Homework7withSolutions
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University of California, Davis**We aren't endorsed by this school
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ECE EEC 230
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Electrical Engineering
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Dec 25, 2024
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UNIVERSITY OF CALIFORNIA, DAVISDepartment of Electrical and Computer Engineering EEC230 Electromagnetics Fall 2023 Homework #7 1.The general solution for the longitudinal fields in a cylindrical waveguide is given by π΄π΄π§π§(ππ,ππ) = [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)]β π½π½ππ(ππππππ)where π΄π΄π§π§= {πΈπΈπ§π§,π»π»π§π§}for TMz and TEz waves, respectively; A and B are constant to be determined; and ππππand ππare wavevector components along the ππand ππdirections. Demonstrate that ππmust be an integer number. Hint: Keep in mind that a cylindrical waveguide is periodic in ππ. Solution: We knot that the waveguide is periodic in ππ, i.e., π΄π΄π§π§(ππ,ππ) =π΄π΄π§π§(ππ,ππ+ 2ππβ). Therefore [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)]β π½π½ππ(ππππππ) =οΏ½π΄π΄cosοΏ½ππ(ππ+ 2ππβ)οΏ½+π΅π΅sinοΏ½ππ(ππ+ 2ππβ)οΏ½οΏ½ β π½π½ππ(ππππππ)Taking advantage of trigonometric relations, we find that [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)] = π΄π΄[cos(ππππ) cos(ππ2ππβ)βsin(ππππ) sin(ππ2ππβ)] +π΅π΅[sin(ππππ) cos(ππ2ππβ)βcos(ππππ) sin(ππ2ππβ)]To enforce this condition, it is required that cos(ππ2ππβ) = 1and sin(ππ2ππβ)=0. The only possibility then is that nmust be an integer. If that is the case, we find that [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)] =π΄π΄[cos(ππππ)β1β0] +π΅π΅[sin(ππππ)β1β0] = [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)]2.The general solution for the longitudinal fields in a cylindrical waveguide is given by π΄π΄π§π§= [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)]β π½π½ππ(ππππππ)where π΄π΄π§π§= {πΈπΈπ§π§,π»π»π§π§}for TMz and TEz waves, respectively; A and B are constant to be determined; and ππππand ππare wavevector components along the ππand ππdirections. Demonstrate that this expression can be simplified to π΄π΄π§π§=πΊπΊcos(ππππ)β π½π½ππ(ππππππ)or to π΄π΄π§π§=πΉπΉsin(ππππ)β π½π½ππ(ππππππ)by choosing an adequate coordinate origin on ππ. Hint: Keep in mind that variables A or B may be related to the origin of the coordinate system in ππ. Solution: Let us consider that the constant A and B acquired a desired value, for instance: π΄π΄=πΉπΉcos(ππππ0)and π΅π΅=βπΉπΉsin(ππππ0), where ππ0is an arbitrary azimuthal point. Then, we find that π΄π΄π§π§= [π΄π΄cos(ππππ) +π΅π΅sin(ππππ)]β π½π½ππ(ππππππ) = [πΉπΉcos(ππππ0) cos(ππππ)β πΉπΉsin(ππππ0) sin(ππππ)]β π½π½ππ(ππππππ)=πΉπΉsinοΏ½ππ(ππ β ππ0)οΏ½In this last expression, F is associate to the modal amplitude and ππ0is the coordinate origin, which can be set to zero. Therefore, π΄π΄π§π§=πΉπΉsin(ππππ)β π½π½ππ(ππππππ)
3.Let us a consider a sectorial waveguide with radius of 1 cm, as the shown in the figure. The walls are metallic and can be considered as perfect electric conductor (PEC). a)Derive the longitudinal fields for TE and TM modes. b)Obtain the cut-off frequencies of the 5 first propagative modes. c)Answer to questions a) and b) assuming now that the waveguide is as shown in the next figure. a)The longitudinal fields can be obtained following the same procedure as with the cylindrical waveguide, i.e. Note that the constant D has already be assumed to be zero, because the field cannot be singular at the origin. TM modes
Note that we have not consider k=0, because then the longitudinal electric field would be zero for all angles. The final longitudinal field yields: TE Modes Note that in this case k=0 is a valid solution, since it does not force the longitudinal magnetic field to be zero in all cases. Then, this field yields: b)The cut-off frequencies are given by:
Which can be re-arranged as: c)For the second waveguide, we obtain the same results as in a) and b), because both structures share boundary conditions, and therefore they governed by the same expressions of the electromagnetic fields.
