高频电子线路张肃文pdf版

页码，1/5 3 1解：f = 1MHz 1 × 10 6 f0 = = 100 2 Δf 0 .7 10 × 10 302 Δf 0 .7 = 1 × 10 6 990 × 10 3 = 10(kHz) Q=取R = 10Ω 100 × 10 QR 则L = = = 159(µH) ω0 2 × 3.14 × 10 6 1 1 C= 2 = = 159(pF) 6 2 ω0 L ( 2 × 3.14 × 10 ) × 159 × 10 63 2 解：( 1 )当ω01 = ( 2 )当ω01 = ( 3 )当ω01 = 1 或ω 02 = L1C1 1 或ω 02 = L1C1 1 或ω 02 = L1C1 1 时，产生并联谐振。 L2C 2 1 时，产生串联谐振。 L2 C2 1 时，产生并联谐振。 L2 C 2L 1 1 ) R 2 + + jω0 LR( 1 2 ) R2 + L jω0 C C ω0 LC C =R 3 3证明：Z = = = 1 1 2R R + jω0 L + R + ) 2 R + jω0 L( 1 2 jω0C ω0 LC (R + jω0 L)(R + 3 4 解：1)由 (15 + C ) × 16052 = (450 + C )5352得C = 40 pF由 (12 + C ) × 16052 = (100 + C )5352 得C = -1 pF (不合理舍去 ) 故采用后一个。 1 1 2 )L = 2 = = 180(μH ) ω0 (C + C ′) (2 × 3.14 × 535 × 103 )2 × (450 + 40 ) × 10 12 3)LCC’mhtml:mk:@MSITStore:C:\Documents%20and%20Settings\Owner\桌面\高频电子线... 2011-3-7

页码，2/53 5解：Q0=1 1 = = 212 ω0 C0 R 2 × 3.14 × 1.5 × 10 6 × 100 × 10 -12 × 52 0L0 =1 1 = = 112( μH ) ω C0 (2 × 3.14 × 1.5 × 10 6 )2 100 × 10 -12 Vom 1 × 10 -3 = = 0.2(mA) 5 R = VCom = Q0VSm = 212 × 1 × 10 - 3 = 212(mV )I om = VLom3 6解：L =1 1 = = 253( μH ) 2 2 ω0 C 2 × 3.14 × 10 6 × 100 × 10 12 V 10 = 100 Q0 = C = VS 0.1()C CX 1 1 = 2 = = 100( pF ) → C X = 200 pF 2 C + C X ω0 L 2 × 3.14 × 10 6 × 253 × 10 6()RX =ω0 L ω0 L 2 × 3.14 × 10 6 × 253 × 10 6 2 × 3.14 × 106 × 253 × 10 6 = = 47.7(Ω ) Q Q0 2.5 0.1 100 1 1 = 47.7 j = 47.7 j 796(Ω ) 6 ω 0C X 2 × 3.14 × 10 × 200 × 10 12Z X = RX j3 7 解：L = Q0 =1 1 = = 20.2( μH ) 2 ω0 C 2 × 3.14 × 5 × 10 6 × 50 × 10 12()2 Δf 100 2 × (5.5 5) × 10 6 20 ξ = Q0 = × = f0 3 5 × 106 3 ′ = 2 × 2 f 0.7， 则Q′ ′ 因2Δ f 0.7 0 = 0.5Q 0，故 R = 0.5R， 所以应并上 21kΩ电阻。3 8证明： 4 πΔf 0 .7 C = 2 πf 0 C ωC = 0 = g∑ f 0 2 Δf 0 .7 Qf0 5 × 10 100 = = 3 2 Δf 0 .7 150 × 10 36mhtml:mk:@MSITStore:C:\Documents%20and%20Settings\Owner\桌面\高频电子线... 2011-3-7

页码，3/53 9 解：C = Ci + f0 =(C2 + C0 )C1C2 + C 0 + C1=5+(20 + 20)2020 + 20 + 20= 18.3( pF )1 1 = = 41.6(MHz ) 2π LC 2 × 3.14 0.8 × 10 6 × 18.3 × 10 12 L = 100 × C12 0.8 × 10 6 = 20.9(kΩ ) (20 + 20) × 10 122 2RP = Q0 C 2 + C0 + C1 20 + 20 + 20 R0 = 10 20.9 R∑ = Ri RP × 5 = 5.88(kΩ ) C1 20 QL = 2 Δf 0.7 R∑ 5.88 × 103 = 28.2 = ω0 L 2 × 3.14 × 41.6 × 10 6 × 0.8 × 10 6 f 0 41.6 × 106 = = = 1.48(MHz ) 28.2 QL3 12解： 1)Z f1 = 0 3)Z f1 = R 2 )Z f1 = 0mhtml:mk:@MSITStore:C:\Documents%20and%20Settings\Owner\桌面\高频电子线... 2011-3-7

