Circular Waveguides in ANSYS Module

Welcome back to this ANSYS module on designing waveguides using the ANSYS electronic desktop student version, AEDT for short. And in that previous module we analyzed a rectangular waveguide using the AEDT HFSS simulation and we looked at various results. And in today's module our focus will be on the governing equations of a circular waveguide. I'm going to change up the geometry. And we'll derive that using Maxwell's equations. So a circular waveguide has a circular cross-section. And in general, this cross-section is in the xy-plane and is uniform along that z-axis. So let's define its radius as A. And a typical waveguide cavity can be empty or filled with some dielectric. with some lossless dielectric material and with some relative permittivity epsilon and permeability mu. The outside walls of that circular waveguide can be a PEC, perfect electroconductor, or metal, or some metal, or some conductor. And like that rectangular waveguide, the circular waveguide is made up of a single conductor. Again, no TEM mode exists. electric mode only TE and TM transverse electric and transfer magnetic and derive these modes let's start with Maxwell's equations for a source free region and in phasor format we have where the electric and the magnetic fields have components along x y and z axis in the Cartesian coordinate system and since the waveguide geometry is identical to a long cylinder will represent both the electric and the magnetic fields in terms of the cylindrical coordinates where we have rho and phi are related to x and y in that cortesian coordinate system as and row can range from 0 to a while 5 ranges from 0 to 2 pi and this representation also results in the simplification of the resulting equations here we have E rho, E phi, H rho, and H phi. And these are the transverse components for the cylindrical system. And EZ and HZ are the longitudinal components to the wave propagation. And the fields are propagating along that z-axis, same as in the rectangular waveguide. The variation along the z-axis will also be in terms of the exponential of j times beta z. And beta is the propagation constant for the medium. as the dielectric is assumed to be lossless again. Now let's expand Maxwell's equations. We'll expand in terms of the individual components and express the transverse components in terms of the longitudinal components. And they yield these results. And here, kc is a cut-off wave number and it's defined by this equation. And k is the wave number in that medium, in that dielectric. And it's given by omega times the square root of mu times epsilon, or 2 times pi divided by lambda. So let's find the solution for the longitudinal components of the fields. And then derive the transverse components from the launch. First, let us consider a general function phi to represent the longitudinal component for deriving the general solution. The longitudinal means lengthwise, so in our case lengthwise in the z direction, which is in the direction of the propagation. And for the transverse electric case, this function is equal to H as E equals zero. And for the TM case to transverse, this function is equal to H as E equals zero. versus magnetic case, this function is equal to E of z as H of z equals zero. And the wave equation in terms of this function is given by an expanding that del squared operator in terms of cylindrical coordinates yields this equation. And if the function can be expressed as the product of some functions in terms of the individual cylindrical coordinates, then by using the separation of variable methods again, that partial differential equation, that PDE, can be written in terms of normal differential equations, shown here. Divide the equation with the function, yields, And given that the variation along the z-axis is expressed in terms of the exponential of j times beta times z, we can now represent the z function as z of z equals c1 times the exponential of j times beta times z. And c1 is the amplitude constant. And the double derivative of z with respect to z is given as a negative. of the square of the beta z and this can be rewritten as shown here and substituting this value in the equation simplifies it to this. Substituting the wave number now results in this and if the differential terms with respect to rho and phi and they're grouped into two sides of the equations then we can rewrite the equation as follows. And each side of the equation only depends on one variable. And the solution for this differential equation exists only if the differential equation is each side is equal to some constant. So let's define the separation constant as k rho and k phi and these are related to the wave number as shown here. And the right-hand side term in the equation it can be expressed as follows. So the wave equation is modified to this And the general equation for this differential equation is given by where c2 and c3 are the integration constants. And this function must be periodic in phi, which leaves k. k phi is equal to some integer n. And so the equation becomes, now substitute the value of k phi in the wave equation gives us this. And this differential equation is now reduced to Bessel's differential equation and whose general solution is given in the form where j n of k c times rho and y n of k c times rho are the Bessel functions of the first and second kind respectively and c 4 and c 5 are some constants and since the Bessel function of the second kind is not applicable from cylindrical waveguides gives us C5 must be zero and therefore the solution to the general function for a cylindrical waveguide is given as and to simplify the expression the products of the constants are replaced by A and B. Using the boundary conditions for the TE case the solution for the longitudinal magnetic field component H of Z is given here The cutoff wave number is determined by applying the boundary condition that the tangential component of the electric field is zero at the conductor surface. And since E of Z is zero, the other tangential components of the electric field E phi of rho phi and Z at rho equal to A equals zero. Here's the modified equation between E phi and H of Z. The transfer's field along phi for H of Z is given by this. Here Jn'of Kc times rho is the derivative of Jn with respect to rho. For E phi to be zero, rho has to be equal to A. Jn of Kc times A should be zero, which gives the cutoff wave number for the mode as shown. Your PnM'is the mth root of Jn prime. And for the TM case the solution for the longitudinal electric field component E of Z is shown here. Now we can directly apply the same boundary conditions to this longitudinal component. which gives the cutoff wave number here. And here Pnm is the mth root of Jn of Kc times rho. And the values of m and n are defined the mode of the field and denotes the number of full cycle variations along the circumference and M denotes the number of half cycle variations along the radial direction. And with the value of M must always be greater than or equal than or equal to one in the transverse plane. And to define the TE and TM modes we use NMM as indices. It's a way of categorizing the modes TENM and TMNM. And usually we follow a standard convention for numbering the modes. We use the mode number along the circumference first and then the mode number along the radial. direction. That's the normal way of doing it. The equation relating the propagation constant to the wave number, the cutoff wave numbers, can be derived here. And substituting the value for Kc for both the modes gives us this. And as the wave number depends on the frequency, this equation can be rewritten to define F of C, the cutoff frequency. and this depends on the values of m and n. Again, the cutoff frequency is different for different modes. Here's a question. What's the minimum value for m and n for the Te and Tm modes of a cylindrical waveguide? As we discussed earlier, m represents the number of roots for the Bessel's function, for Bessel's first function, and that's why we typically start with m equals one. So let's look at the longitudinal wave. components. Pay attention to the terms containing n. These fields are non-zero even if n equals zero. So the value of n starts with zero. If we want to find the dominant mode, the mode with the lowest cutoff frequency, we need to know the values of the roots for the Bessel's function. Usually the TE11 mode has the lowest cutoff frequency. because it has the smallest value of the root, of Bessel's root. And the next modes are TM01 and TE21. These are the governing equations of a circular waveguide. And in order to design waveguides, it's important to have a good understanding of the equations involved. In the upcoming module, we'll be simulating a circular waveguide in ANSYS Electronic Desktop HFSS, which is a very important part of the system. instead of performing an exact analysis using Maxwell's equations, we'll be using a physical model geometry. So thank you for watching this module. And to find more information on AEDT, ANSYS Electronic Desktop, HFSS, or for any of the ANSYS simulation tools, please go to ansys.com forward slash courses today.

