Electromagnetic Waves and Antennas combined - Chapter 2 potx

... expressions: A 2 = A 2 +τ s AB cos φ = A 2 −B 2 τ 2 s 1 −τ 2 s = 1 2  A 2 +B 2 +sD  = A 2 B 2 λ s B 2 = B 2 −τ s AB cos φ = B 2 −A 2 τ 2 s 1 −τ 2 s = 1 2  A 2 +B 2 −sD  = A 2 B 2 λ −s (2. 13.4) 2. 13. Problems ... OD, are given by: A  =  1 2 (A 2 +B 2 )+ s 2  (A 2 −B 2 ) 2 +4A 2 B 2 cos 2 φ B  =  1 2 (A 2 +B 2 )−...

Ngày tải lên: 13/08/2014, 02:20

... +ρ 2 2 ρ 2 3 +2 2 ρ 3 cos 2k 2 l 2 = ρ 2 1 This can be solved for cos 2k 2 l 2 : cos 2 k 2 l 2 = ρ 2 1 (1 +ρ 2 2 ρ 2 3 )−(ρ 2 2 +ρ 2 3 ) 2 2 ρ 3 (1 −ρ 2 1 ) Using the identity, cos 2 k 2 l 2 = 2 cos 2 k 2 l 2 −1, ... condition |Γ 2 |= |ρ 1 |, or, |Γ 2 | 2 = ρ 2 1 , which is written as:      ρ 2 +ρ 3 e −2jk 2 l 2 1 +ρ 2 ρ...

Ngày tải lên: 13/08/2014, 02:20

... (7.1.7): k 2 z +k 2 x = k 2 = n 2 k 2 0 k z  2 +k x  2 = k 2 = n 2 k 2 0 Because, k  x = k x = k sin θ = nk 0 sin θ, we may solve for k  z to get: k 2 z = n 2 k 2 0 −k 2 x = n 2 k 2 0 −k 2 x = ... requires that the numerator of (7. 12. 13) be positive, or, (1 +β 2 )cos θ 2 ≥ 0  cos θ ≥ 2 1 +β 2  θ ≤ acos  2 1 +β 2  (7. 12. 22) Because 2 β...

Ngày tải lên: 13/08/2014, 02:20

... k 2 0 n 2 1 −β 2 −α 2 c = k 2 0 n 2 2 −β 2 ⇒ k 2 c = k 2 0 n 2 1 −β 2 α 2 c = β 2 −k 2 0 n 2 2 (9.11.3) Similarly, Eqs. (9.11 .2) read: ∂ 2 x H z (x)+k 2 c H z (x)= 0 for |x|≤a ∂ 2 x H z (x)−α 2 c H z (x)= ... R s a 2  |H 0 | 2 +|H 1 | 2  +R s b|H 0 | 2 = R s a 2  |H 0 | 2 +|H 1 | 2 + 2b a |H 0 | 2  Using |H 0 | 2 +|H 1 | 2 =|E 0 |...

Ngày tải lên: 13/08/2014, 02:20

... constants: B 2 1 −4|C 1 | 2 = B 2 2 −4|C 2 | 2 = 4|S 12 S 21 | 2 (K 2 −1) |C 1 | 2 =|S 12 S 21 | 2 +  1 −|S 22 | 2  D 1 |C 2 | 2 =|S 12 S 21 | 2 +  1 −|S 11 | 2  D 2 (13.5 .2) 534 13. S-Parameters For ... −|S 11 | 2 |S 22 −ΔS ∗ 11 |+|S 12 S 21 | (Edwards-Sinsky stability parameter) μ 2 = 1 −|S 22 | 2 |S 11 −ΔS ∗ 22 |+|S 12 S...

Ngày tải lên: 13/08/2014, 02:20

... 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 ... x  = ⎡ ⎢ ⎢ ⎢ ⎣ τ  x  y  z  ⎤ ⎥ ⎥ ⎥ ⎦ ,L= ⎡ ⎢ ⎢ ⎢ ⎣ γ 00−γβ 0100 0010 −γβ 00 γ ⎤ ⎥ ⎥ ⎥ ⎦ (H .2) Such transformations leave the quadratic form (c 2 t 2 −x 2...

Ngày tải lên: 13/08/2014, 02:20

... energies: ¯ w mech = 1 2 Re  N  1 2 m|v| 2 + 1 2 mω 2 0 |x| 2  = 1 4 Nm(ω 2 +ω 2 0 )|x| 2 = 1 4 Nm(ω 2 +ω 2 0 )e 2 |E| 2 /m 2 (ω 2 0 −ω 2 ) 2 = 1 4  0 |E| 2  ω 2 p (ω 2 +ω 2 0 ) (ω 2 0 −ω 2 ) 2  where we used the definition (1.11.6) of the plasma frequency. It follows that Eq. (1.16 .2) can ... integrals, for which you may assume that 0 &...

Ngày tải lên: 13/08/2014, 02:20

... gain case: n = 1 + fω 2 p /2 ω 2 1 −ω 2 +jωγ + fω 2 p /2 ω 2 2 −ω 2 +jωγ n g = 1 + fω 2 p (ω 2 +ω 2 1 ) /2 (ω 2 1 −ω 2 +jωγ) 2 + fω 2 p (ω 2 +ω 2 2 ) /2 (ω 2 2 −ω 2 +jωγ) 2 (3.9.8) The two peaks ... parts: Re (n g ) = 1 + fω 2 p (ω 2 +ω 2 r )  (ω 2 −ω 2 r ) 2 −ω 2 γ 2   (ω 2 −ω 2 r ) 2 +ω 2 γ 2  2 Im(n g ) = fω 2 p γω...

Ngày tải lên: 13/08/2014, 02:20

... = n 2 1 +n 2 3 −N 2 N 2 (4.7.11) sin 2 θ = 1 − n 2 1 N 2 1 − n 2 1 n 2 3 , cos 2 θ = 1 − n 2 3 N 2 1 − n 2 3 n 2 1 (4.7. 12) cos 2 θ − n 2 1 N 2 =− n 2 1 n 2 3 sin 2 θ, sin 2 θ − n 2 3 N 2 =− n 2 3 n 2 1 cos 2 θ (4.7.13) Using ... relationships: sin 2 θ 1 − n 2 1 N 2 + cos 2 θ 1 − n 2 3 N 2 = 1 (4.7.9) n 3 n 1 cos 2 θ + n 1 n...

Ngày tải lên: 13/08/2014, 02:20

... have: Z 1 = η 2 1 Z 2 = η 2 1 η 2 2 /Z 3 = η 2 1 η 2 2 Z 3 = η 2 1 η 2 2 η b Inserting this into Γ 1 = (Z 1 −η a )/(Z 1 +η a ), we obtain: Γ 1 = η 2 1 η b −η 2 2 η a η 2 1 η b +η 2 2 η a The two ... η b . If η 1 is half- and η 2 quarter-wavelength, then, Z 1 = Z 2 = η 2 2 /Z 3 = η 2 2 /η b . And, if the quarter-wavelength is first and the half-wavelength se...

Ngày tải lên: 13/08/2014, 02:20