第五章習(xí)題解答5求下列函數(shù)的幅頻特性和相頻特性

5-1 A41）G(s)

20s4G(s)

2022()2）G(s)

1022()90G(s)

10s

G(s)

G(s)

G(s)

G(s)2(sG(s)

(s2)(5s2)(s 答案1 2 3(c 1(1）G(s)

g1=tf(([1]),conv([0.11],[0.22G(s)

s(0.2sg1=tf(([2]),conv([10],[0.2G(s)

s2(0.2sg1=tf(([2]),conv([100],[0.2G(s) s(2sG(s) G(s)Ks

1s

3s其 又40

2 2

1τ1=11 20lgK=L(1) G(s)

ss(s1)(0.0625s

s(s

Ｓ（Ｓ＋２（Ｓ＋３）Ｋ（Ｓ＋１）Ｓ２（Ｓ＋４（Ｓ＋５）

G(j)H(js) j(j2)(jK242

φ(ω)=90 0,A(),()90,,A()0,()270,又：202202X0G(j)H(js)

K

K5(62)j(625

令I(lǐng)m=0，即62 ReK

5

25

G(j)H(js) K(j(j)2(j4)(j

KK222162

50,A(),()180,,A()0,()270,[ω=0.1代入，Φ(ω)=-180°+(5.71°-1.43°-1.15°)=-180°+3.13°又：2020X0K j

(j1)(2029G(j)H(js)

2029

(202)281 令 (20292)(2029)jK(2082)(112) 令 K

K

208

K

K

12

Re

11(20

11

1199 G(s) s(0.5s1)(0.1s如圖，11/0.52s1，

1/0.120lg140lgc120lgK20 1(c1)2 c1 )==-解：開環(huán)頻率特性Gj

G(s)K

KG(j)

兩個(gè)特殊點(diǎn)①ω=0+②ω

G(j)G(j)

G(j)G(j)G(j)j(

Kj(1jT)(1jT1) 1) Kj1jTjT2TT Kj12TTTTT1 T1

1

1T2 T2

)T2)T2 Re[G(j)]T1240000(sG(s)

s(s1)(s2)(s100)(s K3 TG(s)H(s) KT1

T1K

T1T21.2 A( KT 11G(j)0.2(j)3(j)2200,A(),()90,,A()0,()270,G(j) 此時(shí)Gj)

5-15.單位反饋系統(tǒng)的開環(huán)G(s)ss

求相

1 L()20

(1

(5 (520lg100/5c

(518090arctg(c)arctg()arctg(c 0PhasePhase

Gm=InfdB(atInfrad/sec),Pm=57.4deg(at19.3

Frequencyg1=tf(([1/51]),conv([110],[1/50-0ω0-ω- --01ω3-1)G(j) j2）G(j) (j1)(j 10(j3）G(j) (j)2(0.1jG(j)G(j)

G(j)H(js)

0.04210.00042

0,A(),()90,,A()0,()270,又：202202X0() 5100.0425010.0004250nyquist(tf([20],conv(conv([10],[0.21]),[0.02

20.04210.00042()180arctg0.20,A(),()180,,A()0,()360,[ω=0.1代入，Φ(ω)=-180°-1.14°-0.11°=-180°-1.25°:0

包圍（-1，j0）點(diǎn)2圈，因此N=-2，系統(tǒng)不穩(wěn)定。nyquist(tf([40],conv(conv([100],[0.21]),[0.021])))G(j) ，()90tg1(0.2)0，A()，()90，A()0，()270令(90tg1(0.2tg1(2 5-19.已知某二階系統(tǒng)的阻尼0.4，求該系統(tǒng)的超調(diào)量pMrrpe 121M121 nn本題G(s)

210.012 ，()90210.0120，A()，()90，A()0，()270()90tg1tg1(0.1) 5-21.（交通大學(xué)2003本題15分G(s)4(s s2(s G(j)2(j02)()180arctan/0.5arctan/0，A()，()180，A()0，()180---- - - - - - 5-22.（交通大學(xué)2003本題20分G(s)

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