And we'll derive that using Maxwell's equations. So a circular waveguide has a circular cross-section. And in general, this cross-section is in the xy-plane and is uniform along that z-axis.

So let's define its radius as A. And a typical waveguide cavity can be empty or filled with some dielectric. with some lossless dielectric material and with some relative permittivity epsilon and permeability mu. The outside walls of that circular waveguide can be a PEC, perfect electroconductor, or metal, or some metal, or some conductor.

And like that rectangular waveguide, the circular waveguide is made up of a single conductor. Again, no TEM mode exists. electric mode only TE and TM transverse electric and transfer magnetic and derive these modes let's start with Maxwell's equations for a source free region and in phasor format we have where the electric and the magnetic fields have components along x y and z axis in the Cartesian coordinate system and since the waveguide geometry is identical to a long cylinder will represent both the electric and the magnetic fields in terms of the cylindrical coordinates where we have rho and phi are related to x and y in that cortesian coordinate system as and row can range from 0 to a while 5 ranges from 0 to 2 pi and this representation also results in the simplification of the resulting equations here we have E rho, E phi, H rho, and H phi. And these are the transverse components for the cylindrical system. And EZ and HZ are the longitudinal components to the wave propagation.

And the fields are propagating along that z-axis, same as in the rectangular waveguide. The variation along the z-axis will also be in terms of the exponential of j times beta z. And beta is the propagation constant for the medium. as the dielectric is assumed to be lossless again. Now let's expand Maxwell's equations.

We'll expand in terms of the individual components and express the transverse components in terms of the longitudinal components. And they yield these results. And here, kc is a cut-off wave number and it's defined by this equation. And k is the wave number in that medium, in that dielectric. And it's given by omega times the square root of mu times epsilon, or 2 times pi divided by lambda.

So let's find the solution for the longitudinal components of the fields. And then derive the transverse components from the launch. First, let us consider a general function phi to represent the longitudinal component for deriving the general solution. The longitudinal means lengthwise, so in our case lengthwise in the z direction, which is in the direction of the propagation.

And for the transverse electric case, this function is equal to H as E equals zero. And for the TM case to transverse, this function is equal to H as E equals zero. versus magnetic case, this function is equal to E of z as H of z equals zero.