4.Let us a consider a sectorial waveguide as the one shown in the figure. The radius is a=1cm and the internal angle of the waveguide cross section is 45 degrees. Figure 1. A sectorial waveguide a) Derive the expression that describes the electric field supported by the waveguide for TM modes. Obtain the expression that describe the waveguide cut-off frequency. What is the fundamental TM mode supported by this waveguide? Obtain its cut-off frequency. Since we are dealing with TM modes: Then, we impose boundary conditions: ππ= 0βEz must be finite. Therefore, D=0. ππ= 0βπΈπΈπ§π§=π΄π΄+π΅π΅ β0β π΄π΄= 0βππ=ππ4βNote that p=0 is not a valid solution, as it would lead to a constant Ez without any spatial dependence. We note that . Then, the longitudinal electric field yields with We also impose the boundary conditions at: ππ=ππ βThen, the cut-off frequency yields:
Applying the transverse-longitudinal decomposition, we obtain the transverse electric field components as: The fundamental mode is the TM41, with a cut-off frequency Note that v=1, n=4. b) Let us a consider a section of 5cm of this sectorial waveguide. Both ends have been short-circuited using metallic plates. Obtain the expressions that describe the electric field of the resonant TM modes. We impose boundary condition at z=0 and z=d. The boundary condition can be imposed on the transverse component of the electric field. For instance, we can operate with πΈπΈππ. Then:
Similarly, we obtain that And the longitudinal expression yields: Finally, we need to determine the propagation constant. To this purpose, we impose the boundary condition at z=d=5cm: c) Derive the expression of the TM modes resonant frequency. Obtain the resonant frequency of the mode TM411. The structure will resonate when Then:
We can then isolate the expression from the cut-off frequency as: Finally:
5. Let us a consider a waveguide that possesses the cross-section shown in the figure. The waveguide has PEC walls and is filled with air. a. Derive the expression of the longitudinal electric field for all TM modes. We start from the solution of the longitudinal electric field in cylindrical coordinates: For TM modes, Hz=0 and Then, we apply boundary conditions on the waveguide walls: We note that n cannot be equal to zero. Then: with The ratio between these equations yields Which can be rewritten as
We note that this equation does not admit an analytical solution. However, we can simplify it as: Then, the expression of the longitudinal field yields: Where b. Calculate the cut-off frequency of the first supported TM mode. To this purpose, please use the data shown in the figure.
We can obtain the cut-off frequency through the transverse wavenumber as Considering the tables shown in the figure, and that yields to the lower wavenumber We find that Finally
6. Derive the expressions that describe the electric and magnetic fields as well as the quality factor of a cylindrical resonator (PEC wall, filled with air; dimensions: radius βaβ and length βlβ) in which the mode TE011 is excited. The following expressions might be useful: We can start from the expressions that describe the fields for TE modes in a cylindrical waveguide: Where: In the case of the TE01 mode: We note that the waveguide is ended with metals. This imposes a boundary condition for the transverse electric field at z=0 and z=l. The transverse components of the electric field are πΈπΈππand πΈπΈππ. We select the πΈπΈππcomponent because πΈπΈππ= 0. In the particular case of the TE01 mode, this electric field component yields: We apply now boundary conditions at z=0: then: In addition, the longitudinal expression of the magnetic field yields:
Imposing boundary condition, we get that . Then: Finally, we need to operate with the π»π»ππcomponent. We note that this expression depends on the propagation constant π½π½. Therefore, the expression of this field component will be different for waves propagating forward and backward: In addition, we know that . The field component then reads The second part of the problem requires calculating the quality factor of the resonator: We select to calculate the electric energy because it only has one component. Then: The integrals yield And we arrive to: The next step is to calculate the power dissipated on the walls. Note that since the resonator is empty there
are not losses in the dielectric. We can consider the integrals individually. On the top and bottom covers: On the lateral walls: Thus, the total dissipated power yields: Finally, the quality factor is: