Electromagnetic Waves and Antennas combined - Chapter 6 potx

... Structures r= -0 .1000 -0 .2000 -0 .4000 0.5000 A= 1.0000 1.0000 1.0000 1.0000 -0 .1000 -0 .1200 -0 .2000 0 -0 . 064 0 -0 .1000 0 0 -0 .0500000 B= -0 .1000 -0 .2000 -0 .4000 0.5000 -0 .1880 -0 . 360 0 0.5000 0 -0 .3500 0.5000 ... [1, -0 .1, -0 . 064 , -0 .05]; b = [-0 .1, -0 .188, -0 .35, 0.5]; [r,A,B] = bkwrec(a,b); n = r2n(r); r...

Ngày tải lên: 13/08/2014, 02:20

... other. Examples of such waves are the evanescent waves in total internal reflection, various guided-wave problems, such as surface waves, leaky waves, and traveling-wave antennas. The most famous ... still a right-handed vector system, but {E 0 , H 0 , k} will be a left-handed system. This is the reason why Veselago [3 76] named such media left-handed media. † In a lossy negative-ind...

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... as p-polarization, π-polarization, or TM po- larization, the electric fields lie on the plane of incidence and the magnetic fields are Fig. 7.1.1 Oblique incidence for TM- and TE-polarized waves. 7.1. ... real-parts in the right-hand side of Eq. (7. 16. 8) are zero, showing that |ρ TE |=|ρ TM |=1 and there is no (time-averaged) power flow into the right medium. For magnetic media, includ...

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... 9 .6. 1 for the C-band and X-band waveguides, WR-159 and WR-90.  Example 9.8.2: WR-159 Waveguide. Consider the C-band WR-159 air-filled waveguide whose characteristics were listed in Table 9 .6. 1. ... 0.40 6. 56 8.20 12.50 X 250 kW 0.110 WR -6 2 0 .62 2 0.311 9.49 11.90 18.00 Ku 140 kW 0.1 76 WR-42 0.42 0.17 14.05 17 .60 26. 70 K 50 kW 0.370 WR-28 0.28 0.14 21 .08 26. 40 40.00 Ka 27...

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... 0. 866 6∠50.80 o ,r 3 = 0.3893 % load and source stability circles: [cL,rL] = sgcirc(S,’l’); % c L = 2. 160 8 ∠50.80 o ,r L = 1.2 965 [cG,rG] = sgcirc(S,’s’); % c G = 1.74 56 171 .69 o ,r G = 0.8 566 smith; ... circle 13.13. Problems 569 13.7 Computer Experiment. The Hewlett-Packard ATF- 361 63 pseudomorphic high electron mo- bility transistor (PHEMT) has the following S- and noise pa...

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... length dualband - two-section dual-band Chebyshev impedance transformer dualbw - two-section dual-band transformer bandwidths stub1 - single-stub matching stub2 - double-stub matching stub3 - triple-stub ... 4 dielectric slab, 162 , 3 06 half-wave, 165 quarter-wave, 165 reflectionless, 165 dielectric waveguides, 3 86 diffraction integrals, 960 diffraction theory, 67...

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... formula [147], where λ and λ i are in units of μm: n 2 = 1 + 0 .69 6 166 3 λ 2 λ 2 −(0. 068 4043) 2 + 0.40794 26 λ 2 λ 2 −(0.1 162 414) 2 + 0.8974794 λ 2 λ 2 −(9.8 961 61) 2 (1.11. 16) 1.12 Conductors The ... losses) (1.9 .6) where J tot = J + jωD is the total current in the right-hand side of Amp ` ere’s law and accounts for both conducting and dielectric losses. The time-averaged v...

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... , dn dλ =− 11.98× 10 −3 μm −1 , d 2 n dλ 2 =−4.24×10 −3 μm −2 (3 .6. 13) resulting in the group index n g = 1. 463 and group velocity v g = c/n g = 0 .68 4c. Using (3 .6. 10) and (3 .6. 11), the calculated values of D and k  are: D = 21.9 ps km ... quadrature weights and points h = - wp^2 * tf * J1over(wp*sqrt(x.^2 - tf^2)) .* exp(j*w0*(t(i)-x)); Ez(i) = exp(j*w0*(t(i)-tf))...

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... k + and backward with k − , and the reverse is true of the E − (z) component. The forward-moving component of E + and the backward-moving component of E − , that is, E R+ and E R− , are 1 36 4. ... determined. To obtain both the TE and TM solutions, we assume initially that E has all its three components and rewrite Eq. (4 .6. 16) component-wise. Using Eq. (4 .6. 5) and noting...

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... half-wave, (c) layer-1 is quarter- and layer-2 half-wave, and (d) layer-1 is half- and layer-2 quarter-wave. Show that the reflection coefficient at interface-1 is given by the following expressions ... (5.2.3) and (5.2.4). 5.3. Reflected and Transmitted Power 159 5.3 Reflected and Transmitted Power For waves propagating in the z-direction, the time-averaged Poynting vector has only...

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