(完整版)CH6习题及答案
结构化学-Ch6-复习习题-杨媛
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3.分裂能（△o 或10Dq)(P186)： 金属原子或离子的5个d轨道在球形场作用 下分裂成2个高能级eg轨道和3个低能级t2g 轨道，高能的d轨道与低能的d轨道的能量 之差即是分裂能。
Eeg=6Dq(或0.6Δ0) Et2g=-4Dq(或-0.4Δ0)
eg
６ q Ｄ
CFSE1-CFSE2=4Dq-(24Dq-2P)=-20Dq+2P
=2（-Δ0+P ）<0 （Δ0 > P） ∴Co3+采用t2g6 的排布方式，低自旋，没有未 配对电子，反磁性。
P203-8. 试判断下列两组配位化合物顺磁性大小的次序： (1) A. [Co(NH3)6]3+ B. [Co(NH3)6]2+ C. [Co(NO2)6]3D. [Co(CN)6]4 A. NH3为中场配体，Co3+（d6）为强场离子，总体上 [Co(NH3)6]3+属强场配合物，d电子处于低自旋，组态为 t2g6eg0，没有未配对电子。 B. NH3为中场配体，Co2+（d7）为弱场离子，总体上 [Co(NH3)6]2+属弱场配合物，d电子处于高自旋，组态为 t2g5eg2，有3个未配对电子。 C.NO2-属于强场配体, [Co(NO2)6]3- 属强场配合物，d电 子处于低自旋，组态为t2g6eg0，没有未配对电子。 D. CN-属于强场配体， [Co(CN)6]4-属强场配合物，d电 子处于低自旋，组态为t2g6eg1，有1个未配对电子。 ∴顺磁性大小：B>D>A=C
P202-7. 已知[Co(NH3)6]2+的Δ0 < P，而[Co(NH3)6]3+的 Δ0 > P ，试解释此区别的原因，并用稳定化能推算出 二者的d电子构型和磁性。 解：Co3+比Co2+价态高，ΔCo3+ > ΔCo2+ 电子成对能不变，所以出现题目中出现情况。 [Co(NH3)6]2+中 Co2+的d7有两种排布方式： ①t2g5eg2 ② t2g6eg1 ①CFSE=-[5(-4Dq)+26Dq]=8Dq ②CFSE=-[6(-4Dq)+16Dq+3P-2P]=18Dq-P
统计部分(ch2-ch6)练习答案
数理统计部分（ch2-ch6）练习答案1．4321,,,ξξξξ独立，)1,0(~N i ξ，4,3,2,1=i 。
)2(~)(2243221χξξξξη+++=k ，则k=1/3；242321ξξξξ++服从的分布是t(2)解：（1）)3,0(~432N ξξξ++，)1,0(~3432N ξξξ++∴，)1(~3)(22432χξξξ++∴，k =1/3（2）)2,0(~21N ξξ+，)1,0(~221N ξξ+与)2(~22423χξξ+独立，)2(~2/2/)(242321242321t ξξξξξξξξ++=++∴2．n X X X ,,,21 是来自总体),(2σμN 的样本，∑==n i i X n X 11，∑=-=n i i X X n S 122*)(1，2**S S =则1/*--n S X μ的分布是)1(-n t 。
解：∑=--=ni i X X n S 122)(11是样本方差,则2*2)1(nS S n =-，)1(~/1/*--=--n t nS X n S X μμ 3．已知74.4)5,10(05.0=F ，)10,5(~F ξ，95.0}{=≥x P ξ，则x = 0.21 。
解：21.074.41)5,10(1)10,5(05.095.0====F F x4．n X X X ,,,21 是来自总体],0[θU 的样本，则参数θ的矩法估计量是X 2；极大似然估计量是},,,max{21n X X X ⋅⋅⋅。
5．n X X X ,,,21 是总体X 的样本，X 有分布密度⎩⎨⎧≤>=+-)(0)()()1(c x c x x c x f θθθ，其中c>0是已知常数，1>θ是未知参数。
求θ的矩法估计量。
解：dx x xf EX ⎰+∞∞-=)(dx xc x c)1(+-+∞⎰⋅=θθθθθ--=1c ,θθˆ1ˆ--=∴c X ，θ的矩法估计量：cX X-=θˆ。
Ch6 微型计算机原理与接口技术 答案
解：16位地址信号通过译码电路与74LS244芯片连接。其连接如下图所示。
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断点的逆过程。即CPU会自动地将堆栈内保存的断点信息弹出到IP、CS和FLAG中，保证被中断 的程序从断点处继续往下执行。 6.11 CPU满足什么条件能够响应可屏蔽中断？
CH6压电式传感器(含答案)《传感器与检测技术(第2版)》习题及解答.docx
第6章压电式传感器一、单项选择题1、对石英晶体，下列说法正确的是（）0A.沿光轴方向施加作用力，不会产生压电效应，也没有电荷产生。
B.沿光轴方向施加作用力，不会产生压电效应，但会有电荷产生。
C.沿光轴方向施加作用力，会产生压电效应，但没有电荷产生。
D.沿光轴方向施加作用力，会产生压电效应，也会有电荷产生。
2、石英晶体和压电陶瓷的压电效应对比正确的是（）A.压电陶瓷比石英晶体的压电效应明显，稳定性也比石英晶体好B.压电陶瓷比石英晶体的压电效应明显，稳定性不如石英晶体好C.石英晶体比压电陶瓷的压电效应明显，稳定性也比压电陶瓷好D.石英晶体比压电陶瓷的压电效应明显，稳定性不如压电陶瓷好3、两个压电元件相并联与单片时相比说法正确的是（）A.并联时输出电压不变，输出电容是单片时的一半B.并联时输出电压不变，电荷量增加了 2倍C.并联时电荷量增加了 2倍，输出电容为单片时2倍D.并联时电荷量增加了一倍，输出电容为单片时的2倍4、两个压电元件相串联与单片时相比说法正确的是（）A.串联时输出电压不变，电荷量与单片时相同B.串联时输出电压增大一倍，电荷量与单片时相同C.串联时电荷量增大一倍，电容量不变D.串联时电荷量增大一倍，电容量为单片时的一半5、用于厚度测量的压电陶瓷器件利用了（）原理。
A.磁阻效应B.压阻效应C.正压电效应D.逆压电效应6、压电陶瓷传感器与压电石英晶体传感器的比较是（）。
A.前者比后者灵敏度高B.后者比前者灵敏度高C.前者比后者性能稳定性好D.前者机械强度比后者的好7、压电石英晶体表面上产生的电荷密度与（）。
A.晶体厚度成反比C.作用在晶片上的压力成正比8、压电式传感器目前多用于测量（A.静态的力或压力C.位移B.晶体面积成正比D.剩余极化强调成正比）oB.动态的力或压力D.温度A.不产生压电效应B. 产生逆向压电效应C. 产生横向压电效应D. 产生纵向压电效应关于压电式传感器中压电元件的连接，以下说法正确的是（二、多项选择题1、 压电晶体式传感器其测量电路常采用（）。
概论论与数理统计课本答案CH6 ans
概率论第六章习题解答习题6.11． 求下列总体分布中参数的矩估计：（1）21,01,(;)0,,x x f x θθθ+−≤≤⎧=⎨⎩其他 其中θ < 1；（2）f (x ; p ) = p (1 − p ) x − 1，x = 1, 2, …；其中0 < p < 1；（3）1211221e ,,(;,)0,,x x f x θθθθθθ−−⎧⎪≥=⎨⎪⎩其他 其中−∞ < θ 1 < +∞，θ 2 > 0． 解：（1）因11320021211E()(21)d ()323226X x x x x x θθθθθθθ−−=+−=+=+=+∫，有θ = 6 E (X ) − 3，故θ 的矩估计为ˆ63X θ=−； （2）因1121111d d d 11E()(1)d d d 1(1)x x x x x x x x q X x p p p x qp q p q p p q q q q p q ∞∞∞∞−−====⎛⎞=⋅−=⋅=====⎜⎟−−⎝⎠∑∑∑∑， 故1E()p X =，p 的矩估计为1ˆpX=； （3）因∫∫∫∞+−−+∞−−∞+−−∞+−−+−=−⋅=⋅=121121121121d eede)1(d e1)(E 2θθθθθθθθθθθθθx x x x x X x x x x212121121eeθθθθθθθθθ+=−−=+∞−−+∞−−x x x ，且∫∫∫∞+−−+∞−−∞+−−∞+−−⋅+−=−⋅=⋅=121121121121d 2eede)1(d e1)(E 22222θθθθθθθθθθθθθx x x x x x X x x x x22212122122222)(E 2d e12e121121θθθθθθθθθθθθθθ++=+=⋅+−=∫∞+−−+∞−−X x x x x x ， 则2222122212122)(22)](E [)(E )(D θθθθθθθ=+−++=−=X X X ，即)(D 2X =θ，)(D )(E 1X X −=θ，故θ 1和θ 2的矩估计为n S X −=1ˆθ，nS =2ˆθ． 