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18 october 2010 0 02 04 06 h t hong

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Unformatted text preview: ùng caøng nhanh. Th Thôøi gian quaù ñoä cuûa heä quaùn tính baäc 1 laø: t qñ ⎛1⎞ = T ln⎜ ⎟ ⎝ε ⎠ vôùi ε = 0.02 (tieâu chuaån 2%) hoaëc ε = 0.05 (tieâu chuaån 5%) 18 October 2010 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 17 Ñaù Ñaùp öùng quaù ñoä Quan heä giöõa vò trí cöïc vaø ñaùp öùng heä quan tính baäc 1 heä giöa vò trí cöc va ñap ng heä quaùn tính baä Cöïc naèm caøng xa truïc aûo ñaùp öùng cuûa heä quaùn tính baäc 1 caøng nhanh, thôøi gian quaù ñoä caøng ngaén. thôi cang ngan Im s y(t) K Re s 0 t 0 Giaûn ñoà cöïc –zero cuûa khaâu quaùn tính baäc 1 18 October 2010 Ñaùp öùng quaù ñoä cuûa khaâu quaùn tính baäc 1 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 18 Ñaù Ñaùp öùng quaù ñoä Heä dao ñoäng baäc 2 dao ñoä baä R(s) K T 2 s 2 + 2ξTs + 1 Y(s) Haøm truyeàn heä dao ñoäng baäc 2: 2 K Kω n G(s) = 2 2 =2 2 T s + 2ξTs + 1 s + 2ξωn s + ωn 1 (ωn = , 0 < ξ < 1) T Heä dao ñoäng baäc 2 coù caëp cöïc phöùc: p1, 2 = −ξωn ± jωn 1 − ξ 2 Ñaùp öùng quaù ñoä: ⇒ 2 Kω n 1 Y ( s ) = R ( s )G ( s ) = . 2 2 s s + 2ξωn s + ωn [ ⎧ e −ξωnt ⎪ y (t ) = K ⎨1 − sin (ωn 1 − ξ 2 )t + θ ⎪ 1− ξ 2 ⎩ 18 October 2010 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ ⎫ ⎪ (cosθ = ξ ) ⎬ ⎪ ⎭ 19 Ñaù Ñaùp öùng quaù ñoä Heä dao ñoäng baäc 2 (tt) dao ñoä baä (tt) Im s cos θ= ξ jω n 1 − ξ 2 ωn −ξωn θ y(t) Re s (1+ε).K K (1−ε).K 0 − jω n 1 − ξ 2 0 Giaûn ñoà cöïc –zero cuûa khaâu dao ñoäng baäc 2 18 October 2010 t tqñ Ñaùp öùng quaù ñoä cuûa khaâu dao ñoäng baäc 2 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 20 Ñaù Ñaùp öùng quaù ñoä Nhaä xet ve heä dao ñoä baä Nhaän xeùt veà heä dao ñoäng baäc 2 Heä dao ñoäng baäc 2 coù caëp cöïc phöùc, ñaùp öùng quaù ñoä coùù daïng dao ñoäng vôùi bieân ñoä giaûm daàn. Neáu ξ = 0, ñaùp öùng cuûa heä laø dao ñoäng khoâng suy giam ôi tan so giaûm vôùi taàn soá ωn ⇒ ωn goïi laø taàn soá dao ñoäng töï nhieân. Neáu 0< ξ <1, ñaùp öùng cuûa heä laø dao ñoäng vôùi bieân ñoä giaûm daàn ⇒ ξ goïi laø heä soá taét (hay heä soá suy giaûm), ξ caøn...
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