王划一-自动控制原理-5-3稳定裕度
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第五章 频率法5-1用时域与频域法分析设计和设计系统的主要区别是什么? 5-2用时域法分析和设计系统的主要优点是什么? 5-3奈氏稳定判据的本质是什么?5-4何谓幅值裕度与相位裕度,并举例说明之。
5-5试述二阶系统闭环频率特性与时域中阶跃相应之间的关系。
5-6试定性叙述伯德图各段与时域指标之间的对应关系。
5-7已知单位反馈系统的开环传递函数为 W K (s)=110+s当系统的给定信号为 (1))30sin()(01+=t t x r(2) )452cos(2)(02-=t t x r(3))452cos(2)30sin()(03--+=t t t x r求系统的稳态输出。
解:5-7(1)系统的闭环传递函数为1110)(1)()(+=+=s s W s W s W K K B因为)30sin()(0+=t t x r )30(0)(+=t j r ej X ω 02.511arctan29054.012110)(j j B eej W --=+=ωωω)2.530(09054.0)()()(-+==t j B r c ej W j X j X ωωω所以)8.24sin(9054.0)(0+=t t x c 解:5-7(2)系统的闭环传递函数为1110)(1)()(+=+=s s W s W s W K K B因为)452cos(2)(0-=t t x r 化为正弦表达形式则)452sin(2)(0+=t t x r )452(02)(+=t j r ej X ω 3.1011arctan28944.012110)(j j B eej W --=+=ωωω)3.10452(07888.1)()()(-+==t j B r c ej W j X j X ωωω所以)7.342sin(7888.1)(0+=t t x c解:5-7(3)根据叠加原理,系统的输出为5-7(1)-5-7(2))7.342sin(7888.1)8.24sin(9054.0)(0+-+=t t t x c5-8绘出下列各传递对应的幅相频率特性。
《自动控制原理》(第2版)习题答案第2章2-1 (1)t e t e tt23sin 3123cos 122--+- (2)6 + 3t(3))334(322+++---t t e e t t(4)t t ωωωsin 1132-2-2 (1)2351853t t e e --+- (2)t e 2-(3)t e a b t ae n t nn n t n n ωωζωωζωζωsin cos --++ (4)t a Aa t a A e b a A at ωωωωωωωsin cos 222222++++⎪⎭⎫ ⎝⎛++- 2-3 (a ))()()(2110f f ms f s X s X i ++= (b )212110)()()(k k s k k f fs k s X s X i ++= 2-4 (a ))()()(t u t kx t xm =+ (b ))()()(2121t u t x k k k k t xm =++ 2-5 (a ))()()()()(2212121t u R dtt du C R R t u R R dt t du C R R r r c c +=++ (b ))()()()()()(22121221t u R t u R R dt t du C R R L dt t u d LC R r c c c =++++ 2-6 252312)14(100)()(2+++=s s s s R s C 2523125231210)()(22++++⋅=s s s s s R s E 2-7 t t e e t c 2241)(--+-=2-8 )1)(2(23)(+++=s s s s G t t e e t h ---=24)(2-9 (a )1)(1)()(32213+++⋅-=s R R C s CR R R s U s U r c (b )13221)()()(R R R s R CR s U s U r c ++-= 2-10 (a )))((1)()(432121G G G G G G s R s C -+++= (b ))(1)1()()(21221H H G G G s R s C -++= (c )331311321332123113211)()(H G H G H G G G G H G G H G G H G G G G s R s C ++++++= 2-11 (a )32211)()(G G G G s R s C ++= (b )HG H H G s R s C 111)1()()(+--= (c )121223121)()()(H G G H G G G G s R s C +++= 2-12 (a )))((1)1()()(23111232123111134321H G H G H H G G G H G H G H G G G G G G s R s C --++++++= ))((1)1(1)()(2311123212311123423H G H G H H G G G H G H G H H G G H G s R s E --++++-+⋅= (b )21212121312)()(G G G G G G G G s R s C ++-++-= 21212131)1(1)()(G G G G G G s R s E ++-+⋅= 2-13 (a )12121211)()(H G G G G G G s R s C ++= 121211211)1(1)()(H G G G G H G G s R s E +++⋅= 12121231211)1(1)()(H G G G G G G H G G s D s C ++++⋅-= 12121231211)1(1)()(H G G G G G G H G G s D s E ++-+⋅= (b )434242143421)()(G G G G G G G G G G G s R s C ++++= 434242111)()(G G G G G G G s R s E ++-= 434241)()(G G G G G s D s C ++= 434241)()(G G G G G s D s E ++-=2-14 (a )))((1)(23113343321231134321H G H G H G G H G G G H G H G G G G G G s G -+++-++= (b )3541432326543211)(H G G H G G H G G G G G G G G s G +-+=(c ) 15.1(d )))((1)1()(ch af ehgf ch gb af gb ed abcd s G +----++=第3章 3-1 60070600)(2++=s s s G ζ = 1.43 ωn = 24.5 3-2 ζ = 0.6 ωn = 2 σ % = 9.48% t p = 1.96 t r = 1.38 t s = 2.53-3 ζ = 0.5 ωn = 1σ % = 16.3% t p = 2.42 t r = 3.63 t s = 63-4 (1)12211)(KT s T s T s ⋅++=φ (2))5)(2()3(15)(+++=s s s s φ (3)25.06.04016.0)3.0(40)(22++=++=s s s s φ 3-5 (1)ζ = 0.1 ωn = 5)26.8497.4sin(005.11)(5.0︒+-=-t e t c tσ % = 72.9% t s = 6ζ = 0.1 ωn = 10)26.8494.9sin(005.11)(︒+-=-t e t c tσ % = 72.9% t s = 3ζ = 0.1 ωn = 1)26.84994.0sin(005.11)(1.0︒+-=-t e t c tσ % = 72.9% t s = 30(2)ζ = 0.5 ωn = 5)6033.4sin(155.11)(5.2︒+-=-t e t c tσ % = 16.3% t s = 1.23-6 (1)图1 ζ = 0.1 ωn = 1 σ % = 72.9% t s = 30图2 ζ = 0.5 ωn = 1图2 ζ = 0.0745 ωn = 0.7453-7 (1)K h = 0.216(2)10)101()1(10)(21++++=s K s s K s h h φ 10)101(10)(22+++=s K s s h φ (3)474.01015.0 16.310====ζωnσ % = 18.4% t p = 1.128 t s = 2(4)原系统 ζ = 0.158 ωn = 3.16 σ % = 60.5% t p = 1.006 t s = 6加入比例微分和速度反馈后,阻尼比增大,使系统超调量减小,响应速度加快。