{Zl = Rl Ø!>6|÷‚©Ùž„IĘN\ƒ " ù«{|¡£Ä�gŽéu‡Åó§¥�Ù§¯KkéŒ�éu" ���777ÅÅÅDDDÑÑт‚‚ IOG�))K1á´§Zl = 0§Xã2¤«" Ù§G�))Zl = jXl§ ��ݧ ∆Z = 1 4 λg − λg 4pi ϕl 1 Figu"/> {Zl = Rl Ø!>6|÷‚©Ùž„IĘN\ƒ " ù«{|¡£Ä�gŽéu‡Åó§¥�Ù§¯KkéŒ�éu" ���777ÅÅÅDDDÑÑт‚‚ IOG�))K1á´§Zl = 0§Xã2¤«" "/>

c5

2011-10-26 30页 pdf 691KB 64阅读

用户头像

is_055469

暂无简介

举报
c5 DÑ‚Ý ) þ˜ù·‚éu�7ÅDтÚ17ÅDтÚ? IOG�Ú��Ý∆z�Vg"3�7ÅDт¥§rá´óŠG �ŠIOG�¶��aq§317ÅG�¥§Kr�K1>{Zl = Rl Ø!>6|÷‚©Ùž„IĘN\ƒ " ù«{|¡£Ä�gŽéu‡Åó§¥�Ù§¯KkéŒ�éu" ���777ÅÅÅDDDÑÑт‚‚ IOG�))K1á´§Zl = 0§Xã2¤«" Ù§G�))Zl = jXl§ ��ݧ ∆Z = 1 4 λg − λg 4pi ϕl 1 Figure 1: ...
c5
DÑ‚Ý ) þ˜ù·‚éu�7ÅDтÚ17ÅDтÚ? IOG�Ú��Ý∆z�Vg"3�7ÅDт¥§rá´óŠG �ŠIOG�¶��aq§317ÅG�¥§Kr�K1>{Zl = Rl < Z0ŠIOG�§Ù§G�´3IOG�þ\ ˜‡��Ý∆z (note: ∆zŒ�ŒK)"��ª�>Ø!>6|÷‚©Ùž„IĘN\ƒ " ù«{|¡£Ä�gŽéu‡Åó§¥�Ù§¯KkéŒ�éu" ���777ÅÅÅDDDÑÑт‚‚ IOG�))K1á´§Zl = 0§Xã2¤«" Ù§G�))Zl = jXl§ ��ݧ ∆Z = 1 4 λg − λg 4pi ϕl 1 Figure 1: ��Ý�Š^" 2 Figure 2: �7Åá´K1Dт>Ø>6©Ù" ½§  ϕl = pi − 2 tan−1 ( X1 Z0 ) ∆z = λg 2pi tan−1 ( Xl Z0 ) Dт{|§ { Z(z′) = jZ0 tanβ(z′ + ∆z) 3 Figure 3: >|K1Dт>Ø©Ù" 4 Figure 4: >|K1Dт{|©Ù" 5 Dт>Ø>6©Ù§  U(z′) = j2U + l e j 12 (ϕl−pi) sinβ(z′ + ∆z) I(z′) = 2I+l e j 12 (ϕl−pi) cosβ(z′ + ∆z) 111777ÅÅÅDDDÑÑт‚‚ IOG����>{§Zl = Rl < Z0§Xã5¤«§ ?¿K1œ¹§Zl = Rl + jXl§ ��ݧ ∆Z = 1 4 λg − λg 4pi ϕl ½  ϕl = pi − [ tan−1 ( X1 Z0 +Rl ) + 2 tan−1 ( X1 Z0 −Rl )] ∆z = λg 2pi [ tan−1 ( X1 Z0 +Rl ) + 2 tan−1 ( X1 Z0 −Rl )] 17ÅDт{|§ Z(z′) = jZ0 1 + jρ tanβ(z′ + ∆z) ρ+ j tanβ(z′ + ∆z) Ù>Ø>6©Ù§  U(z′) = U + l (1− |Γl|)ejβz ′ + j2U+l |Γl|ej 1 2 (ϕl−pi) sinβ(z′ + ∆z) I(z′) = I+l (1− |Γl|)ejβz ′ + 2I+l |Γl|ej 1 2 (ϕl−pi) cosβ(z′ + ∆z) 6 Figure 5: 17Å�>{K1Dт>Ø>6©Ù" 7 Figure 6: ?¿K1Dт>Ø©Ù" 8 Figure 7: ?¿K1Dт{|©Ù" 9 3�Ù¥§·‚òlp�áv:5w–Dт¯K" § 1 DDDÑÑт‚‚ããã���ÝÝÝ ))) c¡®²?