自控課件答案4章

1、- PAGE 15G(jw)1 jw 1 12151w 1w2 11w 1w2 14w2 110w2 jw(1w2)(14w2)(1 wG( jw) w|G(jw0 1ImG(jw)1令I(lǐng)mGjwReG(jw)4545令ReGjw求得 wImG(jw) 5 5 -w（4）G(jw)11w2 111w2 1(jw)2(1 jw)(12w w|G(j 12求出 w122 2 此時(shí) ImG(jw)2 2 ww02 3w（6）G(jw)H(jw) 1 (jw)2(4(jw)2 0.8jw1 w2(4w2 0.8jw1 (4w4 w2) (1 j5w)(4w4 w2) j0.8w3 (4w4 w2)2

2、(0.8w3)2 w2 j(20w5 w4(4w21)2w2(14w2)2120w2 j w (14w2)2 |G(jw)w2 (14w2)2 1 (14w2)2arctan5w1801w 2G(jw)arctan5w180180 0.8w w 4w2 1w w|G(j w0, G(j)H(j) 令I(lǐng)mG( j)H( j) 0 令-（8）G(jw)H(jw) 5(0.5jw1) jw(0.25jw1)(jw1)5 (0.5w)2 |G(jw)H(jw)w (0.25w)2 1 w2 wG( jw) j |G(jw)H( jw j 1|G(jw)H( j 2 G(jw)H(jw) 5(0.5jw

3、1)(10.25jw)(12 jw 11w2 5(0.5jw1)(10.25jw)(1 jw) jw(0.252w2 1)(w2 1) 5j(10.5jw)(10.25jw)(1 jw) w(0.252w2 1)(w2 1) 5j(0.625w2 1) j(0.125w3w(0.252w21)(w2(0.25w0.125w3) j(0.625w2 w(0.252w21)(w2令w無解 與實(shí)軸無交2令2w0, Gjw)Hjw1.25 jww -w4-解G(jw)H(jw) K(1Ta jw)(1Tb 1(jw)2(1T 1K 1T 2w2 1T |G(jw)H(jw)| 1w2 1T1 w0時(shí)

4、|Gjw)Hjw| j 1w時(shí) |GjwH( jG( jw) j G(jw)H(jw) K(1Ta jw)(1Tb jw)(1T1 1w2(1T2w21K(1TT w2) j(T T )w(1T a 1w2(1T2w21K1TT w2 (T T )Tw2 j(1TT w2)TwT a a 1w2(1T2w21K1(TT TT TT )w2K(T T T)wTTTw2 a b a ja b w2(1T2w2w2(1T2w2 Tb T1 0 (T T )wTw(1TT w2G(jw)H(jw)arctan a 1801TT w2 (T T )Ta (T T T)wTTTw2arctan a b 1

5、801TT (T T )Ta 當(dāng)TT (T T)Tw21時(shí)G( jw) j a 11 TaTb Ta Tb)T1時(shí) 幅相曲線與虛軸無交點(diǎn)（2）T1 Ta 0T1 Tb 0 e ImGHTTTw2T T T a b w2 T1 (Ta Tb當(dāng)T1 Ta Tb時(shí) 曲線與負(fù)實(shí)軸有交點(diǎn)w0時(shí) Gj0)Hj0 w w 當(dāng)T1 Ta Tb4-解G(jw)Kjw(1 jT1w)(1 由G(S可知 w w-1 w wKw 1T2w2Kw 1T2w2 1T 12G(jw)arctanT1warctanG(jw) Kjw(1 jT1w)(1 w2(1T2w2)(1T 2w2K(T T K(TT w2 j1 (1T

6、2w2)(1T2w2w(1T2w2)(1T2w2求幅相曲線與實(shí)軸交點(diǎn) 此時(shí)Gjw1令I(lǐng)mG(jw)=0， 得w1此時(shí) ReGjw KT T ReGjw(1 0.02)(1 N PN （2）要使系統(tǒng)穩(wěn)定必須有即 KTT T T 1 4-（a）N N N N N PN N N N N PN N N N N PN N N N N N N N N N 4-w w w3 20lgK 20lg100-w1- 2w131 2000(S 600( 6G(s) S(S2 4S S 1 S2 SS( 1 S2 SGj6 w2 jw 24. w2 20 lK 20 lg 016w6(w) w1 1 w90 180

7、arctanw 1 w2 1-w16- 90 451w6 2000(S 600( 6（5）G(s) S(S2 4S S 1 S2 SS( 1 S2 SjwGj 1 w2 jw 24. w2 20 lK 20 lg 01(w) w1 1 w90 180 arctanw 1 w2 1-w16- 1813w96解G(jw)H(jw) ( jw)2轉(zhuǎn)折頻率 w8.20lgK 20lg50 -w1- w1w1(1 1 w2)2 (11w1(1 1 w2)2 (17當(dāng)|Gjw)Hjw|1w 1G(jw)H(jw)90 arctan1 1 則Gjw)Hjw|G(jw)H( j S 1 S2 1S5.87能得到30 度解G(jw)H(jw) (jw)(jw1)(jw則|G(jw)H(jw)1w 0.36w wc G(jw )H(jw )90 arctanarctan 1w g0.36w g當(dāng)Gjwg)Hjwg1801w g0.36w gKg |G(jwg )H(jwg )4-G(S) Ks20lgk G(S) sk K(2S G(S) G(jw)S2(0.5SK( j2ww2(j0.5ww0.5時(shí)L(w20lgk40lg當(dāng)0.5 w 2L(w) 20lgk 40lgw20lgw 2L(w) 20lgk 40lgw20lg2w20lgw=1L(1) 20lgk 40lg120lg2*1 k G(S