chapter6第五版

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计算机系统结构(第五版)第6章

【例 6-1】计算D=A×(B+C)，其中A、B、C、D都是有N个元素的
如果采用逐个求D向量元素的方法，即访存取ai、bi、ci 元素求di，再取ai+1、bi+1、ci+1求di+1，则这种处理方式称为横向(水平)处理方式。
6.1.2
向量流水处理机的结构因具体机器的不同而不同。图6 - 1只画出了CRAY-1中央处理机中有关向量流水处理部分的简图。
所谓Vi冲突，指的是并行工作的各向量指令的源向量或结果向量使用了相同的Vi。所谓功能部件冲突，指的是同一个功能部件被要求并行工作的多条向量指令所使用。
第一、二条指令无任何冲突，可以并行执行。第三条指令与第一、二条指令出现Vi冲突，存在先写后读数相关，本来是不能并行执行的，但若能把第一、二条指令的结果分量直接链接进第三条指令所用的功能部件，那第三条指令就能与第一、二条指令在大部分时间内并行。它们的链接过程如图6 - 3所示。
(V061)=(V053)+(V161) =(V15)+(V113)+(V121)+(V129)+(V137) +(V145)+(V153)+(V161)
第八部分(结果部分)
(V062)=(V054)+(V162) =(V16)+(V114)+(V122)+(V130)+(V138) +(V146)+(V154)+(V162)
(V05
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(V00
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(V15
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0(V15
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(V06 ) (V00 ) (V16 ) 0(V16 )
(V07 ) (V00 ) (V17 ) 0(V17 )

电路第五版邱关源第6章(新版)

3.并联的等效电感并联的等效电感
串联的等效电感等于各个电感之和；串联的等效电感等于各个电感之和；并联等效电感的倒数等于各个电感倒数之和
本章小结
• 电容和电感元件的性质，作用电容和电感元件的性质， • 电容和电感元件特性（库伏特性和伏安特电容和电感元件特性（性） • 电容和电感元件的功率及能量 • 电容和电感串并联的等效电容
• 关于电感元件的结论：关于电感元件的结论：
（1）电感元件是一个无源元件，动态元件，具有）电感元件是一个无源元件，动态元件，记忆”功能的元件， “记忆”功能的元件，短路直流的作用（2）能储存和释放能量。具有短路直流的作用）能储存和释放能量。具有短路直流
（）注: 1）实际的电感线圈的模型可用线性电感元件和电阻元件串联组合。件串联组合。（2）非线性电感元件韦安特性不是直线）非线性电感元件韦安特性不是直线, 电感元件（3）L ）元件的参数
第六章：第六章：储能元件
• 本章的重点： 1.电容和电感元件的特性和伏安关系 2.电容和电感的串并联的等效参数 • 本章难点：对电容和电感元件特性的理解
§6-1 电容元件
一、电容元件符号：电容元件符号：二、电容元件特性（库伏特性）电容元件特性（库伏特性）
任何时刻，任何时刻，当电压的参考方向与电容元件极板上存储电荷的方向一致时，电荷q与电压成正比。一致时，电荷与电压 u 成正比。 C
分部电容，三极管极间存在的杂散电容。分部电容，三极管极间存在的杂散电容。 (2)实际电容元件的模型是电容和电阻元件的并联组合。实际电容元件的模型是电容和电阻元件的并联组合。实际电容元件的模型是电容和电阻元件的并联组合
(3)非线性电容元件库伏特性不是直线，如变容非线性电容元件库伏特性不是直线，

同济大学高等数学第六版第五章课后习题答案(包括5.1,5.2,5.3,5.4,总习题五)