2． 求下列总体分布中参数的极大似然估计：（1）f (x ; θ ) = θ (1 − θ ) x − 1，x = 1, 2, …；其中0 < θ < 1； （2）λλλ−=e !);(x x f x，x = 0, 1, 2, …；其中λ > 0；（3）222)(ln 2eπ21),;(σµσσµ−−=x xx f ，x = 0；其中−∞ < µ < +∞，σ > 0．解：（1）nx nx x x n ni i n x f x f x f L −−−−∑−=−−⋅−===121)1()1()1()1();();();()(11121θθθθθθθθθθθθ""，即)1ln()(ln )(ln 1θθθ−−∑+==n x n L ni i ，令011)(1d )(ln d 1=−−⋅−∑+⋅==θθθθn x n L n i i ，得xx nni i11==∑=θ， 故θ 的极大似然估计为X1ˆ=θ； （2）λλλλλλλλλλλλn n x n x x x n x x x x x x x f x f x f L ni in−−−−∑=⋅===e !!!e !e !e !);();();()(212121121"""，即λλλn x x x x L n ni i −−⋅∑==)!!!ln(ln )(ln 211"，令01d )(ln d 1=−⋅∑==n x L n i i λλλ，得x x n ni i ==∑=11λ， 故λ 的极大似然估计为X =λˆ； （3）),;(),;(),;(),(222212σµσµσµσµn x f x f x f L "=212222222212)(ln 212)(ln 2)(ln 22)(ln 1e)π2(1eπ21eπ21eπ21σµσµσµσµσσσσ∑===−−−−−−−−ni i n x nnx nx x x x x x x x ""，即21221222)(ln )ln()ln π2(ln 2),(ln σµσσµ∑−−−+−==ni i n x x x x nL "，令0ln 2)1()(ln 2),(ln 21212=−∑=∑−⋅−−=∂∂==σµσµµσµn x x L ni i ni i ，得∑==ni i x n 1ln 1µ，再令02)(ln 12),(ln 412222=∑−+⋅−=∂∂=σµσσσµni i x n L ，得∑−==n i i x n 122)(ln 1µσ， 故µ和σ 2的极大似然估计为∑==n i i X n 1ln 1ˆµ，∑−===∧∑n i n i i i X n X n 1212)ln 1(ln 1σ． 3． 设总体X 的密度函数为⎩⎨⎧<<+=,,0,10,)1();(其他x x x f θθθ求参数θ 的极大似然估计与矩法估计，并看看它们是否一致？今获得样本观测值为0.4, 0.7, 0.27, 0.55,0.68, 0.31, 0.45, 0.83．试分别求θ 的极大似然估计值与矩估计值．解：因121212()(;)(;)(;)(1)(1)(1)(1)()n n n n L f x f x f x x x x x x x θθθθθθθθθθθθ==+⋅++=+"""，即ln L (θ ) = n ln (θ + 1) + θ ln (x 1 x 2 … x n )，令12d ln ()1ln()0d 1n L n x x x θθθ=⋅+=+"， 则12111ln()ln nn ii nnx x x x θ==−−=−−∑"，故θ 的极大似然估计为1ˆ1ln nii nX θ==−−∑；因1211E()(1)d (1)22xX x x x θθθθθθθ++=⋅+=+⋅=++∫，有2E()11E()X X θ−=−，故θ 的矩法估计为21ˆ1X Xθ−=−； 显然参数θ 的极大似然估计与矩法估计不一致；又因样本观测值为0.4, 0.7, 0.27, 0.55, 0.68, 0.31, 0.45, 0.83，有1(0.40.70.83)0.523758x =+++="，故θ 的极大似然估计值为8ˆ10.3982ln 0.4ln 0.7ln 0.83θ=−−=+++"，θ 的矩估计值为20.523751ˆ0.099710.52375θ×−==−． 习题6.21． 设容量为3的随机样本X 1 , X 2 , X 3取自概率密度函数为1,0,(;)0,,x f x θθθ−⎧<<⎪=⎨⎪⎩其他的总体．证明1(1)ˆ4X θ=和2(3)ˆ43X θ=都是θ 的无偏估计量． 证：总体X 的分布函数为0,0,(;),0,1,,x x F x x x θθθθ<⎧⎪⎪=≤<⎨⎪≥⎪⎩则容量为3的样本的最小顺序统计量X (1) 的分布函数和密度函数为33(1)0,0,(;)1[1(;)]11,0,1,,x x F x F x x x θθθθθ<⎧⎪⎪⎛⎞=−−=−−≤<⎨⎜⎟⎝⎠⎪⎪≥⎩23(1)(1)3(),0,(;)(;)0,x x f x F x θθθθθ⎧−<<⎪′==⎨⎪⎩其他且最大顺序统计量X (3) 的分布函数和密度函数为33(3)0,0,(;)[(;)],0,1,,x x F x F x x x θθθθθ<⎧⎪⎪⎛⎞==≤<⎨⎜⎟⎝⎠⎪⎪≥⎩23(3)(3)3,0,(;)(;)0,x x f x F x θθθθ⎧<<⎪′==⎨⎪⎩其他得234222321(1)33300031212ˆE()4E()4()d (2+)d 2+234x x x X x x x x x x x θθθθθθθθθθθθθ⎛⎞==⋅−=−=−=⎜⎟⎝⎠∫∫，2432(3)33300044344ˆE()E()d d 334x x X x x x x θθθθθθθ==⋅==⋅=∫∫，故1(1)ˆ4X θ=和2(3)ˆ43X θ=都是θ 的无偏估计量． 2． 设总体X 服从伯努利分布B (1, p )，p 为未知参数（0 < p < 1）．样本X 1 , …, X n 来自于X ．（1）证明：当n = 1时，p 2不存在无偏估计；（2）若n ≥ 2，求p 2的一个无偏估计量． 解：（1）当n = 1时，样本X 1的概率分布为101~1X p p ⎛⎞⎜⎟−⎝⎠， 则任何统计量T = T (X 1)的数学期望为E (T ) = T (0) ⋅ (1 − p ) + T (1) ⋅ p = T (0) + [T (1) − T (0)] ⋅ p ≠ p 2， 故当n = 1时，p 2不存在无偏估计；（2）若n ≥ 2，有样本均值11n i i X X n ==∑，样本方差2211()1n i i S X X n ==−−∑， 则E()E()X X p ==，22E()D()(1)S X p p p p ==−=−，即222E()E()E()X S X S p −=−=， 故2X S −是p 2的一个无偏估计量．3． 设从均值为µ ，方差为σ 2（> 0）的总体X 中分别抽取容量为n 1 , n 2的两个独立样本，样本均值分别为1X 和2X ．试证：对于任意满足条件a + b = 1的常数a 和b ，12ˆaX bX µ=+都是µ 的无偏估计，并确定a 、b 使方差ˆD()µ达到最小． 解：因12E()E()X X µ==，211D()X n σ=，222D()X n σ=，有12ˆE()E()E()()a X b X a b a b µµµµ=+=+=+，故当a + b = 1时，ˆE()µµ=，12ˆaX bX µ=+都是µ 的无偏估计； 又22222222222121112121212()2(1)ˆD()D()D()(1)n n a n a n a a a X b X a a n n n n n n σσµσ⎡⎤+−+−=+=⋅+−⋅=+=⎢⎥⎣⎦， 令212112ˆ2()2d D()0d n n a n a n n µσ+−==，得112n a n n =+，且2212212ˆ2()d D()0d n n n n a µσ+=>，故当112n a n n =+，2121n b a n n =−=+时，方差ˆD()µ达到最小． 4． 设X 1 , X 2 , X 3 , X 4是来自均值为θ 的指数分布的样本，其中θ 未知．证明下列三个估计量1123411()()36T X X X X =+++，212341(6543)10T X X X X =+−+，T 3 = 2 X 1 − X 2 + 3 X 3 − 3 X 4 ，均为θ的无偏估计量，并说明上述估计量中哪个最有效．