高频电子线路习题答案（第五版）张肃文(2)

4?9解：p1?gp?N235??0.25N1320p2?N455??0.25N132011??37.2?μS?6?6ω0Q0L2π?10.7?10?100?4?1011?6?6?200?10??2860?10?228.5?μS?22442g??gp?p12goe?p2gie?37.2?10?6?Avo?p1p2yfegΣ0.25?0.25?45?10?3??12.3?6228.5?102Apo?Avo?12.32?151.3QL?2Δf0.711??16.3?66?6gΣω0L228.5?10?2π?10.7?10?4?10f010.7?106???0.657?MHZ?QL16.3?2?2?QL??16.3??K??1???1???1.43?Q??100?0???fe??re?54o?88.5oξ?tan?tan??2.95222gp?p2gie37.2?10?6?0.252?200?10?6g??3008.8?μS?L?22p10.25yfeyre?gs?gie??goe?g?1?ξ2?2?2860?10?6?200?10?6?3008.8?10?6??1?2.952?L??S????145?10?30.31?10?3

?1?gp?4?10解：11??0.037?mS?6?6Q0ω0L100?2?3.14?10.7?10?4?1012?p12goe?p2gie??0.037?0.1?0.32?0.082?0.32?0.15??0.158?mS?R50.3?0.3?382?4.22??21.780.158g??gp?Avo?p1p2yfeg??2?2Δf0.7?ω0Lg?f0?2?3.14??10.7?106?2?4?10?6?0.158?10?3?454.4?kHz??3??Avo?4??Avo?4?21.784?225025.38?4??2Δf0.7?4??5?2Δf0?.7?142?1?2Δf0.7?2?1?454.4?197.65?kHz??1044.6?kHz?14142Δf0.72?12Δf0?.7?2Δf0.7?1044.6?454.4?590.2?kHz???AvoAvo2Δf0.721.78?454.4??9.47?2Δf0.71044.6??4??Avo??4?9.474?8042.66?Avo??4?225025?Avo?4??Avo.38-8042.66?216982.724?11解：C??C?p12Coe?500?0.32?18?501.62?pF?L?1

?2πf0?C?2?1?2?3.14?1.5?10??62?501.62?10?12?22.5?μH?

Kr0.1?1.9不能满足?Avo?S?4?14解：yfe2.5?0Cre26.42?36.42?7.742.5?0.3

4?17解：L1?11??118?μH?223?12ω0C1?2π?465?10??1000?10118118118L36?L2?L34?L56???60??13.5?120737373C12?C1?Co?1000?4?1004?pF??13.5?C36?C2?pCi?1000????40?1004?pF??74.5?ω0C12π?465?103?1000?10?12?6g12?go??20?10??49?μS?Q0100222ω0C2?13.5?2π?465?103?1000?10?12?3g36?pgi????49?μS???0.62?10?Q0100?74.5??初、次级回路参数相等。若为临界耦合，即??1，则13.5?31??40?10p1p2yfe74.5Avo???74g2?49?10?6222ω0C122π?465?103?1004?10?12QL???60?6g1249?102Δf0.7f0465?103?2?2??10.9?kHZ?QL60

Kr0.1?3.1624?20解:vn?4kTRΔfn?4?1.38?10?23?290?1000?107?12.65?μV?2in?4kTGΔfn?4?1.38?10?23?290?10-3?107?12.65?nA?22224?21解：?vn?vn1?vn2?vn3?4kT1R1Δfn?4kT2R2Δfn?4kT3R3Δfn

?4k?T1R1?T2R2?T3R3?Δfn?4kT?R1?R2?R3?Δfn?T?T1R1?T2R2?T3R3R1?R2?R32222又?in?in1?in2?in3?4kT1G1Δfn?4kT2G2Δfn?4kT3G3Δfn?4k?T1G1?T2G2?T3G3?Δfn?4kT?G1?G2?G3?Δfn?T?T1G1?T2G2?T3G3R1R2T3?R2R3T1?R3R1T2?G1?G2?G3R1R2?R2R3?R3R1