习题5-11. 利用定积分定义计算由抛物线y =x 2+1, 两直线x =a 、x =b (b >a )及横轴所围成的图形的面积.解第一步: 在区间[a , b ]内插入n -1个分点i n a b a x i -+=(i =1,2, ⋅⋅⋅, n -1), 把区间[a , b ]分成n 个长度相等的小区间, 各个小区间的长度为: na b x i -=∆(i =1, 2, ⋅⋅⋅, n ).第二步: 在第i 个小区间[x i -1, x i ] (i =1, 2, ⋅⋅⋅, n )上取右端点i n a b a x i i -+==ξ, 作和n a b i n a b a x f S ni i i n i n -⋅+-+=∆=∑∑==]1)[()(211ξ∑=+-+-+-=ni i na b i n a b a a n a b 12222]1)()(2[ ]6)12)(1()(2)1()(2[)(222n n n n n a b n n n a b a na n a b +++⋅-++⋅-+-= ]16)12)(1()()1)(()[(222+++-++-+-=nn n a b n n a b a a a b . 第三步: 令λ=max{∆x 1, ∆x 2, ⋅⋅⋅, ∆x n }na b -=, 取极限得所求面积∑⎰=→∆==ni i i ba x f dx x f S 1)(lim )(ξλ]16)12)(1()()1)(()[(lim 222+++-++-+-=∞→nn n a b n n a b a a a b n a b a b a b a b a a a b -+-=+-+-+-=)(31]1)(31)()[(3322.2. 利用定积分定义计算下列积分:(1)xdx ba⎰(a <b );解取分点为i na b a x i -+=(i =1, 2, ⋅⋅⋅, n -1), 则n a b x i -=∆(i =1, 2,⋅⋅⋅, n ). 在第i 个小区间上取右端点i na b a x i i -+==ξ(i =1, 2, ⋅⋅⋅, n ). 于是∑∑⎰=∞→=∞→-⋅-+=∆=ni n n i i i n ba n ab i n a b a x xdx 11)(lim lim ξ )(21]2)1()()([lim )(22222a b n n n a b a b a a b n -=+-+--=∞→. (2)dx e x ⎰10.解取分点为n i x i =(i =1, 2, ⋅⋅⋅, n -1), 则nx i 1=∆(i =1, 2, ⋅⋅⋅, n ). 在第i 个小区间上取右端点ni x i i ==ξ(i =1, 2, ⋅⋅⋅, n ). 于是) (1lim 1lim 21110n n n n n n i n i n xe e e n n e dx e +⋅⋅⋅++==∞→=∞→∑⎰1)1(]1[lim1])(1[1lim 11111-=--=--⋅=∞→∞→e e n e e ee e nnnn nnn n n .3. 利用定积分的几何意义, 说明下列等式: (1)1210=⎰xdx ;解⎰12xdx 表示由直线y =2x 、x 轴及直线x =1所围成的面积,显然面积为1. (2)41102π=-⎰dx x ;解⎰-121dx x 表示由曲线21x y -=、x 轴及y 轴所围成的四分之一圆的面积, 即圆x 2+y 2=1的面积的41:414112102ππ=⋅⋅=-⎰dx x .(3)⎰-=ππ0sin xdx ;解由于y =sin x 为奇函数, 在关于原点的对称区间[-π, π]上与x 轴所夹的面积的代数和为零, 即⎰-=ππ0sin xdx .(4)⎰⎰=-2022cos 2cos πππxdx xdx .解⎰-22cos ππxdx 表示由曲线y =cos x 与x 轴上]2,2[ππ-一段所围成的图形的面积. 因为cos x 为偶函数, 所以此图形关于y 轴对称. 因此图形面积的一半为⎰20cos πxdx , 即⎰⎰=-2022cos 2cos πππxdx xdx .4. 水利工程中要计算拦水闸门所受的水压力, 已知闸门上水的压强p (单位面积上的压力大小)是水深h 的函数, 且有p =9⋅8h (kN/m 2). 若闸门高H =3m , 宽L =2m , 求水面与闸门顶相齐时闸门所受的水压力P .解建立坐标系如图. 用分点i nH x i =(i =1, 2, ⋅⋅⋅, n -1)将区间[0,H ]分为n 分个小区间, 各小区间的长为n H x i =∆(i =1, 2, ⋅⋅⋅, n ).在第i 个小区间[x i -1, x i ]上, 闸门相应部分所受的水压力近似为∆P i =9.8x i L ⋅∆x i . 闸门所受的水压力为∑∑=∞→=∞→⋅=∆⋅⋅=ni n ni i i n n H i n H L x L x P 11lim 8.98.9lim 228.42)1(lim 8.9H L nn n H L n ⋅=+⋅=∞→.将L =2, H =3代入上式得P =88.2(千牛). 5. 证明定积分性质: (1)⎰⎰=babadx x f k dx x kf )()(;证明∑⎰=→∆=n i iibax kf dx x kf 1)(lim )(ξλ⎰∑=∆==→bani iidx x f k x f k )()(lim 1ξλ.(2)a b dx dx baba-==⋅⎰⎰1.证明a b a b x x dx ni in i iba-=-=∆=∆⋅=⋅→=→=→∑∑⎰)(lim lim 1lim 1011λλλ. 6. 估计下列各积分的值: (1)⎰+412)1(dx x ;解因为当1≤x ≤4时, 2≤x 2+1≤17, 所以 )14(17)1()14(2412-⋅≤+≤-⋅⎰dx x ,即 51)1(6412≤+≤⎰dx x .(2)⎰+ππ4542)sin 1(dx x ;解因为当ππ454≤≤x 时, 1≤1+sin 2x ≤2, 所以)445(2)sin 1()445(14542ππππππ-⋅≤+≤-⋅⎰dx x , 即 ππππ2)sin 1(4542≤+≤⎰dx x .(3)⎰331arctan xdx x ;解先求函数f (x )=x arctan x 在区间]3 ,31[上的最大值M 与最小值m .21arctan )(x x x x f ++='. 因为当331≤≤x 时, f '(x )>0, 所以函数f (x )=x arctan x 在区间]3 ,31[上单调增加. 于是3631arctan 31)31(π===f m ,33arctan 3)3(π===f M .因此 )313(3arctan )313(36331-≤≤-⎰ππxdx x ,即 32arctan 9331ππ≤≤⎰xdx x .(4)⎰-022dx e xx.解先求函数xxe xf -=2)(在区间[0, 2]上的最大值M 与最小值m . )12()(2-='-x e x f xx, 驻点为21=x .比较f (0)=1, f (2)=e 2, 41)21(-=e f , 得41-=e m , M =e 2. 于是)02()02(22412-⋅≤≤-⎰--e dx e e xx,即 41022222---≤≤-⎰e dx dx ee xx .7. 设f (x )及g (x )在[a , b ]上连续, 证明:(1)若在[a , b ]上, f (x )≥0, 且0)(=⎰ba dx x f , 则在[a ,b ]上f (x )≡0;证明假如0/)(≡x f , 则必有f (x )>0. 根据f (x )在[a , b ]上的连续性, 在[a , b ]上存在一点x 0, 使f (x 0)>0, 且f (x 0)为f (x )在[a , b ]上的最大值.再由连续性, 存在[c , d ]⊂[a , b ], 且x 0∈[c , d ], 使当x ∈[c , d ]时,2)()(0x f x f >. 于是⎰⎰⎰⎰++=bdd cc abadx x f dx x f dx x f dx x f )()()()(0)(2)()(0>-≥≥⎰c d x f dx x f dc. 这与条件0)(=⎰badx x f 相矛盾. 因此在[a , b ]上f (x )≡0.(2)若在[a , b ]上, f (x )≥0, 且0/)(≡x f , 则0)(>⎰badx x f ;证明证法一因为f (x )在[a , b ]上连续, 所以在[a , b ]上存在一点x 0, 使f (x 0)>0, 且f (x 0)为f (x )在[a , b ]上的最大值.再由连续性, 存在[c , d ]⊂[a , b ], 且x 0∈[c , d ], 使当x ∈[c , d ]时,2)()(0x f x f >. 于是 ⎰⎰>-≥≥b a d c c d x f dx x f dx x f 0)(2)()()(0.证法二因为f (x )≥0, 所以0)(≥⎰badx x f . 假如0)(>⎰badx x f 不成立. 则只有0)(=⎰badx x f ,根据结论(1), f (x )≡0, 矛盾. 因此0)(>⎰badx x f .(3)若在[a , b ]上, f (x )≤g (x ), 且⎰⎰=bab adx x g dx x f )()(, 则在[a , b ]上f (x )≡g (x ).证明令F (x )=g (x )-f (x ), 则在[a , b ]上F (x )≥0且0)()()]()([)(=-=-=⎰⎰⎰⎰bab ab ab adx x f dx x g dx x f x g dx x F ,由结论(1), 在[a , b ]上F (x )≡0, 即f (x )≡g (x ).4. 根据定积分的性质及第7题的结论, 说明下列积分哪一个的值较大:(1)⎰12dx x 还是⎰13dx x ？解因为当0≤x ≤1时, x 2≥x 3, 所以⎰⎰≥13102dx x dx x .又当0<x <1时, x 2>x 3, 所以⎰⎰>13102dx x dx x .(2)⎰212dx x 还是⎰213dx x ？解因为当1≤x ≤2时, x 2≤x 3, 所以⎰⎰≤213212dx x dx x .又因为当1<x ≤2时, x 2<x 3, 所以⎰⎰<213212dx x dx x .(3)⎰21ln xdx 还是⎰212)(ln dx x ？解因为当1≤x ≤2时, 0≤ln x <1, ln x ≥(ln x )2, 所以⎰⎰≥21221)(ln ln dx x xdx .又因为当1<x ≤2时, 0<ln x <1, ln x >(ln x )2, 所以⎰⎰>21221)(ln ln dx x xdx .(4)⎰1xdx 还是⎰+1)1ln(dx x ？解因为当0≤x ≤1时, x ≥ln(1+x ), 所以⎰⎰+≥11)1ln(dx x xdx .又因为当0<x ≤1时, x >ln(1+x ), 所以⎰⎰+>110)1ln(dx x xdx .(5)⎰10dx e x 还是⎰+1)1(dx x ？解设f (x )=e x -1-x , 则当0≤x ≤1时f '(x ) =e x -1>0, f (x )=e x -1-x 是单调增加的. 因此当0≤x ≤1时, f (x )≥f (0)=0, 即e x ≥1+x , 所以⎰⎰+≥110)1(dx x dx e x .又因为当0<x ≤1时, e x>1+x , 所以⎰⎰+>110)1(dx x dx e x .习题5-21. 试求函数⎰=xtdt y 0sin 当x =0及4π=x 时的导数.解 x tdt dx dy x sin sin 0=='⎰, 当x =0时, y '=sin0=0;当4π=x 时, 224sin =='πy . 2. 求由参数表示式⎰=t udu x 0sin , ⎰=tudu y 0cos 所给定的函数y对x 的导数.解 x '(t )=sin t , y '(t )=cos t , t t x t y dx dy cos )()(=''=.3. 求由⎰⎰=+xyt tdt dt e 00cos 所决定的隐函数y 对x 的导数dxdy. 解方程两对x 求导得 0cos =+'x y e y ,于是 y ex dx dy cos -=. 4. 当x 为何值时, 函数⎰-=xt dt te x I 02)(有极值？解 2)(x xe x I -=', 令I '(x )=0, 得x =0.因为当x <0时, I '(x )<0; 当x >0时, I '(x )>0, 所以x =0是函数I (x )的极小值点. 5. 计算下列各导数: (1)⎰+2021x dtt dxd ;解 dx du dt t du d dt t dx d u ux x ⋅+====+⎰⎰=02021122令421221x x x u +=⋅+=.