证：因总体X 服从均值为θ 的指数分布，即X ~ e (1/θ )，有E (X ) = θ ，D (X ) = θ 2 ，则112341111E()[E()E()][E()E()]()()3636T X X X X θθθθθ=+++=+++=，2123411E()[6E()5E()4E()3E()](6543)1010T X X X X θθθθθ=+−+=+−+=，E (T 3) = 2 E (X 1) − E (X 2) + 3 E (X 3) − 3 E(X 4) = 2θ − θ + 3θ − 3θ = θ ， 故T 1 , T 2 , T 3均为θ 的无偏估计量；又222221123411115D()[D()D()][D()D()]()()93693618T X X X X θθθθθ=+++=+++=， 22222212341143D()[36D()25D()16D()9D()](3625169)10010050T X X X X θθθθθ=+++=+++=，D (T 3) = 4 D (X 1) + D (X 2) + 9 D (X 3) + 9 D (X 4) = 4θ 2 + θ 2 + 9θ 2 + 9θ 2 = 23θ 2 ， 显然D (T 1) < D (T 2) < D (T 3)， 故T 1最有效，T 2其次，T 3最差．5． 设ˆθ是参数θ 的无偏估计量，且ˆD()0θ>，试证：2ˆ()θ不是θ 2的无偏估计量． 证：因ˆθ是参数θ 的无偏估计量，即ˆE()θθ=，有2222ˆˆˆˆE[()]()[E()]()D D θθθθθθ=+=+>， 故2ˆ()θ不是θ 2的无偏估计量． 习题6.31． 随机地从一批零件中抽取10个，测得其长度（单位：cm ）为：2.13, 2.14, 2.12, 2.13, 2.11, 2.15, 2.14, 2.13, 2.12, 2.13．假设该批零件的长度服从正态分布N (µ , σ 2)，试求总体均值µ 的置信系数为95%的置信区间：（1）若已知σ = 0.01；（2）若σ 未知． 解：（1）单个正态总体，已知σ ，估计µ ，总体均值µ 的点估计为X，枢轴量为~(0,1)X U N =，置信系数1 − α = 0.95，置信区间为/2/2(X u u αα−+，因1(2.13 2.14 2.13) 2.1310x =+++="，σ = 0.01，n = 10，u 0.025 = 1.96， 故µ 的置信系数95%的置信区间为(2.13 1.96 2.13 1.96(2.1238,2.1362)−+=；（2）单个正态总体，未知σ ，估计µ ，总体均值µ 的点估计为X，枢轴量为~(1)X T t n =−，置信系数1 − α = 0.95，置信区间为/2/2((1)(1)X t n t n αα−−+−，因1(2.13 2.14 2.13) 2.1310x =+++="， 222221[(2.13 2.13)(2.14 2.13)(2.13 2.13)]0.01159s =−+−++−="，n = 10，t 0.025 (9) = 2.2622，故µ 的95%置信区间为(2.13 2.2622 2.13 2.2622(2.1217,2.1383)−+=．2． 为估计制造某件产品所需的单件平均工时（单位：小时），现制造了五件，记录所需工时为：10.5, 11, 11.2, 12.5, 12.8．设制造单件产品所需工时服从正态分布，试求单件平均工时的置信系数95%的置信区间．解：单个正态总体，未知σ ，估计µ ，总体均值µ 的点估计为X，枢轴量为~(1)X T t n =−，置信系数1 − α = 0.95，置信区间为/2/2((1)(1)X t n t n αα−−+−，因1(10.51112.8)11.65x =+++="，222221[(10.511.6)(1111.6)(12.811.6)]0.99754s =−+−++−="，n = 5，t 0.025 (4) = 2.7764，故µ 的95%置信区间为(11.6 2.7764 2.7764(10.3615,12.8385)−+=．3． 设有两台机床用来生产规格相同的铝合金薄板．随机选取每台机床轧制的产品若干张，测得它们的厚度（单位：cm ）如下：机器I ：0.243, 0.238, 0.248, 0.245, 0.236, 0.241, 0.239， 机器II ：0.261, 0.254, 0.255, 0.257, 0.253, 0.250，设两台机床所生产的薄板的厚度服从方差相等的正态分布．试给出两台机床生产的铝合金薄板平均厚度差的置信系数为95%的置信区间．解：两个正态总体，未知22,x y σσ（但22x yσσ=），估计µ x −µ y ，均值差µ x −µ y 的点估计为X Y −，枢轴量为()()~(2)X Y T t n m µµ−−−=+−， 置信系数1 − α = 0.95，置信区间为(， 因1(0.2430.2380.239)0.24147x =+++="，1(0.2610.2540.250)0.2556y =+++="， 222221[(0.2430.2414)(0.2380.2414)(0.2390.2414)]0.00426x s =−+−++−="，222221[(0.2610.255)(0.2540.255)(0.2500.255)]0.00375ys =−+−++−="， n = 7，m = 6，t 0.025 (11) = 2.2010， 故µ 的95%置信区间为(0.24140.255 2.2010(0.0185,0.0087)−±=−−．4． 由容量为15，取自正态总体N (µ , σ 2)的随机样本算得23.2, 4.24x s ==，确定σ 2和σ 的置信系数90%的置信区间．解：单个正态总体，估计σ 2，总体方差σ 2的点估计为S 2，枢轴量为2222(1)~(1)n S n χχσ−=−，置信系数1 − α = 0.90，置信区间为2222/21/2(1)(1)(,)(1)(1)n S n S n n ααχχ−−−−−，因s 2 = 4.24，n = 15，20.05(14)23.685χ=，20.95(14) 6.571χ=，故σ 2的90%置信区间为14 4.2414 4.24(,(2.5062,9.0336)23.685 6.571××=； σ 的90%置信区间为(1.5831,3.0056)=．5． 设有两个化验员A 和B 独立对某种聚合物中的含氯量用同一种方法各做了10次测定，其测定值的方差分别为220.512,0.665ABs s ==．假定各自的测定值均服从正态分布，方差分别为2Aσ和2Bσ，求22ABσσ的置信系数为0.90的置信区间．解：两个正态总体，估计22A B σσ，方差比22A Bσσ的点估计为22A B S S ，枢轴量为2222~(1,1)A AB B S F F n m S σσ=−−，置信系数1 − α = 0.90，置信区间为2222/22222/21/2/2111(,)(,(1,1))(1,1)(1,1)(1,1)A A A A B B B BS S S S F m n F n m F n m F n m S S S S αααα−⋅⋅=⋅⋅−−−−−−−−，因220.512,0.665A B s s ==，n = 10，m = 10，F 0.05 (9, 9) = 3.18，故22A Bσσ的置信系数为0.90的置信区间为0.51210.512(, 3.18)(0.2421,2.4484)0.665 3.180.665××=．6． 设枪弹的速度（单位：米/秒）服从正态分布．为了比较两种枪弹的速度，在相同的条件下进行了速度测定．算得数据如下：枪弹甲：m = 110，2810x =，s x = 121.41；枪弹乙：n = 100，2682y =，s x = 105.06．试求这两种枪弹的平均速度之差的置信系数近似为95%的置信区间．解：两个正态总体，未知22,x yσσ（大样本），估计µ x −µ y ，均值差µ x −µ y 的点估计为X Y −，大样本情形下枢轴量为()()~(0,1)X Y T N µµ−−−=，置信系数1 − α = 0.95，置信区间为(，因m = 110，2810x =，s x = 121.41，n = 100，2682y =，s x = 105.06，u 0.025 = 1.96，故µ x −µ y 的95%置信区间为(28102682 1.96(97.36,158.64)−±=．复习题六1． 设X 1 , …, X n 为来自总体X 的样本，X 的概率密度函数为22(),0,(;)0,,x x f x θθθθ⎧−<<⎪=⎨⎪⎩其他 其中θ（> 0）是未知参数．试求参数θ 的矩估计量． 解：因3323222002212E()()d ()()23233X x x x x x θθθθθθθθθθ=⋅−=−=−=∫，有θ = 3 E (X )，故θ 的矩估计为ˆX θ=． 注：此题有误，密度函数非零取值范围应为0 < x < θ ．2． 伯莱托（Pareto ）分布是常用于研究收入的模型，其分布函数为⎪⎩⎪⎨⎧<≥⎟⎠⎞⎜⎝⎛−=,,0,,1),;(111212θθθθθθx x x x F 其中θ 1 > 0，θ 2 > 0．