dB L(ω) 5 -20 -40 0 1 φ -900 -1800 1T3?1T1?1T2ω1 ω2 -60 ω ω3 -40 1T3?1T1?1T2K ω
解（2）WK(s)?500s(s?s?100)2?52100ss?s?100?5ss21100?s100s?1
幅频A(?)????1???25???????210?100?????22, ,A(10)??6A(?)?0
???0100??10??90?arctan2??1?2?10相频?(?)??????900?1800?arctan100??102???1?210?
?(10)??1800,?(?)??2700
幅相曲线
jQ(ω) 25 P(ω)
对数频率曲线:交接频率为?1?10和，并且20lg0.5??6dB
dB L(ω) 14 0 -6 1 -20 ω 5 10 -60
5-11 用奈氏稳定判据判断下列反馈系统的稳定性，各系统开环传递函数如下： （1）WK(s)?（2）WK(s)?（3）WK(s)?K(T3s?1)s(T1s?1)(T2s?1)(T3?T1?T2)
10s(s?1)(0.2s?1)100(0.01s?1)s(s?1)
解：(1) WK(s)?K(T3s?1)s(T1s?1)(T2s?1)??s??K??1????3??s??s????s??1?1???????1??2?
其中：?1?1T1，?2?1T2，?3???1????????22?1T3，
K幅频A(?)??????????32?????1???1????1??2, A(?)?0
相频?(?)??900?arctan因为(T3?T1?T2) 所以?(?)??900?arctanT3??arctanT1??arctanT2?
T3??arctanT1??arctanT2???1800
并且?(0)??900，?(?)??1800 幅相曲线
jQ(ω) P(ω)
对数频率曲线:交接频率为?1?1T1??2?1T2??3?1T3
dB L(ω) -20 -40 0 1 φ -900 -1800 ω ω3 ω2 -20 -40 -60 ω1 K ω
解（2）WK(s)?10s(s?1)(0.2s?1)?10?s?s?s?1???1??5?
幅频A(?)?210???1??????15??2, A(?)?0
相频?(?)??900?1800?arctan??arctan?5
并且?(0)??2700，?(?)??2700 幅相曲线
jQ(ω) P(ω)
对数频率曲线:交接频率为?1?1
?2?5
dB L(ω) -20 20 6 0 1 φ -1800 -2700 ω -40 -60 5 ω 10
?s?100??1?100(0.01s?1)?100??解（3）WK(s)?s(s?1)s(s?1)
100幅频A(?)???????1?100?2???12, A(?)?0
相频?(?)??900?1800?arctan??arctan?100
并且?(0)??2700，?(10)??1800?(?)??900 幅相曲线
jQ(ω) P(ω)
对数频率曲线:交接频率为?1?1
?2?100