R(z)} r C^2 k_c^2 J_m’^2(k_c r)+\frac{m^2}{r}C^2 J_m^2(k_c r) dr\)
\(I=2\pi C^2 \int_0^{R(z)} r k_c^2 J_m’^2(k_c r)+\frac{m^2}{r} J_m^2(k_c r) dr\)
Let \(x=k_c r\) then \(dx=k_c dr\)
\(I=2\pi C^2 \int_0^{P’_{mn}} x J_m’^2 (x)+\frac{m^2}{x} J_m^2 (x) dx \)
The right side
\(I=k_c^2\iint_A \phi \phi^* da=k_c^2 \int_0^{2 \pi}d\theta \int_0^{R(z)}C^2 J_m^2(k_c r) r dr\)
\(=2\pi C^2 k_c^2\int_0^{P’_{mn}} J_m^2(x) \frac{1}{k_c^2}x dx\)
\(=2\pi C^2 \int_0^{P’_{mn}} J_m^2(x) x dx=\pi C^2 (P’^2_{mn}-m^2)J_m^2(P’_{mn})\)
Compare the two side, so
\(\int_0^{p_{mn}’}[J_m’^2(x)+\frac{m^2}{x^2}J_m^2(x)]xdx=\frac{1}{2}(p_{mn}’^2-m^2)J_m^2(p_{mn})\)