若随机样本X 1 , …, X n 取自该分布，求θ 1与θ 2的极大似然估计量．解：伯莱托分布的密度函数为⎪⎩⎪⎨⎧<≥⋅=′=+,,0,,),;(),;(11112212122θθθθθθθθθθx x x x F x f则1211211212121112212122112122222222)(),;(),;(),;(),(++++=⋅==θθθθθθθθθθθθθθθθθθθθθθθθn n nnn x x x x x x x f x f x f L """，即ln L (θ 1, θ 2) = n ln θ 2 + n θ 2 ln θ 1 − (θ 2 + 1) ln (x 1 x 2 …x n )，显然θ 1越大，ln L (θ 1, θ 2) 就越大，且x i ≥ θ 1，故θ 1的极大似然估计量为)1(11},,min{ˆX X X n =="θ； 令0)ln(ln 1),(ln 2112221=−+⋅=∂∂n x x x n n L "θθθθθ，得111212ln ln 11ln )ln(θθθ−=−=∑=ni i n x n n x x x n "， 故θ 2的极大似然估计量为)1(12ln ln 11ˆX X n ni i −=∑=θ．3． 设总体X 的概率密度为2231/224πe ,0,(;)0,0,xx x f x x ααα−−−⎧⎪>=⎨≤⎪⎩ 试求参数α 的矩估计和极大似然估计，并证明矩估计量是无偏的． 解：因222231/2211/2200E()4πe d 2π(1)de xxX x x x x αααα+∞+∞−−−−−−=⋅=⋅−∫∫22222211/2211/221/21/2002πe2πe d()0(2πe )2πx x x x x ααααααα+∞+∞+∞−−−−−−−−−=−+=+−=∫，故α=，α的矩估计为ˆXα=；因2222221231/2231/2231/221212()(;)(;)(;)4πe4πe4πenx x x n n L f x f x f x x x x αααααααααα−−−−−−−−−==⋅""2213/22124π()eni i x n n n n x x x αα=−−−∑="，即212211ln ()ln 43ln ln π2ln()2nn i i n L n n x x x x ααα==−−+−∑"，令231d ln ()1230d n i i L n x αααα==−⋅+=∑，得α=，故α的极大似然估计为ˆα= 4． 设总体X 的密度函数为||1(;)e ,2x f x x λλ−−=−∞<<+∞，试求参数λ（−∞ < λ < +∞）的极大似然估计量．解：112||||||||121111()(;)(;)(;)e e e e 2222ni n i x x x x n n L f x f x f x λλλλλλλλ=−−−−−−−−∑==⋅=""，即1ln ()ln 2||ni i L n x λλ==−−−∑，设顺序统计量为x (1) , x (2) , …, x (n )，并且记x (0)为−∞，x (n + 1)为+∞，不妨设x (k ) ≤ λ < x (k + 1)，k = 0, 1, …, n − 1, n ， 则1111ln ()ln 2()()ln 2()kn k ni i i i i i k i i k L n x x n k x x n k λλλλλ==+==+=−−−−−=−−+−+−∑∑∑∑11ln 2(2)kni i i i k n n k x x λ==+=−+−+−∑∑，若2n k <，有n − 2k < 0，ln L (λ )关于λ 单调增加，若2nk >，有n − 2k < 0，ln L (λ )关于λ 单调减少， 当n 为偶数时，取2nk =，ln L (λ )在()()221n n x x λ+≤<时达到最大，（由连续性知()21n x λ=时也达到最大），故当n 为偶数时，λ 的极大似然估计量ˆλ为区间()()221[,]n n X X+上的任何值；当n 为奇数时，取12n k −=，ln L (λ )在()()1122n n x x λ−+≤<时单调增加，取12n k +=，ln L (λ )在()()1322n n x x λ++≤<时单调减少，即ln L (λ ) 在()12n x λ+=时达到最大，故当n 为奇数时，λ 的极大似然估计量()12ˆn X λ+=．5． 设总体X ~ N (µ , σ 2)，X 1 , …, X n 是X 的样本，X 为样本均值，求常数c 和d ，使∑−=+−1121)(n i i i X X c 与∑=−ni i X X d 1||分别为σ 2和σ 的无偏估计．解：因E (X i ) = µ ，2222)](E [)(D )(E µσ+=+=i i i X X X ，则∑∑∑−=++−=++−=+−+=−+=⎥⎦⎤⎢⎣⎡−1112211112211121)](E )(E 2)(E )(E [)2(E )(E n i i i i i n i i i i i n i i i X X X X c X X X X c X X c221122222)1(22)1(]2)()[(σσµµσµσ−=⋅−⋅=−+++=∑−=n c n c c n i ，故当)1(21−=n c 时，21121)(E σ=⎥⎦⎤⎢⎣⎡−∑−=+n i i i X X c ，∑−=+−−1121)()1(21n i i i X X n 为σ 2的无偏估计； 因∑∑≠=−−=−=−ij j i n j j i i X n X n n X n X X X 1111，有i X X −服从正态分布，且E()E()E()0i i X X X X µµ−=−=−=，22222(1)1(1)11D()D()D()(1)i i jj i n n n X X X X n n n n n n σσ≠−−−−=+=+⋅−=∑， 则21~(0,)i n X X N n σ−−~(0,1)X N ，记X Y =Y ~ N (0, 1)，则22222200E(||)||d 2d 2y y y Y y yy y+∞−−−+∞+∞−∞===−=∫∫即E(||)i X X −=，11E(||)E(||)n ni i i i d X X d X X d n ==−=−=⋅∑∑，故当d =时，1E ||ni i d X Xσ=⎡⎤−=⎢⎥⎣⎦∑1||n i X X =−为σ 的无偏估计．。
新编英语教程6第三版练习册答案
新编英语教程6第三版练习册答案【篇一：新编英语教程第六册练习册paraphrase答案】nothing in life is more exciting and rewarding than the sudden flash of light that leaves you a changed person--not only changed, but changed for the better.the most inspiring and gratifying fact of life is the unexpected spark of enlightenment that makes you different and a better person than before.2. he came across the street, finally, muffled in his ancient overcoat, shapeless felt hat pulled down over his bald head, looking more like an energetic gnome than an eminent psychiatrist.at last he walked over from the other side of the street,wrapped in his old-fashioned overcoat, his bald head coveredby a shapeless felt hat. he looked like a dwarfish old man fullof energy rather than a well-known psychiatrist.3. the woman who spoke next had never married because of a sense of obligation to her widowed mother; she recalledbitterly all the marital chances she had let go by.the next speaker on the tape was a woman who had remained single because she thought she was obliged to take care of her mother who was a widow. she still remembered and told others miserably about all the chances of marriage she had missed.4. in the end, if you let it become a habit, it can become a real roadblock, an excuse for not trying any more.eventually, if you form a habit of saying “if only”, the phrasecan really turn to an obstruction, providing you with an excuse for giving up trying anything at all.5. ... you never got out of the past tense. not once did you mention the future.