(2)⎰+32411x x dt tdx d ; 解 ⎰⎰⎰+++=+323204044111111x x x x dt t dx d dt t dx d dt t dx d ⎰⎰+++-=3204041111x x dt t dx d dt t dx d )()(11)()(11343242'⋅++'⋅+-=x x x x 12281312x x x x +++-=. (3)⎰x xdt t dx d cos sin 2)cos(π. 解⎰⎰⎰+-=x x x x dt t dx d dt t dx d dt t dx d cos 02sin 02cos sin 2)cos()cos()cos(πππ))(cos cos cos())(sin sin cos(22'+'-=x x x x ππ )cos cos(sin )sin cos(cos 22x x x x ππ⋅-⋅-= )sin cos(sin )sin cos(cos 22x x x x πππ-⋅-⋅-= )sin cos(sin )sin cos(cos 22x x x x ππ⋅+⋅-= )sin cos()cos (sin 2x x x π-=6. 计算下列各定积分: (1)⎰+-adx x x 02)13(;解a a a x x x dx x x a a+-=+-=+-⎰230230221|)21()13(. (2)⎰+2142)1(dx x x ;解 852)11(31)22(31|)3131()1(333321332142=---=-=+---⎰x x dx x x . (3)⎰+94)1(dx x x ;解 94223942194|)2132()()1(x x dx x x dx x x +=+=+⎰⎰ 6145)421432()921932(223223=+-+=.(4)⎰+33121x dx; 解 66331arctan 3arctan arctan 13313312πππ=-=-==+⎰x x dx . (5)⎰--212121x dx ; 解3)6(6)21arcsin(21arcsin arcsin 1212121212πππ=--=--==---⎰x x dx .(6)⎰+ax a dx3022; 解 aa a axa x a dx a a 30arctan 13arctan 1arctan 1303022π=-==+⎰.(7)⎰-1024x dx ; 解 60arcsin 21arcsin 2arcsin 410102π=-==-⎰x x dx . (8)dx x x x ⎰-+++012241133; 解 013012201224|)arctan ()113(1133---+=++=+++⎰⎰x x dx x x dx x x x 41)1arctan()1(3π+=----=.(9)⎰---+211e x dx;解 1ln 1ln ||1|ln 12121-=-=+=+------⎰e x x dx e e . (10)⎰402tan πθθd ;解4144tan )(tan )1(sec tan 4040242πππθθθθθθπππ-=-=-=-=⎰⎰d d .(11)dx x ⎰π20|sin |;解⎰⎰⎰-=ππππ2020sin sin |sin |xdx xdx dx xπππ20cos cos x x +-==-cos π +cos0+cos2π-cos π=4.(12)⎰2)(dx x f , 其中⎩⎨⎧>≤+=12111)(2x x x x x f .解 38|)61(|)21(21)1()(2131022121020=++=++=⎰⎰⎰x x x dx x dx x dx x f . 7. 设k 为正整数. 试证下列各题: (1)⎰-=ππ0cos kxdx ;证明⎰--=--==ππππππ0)(sin 1sin 1|sin 1cos k kk k kx k kxdx . (2)⎰-=ππ0sin kxdx ;证明 )(cos 1cos 1cos 1sin ππππππ-+-=-=--⎰k kk k x k k kxdx0cos 1cos 1=+-=ππk k k k .(3)⎰-=πππkxdx 2cos ;证明 πππππππ=+=+=---⎰⎰|)2sin 21(21)2cos 1(21cos 2kx kx dx kx kxdx . (4)⎰-=πππkxdx 2sin .证明 πππππππ=-=-=---⎰⎰|)2sin 21(21)2cos 1(21sin 2kx kx dx kx kxdx . 8. 设k 及l 为正整数, 且k ≠l . 试证下列各题: (1)⎰-=ππ0sin cos lxdx kx ;证明 ⎰⎰----+=ππππdx x l k x l k lxdx kx ])sin()[sin(21sin cos0])cos()(21[])cos()(21[=----++-=--ππππx l k l k x l k l k . (2)⎰-=ππ0cos cos lxdx kx ;证明 ⎰⎰---++=ππππdx x l k x l k lxdx kx ])cos()[cos(21cos cos 0])sin()(21[])sin()(21[=--+++=--ππππx l k l k x l k l k . (3)⎰-=ππ0sin sin lxdx kx .证明 ⎰⎰----+-=ππππdx x l k x l k lxdx kx ])cos()[cos(21sin sin . 0])sin()(21[])sin()(21[=--+++-=--ππππx l k l k x l k l k . 9. 求下列极限: (1)xdt t xx ⎰→020cos lim ;解 11cos lim cos lim20020==→→⎰x xdt t x xx .(2)⎰⎰→xt xt x dttedt e 0220022)(lim.解 22222202200)(2lim )(limx xt x t x xt xt x xedt e dt e dtte dt e '⋅=⎰⎰⎰⎰→→222220202lim 2lim x x t x x x x t x xe dte xe edt e ⎰⎰→→=⋅=2212lim 22lim 2020222=+=+=→→xe x e e x x x x x .10. 设⎩⎨⎧∈∈=]2 ,1[]1 ,0[)(2x x x x x f . 求⎰=x dt t f x 0)()(ϕ在[0, 2]上的表达式, 并讨论ϕ(x )在(0, 2)内的连续性.解当0≤x ≤1时, 302031)()(x dt t dt t f x xx ===⎰⎰ϕ;当1<x ≤2时,6121212131)()(2211020-=-+=+==⎰⎰⎰x x tdt dt t dt t f x xxϕ.因此 ⎪⎩⎪⎨⎧≤<-≤≤=21 612110 31)(23x x x x x ϕ.因为31)1(=ϕ, 3131lim )(lim 30101==-→-→x x x x ϕ, 316121)6121(lim )(lim 20101=-=-=+→+→x x x x ϕ, 所以ϕ(x )在x =1处连续, 从而在(0, 2)内连续.11. 设⎩⎨⎧><≤≤=ππx x x x x f 或000sin 21)(. 求⎰=xdt t f x 0)()(ϕ在(-∞, +∞)内的表达式. 解当x <0时,00)()(0===⎰⎰xxdt dt t f x ϕ;当0≤x ≤π时,21cos 21|cos 21sin 21)()(000+-=-===⎰⎰x t tdt dt t f x xxxϕ;当x >π时, πππϕ000|cos 210sin 21)()(t dt tdt dt t f x x x-=+==⎰⎰⎰10cos 21cos 21=+-=π.因此 ⎪⎩⎪⎨⎧≥≤≤-<=ππϕx x x x x 10)cos 1(2100)(.12. 设f (x )在[a , b ]上连续, 在(a , b )内可导且f '(x )≤0,⎰-=x a dt t f ax x F )(1)(. 证明在(a , b )内有F '(x )≤0.证明根据积分中值定理, 存在∈[a , x ], 使))(()(a x f dt t f xa-=⎰ξ.于是有)(1)()(1)(2x f ax dt t f a x x F x a -+--='⎰ ))(()(1)(12a x f a x x f a x ----=ξ )]()([1ξf x f ax --=. 由f '(x )≤0可知f (x )在[a , b ]上是单调减少的, 而a ≤ξ≤x , 所以f (x )-f (ξ)≤0. 又在(a , b )内, x -a >0, 所以在(a , b )内 0)]()([1)(≤--='ξf x f ax x F .习题5-31. 计算下列定积分: (1)⎰+πππ2)3sin(dx x ;解0212132cos 34cos )3cos()3sin(22=-=+-=+-=+⎰ππππππππx dx x . (2)⎰-+123)511(x dx ;解51251110116101)511(2151)511(22122123=⋅+⋅-=+-⋅=+-----⎰x x dx . (3)⎰203cos sin πϕϕϕd ;解⎰⎰-=20323sin cos cos sin ππϕϕϕϕϕd s d410cos 412cos 41cos 4144204=+-=-=πϕπ. (4)⎰-πθθ03)sin 1(d ;解⎰⎰⎰+=-πππθθθθθ0203cos sin )sin 1(d d d⎰-+=ππθθθ020cos )cos 1(d34)cos 31(cos 03-=-+=πθθππ.(5)⎰262cos ππudu ;解 22262622sin 4121)2cos 1(21cos ππππππππu u du u udu +=+=⎰⎰ 836)3sin (sin 41)62(21-=-+-=πππππ.(6)dx x ⎰-2022;解tdt t dx x t x cos 2cos 2 220sin 222⎰⎰⋅======-=π令dt t ⎰+=2)2cos 1(π2)2sin 21(20ππ=+=t t . (7)dy y ⎰--22228;解dy y dy y ⎰⎰---=-2222224228⎰-=⋅=====44sin 2cos 2cos 22ππxdx x x y 令⎰-+=44)2cos 1(22ππdx x)2(2)2sin 21(224+=+=-πππy x . (8)⎰-121221dx x x ; 解⎰⎰⋅=====-=242sin 12122cos sin cos 1ππtdt tt dx x x tx 令 41)cot ()1sin 1(2422πππππ-=--=-=⎰t t dt t. (9)⎰-adx x a x 0222;解⎰⎰⋅⋅=====-=2022sin 0222cos cos sin πtdt a t a t a dx x a x t a x a令⎰=20242sin 4πtdt a 164sin 328)4cos 1(84204204204ππππa t a t a dt t a =-=-=⎰. (10)⎰+31221xx dx ;解⎰⎰⋅⋅=====+=3422tan 3122sec sec tan 11ππtdt t t x x dx tx 令 3322sin 1sin cos 3432-=-==⎰ππππt dt t t .(11)⎰--1145xxdx ; 解 61)315(81)5(81451313324511=--=-======-⎰⎰=--u u du u x xdx u x 令. (12) ⎰+411xdx;解 221112141=⋅+=====+⎰⎰=udu u x dx u x 令 )32ln 1(2)|1|ln (2)111(2121+=+-=+-⎰u u du u .(13)⎰--14311x dx ; 解du u u x dx u x )2(11110211143-⋅-======--⎰⎰=-令 2ln 21)|1|ln (2)111(221021-=-+=-+=⎰u u du u . (14)⎰-a x a xdx 20223; 解 ⎰⎰---=-a a x a d xa x a xdx 2022222022)3(31213 )13(32022-=--=a x a a .(15)dt te t ⎰-1022;解 2110102221021)2(222-----=-=--=⎰⎰e e t d e dt te t tt .