And the wave equation in terms of this function is given by an expanding that del squared operator in terms of cylindrical coordinates yields this equation. And if the function can be expressed as the product of some functions in terms of the individual cylindrical coordinates, then by using the separation of variable methods again, that partial differential equation, that PDE, can be written in terms of normal differential equations, shown here. Divide the equation with the function, yields, And given that the variation along the z-axis is expressed in terms of the exponential of j times beta times z, we can now represent the z function as z of z equals c1 times the exponential of j times beta times z.

And c1 is the amplitude constant. And the double derivative of z with respect to z is given as a negative. of the square of the beta z and this can be rewritten as shown here and substituting this value in the equation simplifies it to this.

Substituting the wave number now results in this and if the differential terms with respect to rho and phi and they're grouped into two sides of the equations then we can rewrite the equation as follows. And each side of the equation only depends on one variable. And the solution for this differential equation exists only if the differential equation is each side is equal to some constant. So let's define the separation constant as k rho and k phi and these are related to the wave number as shown here.

And the right-hand side term in the equation it can be expressed as follows. So the wave equation is modified to this And the general equation for this differential equation is given by where c2 and c3 are the integration constants. And this function must be periodic in phi, which leaves k.

k phi is equal to some integer n. And so the equation becomes, now substitute the value of k phi in the wave equation gives us this. And this differential equation is now reduced to Bessel's differential equation and whose general solution is given in the form where j n of k c times rho and y n of k c times rho are the Bessel functions of the first and second kind respectively and c 4 and c 5 are some constants and since the Bessel function of the second kind is not applicable from cylindrical waveguides gives us C5 must be zero and therefore the solution to the general function for a cylindrical waveguide is given as and to simplify the expression the products of the constants are replaced by A and B.

Using the boundary conditions for the TE case the solution for the longitudinal magnetic field component H of Z is given here The cutoff wave number is determined by applying the boundary condition that the tangential component of the electric field is zero at the conductor surface. And since E of Z is zero, the other tangential components of the electric field E phi of rho phi and Z at rho equal to A equals zero. Here's the modified equation between E phi and H of Z.

The transfer's field along phi for H of Z is given by this. Here Jn'of Kc times rho is the derivative of Jn with respect to rho. For E phi to be zero, rho has to be equal to A.

Jn of Kc times A should be zero, which gives the cutoff wave number for the mode as shown. Your PnM'is the mth root of Jn prime. And for the TM case the solution for the longitudinal electric field component E of Z is shown here.

Now we can directly apply the same boundary conditions to this longitudinal component. which gives the cutoff wave number here. And here Pnm is the mth root of Jn of Kc times rho. And the values of m and n are defined the mode of the field and denotes the number of full cycle variations along the circumference and M denotes the number of half cycle variations along the radial direction.

And with the value of M must always be greater than or equal than or equal to one in the transverse plane. And to define the TE and TM modes we use NMM as indices. It's a way of categorizing the modes TENM and TMNM.

And usually we follow a standard convention for numbering the modes. We use the mode number along the circumference first and then the mode number along the radial. direction. That's the normal way of doing it. The equation relating the propagation constant to the wave number, the cutoff wave numbers, can be derived here.

And substituting the value for Kc for both the modes gives us this. And as the wave number depends on the frequency, this equation can be rewritten to define F of C, the cutoff frequency. and this depends on the values of m and n.

Again, the cutoff frequency is different for different modes. Here's a question. What's the minimum value for m and n for the Te and Tm modes of a cylindrical waveguide? As we discussed earlier, m represents the number of roots for the Bessel's function, for Bessel's first function, and that's why we typically start with m equals one.

So let's look at the longitudinal wave. components. Pay attention to the terms containing n. These fields are non-zero even if n equals zero. So the value of n starts with zero.

If we want to find the dominant mode, the mode with the lowest cutoff frequency, we need to know the values of the roots for the Bessel's function. Usually the TE11 mode has the lowest cutoff frequency. because it has the smallest value of the root, of Bessel's root. And the next modes are TM01 and TE21.

These are the governing equations of a circular waveguide. And in order to design waveguides, it's important to have a good understanding of the equations involved. In the upcoming module, we'll be simulating a circular waveguide in ANSYS Electronic Desktop HFSS, which is a very important part of the system.

instead of performing an exact analysis using Maxwell's equations, we'll be using a physical model geometry. So thank you for watching this module. And to find more information on AEDT, ANSYS Electronic Desktop, HFSS, or for any of the ANSYS simulation tools, please go to ansys.com forward slash courses today.

Transcript for:Circular Waveguides in ANSYS Module

Transcript for:
Circular Waveguides in ANSYS Module