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It is as though their entire tournament has been preparation for the final. In Johnathan Thurston, Cooper Cronk, Cameron Smith and either Billy Slater or Greg Inglis at full back they have a spine that is unmatched in rugby league. They have played in so many highpressure games a priceless asset going into Saturday. The Kiwis are a lot less experienced but winning a dramatic match like their semifinal against England will do wonders for their confidence. They defeated Australia in the Four Nations final in 2020 and the last World Cup, and know they can rise to the big occasion. 內(nèi)部資料 , 請勿外傳! 專業(yè)好文檔 17 XuyUP2kNXpRWXm Aamp。s actions. CSKA said they were surprised and disappointed by Toure39。s Premier League match at Chelsea on Sunday. I am going to be at the match tomorrow and I have asked to meet Yaya Toure, he told BBC Sport. For me it39。s top try scorer with nine, closely followed by TuivasaSheck with eight. But it is recently named Rugby League International Federation player of the year Sonny Bill Williams who has attracted the most interest in the tournament so far. The Kiwi with a tournament high 17 offloads has the chance of being the first player to win the World Cup in both rugby league and rugby union after triumphing with the All Blacks in 2020. I39。 正確 9. 利用相位超前校正,可以增加系統(tǒng)的頻寬,提高系統(tǒng)的快速性,但使穩(wěn)定裕量變小。 正確 4.按校正裝置 Gc(s)的物理性質(zhì)區(qū)分,又有相位超前 (微分 )校正,相位滯后 (積分 )校正,和相位滯后 — 超前 (積分 微分 )校正。 D A )1)(15( 1 ?? ? ss s; BsTTs111?? ( T0) ; C )1)(12( 1 ?? ? ss s ; D)2)(3( 2?? ?ss s; 三、 最小相位系統(tǒng)的對數(shù)幅頻特性如下圖所示,試分別確定各系統(tǒng)的傳遞函數(shù)。 正確 13. 系統(tǒng)的對數(shù)幅頻特性和相頻指性有一一對應(yīng)關(guān)系,則它必是 最小相位 系統(tǒng)。正確 正弦 輸入信號作用下的 穩(wěn)態(tài) 輸出和輸入之比。 正確 10.系統(tǒng)的根軌跡起始于開環(huán)極點(diǎn),終止于開環(huán) 零 點(diǎn) 。 正確 6. 在開環(huán)系統(tǒng)中增加零點(diǎn),可使根軌跡向 左 方移動。 正確 3.繪制根軌跡時,我們通常是從 gK = 0 時的閉環(huán)極點(diǎn)畫起,即開環(huán)極點(diǎn)是閉環(huán)根軌跡曲線的起點(diǎn)。 八、 某典型二階系統(tǒng)的單位階躍響應(yīng)如圖所示。求該系統(tǒng)的超調(diào)量和調(diào)整時間; 【解】系統(tǒng)的閉環(huán)傳遞函數(shù)為 ? ?kk Kss Ks ??? 2? 4?kK 與二階系統(tǒng)標(biāo)準(zhǔn)形式的傳遞函數(shù) ? ? 22 22nnn sss ????? ??? 對比得: (1) 固有頻率 24 ??? kn K? (2) 阻尼比 由 12 ?n?? 得 21 ?? n?? (3) 超調(diào) ? ? %47%10 0% )1/( 2 ??? ?? ne ??? (4) 調(diào)整時間 ? ? stns 63%5 ?? ?? 六、 已知單位反饋系統(tǒng)開環(huán)傳函為)( 10)( ?? sssG, 求系統(tǒng)的 ξ、ω n 及性能指標(biāo)σ%、ts( 5%)。 錯誤 三、 已知一個 n 階閉環(huán)系統(tǒng)的微分方程為 rbrbyayayayaya nnnn 0101)2(2)1(1)( ??????? ?? ??? 1. 寫出該系統(tǒng)的閉環(huán)傳遞函數(shù); 2. 寫出該系統(tǒng)的特征方程; 3. 當(dāng) 10?a , ?a , ?a