…you are always thinking of the past, regretting and lamenting. you did not look forward to what you can do in the future at all.6. my, my, said the old man slyly. if only we had come downten seconds sooner, wed have caught that cab, wouldnt we?the old man said to me trickily, using the phrase “if only” on purpose, “if only we’d got here ten seconds earlier, we’d havecaught the cab.” i laughed and understood what he meant. so i foll owed his advice and said, “next time i’ll run faster”.unit 21. moses pleaded a speech defect to rationalize his reluctance to deliver jehovahs edict to pharaoh. moses justified his unwillingness to pass jehovah’s order to pharaoh, saying that he was “slow of speech”.2. yet for all the trouble procrastination may incur, delay can often inspire and revive a creative soul.delay leads to problems. however, in many cases, it can often stimulate the creativity in an artist.3. he notes that speedy action can be embarrassing or extremely costly.he points out that hastiness may give rise to decision which turn out to be humiliating or expensive.4. bureaucratization, which flourished amid the growing burdens of government and the greater complexity of society, was designed to smother policymakers in blankets of legalism, compromise and reappraisal---and thereby prevent hasty decisions from being made.excessive red-tape(官样文章；繁文缛节) developed because public administration was expanding in scope and because society was growing more and more complicated. in this sense, red-tape helped those in charge of policy to be fully engaged in enormous amount of paperwork and judgment, thus making it impossible for an immature decision to result.5. ...many of my friends go through agonies when they face a blank page.…many of my friends have a hard time the moment they attempt to put pen to paper.unit 31. of course, my father is a gentleman of the old school, a member of the generation to whom a good deal of modern architecture is unnerving; but i suspect---i more than suspect, i am convinced---that his negative response was not so much to the architecture as to a violation of his concept of the nature of money.brought up in the old tradition, my father is naturally not prepared to accept the idea of modern architecture; his objection to it, i would assume, indeed i should say i am prettysure, is not a result of his strong dislike of the physical building itself, but rather that of his refusal to change his attitude towards money.2. if a buildings design made it appear impregnable, the institution was necessarily sound, and the meaning of the heavy wall as an architectural symbol dwelt in the prevailing attitude toward money, rather than in any aesthetic theory.if a building was made to look sturdy/invulnerable, it would be accordingly regarded as reliable, and the significance of the thick walls would be measured not by their artistic value, but by their seeming ability to provide a safe location for money.3. in a primitive society, for example, men pictured the world as large, fearsome, hostile, and beyond human control.people in a primitive society, for example, saw the world as an enormous planet full of fear, hatred and disorder.4.the principal function of todays wall is to separate possible undesirable outside air