After the 90 H pulse at the end of t, we get -Iiz cos &2,tI cos IJt +lly sin Q2,tI cos IJt +21Ivl2x cos &2,tI sin TJt:+ 211,2x sin Q2,tI sin Jt1 polarization DQC+zQC SO The phase cycle is set up so that only coherences with coherence order zero during tm are selected i.e., the first term(polarization)and the zQc part of the third term( I1yl2x-211x2y) The first term is a z magnetization which is modulated with the cosine of the chemical shift(and coupling), i.e., it oscillates between +liz and -1lz. This non-equilibrium polarization will cause ar NOE on neighbouring protons I1z cos a,t cos TJt1->-(1-k12)11z cos Q2,tr cos Tt: -k12 12z cos 22,t cos TTr The amount of NoE transfer is, of course, determined by the cross-relaxation rate 12 and the duration of tm. After the final 'H 90 pulsethese z components are converted into detectable signal - (1-k12IIx cos Q2,tI cos TJt -k12 l2x cos 2,tI cos TTr The first term results in a diagonal peak after 2D FT(at Q2,), the second one in a cross-peak at Q2, in F1(from the tI modulation) and at Q2, in F2 (since it is an 2 coherence now ) The sign of the cross peak depends on the sign of the cross-relaxation rate, i.e., the dominance of either Wo or W2(the diagram shows the maximum NOE cross-peak intensity relative to the diagonal system) 05 NOE with a negative NOE effect, the cross-peaks will 025 be negative with a negative diagonal signal (like in the Id difference NOE experiment!) 00 for a positive NOE effect, one gets positive cross-peaks(with a negative diagonal) T.,0 for ootes l, the NOE effect will disappear, and no NOESY cross-peaks will be visible100 After the 90°y 1H pulse at the end of t1 we get –I1z cos W1 t1 cos pJt1 + I1y sin W1 t1 cos pJt1 + 2I1yI 2x cos W1 t1 sin pJt1 + 2I1zI 2x sin W1 t1 sin pJt1 polarization SQC DQC+ZQC SQC The phase cycle is set up so that only coherences with coherence order zero during tm are selected, i.e., the first term (polarization) and the ZQC part of the third term (= I1yI 2x – 2I1xI 2y). The first term is a z magnetization which is modulated with the cosine of the chemical shift (and coupling), i.e., it oscillates between +I1z and –I1z . This non-equilibrium polarization will cause an NOE on neighbouring protons: –I1z cos W1 t1 cos pJt1 ¾® – (1-k12) I1z cos W1 t1 cos pJt1 – k12 I2z cos W1 t1 cos pJt1 The amount of NOE transfer is, of course, determined by the cross-relaxation rate s12 and the duration of tm . After the final 1H 90° pulsethese z components are converted into detectable signals: ¾® – (1-k12) I1x cos W1 t1 cos pJt1 – k12 I2x cos W1 t1 cos pJt1 The first term results in a diagonal peak after 2D FT (at W1 ) , the second one in a cross-peak at W1 in F1 (from the t1 modulation) and at W2 in F2 (since it is an I2 coherence now). The sign of the cross￾peak depends on the sign of the cross-relaxation rate, i.e., the dominance of either W0 or W2 (the diagram shows the maximum NOE cross-peak intensity relative to the diagonal system): - with a negative NOE effect, the cross-peaks will be negative with a negative diagonal signal (like in the 1D difference NOE experiment!) - for a positive NOE effect, one gets positive cross-peaks (with a negative diagonal). - for w0 tc » 1, the NOE effect will disappear, and no NOESY cross-peaks will be visible