页码，4/53 13解：1)L1 = L2 = C1 = C 2 =ρ1 103 = = 159( μH ) ω01 2 × 3.14 × 1062 01 11 1 = = 159( pF ) 2 ω L 2 × 3.14 × 106 × 159 × 10 6 ηR 1 × 20 M = 1 = = 3.18( μH ) ω01 2 × 3.14 × 106()2 )Z f 12 ( ω01 M ) =R2(2 × 3.14 × 10 =6× 3.18 × 10 6 20)2= 20(Ω )ZP = 3)Q1 =L1 159 × 10 6 = (R1 + R f 1 )C1 (20 + 20) × 159 × 10 12 = 25(kΩ ) 2 × 3.14 × 106 × 159 × 10 6 ω01 L1 = 25 = R1 + R f 1 20 + 20 10 6 × 20 f0 f0 = 2 = 2× = 28.2(kHz ) 10 3 Q ρ1 R1 L2 =4 )2 Δf 0 .7 = 2 ′= 5)C 2 ′ ) (ω02 12(2 × 3.14 × 950 × 10 )13 2× 159 × 10 6= 177( pF ) 1 Z 22 = R2 + j ω02 L1 ω C ′ 02 2 1 = 20 + j100 = 20 + j 2 × 3.14 × 106 × 159 × 10 6 6 12 2 × 3.14 × 10 × 177 × 10 Z f1 = ΘR= 3 15解： ∴Rf1(ω01 M )2Z 22=(2 × 3.14 × 10× 3.18 × 10 6 20 + j1006)2= 0.768 j 3.84(Ω )L 159 × 10 6 = = 20( ) = R1 RP C 50 × 103 × 159 × 10 12 =0→ M =0 f0 106 = 2 = 100 2 f 0.7 14 × 103Q= 2mhtml:mk:@MSITStore:C:\Documents%20and%20Settings\Owner\桌面\高频电子线... 2011-3-7

页码，5/53 16 解：1)R f 1 = Rab = 2 )η = 3)Q =(ω01 M )2R2(10 =7× 10 6 ) = 20( ) 52(R(ω01 L )21+ Rf1)(10 =7× 100 × 10 6 ) = 40(k ) 5 + 202ω01 M 10 7 × 10 6 = =2 R1 5ω01 L 107 × 100 × 10 6 = = 200 R1 5 2 f 0.7 1 1 = η 2 + 2η 1 = 2 2 + 2 × 2 1 × = 0.013 f0 Q 200 1 2 f 1+ Q′ f 0 23 17 解： 10 × 103 1+ Q ′ 300 × 10 3 1 1 R= = = 11.8 Qω0C 22.5 × 2 × 3.14 × 300 × 103 × 2000 × 10 12 I = I0 12I = I0=12=1 → Q ′ = 22.5 1.25 2 f 10 × 10 3 + 1 Q 1+ Q 300 × 10 3 f0 Q Q ′ = 30 22.5 = 7.5=12=1 → Q = 30 21 2ω = 串联联谐 L = 375 μH L2C 3 18解： → 1 1 L2 = 125 μ ω = 并联联谐 (L1 + L2 )C mhtml:mk:@MSITStore:C:\Documents%20and%20Settings\Owner\桌面\高频电子线... 2011-3-7

页码，1/6 β0 β0 f 1+ f T β0 β0 f 1+ f T β0 β0 f 1+ f T 24 5解：当f = 1MHz时，β ==50 50 × 10 1+ 250 × 10 6 506 2= 49当f = 20MHz时，β =2=当f = 50 MHz时，β =2= 50 × 20 × 10 1+ 250 × 10 6 506 2= 12.1 50 × 50 × 106 1+ 250 × 10 6 2=54 7解：g b′e = gm = C b′e =1 IE = = 0.754(mS ) 26( β0 + 1) 26 × (50 + 1)β0 = 50 × 0.754 × 10 3 = 37.7(mS ) rb′e 37.7 × 10 3 gm = = 24( pF ) 2πf T 2 × 3.14 × 250 × 10 6a = 1 + rbb′ g b′e = 1 + 70 × 0.754 × 10 3 ≈ 1 b = ωC b′e rbb′ = 2 × 3.14 × 10 7 × 24 × 10 12 × 70 ≈ 0.1 y ie = jωCb′e )(a j …… 此处隐藏：13400字，全部文档内容请下载后查看。喜欢就下载吧 ……