from the controlled conditions of temperature and humidity which we have created inside.today a wall serves mainly as a physical means to protect the desired atmosphere inside from being disturbed by anything unwelcome outside.5. to repeat, it is not our advanced technology, but our changing conceptions of ourselves in relation to the world that determine how we shall build our walls.again, the decisive factor that can influence the design of a wall is not the advancement of science and technology, but our ever-changing attitude towards our place in this world.unit 41. he was a man of exuberant fancy, and, withal, of an authority so irresistible that, at his will, he turned his varied fancies into facts.he was a man rich in whimsies, and intolerant of any act bold enough as to challenge his authority. when his mind caught upon something, absurd as it might be, he would do everything to make sure that it was done in the way he wished.2. when every member of his domestic and political systems moved smoothly in its appointed course, his nature was bland and genial; but whenever there was a little hitch, and some of his orbs got out of their orbits, he was blander and more genialstill, for nothing pleased him so much as to make the crooked straight, and crush down uneven places.when all his subjects behaved in such a manner as they were told to, he could be gentle and kind. and he could even be more so, if anything not conforming to what he expected should occur, because that offered a great chance for him to see the undesirable removed, a thing he was most delighted in doing.3. he could open either door he pleased: he was subject to no guidance or influence but that of the aforementioned impartial and incorruptible chance.he enjoyed total freedom to choose what to do: he was not directed or influenced by anyone as to which door to open. the only thing that was decisive in terms of his fate was the above-mentioned chance, granted to all the accused alike.4. this element of uncertainty lent an interest to the occasion which it could not otherwise have attained.the fact that no one could tell for sure what might happen (to the accused) made this from of trial more attractive than any other form of justice.5. thus the masses were entertained and pleased, and the thinking part of the community could bring no charge of unfairness against this plan; for did not the accused person have the whole matter in his own hands?thus people enjoyed coming here to watch, and those guided by reason in the society could not possibly question the fairness of this form of trial; for was it not the fact that all the accused were given equal chances to make decisions upon their won destiny?unit51. this semi-barbaric king had a daughter as blooming as his most florid fancies, and with a soul as fervent and imperious as his own.this semi-barbaric king had a daughter as exuberant as the wildest of his notions, a daughter who possessed a nature as fierce and tyrannical as his own.2. of course, everybody knew that the deed with which the accused was charged had been done.it was, of course, known to all that he was guilty of the offense of conducting an affair with the princess.3. ...; but the king would not think of allowing any fact of this kind to interfere with the workings of the tribunal, in which he took such great delight and satisfaction.…,even though the king was well aware that the love affair had taken place, he would still refuse to let the normal method of deciding guilt or innocence be disturbed, because he was extremely enthusiastic about his way of setting matters of this kind.4. ...; but gold, and the power of a womans will, had brought the secret to the princess..