(16)⎰+21ln 1e x x dx ; 解 )13(2ln 12ln ln 11ln 1222111-=+=+=+⎰⎰e e e x x d x x x dx . (17)⎰-++02222x x dx ;解 ⎰⎰--++=++022022)1(1122dx x x x dx 2)1arctan(1arctan )1arctan(02π=--=+=-x .(18)⎰-222cos cos ππxdx x ;解 32)sin 32(sin sin )sin 21(2cos cos 2322222=-=-=---⎰⎰ππππππx x x d x xdx x . (19)⎰--223cos cos ππdx x x ;解⎰⎰---=-222223cos 1cos cos cos ππππdx x x dx x x⎰⎰+-=-202sin cos )sin (cos ππxdx x dx x x34cos 32cos 3220230223=-=-ππx x . (20)⎰+π2cos 1dx x .解22cos 2sin 22cos 100=-==+⎰⎰πππx xdx dx x .2. 利用函数的奇偶性计算下列积分: (1)⎰-ππxdx x sin 4;解因为x 4sin x 在区间[-π, π]上是奇函数, 所以0sin 4=⎰-ππxdx x .(2)⎰-224cos 4ππθθd ;解⎰⎰⎰+==-202204224)22cos 1(8cos 42cos 4ππππθθθθθd x d d ⎰++=202)2cos 2cos 21(2πθd x x⎰++=20)4cos 212cos 223(2πθd x x 23)4sin 412sin 23(20πθπ=++=x x . (3)⎰--2121221)(arcsin dx x x ; 解⎰⎰-=--210222121221)(arcsin 21)(arcsin dx xx dx x x ⎰=2102)(arcsin )(arcsin 2x d x324)(arcsin 3232103π==x .(4)⎰-++55242312sin dx x x xx .解因为函数12sin 2423++x x x x 是奇函数, 所以012sin 552423=++⎰-dx x x x x . 3. 证明:⎰⎰-=aaadx x dx x 022)(2)(ϕϕ, 其中ϕ(u )为连续函数.证明因为被积函数ϕ(x 2)是x 的偶函数, 且积分区间[-a , a ]关于原点对称, 所以有⎰⎰-=aaadx x dx x 022)(2)(ϕϕ.4. 设f (x )在[-b , b ]上连续, 证明⎰⎰---=bbbbdx x f dx x f )()(. 证明令x =-t , 则dx =-dt , 当x =-b 时t =b , 当x =b 时t =-b , 于是 ⎰⎰⎰----=--=bb bbb bdt t f dt t f dx x f )()1)(()(,而 ⎰⎰---=-bbbb dx x f dt t f )()(,所以⎰⎰---=bbbbdx x f dx x f )()(.5. 设f (x )在[a , b ]上连续., 证明⎰⎰-+=babadx x b a f dx x f )()(. 证明令x =a +b -t , 则dx =d t , 当x =a 时t =b , 当x =b 时t =a , 于是 ⎰⎰⎰-+=--+=ba baa bdt t b a f dt t b a f dx x f )()1)(()(,而 ⎰⎰-+=-+baba dx xb a f dt t b a f )()(,所以⎰⎰-+=babadx x b a f dx x f )()(.6. 证明: ⎰⎰>+=+11122)0(11x x x x dx x dx . 证明令t x 1=, 则dt t dx 21-=, 当x =x 时xt 1=, 当x =1时t =1, 于是⎰⎰⎰+=-⋅+=+11121122211)1(1111x x x dt t dt t t x dx ,而 ⎰⎰+=+x xdx x dt t 1121121111, 所以 ⎰⎰+=+1112211x x x dx x dx .7. 证明:⎰⎰-=-110)1()1(dx x x dx x x m n nm.证明令1-x =t , 则 ⎰⎰⎰⎰-=-=--=-11110)1()1()1()1(dx x x dt t t dt t t dx x x m n nm nm nm,即⎰⎰-=-11)1()1(dx x x dx x x m n nm.8. 证明: ⎰⎰=ππ020sin 2sin xdx xdx n n .证明 ⎰⎰⎰+=ππππ22sin sin sin xdx xdx xdx n nn,而 ⎰⎰⎰⎰==--=====-=202022sin sin ))((sin sin πππππππxdx tdt dt t xdx nn n tx n 令,所以⎰⎰=ππ20sin 2sin xdx xdx n n.9. 设f (x )是以l 为周期的连续函数, 证明⎰+1)(a adx x f 的值与a无关.证明已知f (x +l )=f (x ).⎰⎰⎰⎰++++=la ll aa a dx x f dx x f dx x f dx x f )()()()(001⎰⎰⎰-+=+ala lldx x f dx x f dx x f 00)()()(,而 ⎰⎰⎰⎰=+=+=====+=+aaalt x la l dx x f dx l x f dt l t f dx x f 0)()()()(令, 所以 ⎰⎰=+la adx x f dx x f 01)()(.因此⎰+1)(a adx x f 的值与a 无关.10. 若f (t )是连续函数且为奇函数, 证明⎰xdt t f 0)(是偶函数;若f (t )是连续函数且为偶函数, 证明⎰xdt t f 0)(是奇函数.证明设⎰=xdt t f x F 0)()(.若f (t )是连续函数且为奇函数, 则f (-t )=-f (t ), 从而 ⎰⎰--======--=-xut xdu u f dt t f x F 0)1)(()()(令)()()(0x F dx x f dx u f xx ===⎰⎰,即⎰=xdt t f x F 0)()(是偶函数.若f (t )是连续函数且为偶函数, 则f (-t )=f (t ), 从而 ⎰⎰--======--=-xut xdu u f dt t f x F 0)1)(()()(令)()()(0x F dx x f dx u f xx -=-=-=⎰⎰,即⎰=xdt t f x F 0)()(是奇函数.11. 计算下列定积分: (1)⎰-10dx xe x ;解11011101121--------=--=+-=-=⎰⎰⎰e ee dx e xexde dx xe x xx xx.(2)⎰e xdx x 1ln ;解 ⎰⎰⎰⋅-==e e e edx xx x x xdx xdx x 0212121121ln 21ln 21ln)1(4141212122+=-=e x e e.(3)⎰ωπω20sin tdt t (ω为常数);解⎰⎰-=ωπωπωωω2020cos 1sin t td tdt t⎰+-=ωπωπωωωω2020cos 1cos 1tdt t t220222sin 12ωπωωωπωπ-=+-=t . (4)⎰342sin ππdx xx ; 解 ⎰⎰⎰+-=-=34334342cot cot cot sin ππππππππxdx x x x xd dx x x 34sin ln 4313ππππx ++⋅-=23ln 21)9341(+-=π. (5)⎰41ln dx x x;解 ⎰⎰⎰⋅-==4141414112ln 2ln 2ln dx x x x x x xd dx xx)12ln 2(442ln 8122ln 84141-=-=-=⎰x dx x.(6)⎰1arctan xdx x ;解 ⎰⎰=1021arctan 21arctan xdx xdx x x d xx x x ⎰+⋅-=10221021121arctan 21 10102)arctan (218)111(218x x x d x --=+--=⎰ππ214)41(218-=--=πππ. (7)⎰202cos πxdx e x ;解⎰⎰⎰-==202202202202sin 2sin sin cos ππππxdx e x e x d e xdx e x xxx⎰⎰-+=+=202202202cos 4cos 2cos 2πππππxdx e x e e x d e e x x x ⎰-+=202cos 42ππxdx e e x所以)2(51cos 202-=⎰ππe xdx e x.(8)⎰212log xdx x ;解 ⎰⎰=2122212log 21log xdx xdx x⎰⋅-=21221222ln 121log 21dx x x x x2ln 432212ln 212212-=⋅-=x . (9)⎰π2)sin (dx x x ;解 ⎰⎰⎰-=-=ππππ20302022sin 4161)2cos 1(21)sin (x d x x dx x x dx x x ⎰⎰-=⋅+-=πππππππ03000332cos 41622sin 412sin 416x xd x xd x x x 462sin 81462cos 412cos 416303003ππππππππ-=+-=+-=⎰x xdx x x . (10)⎰edx x 1)sin(ln ;解法一 ⎰⎰⋅======1ln 1sin )sin(ln dt e t dx x t tx e令.因为⎰⎰⎰-==⋅11011cos sin sin sin tdt e t e tde dt e t t ttt⎰⎰--⋅=-⋅=1101sin cos 1sin cos 1sin tdt e t e e tde e t tt⎰-+⋅-⋅=10sin 11cos 1sin tdt e e e t ,所以)11cos 1sin (21sin 1+⋅-⋅=⎰e e tdt e t . 因此 )11cos 1sin (21)sin(ln 1+⋅-⋅=⎰e e dx x e.解法二⎰⎰⋅⋅-⋅=eeedx xx x x x dx x 1111)cos(ln )sin(ln )sin(ln ⎰-⋅=edx x e 1)cos(ln 1sin⎰⋅⋅-⋅-⋅=eedx xx x x x e 111)sin(ln )cos(ln 1sin⎰-+⋅-⋅=edx x e e 0)sin(ln 11cos 1sin ,故)11cos 1sin (21)sin(ln 1+⋅-⋅=⎰e e dx x e. (11)dx x ee⎰1|ln |;解dx x dx x dx x eee e⎰⎰⎰+-=1111ln ln |ln |⎰⎰-++-=eeeedx dx x x x x 111111ln ln )11(2)1()11(1e e e e e -=---++-=.(12)⎰-1022)1(dx x m(m 为自然数);解⎰⎰+======-201sin 122cos )1(πtdt dx x m t x m令. 根据递推公式⎰⎰--=20220cos 1cos ππxdx n n xdx n n,⎪⎩⎪⎨⎧⋅⋅⋅⋅⋅--⋅--⋅+⋅⋅⋅⋅⋅⋅--⋅--⋅+=-⎰. ,325476 34121, ,2214365 34121)1(1022为偶数为奇数m m m m m m m m m m m m m m dx x m π(13)⎰=πsin xdx x J m m (m 为自然数).解因为⎰⎰---=====-=0)1)((sin )(sin πππππdt t t xdx x m tx m令⎰⎰-=πππ0sin sin tdt t tdt m m,所以 ⎰⎰==πππ0sin 2sin xdx xdx x J m mm⎰⎰=⋅=2020sin sin 22ππππxdx xdx m m (用第8题结果).根据递推公式⎰⎰--=20220sin 1sin ππxdx n n xdx n n ,⎪⎩⎪⎨⎧⋅⋅⋅⋅⋅--⋅--⋅-⋅⋅⋅⋅⋅⋅--⋅--⋅-=. ,325476 45231, ,2214365 452312为奇数为偶数m m m m m m m m m m m m m m J m ππ习题5-41. 判别下列各反常积分的收敛性, 如果收敛, 计算反常积分的值: (1)⎰∞+14x dx ; 解因为3131)31(lim 3131314=+-=-=-+∞→∞+-∞+⎰x x x dx x , 所以反常积分⎰∞+14x dx 收敛, 且3114=⎰∞+x dx . (2)⎰∞+1xdx;解因为+∞=-==+∞→∞+∞+⎰22lim 211x x x dx x , 所以反常积分⎰∞+1x dx发散.