4?18证明：1?Ib1?yieVbe1?yreVce1???1?Ic1?yfeVbe1?yoeVce1???2?Ib2?yieVbe2?yreVce2Ic2?yfeVbe2?yoeVce2????????????????yieVce1?yre?Vcb2?Vce1??yreVcb2??yie?yre?Vce1???3???????????yfeVce1?yoe?Vcb2?Vce1??yoeVcb2??yfe?yoe?Vce1???4??????????????2???3?得Ic2?yfeVbe1??yie?yre?yoe?Vce1?yreVcb2Vce1??Ic2?yreVcb2?yfeVbe1yie?yre?yoe????????5?Ic2?yreVcb2?yfeVbe1yie?yre?yoe????5?代入?4?Ic2?yoeVcb2??yfe?yoe?Ic2?2??yfe?yfe?yoe?yieyoe?yreyfe?yoe?Vbe1?Vcb2???6?yie?yre?yfe?2yoeyie?yre?yfe?2yoe?yf?yfe?yfe?yoe??yfeyie?yre?yfe?2yoe由?1?乘?yfe?yoe?与?4?乘yre后相加得Ib1?yfe?yoe??Ic2yre?yie?yfe?yoe?Vbe1?yreyoeVcb2由?6?代入消去Ic2得Ib1?2?yie?yieyre?yieyfe?2yieyoe?yreyfe?yre?yoe?yre??Vbe1?Vcb2yie?yre?yfe?2yoeyie?yre?yfe?2yoe?????2yieyoe?yreyfe?yoeyo???yreyie?yre?yfe?2yoe2yie?yieyre?yieyfe?2yieyoe?yreyfeyi??yieyie?yre?yfe?2yoeyr?yre?yoe?yre?y?y?yre???reoeyie?yre?yfe?2yoeyfe略同理可证2?24?22解：vbn?4kTrb?fn?4?1.38?10?23??273?19??70?200?103?0.226?10?12?V2?2ien?2qIE?fn?2?1.6?10?19?10?3?200?103?0.64?10?16?A2????0?f?1???f?????2?0.95?10?10?1???500?106????62?0.952icn?2qIC?1??0??fn?2?1.6?10?19?10?3??1?0.95??200?103?0.32?10?17?A2?

4?23证明：?fn???0A2?f?dfA2?f0????1204?24解：Fn高?3dB?1.995倍?Fn混?1?Fn?Fn高?f?f0??1??2Q??f0??Fn中?6dB?3.981倍?df??f02Q

Ti60?1??1.207T290F?1Fn中?11.207?13.981?1?n混??1.995???10Ap高KpcAp高Ap高0.2?Ap高20lg1.888?2.76?dB?2sAp高?1.888

4?25解：Fn?PsiPniPPV4RsR?RR11?no???s?2?s?1?sPsoPnoPniApApPoPsPoVs4?Rs?R?RR PsiPniPsIs24GsG?GL?rCL14?26解：Fn????2?1?PsoPnoApPoIs4?Gs?G?GL?rCL?GsAPA?6dB?3.981倍?Fn?FnA?ApB?12dB?15.849倍?

4?27解：A为输入级，B为中间级，C为输出级。F?1FnB?12?14?1?nC?1.7???2ApAApA?ApB3.9813.981?15.8494?28解：不能满足要求。设A前置放大器，B为输入级，C为下一级。PsiPni105F?1F?110?11.995?1Fn??4?FnA?nB?nC?FnA???FnA?8.1PsoPno10ApAApA?ApB1010?0.15?8解：i?kv?k?V0?Vmcos?0t?2

21212???k?V02?2V0Vmcos?0t?Vm?Vmcos22?0t?22??当Vm??V0时，i?k?V02?2V0Vmcos?0t?，该非线性元件就能近似当成线性元件来处理，即当V0较大时，静态工作点选在抛物线上段接近线性部分，然后当Vm很小时，根据泰勒级数原则，可认为信号电压在特性的线性范围内变化，不会进入曲线弯曲部分，故可只取其级数的前两项得到近似线性特性。5?12解：为了使iC中的二次谐波振幅达到最大值，?C应为60o。cos60o?VBZ?VBB1?Vm21?VBB?Vm?VBZ2