27746電力拖動自動控制系統(tǒng)(第4版)習(xí)題解答

1、27746電力拖動自動控制系統(tǒng)（第4版）習(xí)題解答習(xí)題解答2.2解：1000x0.02/(10x0.98) = 2.046?系統(tǒng)允許的靜態(tài)速降為2.04rpm o2.3解：1）調(diào)速范圍（均指額定負載情況下）第8頁共7頁"max = "Omax 一=1500-15 = 1485“min = omta 一加N = 150 -15 = 135"max/min =1485/135 = 112)靜差率 s = Mv/o = 15/15O = lO%2.4解： Ce = (Un -及()為=(220 378x0.023)/1430 = 0.147W/rpmZi = INR/Ce

2、 = 378 X (0.023 + 0.022)/0.1478 = 1 15rpmD = ®S/ AW s) = 1430 x 0.2/l 15x(1- 0.2) = 3.1。=nS/ &?(1s) = 1430x0.3/l 15x(1-0.3) = 5.332.5解：(1)= IN X R/Ce = 305x0.18/0.2 = 274.5r /min(2) 5V = Ahv/h0 = 274.5/(1000 + 274.5) = 21.5%(3) An = a，S/O(1 - s)= 1000x0.05/(20x0.95 = 2.63 尸 / min2.6解：(1) Ud

3、 = K“K、U；/(1 + KpK?) = 2xl5x 8.8/(l + 2xl5x 0.7) = 12V(2) Ud =8.8x2x15 = 26%開環(huán)輸出電壓是閉環(huán)的22倍(3)U； = Ud (1 + KpKsy)!KpKs =12x(l + 2xl5x 0.35)/(2 x15) = 4.6K2.7解： 1) D = nNs/ AnN(-s)10 = 1500x2%/A/jvx98%4? v = 1500x2%/98%xl0 = 3.06 f min2) K = (&7叩 / A/7(.J-1 = 100/3.06-1 =31.72.8解：=(1 + KW% =(1 + 15

4、)x8 = 128如果將開環(huán)放大倍數(shù)提高到30,則速降為:/(1 + K) = 128 /(1 + 30)= 4A3rpm在同樣靜差率要求下，D可以擴大&2必 /&為2 = L937 倍2.9解：n=(UN-INxRt)/Ce=> C. =(220-12.5x 1.5)/1500 = 201.25/1500 = 0.134Vmin/ r=(4-/戶&)/，= lN X & / C = 12.5X3.3/0.134 = 307.836r/minAn*. = nv5/(D(l-5) = 1500x!0%/(20*90%) = 8.33r / min所以， =8

5、.33r/min+(3)(4) =(KpK.M -/)/(Ce(l + K)= KU； /a(l + K)-ldR/(Ce( + K)K = (4a / An”)-1 = 307.836 / 8.33-1 = 35.9551500 = 35.955 x 15 / a (1 + 35.955) - 12.5 x 3.3 /(0.134(1 + 35.955)可以求得,=> a = 0.0096V min/ rK =佇Q = 35.955*034 . w34 KJ a 35*0.0096也可以用粗略算法:U* 15U； % U = coi 9 a = = = 0.01 n 1500=KC/K

6、,a 9K, = 35.955xO.134/(35xO.Ol) = 13.762.12解.Ce =(220-15.6x1.5)/1500 =0.131 IVmin/ r(1)= /v x/ez /Cr = 15.6x33/0.1311 = 392.68" min由=150°/3° = 5°s =/ A/iOmin = 392.68 / (392.68 + 50) = 88.7%(2)0.1 = An/(An + 50)An = 5/0.9 = 5.56r/min(3)f；/q(i + K)-&/d/c(i + K)K = KpaKs/Ce1500

7、 = Kp K： / Q (1 + K) - (4 15.6)/ q (1 + K)K = (% /% )-1 =(297.48/5.56)-1 = 52.5習(xí)題三3. 1解：a = U； / iiN = 15V /1500rp？ = 0.01V / rpm當(dāng) U； = 5V, 轉(zhuǎn)速n = =-= 500/pza O.OW/rpm/3 = = - = 0315V/A/擷 40AU = pld = 0.375 *10 = 3.75V = QU. E +1dl R _ C ,hv +1dl R _ 0.127 * 500 + 10 * 2Ks KsKs20即 =500rz,U” = 5V,U； = U, = 3.75 V" =4.175n(2)堵轉(zhuǎn)時，U：5V,u=U(c")=g$=wK, K, K, 203.2解：1) /? = >- = = 0.2V/A a = % = - = 0.008V / rpmb '