(3)dx e ax ⎰∞+-0(a >0);解因为aa e a e a dx e ax x ax ax11)1(lim 100=+-=-=-+∞→∞+-∞+-⎰, 所以反常积分dx e ax⎰∞+-0收敛, 且adx e ax 10=⎰∞+-.(4)⎰∞+-0ch tdt e pt (p >1);解因为⎰⎰∞++--∞+-+=0)1()1(0][21ch dt e e tdt e t p t p pt1]1111[2120)1()1(-=+--=∞++--p pe p e p t p t p ,所以反常积分⎰∞+-0ch tdt e pt收敛, 且1ch 20-=⎰∞+-p ptdt e pt .(5)⎰∞+-0sin tdt e pt ω(p >0, ω>0);解 ⎰⎰∞+-∞+--=0cos 1sin t d e tdt ept ptωωω⎰∞+-∞+--⋅+-=00)(cos 1cos 1dt pe t t e pt ptωωωω ⎰∞+--=02sin 1t d e p pt ωωω ⎰∞+-∞+--⋅+-=0202)(sin sin 1dt pe t p t e p pt pt ωωωωω⎰∞+--=022sin 1tdt e p pt ωωω, 所以220sin w p tdt e pt +=⎰∞+-ωω. (6)⎰+∞∞-++222x x dx ;解 πππ=--=+=++=++⎰⎰∞+∞-∞+∞-+∞∞-)2(2)1arctan()1(12222x x dx x x dx . (7)dx x x ⎰-121; 解这是无界函数的反常积分, x =1是被积函数的瑕点.11)1(lim 1121102102=+--=--=--→⎰x x dx x x x . (8)⎰-202)1(x dx;解这是无界函数的反常积分, x =1是被积函数的瑕点. 因为⎰⎰⎰-+-=-212102202)1()1()1(x dx x dx x dx , 而+∞=--=-=--→⎰111lim 11)1(110102xx x dx x ,所以反常积分⎰-202)1(x dx 发散. (9)⎰-211x xdx ; 解这是无界函数的反常积分, x =1是被积函数的瑕点.21232121]12)1(32[)111(1-+-=-+-=-⎰⎰x x dx x x x xdx 322]12)1(32[lim 38231=-+--=+→x x x . (10)⎰-e x x dx 12)(ln 1. 解这是无界函数的反常积分, x =e 是被积函数的瑕点.x d x x x dx e eln )(ln 11)(ln 11212⎰⎰-=- 2)arcsin(ln lim )arcsin(ln 1π===-→x x e x e. 2. 当k 为何值时, 反常积分⎰∞+0)(ln kx x dx 收敛? 当k 为何值时, 这反常积分发散? 又当k 为何值时, 这反常积分取得最小值?解当k <1时,+∞=-==∞++-∞+∞+⎰⎰2122)(ln 11ln )(ln 1)(ln k k k x kx d x x x dx ;当k =1时,+∞===∞+∞+∞+⎰⎰222)ln(ln ln ln 1)(ln x x d xx x dx k ;当k >1时,k k k k k x k x d x x x dx -∞++-∞+∞+-=-==⎰⎰12122)2(ln 11)(ln 11ln )(ln 1)(ln .因此当k >1时, 反常积分⎰∞+0)(ln kx x dx 收敛; 当k ≤1时, 反常积分⎰∞+0)(ln kx x dx 发散.当k >1时, 令k k k x x dx k f -∞+-==⎰10)2(ln 11)(ln )(, 则 2ln ln )2(ln 11)2(ln )1(1)(112k k k k k f ------=' )2ln ln 11()1(2ln ln )2(ln 21+---=-k k k . 令f '(k )=0得唯一驻点2ln ln 11-=k .因为当2ln ln 111-<<k 时f '(k )<0, 当2ln ln 11->k 时f '(k )>0, 所以2ln ln 11-=k 为极小值点, 同时也是最小值点, 即当2ln ln 11-=k 时, 这反常积分取得最小值3. 利用递推公式计算反常积分⎰∞+-=0dx e x I x n n .解因为⎰⎰∞+-∞+--==0x n x n n de x dx e x I1010-∞+--∞+-=+-=⎰n x n x n nI dx e x n ex ,所以 I n = n ⋅(n -1)⋅(n -2)⋅⋅⋅2⋅I 1. 又因为 1001=-=+-=-==∞+-∞+-∞+-∞+-∞+-⎰⎰⎰x x x xxe dx e xexde dx xe I ,所以 I n = n ⋅(n -1)⋅(n -2)⋅ ⋅ ⋅2⋅I 1=n !.总习题五1. 填空:(1)函数f (x )在[a , b ]上(常义)有界是f (x )在[a , b ]上可积的______条件, 而f (x )在[a , b ]上连续是f (x )在[a , b ]上可积______的条件;解函数f (x )在[a , b ]上(常义)有界是f (x )在[a , b ]上可积的___必要___条件, 而f (x )在[a , b ]上连续是f (x )在[a , b ]上可积___充分___的条件;(2)对[a , +∞)上非负、连续的函数f (x ), 它的变上限积分⎰xa dx x f )(在[a , +∞)上有界是反常积分⎰+∞a dx x f )(收敛的______条件;解对[a , +∞)上非负、连续的函数f (x ), 它的变上限积分⎰xa dx x f )(在[a , +∞)上有界是反常积分⎰+∞a dx x f )(收敛的___充分___条件;(3)绝对收敛的反常积分⎰+∞a dx x f )(一定______; 解绝对收敛的反常积分⎰+∞a dx x f )(一定___收敛___;(4)函数f (x )在[a , b ]上有定义且|f (x )|在[a , b ]上可积, 此时积分⎰ba dx x f )(______存在. 解函数f (x )在[a ,b ]上有定义且|f (x )|在[a , b ]上可积, 此时积分⎰ba dx x f )(___不一定___存在.2. 计算下列极限: (1)∑=∞→+n i n nin 111lim ;解 )122(32)1(32111lim 103101-=+=+=+⎰∑=∞→x dx x n i n n i n . (2)121lim+∞→+⋅⋅⋅++p pp p n nn (p >0);解 11111])( )2()1[(lim 21lim101101+=+==⋅⋅⋅⋅++=+⋅⋅⋅+++∞→+∞→⎰p x p dx x n n n n n nn p p p p p n p p p p n . (3)nn nn !lnlim ∞→; 解 ]ln 1)ln 2ln 1(ln 1[lim !lnlim n n nn n n n n nn ⋅-+⋅⋅⋅++=∞→∞→nn n n n n 1)]ln (ln )ln 2(ln )ln 1[(ln lim ⋅-+⋅⋅⋅+-+-=∞→⎰=⋅+⋅⋅⋅++=∞→10ln 1)ln 2ln 1(ln lim xdx n n n n n n1)ln ()ln (10101010-=-=-=⎰xx x dx x x .(4)⎰-→xaa x dt t f a x x )(lim, 其中f (x )连续; 解法一 )()(lim )(lima af xf dt t f a x x a xa a x ==-→→⎰ξξ (用的是积分中值定理). 解法二 )(1)()(lim)(lim )(lim a af x xf dt t f ax dt t f x dt t f a x x xa ax xa ax xa a x =+=-=-⎰⎰⎰→→→ (用的是洛必达法则). (5)1)(arctan lim202+⎰+∞→x dtt xx .解 4)(arctan 1lim 1)(arctan lim 1)(arctan lim22222202π=+=+=+∞→+∞→+∞→⎰x x x x x x x dtt x x xx . 3. 下列计算是否正确, 试说明理由: (1)⎰⎰----=-=+-=+111111222)1arctan ()1(1)1(1πx xx d x dx ;解计算不正确, 因为x 1在[-1, 1]上不连续. (2)因为⎰⎰--++-=++111122111t t dt tx x x dx , 所以⎰-=++11201x x dx .解计算不正确, 因为t1在[-1, 1]上不连续.(3)01lim 122=+=+⎰⎰-∞→+∞∞-A A A dx x xdx x x . 解不正确, 因为⎰⎰⎰⎰-+∞→+∞→+∞∞--∞→+≠+++=+A A A b b a a dx x xdx x x dx x x dx x x 2020221lim 1lim 1lim 1. 4. 设p >0, 证明⎰<+<+10111px dx p p. 证明 p pp pp p px xx x x x x->+-=+-+=+>11111111. 因为⎰⎰⎰<+<-1010101)1(dx x dxdx x pp,而 110=⎰dx , pp p x x dx x p p+=+-=-+⎰1)1()1(10110, 所以⎰<+<+10111p xdx p p. 5. 设f (x )、g (x )在区间[a , b ]上均连续, 证明: (1)⎰⎰⎰⋅≤ba ba ba dx x g dx x f dx x g x f )()(])()([222;证明因为[f (x )-λg (x )]2≥0, 所以λ2g 2(x )-2λ f (x )g (x )+f 2(x )≥0, 从而 0)()()(2)(222≥+-⎰⎰⎰ba ba ba dx x f dx x g x f dx x g λλ.上式的左端可视为关于λ的二次三项式, 因为此二次三项式大于等于0, 所以其判别式小于等于0, 即0)()(4])()([4222≤⋅-⎰⎰⎰ba b a b a dx x g dx x f dx x g x f ,亦即 ⎰⎰⎰⋅≤ba ba ba dx x g dx x f dx x g x f )()(])()([222. (2)()()()212212212)()()]()([⎰⎰⎰+≤+b ab a b a dx x g dx x f dx x g x f , 证明⎰⎰⎰⎰++=+ba b a b a b a dx x g x f dx x g dx x f dx x g x f )()(2)()()]()([222 212222])()([2)()(⎰⎰⎰⎰⋅++≤ba ba ba ba dx x g dx x f dx x g dx x f ,又()2212212212222])([])([])()([2)()(⎰⎰⎰⎰⎰⎰+=⋅++ba ba ba ba ba ba dx x gdx x fdx x gdx x f dx x g dx x f,所以()()()212212212)()()]()([⎰⎰⎰+≤+b ab a b a dx x g dx x f dx x g x f . 6. 设f (x )在区间[a , b ]上连续, 且f (x )>0. 证明⎰⎰-≥⋅b a baa b x f dxdx x f 2)()()(. 证明已知有不等式⎰⎰⎰⋅≤ba ba ba dx x g dx x f dx x g x f )()(])()([222, 在此不等式中, 取)(1)(x f x f =, )()(x f x g =, 则有⎰⎰⎰⋅≥⋅⋅ba ba ba dx x f x f dx x f dx x f 222])(1)([])(1[])([,即⎰⎰-≥⋅ba baa b x f dxdx x f 2)()()(. 7. 计算下列积分: (1)⎰++2cos 1sin πdx xxx ;解 20202020220)cos 1ln()2(tan cos 1)cos 1(2cos2cos 1sin πππππ⎰⎰⎰⎰+-=++-=++x xxd x x d dx x x dx xxx⎰=++=+-=2022022ln 2cos ln 222ln 2tan )2tan (πππππx dx x xx .(2)⎰+40)tan 1ln(πdx x ;解 ⎰⎰+=+4040cos )4sin(2ln)tan 1ln(πππdx xx dx x ⎰⎰⎰-++=404040cos ln )4sin(ln 2ln ππππxdx dx x dx .令,4u x =-π则。