…; but because she had the money, and above all, because her determination was so irresistible, the princess was able to get access to the secret.5. he understood her nature, and his soul was assured that she would never rest until she had made plain to herself this thing, hidden to all other lookers-on, even to the king.he knew her so well that he was perfectly positive that she would never cease to search for the secret, which remained unknown to all other spectators, even to the king himself.unit 61. there seems to be a general assumption that brilliant people cannot stand routine; that they need a varied, exciting life in order to do their best.it is generally believed that a colorless life can freeze a creative mind, and that only a colorful life can inspire a man to creative work.2. the outstanding characteristic of mans creativeness is the ability to transmute trivial impulses into momentous consequences.one of the wonders human creativity works is that man can make full use of even insignificant feelings to produce far-reaching results.3. an eventful life exhausts rather than stimulates.a life full of diversions stops man’s creativity instead of activating it.4. it is usually the mediocre poets, writers, etc.,who go in search of stimulating events to release their creative flow.only literary artists of an average type rely on excitements in life as a source for their creative work./ great poets, writers, etc., create works of art out of trivial and common subject.5. people who find dull job unendurable are often dull people who do not know what to do with themselves when at leisure. people who are unable to see how to be patient with repetitious work are usually those who are unable to see where to find fun in life when it comes to relaxation.【篇二：新编英语教程6 练习与答案】txt>practice testsforadvanced english(2)主编张华鸿第五、六册本书的主要特点：1．2．前言编写本书的目的：目前英语专业三年级所使用的由上海外国语大学李观仪教授主编的〈新编英语教程〉紧扣精读课文编写练习，实用性、针对性强。
ch6习题及答案
习题6解答判断题：1.二叉树中每个结点有两个子女结点，而对一般的树则无此限制，因此二叉树是树的特殊情形。
( ╳ )2.二叉树就是结点度为2的树。
( ╳ )( (哈工大2000年研究生试题)3.二叉树中不存在度大于2的结点，当某个结点只有一棵子树时无所谓左、右子树之分。
( ╳ ) (陕西省1998年自考试题)4.当k≥1时，高度为k的二叉树至多有21 k个结点。
( ╳ )5.完全二叉树的某结点若无左孩子，则它必是叶结点。
(√)(中科院软件所1997年研究生试题)6.用一维数组存放二叉树时，总是以前序遍历顺序存储结点。
( ╳ )7.若有一个结点是某二叉树子树的中序遍历序列中的最后一个结点，则它必是该子树的前序遍历序列中的最后一个结点。
( ╳ )8.存在这样的二叉树，对它采用任何次序的遍历，结果相同。
(√)(哈工大2000年研究生试题)9.中序线索二叉树的优点之一是便于在中序下查找前驱结点和后继结点。
(√)10.将一棵树转换成二叉树后，根结点没有左子树，( ╳ )(北邮1999年研究生试题。
)11.由树转换成二叉树，其根结点的右子树总是空的。
(√)12.前序遍历森林和前序遍历与该森林对应的二叉树其结果不同。
( ╳ )13.在叶子数目和权值相同的所有二叉树中，最优二叉树一定是完全二叉树。
( ╳ )14.在哈夫曼编码中，当两个字符出现的频率相同时，其编码也相同，对于这种情况应作特殊处理。
( ╳ )15.霍夫曼树一定是满二叉树。
( ╳ )16.树的度是树内各结点的度之和。
( ╳ )17.由二叉树的结点构成的集合可以是空集合。
(√)18.一棵树中的叶子结点数一定等于与其对应的二叉树中的叶子结点数。
( ╳ )选择题：19.树最适合用来表示( C )。
A．有序数据元素 B. 无序数据元素C．元素之间具有分支层次关系的数据 D. 元素之间无联系的数据20.如果结点A有3个兄弟，而且B是A的双亲，则B的度是( D )。
Book6 Unit2课时作业答案
详细答案Book 6 Unit 2Period 1 PreviewI. 词组P 9 Warming up / pre-reading1. make a list of2. recall an enjoyable or unpleasant incident3. make sense4. recite / read / explain sth. to sb.5. rhyming wordsP 10-11 Reading1. give the reader a strong impression2. convey one`s emotions3. nursery rhymes4. delight small children5. be easy to learn and recite6.on fire7.score that goal8. take it easy = take things easy 别着急,慢慢来take one`s time9. 用完，耗尽（及物）(Sb.) run out of sth 用完（不及物）(Sth.) run out 10. make up/ be made up of 11. be brimful of = be full of … 12. translate A into B 13.尤其in particular 14. worth a tryP14-16 Using language1. (Sb. / Sth.) be likely to do sth.2. hold on3. try out试穿try on4. let out II．完成句子.as if she had studied English in EnglandIt’s necessary that 3. he (should) finish his homework first4. It’s required that5. you (should) practice speaking English as often as possible6. that his pay (should) be paid by the hour7. that the work (should) be finished two days ahead of time8. hadn’t missed the trainIII 句型1.Some rhyme(like B) while （然而）others do not(like C).2.It is not a traditional form of English poetry，but is very popular with（受欢迎）English writers.3.With so many different forms of poetry to choose from（从……中选择），students may eventually want to write poems of their own由于有很多的作业要做，他没有时间玩。
