電路原理第十二章

1、12.1 非正弦周期信號非正弦周期信號例ut方波電壓ut鋸齒波it脈沖波形12.2 周期函數分解為傅里葉級數周期函數分解為傅里葉級數 )sin()cos()sin()cos()2sin()2cos()sin()cos()(111011121211110tkbtkaatkbtkatbtatbtaatfkkkkk )cos()cos()2cos()cos()(110121121110kkmkkkmmmtkAAtkAtAtAAtf 00aA 22kkkmbaA kkmkAb sin )arctan(kkkab kkmkAa cos A A0 0 )cos()cos()2cos()cos()(1101

2、21121110kkmkkkmmmtkAAtkAtAtAAtf A A1m1mcos(cos(1 1t +t +1 1 ) ) 2200)(1)(1TTTdttfTdttfTa)()cos()(1)()cos()(1)cos()(2)cos()(211201122101 tdtktftdtktfdttktfTdttktfTaTTTk)()sin()(1)()sin()(1)sin()(2)sin()(211201122101 tdtktftdtktfdttktfTdttktfTbTTTk )sin()cos()sin()cos()2sin()2cos()sin()cos()(111011121

3、211110tkbtkaatkbtkatbtatbtaatfkkkkk 例0Akmk115141312161f(t)Em-Em021ttT/2TmmEtfEtf )()(TtTTt 2200)(100 TdttfTa 0)()cos(2)()cos()()cos(1)()cos()(10112110112011 tdtkEtdtkEtdtkEtdtktfammmk )cos(12)cos(12)()sin(2)()sin()()sin(1)()sin()(1010112110112011 kkEtkkEtdtkEtdtkEtdtkEtdtktfbmmmmmk )5sin(51)3sin(31)

4、sin(4)(111tttEtfm 0k115131AkmmE434mE54mEtf f( (t t) )tf f( (t t) )tf f( (t t) )12.3 有效值、平均值和平均功率有效值、平均值和平均功率 TdtiTI021)cos(110kkkmtkIIi TkkkmdttkIITI01210)cos(1 200201IdtITT 122023222120kkIIIIIII21202)(cos1kkTkmIdttkIT 0)cos(21100 dttkIITkkmT )0)cos()cos(21110qkdttqtkIITqkqmTkm ，（ dtiTITav 01IItTIdt

5、tTIdttITImTmTmTmav898. 0637. 0)sin(4)cos(4)cos(140400 )cos()cos(100100ikkkmukkkmtkIItkUUuip TpdtTP01 kkkIUIUIUIUP coscoscos22211100，2122 kIIUUikukkkmkkmk 12.4 12.4 非正弦電流電路的計算非正弦電流電路的計算 CkXLkXCkLk111 ， VtttttuCRs )9cos(71.15)7cos(20.20)5cos(28.28)3cos(13.47)cos(40.14110:,45. 91,3111111 輸輸入入電電源源為為CkjR

6、UIksk1)()(1 RC+-us0,0,10,0000 PIVUk直直流流分分量量WRIPAAjIVUks02.30539.7208.1045.9301000100,12)1()1()1()1( WRIPAAjIVUks93.1754.4666.715.33033.33033.33,32)3()3()3()3( ,29.1949.3,23.2434.4,21.3258.5)9()7()5(AIAIAI (5)(7 )(9 )95.5256.5536.60PWPWPW最后按時域形式迭加為：最后按時域形式迭加為： Attttti )29.199cos(249. 3)23.247cos(234.

7、 4)21.325cos(258. 5)4 .463cos(266. 7)39.72cos(208.1011111 WPPPPP80.6699310 同理求得：同理求得：Lus+-CRusUm021t Vttus )4cos(151)2cos(3121157411 ，用用結結點點電電壓壓法法有有：次次諧諧波波為為設設負負載載兩兩端端的的第第)(kUk )(1)(11111kskUULjkCjkRLjk 1111)()( LjkCjkRkskUU 令令k=0, 2, 4, , 并代入數據并代入數據,可分別求得可分別求得:VUVUVU12.05.2100)4()2(0 usCRL例：圖示電路中，激

8、勵例：圖示電路中，激勵 u1(t) = u11( 1)+u12( 2)，包含包含 1、 2 兩個頻率分量，且兩個頻率分量，且 1 2 ，要求響應，要求響應 u2(t) 只含只含有有 1 頻率電壓，如何實現？頻率電壓，如何實現？+_u1(t)u2(t)可由下列濾波電路實現：可由下列濾波電路實現：CRC2C3L1+_u1(t)+_u2(t)2121CL )(13211CCL 并聯諧振，開路并聯諧振，開路串聯諧振，短路串聯諧振，短路Vttuts)2cos(216)cos(21512)( +-us(t)L1L2CRuR(t)+-VUs25161512222 中中不不含含基基波波，即即則則，對對并并聯聯諧諧振振與與（)1)(tRuCL 11LC HCL1101000116221 HLCjLjCjLjLj31021221222112 ARUI1200 ARUI16