Understanding NMR Spectroscopy phần 2 potx

... rotating frame, B − 1 appears to be static 2. 6 Strong coupling 2 17 40 503 020 100 –ν 0,A – 1 / 2 J AX –ν 0,B – 1 / 2 J BX –ν 0,A + 1 / 2 J AX –ν 0,B + 1 / 2 J BX –ν 0,B –ν 0,A frequency (Hz) full ... and soft pulses 3–17 (a) (b) (c) -20 -10 10 20 -1 -0.5 0.5 1 -20 -10 0 10 20 0.5 1 -20 -10 10 20 -1 -0.5 0.5 1 M x M y M abs Ω / ω 1 Ω / ω 1 Ω / ω 1 Fig. 3 .26 Plots of...

Ngày tải lên: 14/08/2014, 09:21

... constant R 2 = 1/T 2 and so S(t) becomes S(t) = S 0 exp(it) exp(−R 2 t). (4 .2) The shorter the time T 2 (or the larger the rate constant R 2 ) the more rapidly the signal decays. 4 .2 Fourier transformation Fourier ... lines S(t) will be: S(t) =S 0,1 exp(i 1 t) exp  −t T (1) 2  + S 0 ,2 exp(i 2 t) exp  −t T (2) 2  + S 0,3 exp(i 3 t) exp  −t T (3) 2  . where we...

Ngày tải lên: 14/08/2014, 09:21

... Appellate Bench, 20 00, pp. 557–564; Zaman, 1966, pp. 8– 12. 15 Tabari, n.d., pp. 26 , 27 . 16 University of the Punjab, 1973, 10, p. 1 72. 17 Shariat Appellate Bench, 20 00, pp. 539–543. 28 Understanding ... through compulsion. 2. The maintenance of healthy conditions in the market to ensure its proper functioning. 18 See Holy Qur’ ¯ an, 2: 28 1. 19 See Chapra, 20 00a, pp. 69– 7...

Ngày tải lên: 09/08/2014, 16:21

... intensity 1 2 1 2 D − 1 2  − 1 2 J 12 (1 + sin 2 ) 3–4 1 2 D − 1 2  + 1 2 J 12 (1 − sin 2 ) 1–3 − 1 2 D − 1 2  − 1 2 J 12 (1 − sin 2 ) 2 4 − 1 2 D − 1 2  + 1 2 J 12 (1 + sin 2 ) In this table ... ν 0 ,2 − 1 2 J 13 − 1 2 J 23 5–8 ααβ βββ −ν 0,1 − ν 0 ,2 + 1 2 J 13 + 1 2 J 23 1–7 ααα β αβ −ν 0,1 − ν 0,3 − 1 2 J 12 − 1 2 J 23 2 8 αβ...

Ngày tải lên: 14/08/2014, 09:21

... y; using diagram V: 22 12 2 12 1 2 12 1 12 1 2 II JtII JtI xz JtII xz y zz π ππ → +cos sin The diagrams apply equally well to spin -2; for example −→ +22 12 2 12 1 2 12 2 12 1 2 II Jt II JtI zy JtII zy ... ππ Jt 12 2= i.e. a delay of 1/(2J 12 ). A delay of 1/J 12 causes in- phase magnetization to change its sign: IIIII x JtII t J yz y JtII t J y zz zz 1...

Ngày tải lên: 14/08/2014, 09:21

... () () ( ) 1 32 4 1 1 1 121 1 2 1 324 1 3 2 303 1 1 2 4 02 11 2 2 20 2 1 1 2 403 2 4 2 424 2 3 2 221 () ( ) () ( ) () () () () −−+++ ++ +−− − nWnWnWnWnWn Wn Wn Wn Wn Wn Wn ISI SI SI [ 12] This unpromising ... DQ DQ + () → −+ () + () −+ () + () () + () () 31 22 13 3 1 12 23 1 13 3 1 12 23 1 2 12 1 1 2 23 1 2 3 2 t BtJJt BtJJtI x JtII JtI I x zy zz...

Ngày tải lên: 14/08/2014, 09:21

... () = −       − () −− () [] +− − () +− () 0 12 12 1 exp exp exp exp where RR RRR RRR RRR IISSIS IS IS =− ++ =++ [] =+− [] 22 2 1 1 2 2 1 2 24 σ λλ These definitions ensure that λ 1 > λ 2 . If R I and R S are ... moved and the original pattern is lost. 8 22 γγ 1 2 2 2 12 6 1 r where γ 1 and γ 2 are the gyromagnetic ratios of the two nuclei involved and r 1...

Ngày tải lên: 14/08/2014, 09:21

... tt tttT CPN S i PN i i 12 12 12 1 2 121 222 12 1 12 12 1 2 121 2 ,,, cos exp exp exp ,,, sin exp exp () = () + () = ()( ) − () − () () = () + () [] = ()( ) − ΩΩ ΩΩ (() − () exp tT 22 These cosine and ... S P (t 1 ,F 2 ) StF t tT A D StF StF StF ttTA N NP ii i 12 1 4 1 122 2 12 12 12 1 2 1 122 , * exp exp ,,*, exp exp () = () − () − [] () = () + () = () − () + Ω Ω Tr...

Ngày tải lên: 14/08/2014, 09:21

... simple cycle 0 2 on the first pulse and the receiver. The final 8 step cycle is 1st pulse: 0 1 2 3 2 3 0 1, 2nd pulse: 0 1 2 3 0 1 2 3, 3rd pulse fixed, receiver: 0 0 0 0 2 2 2 2. An alternative ... gradient the observable terms on spin 1 are i 2 1 21 121 2 121 2 22 exp – cos siniIIII xz yz ΩΩ Ω Ω + () () + [] ττ τ [14] where it has been assumed that τ 2 is sufficiently s...

Ngày tải lên: 14/08/2014, 09:21

... I zx JtII zz z yyy yyy zz 2 12 2 123 2 12 2 11 2 2 33 123 123 12 1 2 2 2 2 ΩΩ Ω → → → → → → () ++ () () ++ () π π π 22 4 12 2 12 3 2 1 2 2 23 2 3 23 2 3 12 1 2 13 1 3 II II I I zx JtI ... rotations I II II I I II x tI xz II xz I x tI tI x JtII zy JtII y yy y zz zz zz 1 12 2 12 1 1 2 12 2 11 12 2 11 2 2 12 1 2 12 1...

Ngày tải lên: 14/08/2014, 09:21