Waveguides - NUS/ECE EE4101 Waveguides At high frequencies,...

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Hon Tat Hui Waveguides NUS/ECE EE4101 1 At high frequencies, the loss of electromagnetic waves traveling along transmission lines due to conductor resistance and radiation leakage becomes exceedingly large. To alleviate this problem, hollow waveguides can be used. We will study the rectangular waveguide as a typical example. Waveguides x y a b ε , μ x y ε , μ r z z Rectangular waveguide Circular waveguide
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Hon Tat Hui Waveguides NUS/ECE EE4101 2 General Field Expression inside a Waveguide ( ) z e y x γ = , ~ E E E E 2 2 2 = z Then Helmholtz’s equations: 0 2 2 = + E E k ( ) με ω = k 0 2 2 = + H H k Transverse directions: ( x , y ) or ( r , φ ) Longitudinal direction: z (propagation direction) In general
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Hon Tat Hui Waveguides NUS/ECE EE4101 3 Method of Solution: Express transverse field components E x , E y in terms of longitudinal field component E z Obtain solution for the longitudinal field E z from the wave equation Obtain E x, E y from E z Step 1 Step 2 Step 3
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Hon Tat Hui Waveguides NUS/ECE EE4101 4 In rectangular coordinates: 0 2 2 2 2 2 2 2 = + + + E k z y x 0 2 2 2 2 = + + E k z xy ( ) 0 2 2 2 = + + E E k xy γ ( ) 0 2 2 2 = + + E k xy Similarly, ( ) 0 2 2 2 = + + H H k xy (1a) (1b) 0 2 2 = + E E k
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Hon Tat Hui Waveguides NUS/ECE EE4101 5 H E ωμ j = × H z y x j E E E z y x z y x = ˆ ˆ ˆ x y z x y z H j ω E γ y E H j ω z E y E ~ ~ ~ μ = + = y z x y z x H j ω x E E γ H j ω x E z E ~ ~ ~ = = z x y z x y H j ω y E x E H j ω y E x E ~ ~ ~ = = (2a) (2b) (2c) Note that: ( ) ( ) () ( ) z y x i e y x H z y x H e y x E z y x E z i i z i i , , , ~ , , , ~ , , = = = γ
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Hon Tat Hui Waveguides NUS/ECE EE4101 6 Similarly from E H ωε j = × x y z E j ω H γ y H ~ ~ ~ ε = + (3a) y z x E j ω x H H γ ~ ~ ~ = z x y E j ω y H x H ~ ~ ~ = (3b) (3c) We have
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Hon Tat Hui Waveguides NUS/ECE EE4101 7 Finally from equation sets in (2) and (3), we have: + = y E j ω x H γ k γ H z z x ~ ~ 1 ~ 2 2 ε (4a) + + = x E j ω y H γ k γ H z z y ~ ~ 1 ~ 2 2 (4b) + + = y H j ω x E γ k γ E z z x ~ ~ 1 ~ 2 2 μ (4c) + = x H j ω y E γ k γ E z z y ~ ~ 1 ~ 2 2 (4d) Hence, we can solve the scalar Helmholtz’s equations for E z and H z , and use the above formulas to determine the other components. ~ ~
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Hon Tat Hui Waveguides NUS/ECE EE4101 8 Waveguide Mode Classification It is convenient to first classify waveguide modes as to whether E z or H z exists according to: TEM: E z = 0 H z = 0 TE: E z = 0 H z 0 TM: E z 0 H z = 0 TEM = Transverse ElectroMagnetic TE = Transverse Electric TM = Transverse Magnetic
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Hon Tat Hui Waveguides NUS/ECE EE4101 9 (1) TEM Modes: 0 ~ ~ = = = = z z z z H E H E From the equations in (4), for the existence of non-trivial solutions, the denominators must be zero also. That is, 0 2 2 = + k γ με ω
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This note was uploaded on 10/21/2011 for the course EE 4101 taught by Professor Yeotatsoon during the Spring '11 term at National University of Singapore.

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Waveguides - NUS/ECE EE4101 Waveguides At high frequencies,...

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