Presentation on theme: "1 ENE 428 Microwave Engineering Lecture 8 Rectangular waveguides and cavity resonator."— Presentation transcript:
1 ENE 428 Microwave Engineering Lecture 8 Rectangular waveguides and cavity resonator
2 TM waves in rectangular waveguides Finding E and H components in terms of z, WG geometry, and modes. From Expanding for z-propagating field for the lossless WG gets where
3 Method of separation of variables (1) Assume where X = f(x) and Y = f(y). Substituting XY gives and we can show that for lossless WG.
4 Method of separation of variables (2) Let and then we can write We obtain two separate ordinary differential equations:
5 General solutions Appropriate forms must be chosen to satisfy boundary conditions.
6 Properties of wave in rectangular WGs (1) 1.in the x-direction E t at the wall = 0 E z (0,y) and E z (a,y) = 0 and X(x) must equal zero at x = 0, and x = a. Apply x = 0, we found that C 1 = 0 and X(x) = c 2 sin( x x). Therefore, at x = a, c 2 sin( x a) = 0.
7 Properties of wave in rectangular WGs (2) 2. in the y-direction E t at the wall = 0 E z (x,0) and E z (x,b) = 0 and Y(y) must equal zero at y = 0, and y = b. Apply y = 0, we found that C 3 = 0 and Y(y) = c 4 sin( y y). Therefore, at y = a, c 4 sin( y b) = 0.
8 Properties of wave in rectangular WGs (3) and therefore we can write
9 Every combination of integers m and n defines possible mode for TM mn mode. m = number of half-cycle variations of the fields in the x- direction n = number of half-cycle variations of the fields in the y- direction For TM mode, neither m and n can be zero otherwise E z and all other components will vanish therefore TM 11 is the lowest cutoff mode. TM mode of propagation
TM 11 field lines 10 Side view End view
11 Cutoff frequency and wavelength of TM mode
12 Ex2 A rectangular wg having the interior dimension a = 2.3cm and b = 1cm filled with a medium characterized by r = 2.25, r = 1 Find h, f c, and c for TM 11 mode If the operating frequency is 15% higher than the cutoff frequency, find (Z) TM11, ( ) TM11, and ( g ) TM11. Assume the wg to be lossless for propagating modes.
13 TE waves in rectangular waveguides (1) E z = 0 From Expanding for z-propagating field for a lossless WG gets where
14 TE waves in rectangular waveguides (2) In the x-direction Since E y = 0, then from we have at x = 0 and x = a
15 TE waves in rectangular waveguides (3) In the y-direction Since E x = 0, then from we have at y = 0 and y = b
16 Method of separation of variables (1) Assume then we have
17 Properties of TE wave in x-direction of rectangular WGs (1) 1.in the x-direction at x = 0, at x = a,
18 Properties of TE wave in x-direction of rectangular WGs (2)
19 Properties of TE wave in y-direction of rectangular WGs (1) 2. in the y -direction at y = 0, at y = b,
20 Properties of TE wave in y-direction of rectangular WGs (2) For lossless TE rectangular waveguides,
21 Cutoff frequency and wavelength of TE mode
TE 10 field lines 22 Side view End view Top view
23 A dominant mode for TE waves For TE mode, either m or n can be zero, if a > b, is a smallest eigne value and f c is lowest when m = 1 and n = 0 (dominant mode for a > b)
24 A dominant mode for TM waves For TM mode, neither m nor n can be zero, if a > b, f c is lowest when m = 1 and n = 1
25 Ex1 a) What is the dominant mode of an axb rectangular WG if a < b and what is its cutoff frequency? b) What are the cutoff frequencies in a square WG (a = b) for TM 11, TE 20, and TE 01 modes?
26 Ex2 Which TM and TE modes can propagate in the polyethylene-filled rectangular WG ( r = 2.25, r = 1) if the operating frequency is 19 GHz given a = 1.5 cm and b = 0.6 cm?
27 Rectangular cavity resonators (1) At microwave frequencies, circuits with the dimension comparable to the operating wavelength become efficient radiators An enclose cavity is preferred to confine EM field, provide large areas for current flow. These enclosures are called ‘cavity resonators’. There are both TE and TM modes but not unique. a b d
28 Rectangular cavity resonators (2) z-axis is chosen as the reference. “mnp” subscript is needed to designate a TM or TE standing wave pattern in a cavity resonator.
29 Electric field representation in TM mnp modes (1) The presence of the reflection at z = d results in a standing wave with sin z or coz z terms. Consider transverse components E y (x,y,z), from B.C. E y = 0 at z = 0 and z = d 1) its z dependence must be the sin z type 2) similar to E x (x,y,z).
30 Electric field representation in TM mnp modes (2) From H z vanishes for TM mode, therefore
31 Electric field representation in TM mnp modes (3) If E x and E y depend on sin z then E z must vary according to cos z, therefore
32 Magnetic field representation in TE mnp modes (1) Apply similar approaches, namely 1)transverse components of E vanish at z = 0 and z = d - require a factor in E x and E y as well as H z. 2)factor indicates a negative partial derivative with z. - require a factor for H x and H y f mnp is similar to TM mnp.
33 Dominant mode The mode with a lowest resonant frequency is called ‘dominant mode’. Different modes having the same f mnp are called degenerate modes.
34 Resonator excitation (1) For a particular mode, we need to 1)place an inner conductor of the coaxial cable where the electric field is maximum. 2)introduce a small loop at a location where the flux of the desired mode linking the loop is maximum. source frequency = resonant frequency
35 Resonator excitation (2) For example, TE 101 mode, only 3 non-zero components are E y, H x, and H z. insert a probe in the center region of the top or bottom face where E y is maximum or place a loop to couple H x maximum inside a front or back face. Best location is affected by impedance matching requirements of the microwave circuit of which the resonator is a part.
36 Coupling energy method place a hole or iris at the appropriate location field in the waveguide at the hole must have a component that is favorable in exciting the desired mode in the resonator.
37 Ex3 Determine the dominant modes and their frequencies in an air-filled rectangular cavity resonator for a > b > d a > d > b a = b = d