ØL§Dт�¦)´ÏL‡©§�Ï)\>.^‡5�¤�" 3þ¡?Ø¥®‰·‚˜‡­‡é«µDт�ˆ«A^ь±8(˜ãݏl�Dтã§Ø+´á´!m´½? ¿K1"Dтãå�C†�Š^§ Ý nØTT´L�ù«C†�ÐêÆóä"Ïd§�) Dтã�Ý )g Ž" CCC†††�,˜‡A:´3¦)ž§rüà(Ñ\ÚÑÑ)>.^‡”!˜”"Ïd§¤���(JŒ·Ü?Û>.^‡" Figure 8: Dт˜„){" 10 Figure 9: DтãÝ )" ·‚„´l˜„ÃÑDт§Ñu?1?Ø" dU dz = jωLI dI dz = jωCU (1) æ^LaplaceC†(î‚/`´ü>C†)§  V (s) = ∫ ∞ 0 U(z)e−szdz J(s) = ∫ ∞ 0 I(z)e−szdz (2) y3Ęãݏl�Dтã§3ù˜!§lK1Ñu•> �‹I^z′L«§Xã(10)§éª(1)†>ŠLaplaceC† 11  L ( dU dz ) = sV (s)− U(0) L ( dI dz ) = sJ(s)− I(0) (3) “\ª(2)§k  sV (s)− jωLJ(s) = U(0)−jωCV (s) + sJ(s) = I(0) (4) Œ±)Ñ  V (s) = sU(0) + jωLI(0) s2 + ω2LC J(s) = jωCU(0) + sI(0) s2 + ω2LC (5) 5¿�Laplace_C†  L −1 ( a s2 + a2 ) = sin at L −1 ( s s2 + a2 ) = cos at (6) éª(5)–±Laplace_C†§k 12 Figure 10: Dтã‹I"  U(l) = cosβlU(0) + jZ0 sinβlI(0)I(l) = j 1Z0 sinβlU(0) + cosβlI(0) (7) Ù¥§β = ω √ LC§Z0 = √ L C "q-θ = βl¡>ݧª(7)�Ý /ª´  U(l) I(l)  =  cos θ jZ0 sin θ j 1 Z0 sin θ cos θ  U(0) I(0)  (8) §(8)¡DтãÝ "Œ±`§IP4ù˜Ý §=Œ‰ÑŒÜ©Dтúª"·‚2˜g5¿�í�Ý (8)L§ 13 ¥vk|^?Û>.^‡"�ϏXd§§Œ±·Ü?¿>.^‡" ÖÖÖ???ØØØØØØ 1. òª(8)Šü‡‚5§§…5¿� U(l) I(l) = Z(z) § U(0) I(0) = Zl Kk Z(z) = Z0 Zl + jZ0 tan θ Z0 + jZl tan θ (9) 2. �ª(9)¥Zl = 0§=�7Åá´G�§k Z(z) = jZ0 tan θ (10) �ª(9)¥Zl =∞§=�7Åm´G�§k Z(z) = −jZ0c tan θ (11) �ª(9)¥Zl = jXl§=�7Å?¿G�§k Z(z) = Z0 j(Xl + Z0 tan θ) Z0 −Xl tan θ = jZ0  Xl Z0 + tan θ 1− Xl Z0 tan θ  14 -tanϕl = Xl Z0 §=Œ�Ñ Z(z) = jZ0 tan(θ + ϕl) (12) ùNy ��ƒ �gŽ" 3. ª(8)´Ñ\à^K1àL«"XJ‡L5µK1à^Ñ\àL«§qk U(0) I(0)  =  cos θ jZ0 sin θ j 1 Z0 sin θ cos θ −1  U(l) I(l)  =  cos θ −jZ0 sin θ −j 1 Z0 sin θ cos θ  U(l) I(l)  (13) †c¡Ý ��¬Ü"¢Sþ§L^−θ�“θ=ŒrÑ\ÑÑC† ˜" § 2 DDDÑÑт‚‚ãããAÝÝÝ ���ÊÊÊHHHnnnØØØ ·‚?˜Úí2þãÝ gŽ"3þ¡?