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6.1 复随机过程0()()j t Z t e ω+Φ=，式中0ω为常数，Φ是在(0,2)π上均匀分布的随机变量。
求：（1）[()()]E Z t Z t τ*+和[()()]E Z t Z t τ+；（2）信号的功率谱。
解：(1)()0000[()][]20[()()]1212j t j t j j Z E Z t Z t eed ed e R ωτωπωτωττπτπ*+∞++Φ-+Φ-∞+=Φ=Φ==⎰⎰0000[()][]2[(2)2]2(2)201[()()]212120j t j t j t j t j E Z t Z t eed ed e e d ωτωπωτπωττπππ+∞++Φ+Φ-∞++Φ+Φ+=Φ=Φ=Φ=⎰⎰⎰(2)00()[()]{[()()]}[]2()Z Z j S F R F E Z t Z t F e ωτωττπδωω*==+==-6.26.36.4 已知()a t 的频谱为实函数()A ω，假定ωω>∆时，()0A ω=，且满足0ωω∆，试比较：（1）0()cos a t t ω和0(12)()exp()a t j t ω的傅立叶变换。
（2）0()sin a t t ω和0(2)()exp()j a t j t ω-的傅立叶变换。
（3）0()cos a t t ω和0()sin a t t ω的傅立叶变换。
解：由傅立叶变换的定义可以得到： （1）000001()cos [()()]211()()22FTj t FTa t t A A a t e A ωωωωωωωω←−→-++←−→- 01()2j t a t e ω的傅立叶变换是0()cos a t t ω的傅立叶变换的正频率部分。
（2）000001()sin [()()]21()()22FTj t FTa t t A A j j a t e A jωωωωωωωω←−→--+-←−→- 0()2j tj a t e ω-的傅立叶变换是0()sin a t t ω的傅立叶变换的正频率部分。
（3）0()cos a t t ω和0()sin a t t ω的傅立叶变换[][]()00()sin ()cos sgn F a t t F a t t j ωωω=⋅-⎡⎤⎣⎦6.56.66.7 若零均值平稳窄高斯随机信号()X t 的功率谱密度如题图6.7(1) 试写出此随机信号的一维概率密度函数； (2)写出()X t 的两个正交分量的联合概率密度函数。
题图6.7解：(1) 零均值平稳窄带高斯信号()X t 的正交表达式为 00()()cos ()sin x t i t t q t t ωω=-基于功率谱计算功率得21(0)()22X XX AWP R S d σωωππ+∞-∞====⎰ 所以2()(0,)X X t N σ所以一维概率密度222211()x AW x AWf x ee ππ-⋅-==(2) 又因为()X t 的功率谱关于中心频率0ω偶对称()0qi S ω=即 12()[()()]0qi R E i t q t τ==所以(),()i t q t 彼此正交，做为零均值的高斯信号也彼此独立，且()()()0E X t E i t E q t ===⎡⎤⎡⎤⎡⎤⎣⎦⎣⎦⎣⎦,2222X i qAW σσσπ=== 222221212()()22(,;,)(,)(,)112X iq i q i q i q AWXf i q t t f i t f q t e e AWπσπσ++--===6.8 对于窄带平稳随机过程00()()cos ()sin x t i t t q t t ωω=-，若其均值为零，功率谱密度为0000cos[()/],/2()cos[()/],/20x P S P πωωωωωωωπωωωωωω⎧-∆-≤∆⎪=+∆+≤∆⎨⎪⎩，其它式中0,P ωωω∆>>∆及都是正实常数。
试求(1) x(t)的平均功率； (2) i(t)的功率谱密度；(3) 互相关函数()iq R τ或互谱密度()iq S ω； (4) i(t)与q(t)是否正交或不相关？ 解：（1）()x t 的平均功率：()()2221122cos /22w N X w P wP S d P w d ωωπωωπππ∆∞∆-∞-⋅∆==⋅∆=⎰⎰（2）()x t 是零均值平稳窄带随机信号，所以有：()()()()00,20,2cos ,20,X X i q w S S S S otherP w w otherωωωωωωωπωω∆⎧++-≤⎪==⎨⎪⎩⎧⎛⎫∆⎪ ⎪=≤∆⎝⎭⎨⎪⎩（3）互相关函数()iq R τ或互谱密度()iq S ω因为()X S ω是关于0ω偶对称，故互谱密度()iq S ω为0，互相关函数()iq R τ也为0（4）由()0iq R τ=，所以()i t 与()q t 任意时刻正交。
因为()i t 与()q t 是零均值的，所以()i t 与()q t 是不相关的。
6.96.10 6.11已知零均值窄带平稳噪声00()()cos ()sin X t A t t B t t ωω=-的功率谱密度如题图6.11所示。
画出下列情况下随机过程 ()A t ，()B t 各自的功率谱密度：（1） 01ωω= （2）02ωω= （3）012()/2ωωω=+判断上述各种情况下，过程()A t ，()B t 是否互不相关。
1221题图6.11解：因为()X t 是零均值平稳窄带随机信号,所以有：000()()()()0x x A B S S S S ωωωωωωωω++-⎧<==⎨⎩其它00[()()]()()0x x BA AB j S S S S ωωωωωωωω--+⎧<=-=⎨⎩其它功率谱图形如下： （1）（2）（3）由于()X t 的功率谱不以中心频率0ω偶对称，所以互功率谱密度()BA S ω在三种情况下都不为0, 所以A(t),B(t)相关.6.126.13 同步检波器如下题图6.13所示，输入()X t 为窄带平稳噪声，它的自相关函数为20()cos X X R eβττσωτ-=，0βω。
若另一输入0()sin()Y t A t ωθ=+，其中A 为常数，θ服从(0,2)π上的均匀分布，且与()X t 独立。
求检波器输出()Z t 的平均功率。
题图6.13解： ()()()()()00cos sin X t i t t q t t ωθωθ=+-+令： 由题意知0200[()][sin()]1sin()02E Y t E A t A t d πωθωθθπ=+=+=⎰200020(,)[()()]1sin[()]sin()2cos ()2Y Y R t t E Y t Y t A t A t d A R t πττωτθωθθπωτ+=+=+++==⎰所以()Y t 也是平稳的.()()()()()()2020022220222200()cos 2()2X X X X X XR eS i t q t βττσωτσβωπδωωδωωπωββσβσωωωβωωβ-=∴=⋅*++-⎡⎤⎣⎦+=+→++-+关于偶对称，与正交()()()()()()()()()()()()()()()()()000000000()()()cos sin sin cos sin sin sin sin sin 2cos cos 222M t X t Y t i t t q t t A t Ai t t t Aq t t t A Ai t t q t t ωϕωϕωθωϕωθωϕωθθϕωϕθθϕωϕθ==+-++⎡⎤⎣⎦=++-++=-+++---++⎡⎤⎡⎤⎣⎦⎣⎦经过低通滤波器LPF 后，()()()()()sin cos 22A AZ t i t q t θϕθϕ=---（对于同步检波器，θϕ-，()()2AZ t q t =-）()()()()()()()()()()()()()()2222222222222222()sin cos 22sin cos sin 244404444X X A A E Z t E i t q t A A A E i t q t i t q t A A A A E i t E X t R θϕθϕθϕθϕθϕσ⎡⎤⎡⎤=---⎣⎦⎢⎥⎣⎦⎡⎤=-+---⎢⎥⎣⎦⎡⎤⎡⎤====⎣⎦⎣⎦。