Introduction to Management Science 5th Edition, 课后习题答案 Chapter 6

CHAPTER 6NETWORK OPTIMIZATION PROBLEMSSOLUTION TO SOLVED PROBLEMS6.S1Distribution at Heart BeatsHeart B eats i s a m anufacturer o f m edical e quipment. T he c ompany’s p rimary p roduct i s a device u sed t o m onitor t he h eart d uring m edical p rocedures. T his d evice i s p roduced i n t wo factories a nd s hipped t o t wo w arehouses. T he p roduct i s t hen s hipped o n d emand t o f ourthird-‐party w holesalers. A ll s hipping i s d one b y t ruck. T he p roduct d istribution n etwork i s shown b elow. T he a nnual p roduction c apacity a t F actories 1 a nd 2 i s 400 a nd 250, respectively. T he a nnual d emand a t W holesalers 1, 2, 3, a nd 4 i s 200, 100, 150, a nd 200, respectively. T he c ost o f s hipping o ne u nit i n e ach s hipping l ane i s s hown o n t he a rcs. D ue t o limited t ruck c apacity, a t m ost 250 u nits c an b e s hipped f rom F actory 1 t o W arehouse 1 e ach year. F ormulate a nd s olve a n etwork o ptimization m odel i n a s preadsheet t o d etermine h ow to d istribute t he p roduct a t t he l owest p ossible a nnual c ost.This i s a m inimum-‐cost f low p roblem. T o s et u p a s preadsheet m odel, f irst l ist a ll o f t he a rcs as s hown i n B4:C11, a long w ith t heir c apacity (F4) a nd u nit c ost (G4:G11). O nly t he a rc from F1 t o W H1 i s c apacitated. T hen l ist a ll o f t he n odes a s s hown i n I4:I11 a long w ith e ach node’s s upply o r d emand (L4:L11).The c hanging c ells a re t he a mount o f f low t o s end t hrough e ach a rc. T hese a re s hown i nFlow (D4:D11) b elow, w ith a n a rbitrary v alue o f 10 e ntered f or e ach. T he f low t hrough t he arc f rom F1 t o W H1 m ust b e l ess t han t he c apacity o f 250, a s i ndicated b y t he c onstraint D4<= F4.For e ach n ode, c alculate t he n et f low a s a f unction o f t he c hanging c ells. T his c an b e d one using t he S UMIF f unction. I n e ach c ase, t he f irst S UMIF f unction c alculates t he f low l eaving the n ode a nd t he s econd o ne c alculates t he f low e ntering t he n ode. F or e xample, c onsider the F 1 n ode (I4). S UMIF(From, N odes, F low) s ums e ach i ndividual e ntry i n F low (thechanging c ells i n D 4:D11) i f t hat e ntry i s i n a r ow w here t he e ntry i n F rom (B4:B11) i s t he same a s i n t hat r ow o f N odes (i.e., F 1). S ince I 4 = F 1 a nd t he o nly r ows t hat h ave F 1 i n F rom (B4:B11) a re r ows 4 a nd 5, t he s um i n t he s hip c olumn i s o nly o ver t hese s ame r ows, s o t his sum i s D 4+D5.The g oal i s t o m inimize t he t otal c ost o f s hipping t he p roduct f rom t he f actories t o t he wholesalers. T he c ost i s t he S UMPRODUCT o f t he U nit C osts w ith t he F low, o r T otal C ost = SUMPRODUCT(UnitCost, F low). T his f ormula i s e ntered i nto T otalCost (D13).34567891011JNet Flow=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)The S olver i nformation a nd s olved s preadsheet a re s hown b elow.Thus, F low (D4:D11) i ndicates h ow t o d istribute t he p roduct s o a s t o a chieve t he m inimum Total C ost (D13) o f $58,500.Solver ParametersSet Objective Cell: TotalCost To: MinBy Changing Variable Cells: FlowSubject to the Constraints: D4 <= CapacityNetFlow = SupplyDemand Solver Options:Make Variables Nonnegative Solving Method: Simplex LP34567891011JNet Flow=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)6.S 2 Assessing the Capacity of a Pipeline NetworkExxo 76 i s a n o il c ompany t hat o perates t he p ipeline n etwork s hown b elow, w here e achpipeline i s l abeled w ith i ts m aximum f low r ate i n m illion c ubic f eet (MMcf) p er d ay. A n ew o il well h as b een c onstructed n ear A . T hey w ould l ike t o t ransport o il f rom t he w ell n ear A t o their r efinery a t G . F ormulate a nd s olve a n etwork o ptimization m odel t o d etermine t he maximum f low r ate f rom A t o G .This i s a m inimum-‐cost f low p roblem. A ssociated w ith e ach p ipe i n t he n etwork w ill b e a n arc (or, f or p ipes w hich m ight f low i n e ither d irection, t wo a rcs, o ne i n e ach d irection). T o set u p a s preadsheet m odel, f irst l ist a ll o f t he a rcs a s s hown i n B 5:C19, a long w ith t heir capacity (F5:F19). T hen l ist a ll o f t he n odes a s s hown i n H 5:H11. A ll t he t ransshipment nodes (every n ode e xcept t he s tart n ode A a nd t he e nd n ode G ) w ill b e c onstrained t o h ave net f low = 0 (Supply/Demand = 0). T he start n ode (A) a nd e nd n ode (G) a re l eft unconstrained. W e w ant t o m aximize t he n et f low o ut o f n ode A .The c hanging c ells a re t he a mount o f f low t o s end t hrough e ach p ipe (arc). T hese a re s hown in F low (D5:D19) b elow, w ith a n a rbitrary v alue o f 0 e ntered f or e ach. T he f low t hrough each a rc i s c apacitated a s i ndicated b y t he <= i n E5:E19.For e ach n ode, c alculate t he n et f low a s a f unction o f t he c hanging c ells. T his c an b e d one using t he S UMIF f unction. I n e ach c ase, t he f irst S UMIF f unction c alculates t he f low l eaving the n ode a nd t he s econd o ne c alculates t he f low e ntering t he n ode. F or e xample, c onsider the A n ode (H5). S UMIF(From, N odes, F low) i n I 5 s ums e ach i ndividual e ntry i n F low (the changing c ells i n D 5:D19) i f t hat e ntry i s i n a r ow w here t he e ntry i n F rom (B5:B19) i s t he same a s i n t he e ntry i n t hat r ow o f N odes (i.e., A ). S ince t he o nly r ows t hat h ave A i n F rom (B5:B19) a re r ows 5 a nd 6, t he s um i n t he s hip c olumn i s o nly o ver t hese s ame r ows, s o t his sum i s D 5+D6.4567891011INet Flow=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)The g oal i s t o m aximize t he a mount s hipped f rom A t o G. S ince n odes B t hrough F a retransshipment n odes (net f low = 0), a ny a mount t hat l eaves A m ust e nter G. T hus, maximizing t he f low o ut o f A w ill a chieve o ur g oal. T hus, t he f ormula e ntered i nto t heobjective c ell M aximumFlow (D21) i s =I5.The S olver i nformation a nd s olved s preadsheet a re s hown b elow.Thus, F low (D5:D19) i ndicates h ow t o s end o il t hrough t he n etwork s o a s t o a chieve t he Maximum F low (D21) o f 34 t housand g allons/hour.Solver ParametersSet Objective Cell: MaximumFlow To: MaxBy Changing Variable Cells: FlowSubject to the Constraints: Flow <= CapacityNetFlow = SupplyDemand Solver Options:Make Variables Nonnegative Solving Method: Simplex LP4567891011INet Flow=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)6.S 3 Driving to the Mile -High CitySarah a nd J ennifer h ave j ust g raduated f rom c ollege a t t he U niversity o f W ashington i n Seattle a nd w ant t o g o o n a r oad t rip. T hey h ave a lways w anted t o s ee t he m ile-‐high c ity o f Denver. T heir r oad a tlas s hows t he d riving t ime (in h ours) b etween v arious c ity p airs, a s shown b elow. F ormulate a nd s olve a n etwork o ptimization m odel t o f ind t he q uickest r oute from S eattle t o D enver?This i s a s hortest p ath p roblem. T o s et u p a s preadsheet m odel, f irst l ist a ll o f t he a rcs a s shown i n B 4:C11, a long w ith t heir c apacity (F4). O nly t he a rc f rom F 1 t o W H1 i scapacitated. T hen l ist a ll o f t he n odes a s s hown i n I 4:I11 a long w ith e ach n ode’s s upply o r demand (L4:L11).The c hanging c ells a re t he a mount o f f low t o s end t hrough e ach a rc. T hese a re s hown i nFlow (D4:D11) b elow, w ith a n a rbitrary v alue o f 10 e ntered f or e ach. T he f low t hrough t he arc f rom F 1 t o W H1 m ust b e l ess t han t he c apacity o f 250, a s i ndicated b y t he c onstraint D 4 <= F 4.SeattleGrand JunctionDenverFor e ach n ode, c alculate t he n et f low a s a f unction o f t he c hanging c ells. T his c an b e d one using t he S UMIF f unction. I n e ach c ase, t he f irst S UMIF f unction c alculates t he f low l eaving the n ode a nd t he s econd o ne c alculates t he f low e ntering t he n ode. F or e xample, c onsider the F 1 n ode (I4). S UMIF(From, I 4, F low) s ums e ach i ndividual e ntry i n F low (the c hanging cells i n D 4:D11) i f t hat e ntry i s i n a r ow w here t he e ntry i n F rom (B4:B11) i s t he s ame a s i n I4 (i.e., F1). S ince I 4 = F 1 a nd t he o nly r ows t hat h ave F 1 i n F rom (B4:B11) a re r ows 4 a nd 5, t he s um i n t he s hip c olumn i s o nly o ver t hese s ame r ows, s o t his s um i s D 4+D5.34567891011JNet Flow=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)The g oal i s t o m inimize t he t otal c ost o f s hipping t he p roduct f rom t he f actories t o t he wholesalers. T he c ost i s t he S UMPRODUCT o f t he U nit C osts w ith t he F low, o r T otal C ost = SUMPRODUCT(UnitCost, F low). T his f ormula i s e ntered i nto T otalCost (D13).The S olver i nformation a nd s olved s preadsheet a re s hown b elow.Thus, F low (D4:D11) i ndicates h ow t o d istribute t he p roduct s o a s t o a chieve t he m inimum Total C ost (D13) o f $58,500.Solver ParametersSet Objective Cell: Total Cost To: MinBy Changing Variable Cells: FlowSubject to the Constraints: D4 <= CapacityNetFlow = SupplyDemand Solver Options:Make Variables Nonnegative Solving Method: Simplex LP34567891011JNet Flow=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)=SUMIF(From,Nodes,Flow)-SUMIF(To,Nodes,Flow)。

chapter-6-non-verbal-communication-非言语

Body Behavior
General Appearance and Dress Body Movements: Kinesics Posture Gestures Facial expressions Eye Contact and Gaze Touch Smell paralanguage
What does this gesture mean? French gesture for 'I don't believe you'
Guess what this Iranian gesture means?
No. 1 for me/ Good luck / screw you(滚蛋)
Much of nonverbal communication is universal.
Our emotions and attitudes are reflected in our stance, dress, and form of eye contact.
A handshake
--- welcome / goodbye
Categorization by毕继万（1995）
体态语（body language，包括各种表情、动作、姿态）副语言（paralanguage，包括沉默与非语义声音）客体语（object language，包括皮肤颜色、气味、衣着化妆、家具等）环境语言（environmental language，包括时间，空间、颜色、城市规划以及人对自然的影响等，即影响生理与心理的环境因素）。
1. Nonverbal Communication
Nonverbal codes refer to communicative messages which are not in word form. More than 55% messages are communicated nonverbally!

