幅相曲线
jQ(ω) 1.414 P(ω) 1
对数频率曲线
dB L(ω) 40 0 φ(ω) 1800 900 00 ωn ω ω
?s?25??1?25(0.2s?1)?5?解（10）W(s)?2?2s +2s+1s +2s+1???25???1?5?2
幅频A(?)??1?????2??222,
A(0)?25,A(1)?12.5A(?)?0
?arctan??相频?(?)???arctan??0?5?arctan02?1??2??12??50
??1?180?arctan??12?(0)?0,?(1)??78.7,?(?)??900
幅相曲线
jQ(ω) 25 P(ω)
对数频率曲线:交接频率为?1?1和?2 dB 28 -40 L(ω) -20 ?5，并且20lg25?28dB
0 1 5 ω -20
5-9绘出习题5-8各传递函数对应的对数频率特性。
5-10绘出下列系统的开环传递函数的幅相频率特性和对数频率特性。
（1）WK(s)?（2）WK(s)?（3）WK(s)?K(T3s?1)s(T1s?1)(T2s?1)(1?T1?T2?T3?0)
500s(s?s?100)e?0.2s2
s?1
?s??K??1????3??s??s????s??1?1???????1??2?解：(1) WK(s)?K(T3s?1)s(T1s?1)(T2s?1)?
其中：?1?1T1，?2?1T2，?3?1T3，
因为1?T1?T2?T3?0，所以1??1??2??3
??????32K幅频A(?)???????1????????2???1??22, A(?)?0
?????1???1?相频?(?)??900?arctanT3??arctanT1??arctanT2?
是否小于?1800
这里需要讨论?(?)??900?arctan即arctan令arctanT3??arctanT1??arctanT2?T3??arctanT1??arctanT2?是否小于?900
T3???3，arctanT1???1，arctanT2???2arctanT3??arctanT1??arctanT2???3??1??2
即?1??2??3是否大于900，令?2??3?1??2??3??1????则
tan?1?tan?1?tan?1tan?tan(?1??2??3)?tan(?1??)?
tan(?2??3)?tan??tan?2?tan?31?tan?2tan?3
其中tan将tan?1?T1?，tan?2?T2?，tan?3?T3?
?2??3)?tan??tan?2?tan?31?tan?2tan?3?1?T1?，tan?2?T2?，tan?3?T3?和tan(代入到tan(?1??)?tan?1?tan?1?tan?1tan?中
tan?2?tan?31?tan?2tan?3tan?2?tan?31?tan?2tan?32tan(?1??)?tan?1?tan?1?tan?1tan?T2??T3?1?T2T3?1?T2T3?22tan?1??1?tan?1T1??则
?1?T1?T2??T3?2?T1?1?T2T3?1?T2T3?2??T1?T2?T3?????T??T??
232?T1?1?T2T3?????T??T??2321??T2T3?T1T2?T1T3??即tan(?1??)?T1?1?T2T3??21??T2T3?T1T2?T1T3?????T??T??
232因为T1??1?T2T3?2???T2??T3???0 1．如果1??T2T3?T1T2?T1T3??2即T2T3?T1T3?T1T2?0?0则tan(?1??)?900
1T3?1T1?1T2→T3?T1T2T1?T2因为1?T1?T2?T3?0→
2如果1??T2T3?T1T2?T1T3??2即T2T3?T1T3?T1T2?0?1则tan(?1??)?900
1T3?1T1?1T2→T3?T1T2T1?T2?0因为1?T1?T2?1??)?90?T3?0→
3如果1??T2T3?T1T2?T1T3??2即T2T3?T1T3?T1T2所以当
1T3?1T1?则tan(0
1T3?1T1?1T2?0→T3?T1T2T1?T2因为1?T1?T2?T3?0→
1T2时，
大于?1800
?(?)??90?arctanT3??arctanT1??arctanT2?0当
1T3?1T1?1T2时，
小于?1800
?(?)??90?arctanT3??arctanT1??arctanT2?0并且?(0)??900，?(?)??1800 幅相曲线
jQ(ω) P(ω) 1T31T11T2??
jQ(ω) P(ω) 1T31T11T2??
对数频率曲线:交接频率为1??1?1T1??2?1T2??3?1T3