Ø¥§8(å5´|^DтãÝ rÑ\>Ø>6ÚÑÑ>Ø>6‚5/é Xå5§½ö`§ÏLDтãÝ �C†§rK1>Ø>6C¤Ñ\>Ø>6" ù«gŽŒŠÜn�ÿ2§=¥m�C†Ý ؘ½´Dтã–ùÒ´�ägŽ"˜‡‚5�ä(Network)§Ñ\> Ø>6U1!I1§ÑÑ>Ø>6U2!I2Œ±^Ý [A]éXå5§ù«Ý Ï~¡/=£Ý 0½/~êÝ 0§q¡ /AÝ 0½/ABCDÝ 0"  U1 = A11U2 +A12I2I1 = A21U2 +A22I2 (14) 15 �¤Ý /ª  U1 I1  =  A11 A12 A21 A22  U2 I2  (15) ÖÖÖ555ŸŸŸØØØ 1. ?é5Ÿ XJ1.‡�ä�ÑÑà´1/‡�ä�Ñ\à§K¡ùü‡�ä?é(Cascade)§k U1 I1  = [AI]  U2 I2   U2 I2  = [AII]  U3 I3  (16) KŒ  U1 I1  = [AI][AII]  U3 I3  (17) í2�N‡�ä?é§Ko�[A]Ý �uˆ[A]Ý g¦È= U1 I1  = N∏ i=1 [Ai]  UN+1 IN+1  (18) 2. é¡5Ÿ 16 Figure 11: [A]Ý Figure 12: �ä?é" 17 é¡�ä(~X§ÃÑDт)§k A11 = A22 (19) 3. ÃÑ5Ÿ ÃÑ�䧌  A11§ A22 ∈ RealA12§ A21 ∈ Imagenary (20) 4. p´5Ÿ 3p´�䥧[A]Ý �1�ªŠ�u1§= det[A] = 1 (21) 5. {|C†5Ÿ Zin = A11Zl +A12 A21Zl +A22 (22) § 3 ;;;...[A]ÝÝÝ 1. Gé{| 18  1 Z 0 1  2. ¿é�B  1 0 Y 1  3. Dтã  cos θ jZ0 sin θ j 1 Z0 sin θ cos θ  19 4. ¿éá´{!  1 0 −j 1 Z0 c tanθ 1  5. ¿ém´{!  1 0 −j 1 Z0 tanθ 1  6. Géá´{! 20  1 jZ0tanθ 0 1  7. Gém´{!  1 −jZ0c tanθ 0 1  § 4 AAA^^^ÞÞÞ~~~ Ö~1ØX㫧Zl = 100 + j200, L = 0.1µH, C = 20pF , Z0 = 50Ω , f = 300MHz ¦Ñ\7Å'" 21 Figure 13: ~1>´" Ö)ØòXÚéZ08˜z 22 Z¯l = Zl/Z0 = 2 + j4 Y¯1 = ωCZ0 = 2pi × 3× 108 × 20× 10−12 × 50 = 1.8850 Y¯2 = Z0 ωL = 50 2pi × 3× 108 × 10−7 = 0.2653 θ1 = βl1 = 2pi λ · 0.1λ = 36◦ θ2 = βl2 = 2pi λ · 0.2λ = 72◦ æ^Ý )—kØÄZ¯l§5¿8˜z�DÑãÝ  cos θ j sin θ j sin θ cos θ  23 [A¯] =  1 0 −j1.8850 1 ×  cos 36◦ j sin 36◦ −j sin 36◦ cos 36◦ ×  1 0 −j0.2653 0  ×  cos 72◦ −j sin 72◦ j sin 72◦ cos 72◦  =  −0.26083 j1.09937 j0.39307 −2.17720  Z¯in = A11Z¯l +A12 A21Z¯l +A22 = 0.52166 + j0.05605 3.749480− j0.39307 Γ = Z¯in − 1 Z¯in + 1 = 3.22782− j0.44912 4.27114− j0.39307 |Γ| = 3.25892 4.28442 = 0.76064 ρ = 1 + |Γ| 1− |Γ| = 7.35561 Ö~2ØXã>´L«V+>Npin+P~ì"¦Ñ\7Å'1ž§R1ÚR2ü+f>{��å^‡" Ö)Øæ^Ý 5¦) 24 Figure 14: V+PIN>NP~ì" A¯ =  1 01 R1 1  0 j j 0  1 01 R2 1  = j  1R2 1 1 + 1 R1R2 1 R1  Z¯in = A11Z¯l +A12 A21Z¯l +A22 = 1 + 1 R2 1 + 1 R¯1R¯2 + 1 R1 = 1 Œ��^‡´ R¯1 = 1 + R¯2 U�yP~ìÑ\àš�" 25 § 5 öööSSS ˜!