原版的操作系统_精髓与设计原理_第5版Chapter06

Concurrency: Deadlock and Concurrency:Deadlock andStarvationChapter6Chapter 61Deadlock •Permanent blocking of a set of processes P bl ki f fthat either compete for system resources or communicate with each other•No efficient solutionNo efficient solution•Involve conflicting needs for resources by two or more processes2345Reusable ResourcesReusable Resources •Used by only one process at a time and notU d b l t ti d t depleted by that use•Processes obtain resources that they later release for reuse by other processes •Processors, I/O channels, main and secondary memory, devices, and data structures such as yfiles, databases, and semaphoresDeadlock occurs if each process holds one •Deadlock occurs if each process holds one resource and requests the other6Example of Deadlock Example of Deadlock7Another Example of Deadlock Another Example of Deadlock S i il bl f ll i f •Space is available for allocation of 200Kbytes, and the following sequence of events occurP1P2. . .. . .Request 80 Kbytes;R 60Kb . . .. . .Request 70 Kbytes;R 80Kb Deadlock occurs if both processes Request 60 Kbytes;Request 80 Kbytes;•Deadlock occurs if both processes progress to their second request8Consumable ResourcesConsumable Resources •Created (produced) and destroyedC d(d d)d d d (consumed)•Interrupts, signals, messages, and information in I/O buffersinformation in I/O buffers•Deadlock may occur if a Receive message is blockingMay take a rare combination of events to •May take a rare combination of events to cause deadlock9Example of DeadlockExample of DeadlockD dl k if i i bl ki •Deadlock occurs if receive is blockingP1 . . .P2 . . .. . .Receive(P2); Send(P2, M1);. . .Receive(P1);Send(P1, M2);10Resource Allocation Graphs Resource Allocation Graphs •Directed graph that depicts a state of theDi t d h th t d i t t t f th system of resources and processes11Resource Allocation Graphs Resource Allocation Graphs12Conditions for DeadlockConditions for Deadlock •Mutual exclusionM l l i–Only one process may use a resource at atimeHold and wait•Hold-and-wait–A process may hold allocated resourceswhile awaiting assignment of otherswhile awaiting assignment of others•No preemption–No resource can be forcibly removed form ap gprocess holding it13Conditions for DeadlockConditions for Deadlock •Circular waitCi l it–A closed chain of processes exists, such that eachprocess holds at least one resource needed by theh ld t l t d d b thnext process in the chain1415Possibility of DeadlockPossibility of Deadlock •Mutual ExclusionM l E l ip p•No preemption•Hold and wait16Existence of DeadlockExistence of Deadlock •Mutual ExclusionM l E l ip p•No preemption•Hold and wait•Circular wait17Deadlock PreventionDeadlock Prevention •Mutual ExclusionM l E l i–Must be supported by the operating system •Hold and Wait–Require a process request all of its requiredRequire a process request all of its requiredresources at one time18Deadlock PreventionDeadlock Prevention•No PreemptionN P i–Process must release resource and requestagainOpe g sys e y p ee p p ocess o –Operating system may preempt a process torequire it releases its resources •Circular WaitCircular Wait–Define a linear ordering of resource types19Deadlock AvoidanceDeadlock Avoidance•A decision is made dynamically whether A d i i i d d i ll h h the current resource allocation request will, if granted, potentially lead to ade d ocdeadlock•Requires knowledge of future processtrequest20Two Approaches toDeadlock Avoidance •Do not start a process if its demandsD if i d dmight lead to deadlock•Do not grant an incremental resourcerequest to a process if this allocation request to a process if this allocationmight lead to deadlock21Resource Allocation Denial Resource Allocation Denial •Referred to as the banker’s algorithmR f d h b k’l i hy•State of the system is the current allocation of resources to process •Safe state is where there is at least one S f t t i h th i t l t sequence that does not result in deadlock •Unsafe state is a state that is not safe22Initial State23P2 Runs to Completion24P1 Runs to Completion25P3 Runs to Completion26Unsafe State27Unsafe State28Deadlock Avoidance Logic Deadlock Avoidance Logic29Deadlock Avoidance Logic Deadlock Avoidance Logic30Deadlock AvoidanceDeadlock Avoidance •Maximum resource requirement must be M i i t t b stated in advance•Processes under consideration must be independent; no synchronization requirements•There must be a fixed number ofThere must be a fixed number of resources to allocate•No process may exit while holdingNo process may exit while holding resources31Deadlock Detection Deadlock Detection32Strategies once DeadlockDetected•Abort all deadlocked processesAb t ll d dl k d•Back up each deadlocked process to some previously defined checkpoint, and restart all process–Original deadlock may occur •Successively abort deadlocked processes Successively abort deadlocked processes until deadlock no longer exists •Successively preempt resources until Successively preempt resources until deadlock no longer exists33Selection Criteria DeadlockedProcesses•Least amount of processor timeL f i consumed so far•Least number of lines of output produced so farproduced so far•Most estimated time remaining •Least total resources allocated so far •Lowest priorityLowest priority34Strengths and Weaknesses of theStrategies35Dining Philosophers Problem Dining Philosophers Problem36Dining Philosophers Problem Dining Philosophers Problem37Dining Philosophers Problem Dining Philosophers Problem38Dining Philosophers Problem Dining Philosophers Problem39Dining Philosophers Problem Dining Philosophers Problem40UNIX ConcurrencyMechanisms•PipesPig•Messages•Shared memory•Semaphores•Signals4142Linux Kernel ConcurrencyMechanisms •Includes all the mechanisms found in I l d ll h h i f d i UNIX•Atomic operations execute without interruption and without interference interruption and without interference43Linux Atomic Operations Linux Atomic Operations44Linux Atomic Operations Linux Atomic Operations45Linux Kernel ConcurrencyMechanisms •SpinlocksS i l k–Used for protecting a critical section4647Linux Kernel ConcurrencyMechanisms48Solaris Thread Synchronization Primitives •Mutual exclusion (mutex) locksM l l i()l kp•Semaphores•Multiple readers, single writer (readers/writer) locks(d/it)l k•Condition variables4950。

清华第五版数值分析第6章课件

即：
x x
( ( ( ( ( x1k 1) 1 ( a12 x2k ) a13 x3k ) a14 x4k ) a1n xnk ) b1 ) a11 ( ( ( ( ( x2k 1) 1 (a21 x1k 1) a23 x3k ) a24 x4k ) a2 n xnk ) b2 ) a22
( i 1, 2, , n)
或
i 1 n
x
( k 1) i

bi aij x
j 1
(k ) j

j i 1
a
ij
x
(k ) j
a ii
; i 1, 2, , n
雅可比迭代法的矩阵表示
将系数矩阵分裂为：A
D LU
其中 D diag(a11 , a22 , , ann )
G-S迭代算法描叙
1 输入 A, then, , M . 2 if , b, x0 1 x ( a x a 2.1k M for i 1, 2, 2.1.1s=0， , n 2.1.2 for t ( x0 )i
( k 1) i i 1 ii j 1 ij ( k 1) j
Jacobi迭代算法描述
1 输入 A, b, x0 , , M . n 2 if k M , then 1 xi( k 1) (bi aij x (jk ) ) xi( k ) . 2.1 for i 1, 2, , n aii j 1 2.1.1 s=0, 2.1.2 forj 1, 2, , n
过程建立Jacobi迭代公式，即
a
i 1
n
ij
x j bi , aii 0

语言学教程Chapter6

The second wife will claim the inheritance
belongs to her.
Garden path sentence Sentences that are initially interpreted with a different structure than they actually have.

proposition
L comprehension: discourse interpretation

Schemata and drawing inferences Schema: a pre-existing knowledge structure in
the world

Structural factors: identify constituents of a
sentence and the ways in which they relate to one another

Minimal attachment: structural simplicity guides all initial analyses in sentence comprehension.
Problems: Syntactic ambiguity
a.
Different possible ways that words can be fit into phrases. Ambiguous category of some of the words in the sentence.
b.

The formal approach: structural patterns, including the study of morphological, syntactic, and lexical structure. The psychological approach: language from the view of general systems ranging from perception, memory, attention, and reasoning. The conceptual approach: how language structures (processes & patterns) conceptual content.

高等数学(同济第五版)课后答案第六章

=(3p 2
y−
1 2
y2
−
1 6p
y3)
p −3
p
= 16 3
p2
.
5. 求由下列各曲线所围成的图形的面积;
(1)ρ=2acosθ ; 解:
所求的面积为
A=
1 2
π
∫−2π 2
(2acosθ )2dθ
π
=
4a
2
∫2
0
cos2θdθ
=πa2.
(2)x=acos3t, y=asin3t; 解
a
a
z 20. 利用题 19 和结论, 计算曲线 y=sin x(0≤x≤π)和 x 轴所围
成的图形绕 y 轴旋转所得旋转体的体积.
∫ ∫ .3 解
V = 2π π xsin xdx = −2π π xd cosx = 2π (−xcosx +sin x)π = 2π 2 .
0
0
0
w 21. 计算曲线 y=ln x 上相应于 3 ≤ x≤ 8 的一段弧的长度.
o 解设弦的倾角为 α. 由图可以看出, 抛物线与过焦点的弦所围成的图形的面积为
.c A= A0+ A1.
e 显然当α
=π 2
时,
A1=0;
当α <π 2
时,
A1>0.
因此, 抛物线与过焦点的弦所围成的图形的面积的最小值为
h a ∫ z A0 =2 0
2axdx = 8 3
a
x3 a = 8 a2 . 03
曲线 ρ = 2 sinθ 与 ρ 2 =cos 2θ 的交点 M 的极坐标为 M ( 2 , π ) . 所求的面积为 26
A

大学英语翻译教程第五版 PPT第六讲

6.1 转换成汉语动词
❖ 词类转换是英汉翻译中不可或缺的变通手段，是突破原文词性束缚、避免貌合神离的硬译死译，使译文准确、通顺、易懂的有效方法。
❖并不意味着词类转换可以随心所欲，而必须遵循忠实于原文意义、使译文自然流畅、通顺易懂的原则。
6.1 转换成汉语动词
❖ 英语动词远比汉语动词用得少。一个英语句子（单句）往往只用一个限定性动词作谓语，而在一个汉语句子中却可以出现几个动词。
(41) Experts say this is actually a picture after
Picasso.
6.1.2 介词（与介词短语）转译成动词
❖ 此外，英语中不少的短语介词同样具有动作、行为意义，这些短语介词汉译时，也往往转换成动词。 (42) Before long his thoughts were centred on the use of air as the working fluid, instead of steam. (43) In view of human history this could be a slight enough hope.
LOGO
第六讲翻译技巧（二）：词类转换
❖ 在忠实于原文的前提下，根据译入语的行文习惯，适当进行词类转换（即将原文中属于某种词类的词译成译语的另一类词），能使译文通顺达意、自然流畅。
(1) Those small factories are also lavish consumers, and wasters, of raw materials. 原译：那些小厂还是原材料极大的消耗者和浪费者。
(14) The engine has given a constantly good performance.

