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Basics Concepts Nuclear spins Magnetic field B 0 Energy Sensitivity The NMR transition Larmor Frequency Magnetic field B 1 The rotating frame A pulse!

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Presentation on theme: "Basics Concepts Nuclear spins Magnetic field B 0 Energy Sensitivity The NMR transition Larmor Frequency Magnetic field B 1 The rotating frame A pulse!"— Presentation transcript:

1 Basics Concepts Nuclear spins Magnetic field B 0 Energy Sensitivity The NMR transition Larmor Frequency Magnetic field B 1 The rotating frame A pulse!

2 The energy of the NMR Transition Sensitivity B0B0 E m=+1/2 m=-1/2  E=h 0  NUCLEUS B 0 MAGNETIC FIELD Larmor Frequency The two Zeeman level are degenerate at B 0 =0  E=  (h/2  )B 0 E= - B 0

3 d  /dt=  (t)^B 0 (t)

4 B0B0 M0M0 I=1/2 E =  B 0 E =   mB 0  :m = +½    +½   E   ½   B 0  :m =  ½    ½   E   ½   B 0 M 0 =  = M z M x = M y = 0  E =   B 0 =     h 

5  0 =2    0 =    0  0  Mz=-1/2 h  /2  E=-  B 0 = -  hm z B 0 /2   E=  hB 0 /2  STILL, NO NMR EXPERIMENT Mz=1/2 h  /2 

6 B 1 (t)=  B 1  cos  1 t B 1 is a radiofrequency transmitter A pulse!

7 dM(t)/dt= M(t)^B(t) B(t)= B 0 + B 1 (t) dM(t)/dt= M(t)^B 0 + M(t)^B 1 (t) = M(t)^B 0 + M(t)^|B 1 | cos  1 t (t) Double precession A pulse! Precession around B 0 (z axis) Precession around B 1 (axis defined in the xyplane and rotating at speed  1)

8 Laboratory Frame Nuclear frequency 1= precession frequency of magnetic field B 1 A pulse!

9 The Rotating frame X’,y’,z’ =laboratory frame X,y,z,=rotating frame (rotating at the frequency 1) In the rotating frame,there is no frequency precession for  and the radifrequency B1 is seen as a static magnetic field The static magnetic field B 0 is not observed in the rotating frame

10 Laboratory Frame jam Fly A (Laboratory Frame) Fly B The movement of Fly B as seen by Fly A Rotating Frame Fly B jam The movement of Fly B as seen by Fly A Fly A (Rotating frame)

11 Precession in the laboratory frame  0 dM/dt=M^  BdM/dt=M^  (B-    ) L.F. R.F. at freq.   If  =  0 dM/dt=0 If    0 dM/dt=M^  B 1 =    If  =  0 +B 1 dM/dt=M^  (B 0 +B 1 -  0  ) Rotation! dM/dt=M^  (B +B 1 -  0  ) dM/dt=M^  (B 1 + (  0  )) dM/dt=M^  (B 1 + (  )) B1B1  0

12 Precession in the laboratory frame  0 M xy from any nuclear spin not exactly on resonance, will also precess in the x’y’ plane at the difference frequency  0.

13 Basics (II) Free Induction Decay (FID) Fourier transformation A pulse

14 Free Induction Decay (FID) Observed NMR signal in the time domain Resonance frequencies are acquired as a function of time Common case of observed FIDs t tt

15 What happens? Relaxation. Magnetization disappear from the xy plane because the system goes back to the equilibrium. The observed signal is always an Exponential DECAY. Chemical shift precession. Different spins may have a different resonance frequency. When the resonance frequency is different from that of the field B 1, the signal rotates on the xy plane, with a precession ferquency 1 - 0

16 Thank you, Mr. Fourier! F( ) F(t)

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20 FOURIER TRANSFORMATIONS F( )=  ( 0 ) F( )=A  (sin  )/  centered at 0 F( )=T 2 /1+(2  T 2 ) 2 -i 2  (T 2 ) 2 /1+(2  T 2 ) 2 0 F( )=T 2 /1+(2    T 2 ) 2 -i 2    (T 2 ) 2 /1+(2    T 2 ) 2 0 F(t)=exp(-t/T 2 ) F(t)=exp(-t/T 2 )exp(i2  A )

21 Why bother with FT? FT allows to decompose a function in a sum of sinusoidal function (deconvolution). In NMR FT allows to switch from the time domain, i.e. the signal emitted by the sample as a consequence of the radiofrequency irradiation and detected by the receiving coil to the frequency domain (NMR spectrum) The FT allows to determine the frequency content of a squared function

22 A “real” F.I.D.

23 Excitation pulses A single resonance at Larmor frequency 1= excitation frequency 0 (precession of the rotating frame) Transmitter B 1 A Pulse B1 switched off B1 on. A pulse! The My magnetization is observed by the receiver coil

24 Received signal The received signal 1 is compared with the excitation frequency 0 The resulting signal has observed frequency =0 During acquisition time the signal relaxes (T2) My=exp(t/T2) Time (t) Fourier Transformation Frequency ( ) =0

25 Excitation pulses A single resonance at Larmor frequency 1different from excitation frequency 0 (precession of the rotating frame) Transmitter B1 Pulse B1 switched off B1 on. A pulse! The My magnetization is observed by the receiver coil

26 Received signal The received signal 1 is compared with the excitation frequency 0 The resulting signal has observed frequency obs=( 1- 0) During acquisition time the signal relaxes (T2) My=cos( obs )texp(t/T2) Time (t) Fourier Transformation Frequency ( ) My=cos( obs )texp(t/T2) =0 = obs

27 FT relax. PreparationDetection x y z t2t2 00

28 Typical 1 H NMR Spectrum Absorbance

29 Protein 1 H NMR spectrum: a “real spectrum” Fourier Transformation The NMR signal in the time domain Free Induction Decay A short pulse will excite all spins All spins will relax (all together) during time AQ The FT of FID gives the NMR spectrum


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