delta=beta*1.554*1e-9/(2*pi*3e-7); disp2=23: ps 2/km for ml=1: fix(z/h) DI=exp(( j*(i*w1)+i*0. 5*disp2*(i*w1).2)*h/2) ul=fft(u).*D1 u2=ifft(ul) D2=exp( deltas*(i*w1)+i*0.5*dsp2*(*w1).2)*h/2) vl=fft(v).米D2 v2=ifft(v1)

第一周作业参考答案 3.11011.011(B)=27.375(D)=33.3(O)=1B.6(H) 4.240.75(D)=110000.11(B)=360.6(O)=0.C(H) 5 E3(H)=11100011() E3(H)为原码时,x=-110011(B)=-99(D) E3(H)为反码时,[]原=10011100(B)(x=-11100=-28(D) E3(H)为补码时,[]原=10011101(B),(x=-11101=-29(D) E3(H)为无符号数时,X=E3(H)=227(D) 因为E3(H)的高4位E(H)=14(D)超出了0~9的范围,所以E3(H)不可能为压缩BCD码

一、选择题 A型题 1.d2.c3.e4.e5.d6.a7.e8.c9.c10.b1.d12.a 13.b14.a15.d16.e17.e18.e19.c20.c21.b22d 23.c24.c25.c26.d27.e28.c29.c30.c31.b32.b

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Transmittal No. 96-E1 (9/96) Form FDA 2905a (6/92) INDEX: METHODS–1 Pesticide Analytical Manual Vol. I INDEX: METHODS Index to PAM I Methods, by Chemicals Tested for Recovery Codes: C: complete (recovery); P: partial; S: small; V: variable; R: recovered; NR: not recovered

Department of Electrical Engineering and Computer Science 6.003: Signals and Systems--Fall 2003 PROBLEM SET 9 SOLUTIONS Home Study 1 O&w 9.25(e) We assume that the system we are considering, H(s)takes the following form:

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(一)底物浓度对酶反应速率的影响 用反应初速度ⅴ对底物浓度[S]作图得P355图9-6。 曲线分以下几段 (1)OA段:反应底物浓度较低时ⅴ与[S]成正比,表现为一级反应,v= kISI 根据酶底物中间络合物学说,酶催化反应时,首先和底物结合生成中间 复合物ES,然后再生成产物P,并释放出E E+S P+E OA段上,底物浓度小,酶未被底物饱和,有剩余酶,反应速率取决于 ES浓度,与[S]呈线性关系,V正比于[S]

1 Recurrent Problems 2 Sums 3 Integer Functions 4 Number Theory 5 Binomial Coefficients 6 Special Numbers 7 Generating Functions 8 Discrete Probability 9 Asymptotics