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2 阻感负载阻感负载时a的移相范围
负载阻抗角：j = arctan(L / R) 稳态时负载电流为正弦波，相位滞后于u1的角度为j a =0时刻仍定为u1过零的时刻，a的移相范围应为j ≤ a ≤π
VT1 VT2 u1 io uo R L
u1 O u G1 O u G2 O uo O io O u VT O 图4-2 7/7 阻感负载单相交流调压电路及其波形
电流谐波次数和电阻负载时相同，也只含3、5、7…等次谐波随着次数的增加，谐波含量减少和电阻负载时相比，阻感负载时的谐波电流含量少一些当a 角相同时，随着阻抗角j 的增大，谐波含量有所减少
11/7
单相交流调压电路 a < j 时
u
t
G1
O
G2
a

t t
O
T1
j
O
a
波形分析:
a<j时阻感负载交流调压电路工作波形
4/7
1 电阻负载
原理分析
6.1.1 单相交流调压电路
VT1 VT2 io uo R u1 VT1 VT2 io uo R
u1
u1 O uo O io O uVT O
在 u1的正半周和负半周，分别对 u1 VT1和VT2的开通角a进行控制就可以调节输出电压 t O 负载电压波形是电源电 uo 压波形的一部分，负载电流（也即电源电流） t O 正负半周 a 起始时刻和负载电压的波形相同 io （a =0）均为电压过零时刻，稳态时，正 t O 负半周的a 相等
6.1.1 单相交流调压电路
4．斩控式交流调压电路
原 •设斩波器件（V1或 V负半周 2）导通时间为ton，开关周期为 • u •基本原理和直流斩波电路有类似之处 1 •u 1正半周 T ，则导通比 a = t / T 理 on ，改变a 可调节输出电压分 VD 1 V1 一般采用全控型器析件作为开关器件 i1 斩波控制
60 40 20 3次 5次 7次
I on = U on / R
电流基波和各次谐波标么值随 a 变化的曲线（基准电流为a =0时的有效值）如图所示
0
60
120 180
触发延迟角a/( ° )
电阻负载单相交流调压电路基图4-6 波和谐波电流含量
10/7
6.1.1 单相交流调压电路
阻感负载的情况
无器件导通
1. 6 通 Uab/2
无器件导通
20/7
6.1.2 三相交流调压电路
VT 1 VT 4 VT 1 晶闸管 VT VT 3 VT 6 导通区间 VT 6 VT VT 5 2 5 uab uac ua 2 2 uao ' a) 2 0 a 4 5 2 3 3 3 3 t 1 t 2 t3 VT 5 VT 1 VT 3 VT 5 晶闸管 VT 6 VT 2 VT 4 VT 6 导通区间 uab ua uac 2 2 4 5 uao ' 3 3 b) 0 2 2 a 3 3 t1 t2 t3 VT 5 VT 5 VT 1 VT 1 VT 3 VT 3 VT 5 VT 5 晶闸管导通区间 VT 4VT 6 VT 6 VT 2 VT 2 VT 4 VT 4 VT 6 uab ua uac 2 uao ' 2 5 3 c) 0 2 4 2 3 3 3
交流调功电路:
以交流电周期为单位控制晶闸管通断，改变通断周期数的比，调节输出功率的平均值
交流电力电子开关:
并不着意调节输出平均功率，而只是根据需要接通或断开电路。 3/7
交流调压电路的应用：
灯光控制（如调光台灯和舞台灯光控制）
异步电动机软起动异步电动机调速
供用电系统对无功功率的连续调节
在高压小电流或低压大电流直流电源中，用于调节变压器一次电压
VT 1 VT 2 u1
io uo R

u1 O uo O io

负载电流有效值
U U oo II = = oo R R
2
t
晶闸管电流有效值
t
IT =
2 2 U sin t U1 1 1 a sin 2a 1 sin t U 1I T = 2U 1 a sin d t = ( 1 2a ) 1 1 a d t = ( 1 2 R R 2 2 ) a O 2 R R 2 2
第6章交流电力控制电路和交交变频电路
6.1 交流调压电路 6.1.1 单相交流调压电路 6.1.2 三相交流调压电路 6.2 其他交流电力控制电路 6.2.1 交流调功电路 6.2.2 交流电力电子开关 6.3 交交变频电路 6.3.1 单相交交变频电路 6.3.2 三相交交变频电路
概述
交流-交流变流电路

t
u VT

UoI Io Uo P U P = = o o = Uo = = S =U I =U = 1I o 1 S U1 U1 o
功率因数
1 a a 1 sin 2 a sin 2a 2 2
O
t
电阻负载单相交流调 6/7 压电路及其波形
6.1.1 单相交流调压电路
改变频率，大多不改变相数，也有改变相数的
先把交流整流成直流，再把直流逆变成另一种频率或可变频率的交流,间接变频电路
2/7
6.1 交流调压电路
交流电力控制电路的结构
两个晶闸管反并联后串联在交流电路中，控制晶闸管就可控制交流电力
交流电力控制电路的类型
交流调压电路:
每半个周波控制晶闸管开通相位，调节输出电压有效值
uVT
t
t
t
t
O
t
5/7
6.1.1 单相交流调压电路
数量关系
负载电压有效值
2 1 1 11 aa 2 dt = U o =U = 2U 12sin t d U sin 2a 1U U sin t t = sin 2 a a o 1 12 a 2
16/7
6.1.2 三相交流调压电路三相三线，电阻负载时的情况
任一相导通须和另一相构成回路
电流通路中至少有两个晶闸管，应采用双脉冲或宽脉冲触发
触发脉冲顺序和三相桥式全控整流电路一样，为 VT1~ VT6,依次相差60° 相电压过零点定为a的起点， a角移相范围是0°~ 150°
17/7
a =30°，分析a相波形。
为交流调功电路控制对象时间常数很大，以周波数为单位控制即可通常晶闸管导通时刻为电源电压过零的时刻，负载电压电流都是正弦波，不对电网电压电流造成通常意义的谐波污染
u1 V2 VD 2 V3 VD 3 VD 4 uo V4 R L
续流通道
斩波控制
图4-7 斩控式交流调压电路
续流通道
图4-7
13/7
6.1.1 单相交流调压电路
单相--斩控式交流调压电路波形
图4－8给出了电阻负载时斩控式交流调压电路的负载电压Uo和电源电流i1波形。
电源电流的基波分量和电源电压同相位，即位移因数为 1。电源电流不含低次谐波，只含和开关周期T有关的高次谐波。功率因数接近1。
图4-8 电阻负载斩控式交流调压电路波形
14/7
6.1.2 三相交流调压电路
根据三相联结形式的不同，三相交流调压电路具有多种形式
ia VT 1 ua VT 3 n ub uc a) ua VT 5 a VT 4 b VT 6 c VT 2 负载 uc b) 负载 ia a Ua0' ia n' n ua b ub c 负载 a
(1) 0°≤ a <60°：三管导通与两管导通交替，每管导通 180°－a 。但a =0°时一直是三管导通
(2) 60°≤ a <90°：两管导通每管导通120° (3) 90°≤ a <150°：两管导通与无晶闸管导通交替，每相每管导通150- a，两管共导通角度为300°－2 a
21/7
2
=
U1 sin cos( 2a j ) cosj 2 Z
负载电流有效值
I 0 = 2 IVT
8/7
6.1.1 单相交流调压电路
3.单相交流调压电路的谐波分析
电阻负载的情况
波形正负半波对称，所以不含直流分量和偶次谐波
式中:
a1 =
an =
u o ( t ) =
n =1, 3, 5 ,
T2
t
– 过渡过程和带 R-L 负载的单相交流电路在 ωt =a i o – 触发 VT2 时， ioL 尚未过零， VT1仍导通， VT2 不通 –– VT1 提前通，被过充电，放电时间延长， – 衰减过程中， VT1 导通时间渐短， VT2 的导通 i o 由两个分量组成：正弦稳态分量、指数衰减分量 –方程式 (6-5) 和(6-6)所得io 表达式仍适用，只是 a≤ωt <∞ – 稳态的工作情况和 a = j 时完全相同 ( a < j ) 时合闸的过渡过程相同过零后， VT2 开通， VT2 导通角小于 π VT1 的导通角超过π 12/7 时间渐长
(a

n
cos n t bn sin n t )
2U 1 sin 2a 2 ( a ) 2
2U 1 (cos 2a 1) 2
b1 =
2U 1 1 1 cos( n 1 ) a 1 cos( n 1 ) a 1 n 1 n 1
0. 6
t t t t
t
t
6.1.1 单相交流调压电路
数量关系
Uo = 1
– 负载电压有效值：