TRNG I HC BCH KHOA KHOA C KH B MN C IN T THI GIA HC K MN NG LC HC V IU KHINNgy thi: 15/04/2011 Thi gian thi: 45 pht Sinh vin khng c s dng ti liu Cu 1: ( 3,0 ) Cho h thng hi tip m nh hnh v ( )2CKG ss; 1( )( 1)( 3)Gss s Xc nh iu kin ca K h thng n nhCu 2: (7,0 ) V qu o nghim s ca h thng khi K = 0 + 0( s) =K( s 1)s( s 2 ) ( s2 + 3s +10 ) bit: _s = 1.28 0.86js = - 0 .94 1.38j Ch nhim b mn Ging vin ra thi TS. Nguyn Duy Anh P N THI MN NG LC HC V IU KHIN Ngy thi: 15/04/2011 Thi gian: 45 pht Cu 1: (3,0 ) im Xc nh iu kin ca K h thng n nh: Phng trnh c trng ca h: 1 + 0c( s) 0( s) = 0 1 +Ks + 2 1( s 1) ( s + 3)= 0 s3 +4s2 + s +( K 6) = 0 Lp bng Routh s311s24K 6s1 10K40s0K 6iu kin h thng n nh _10 K4> 0K 6> 0 6< K < 10 Cu 2: ( 7,0 ) Phng trnh c tnh h thng: 1 +K( s 1 )s( s 2) ( s2 + 3 s +10)= 0 ( 1)Cc cc:p1 = 0, p2 = 2, p3,4 = 1 .5 2.78]Cc zero: z1 = 1 Tin cn: o =( 2l + 1 ) nn m=( 2l + 1 ) n41=n3( l = 0)n3 ( l = 1 )n ( l = 1) 0A =Zcc Zzcon m=0 +2 +( 1 .5 + 2.78]) +( 1.5 2 .78 ]) 13= 23 im tch nhp: ( 1) K = ( s4 + s3 +4s2 20 s)s 1 JKJs=3 s4 +2 s3 s2 +8s 20( s 1 )2 Do JKJs= 0 _s = 1 .28 0 .86] s = 0 .94 1.38](Khng c im tch nhp) Giao im qu o nghim s trc o: ( 1) s4 + s3 + 4s2 +( K 20) s K = 0 ( 2 ) Thay s = ] :o ( 2 ) : 4 ]3 42 +( K 20) ] K = 0 Suy ra: 4 4 2 K = 03 + ( K 20) = 0] = 0K = 0] = 2.7 6K = 27.62] = 1 .62]K = 17.38 Vy giao im cn tm: s = 2.7 6] H s khuch i gii hn l Kgh = 27.62 Gc xut pht ca QNS ti cc phc P3: 03 = 18 0 + [1 ( [2 + [3 + [4) = 3 7.8 5