㫏Ý/Å�H¡�U/$���>´§x´8˜z>|§b´8˜z�B§®µx = 2§b = 1"eà�š�K 1§=zl = 1§¯µθ ÛŠžUþDсZº LaplaceC† 1. LaplaceC†�ê5Ÿ L [f ′(t)] = sL [f(t)]− f(0) (23) Öy²ØdLaplaceC†½Â L [f ′(t)] = ∫ ∞ 0 f ′(t)e−stdt = ∫ ∞ 0 e−stdf(t) = f(t)e−st|∞0 + s ∫ ∞ 0 f(t)e−stdt LaplaceC†^‡ 26 Figure 15: öS1>´" 27 lim l→∞ f(t)e−st = 0 (24) Ïd§k L [f ′(t)] = sL [f(t)]− f(0) 2. ‚5§|¦) sV (s)− jωLJ(s) = U(0)− jωCV (s) + sJ(s) = I(0) D = ∣∣∣∣∣∣ s −jωL−jωC s ∣∣∣∣∣∣ = s2 + ω2LC DV = ∣∣∣∣∣∣ U(0) −jωLI(0) s ∣∣∣∣∣∣ = sU(0) + jωLI(0) DI = ∣∣∣∣∣∣ s U(0)−jωC I(0) ∣∣∣∣∣∣ = jωCU(0) + sI(0) ��� 28 V (S) = sU(0) + jωLI(0) s2 + ω2LC I(S) = jωCU(0) + sI(0) s2 + ω2LC 3. Laplace_C†  L −1 ( a s2+a2 ) = sin at L −1 ( S s2+a2 ) = cos at Öy²ØŠâ½Â L −1 ( a s2 + a2 ) = ∫ ∞ 0 a s2 + a2 estds = j 2 ∫ ∞ 0 [ 1 s+ ja − 1 s− ja ] estds Ù¥ ∫ ∞ 0 1 s+ ja estds = e−jat ∫ ∞ −ja 1 s+ ja e(s+ja)d(s+ ja) = e−jate(s+ja)t|∞−ja = −e−jat∫ ∞ 0 1 s− jae stds = ejat ∫ ∞ ja 1 s− jae (s−ja)d(s− ja) = ejate(s−ja)t|∞ja = −ejat u´ 29 L −1 ( a s2 + a2 ) = 1 2j (ejat − e−jat) = sin at ��aq/§k L −1 ( a s2 + a2 ) = cos at 4. ÃÑDтã)  ·A_C†úª§­#�Ñ  V (s) = sU(0) + j √ L C · ω√LCI(0) s2 + ω2LC J(s) = j √ L C · ω√LCI(0) + SI(0) s2 + ω2LC (25) -Z0 = √ L C §β = ω √ LC§Š_C†k  U(z) = cosβ zU(0) + jZ0 sinβzI(0)I(z) = j 1Z0 sinβzU(0) + cosβzI(0) (26) 30
/
本文档为【c5】,请使用软件OFFICE或WPS软件打开。作品中的文字与图均可以修改和编辑, 图片更改请在作品中右键图片并更换,文字修改请直接点击文字进行修改,也可以新增和删除文档中的内容。
[版权声明] 本站所有资料为用户分享产生,若发现您的权利被侵害,请联系客服邮件isharekefu@iask.cn,我们尽快处理。 本作品所展示的图片、画像、字体、音乐的版权可能需版权方额外授权,请谨慎使用。 网站提供的党政主题相关内容(国旗、国徽、党徽..)目的在于配合国家政策宣传,仅限个人学习分享使用,禁止用于任何广告和商用目的。

历史搜索

    清空历史搜索