basics concepts nuclear spins magnetic field b 0 energy sensitivity the nmr transition larmor...
Post on 22-Dec-2015
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Basics Concepts
• Nuclear spins
• Magnetic field B0
• Energy
• Sensitivity
• The NMR transition
• Larmor Frequency
• Magnetic field B1
• The rotating frame
• A pulse!
The energy of the NMR Transition
• Sensitivity
B0
E
m=+1/2
m=-1/2
E=h0
NUCLEUS
B0 MAGNETIC FIELD
Larmor Frequency
The two Zeeman level are degenerate at B0=0
E=(h/2)B0
E= -•B0
B0
M0
I=1/2 E = B0
E = mB0
: m = +½
+½
E½B0
: m = ½½
E½B0
kT
E
eP
P
M0 = = Mz
Mx = My = 0
E = B0 = h
dM(t)/dt= M(t)^B(t) B(t)= B0 + B1(t)dM(t)/dt= M(t)^B0 + M(t)^B1 (t) = M(t)^B0 + M(t)^|B1| cos1t (t)
Double precession
A pulse!
Precession around B0 (z axis)Precession around B1 (axis defined in the xyplane and rotating at speed 1)
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 B0 is not observed in the rotating frame
Laboratory Frame
jam
Fly A (Laboratory Frame)
Fly B The movement of Fly B as seen by Fly A
Rotating Frame
Fly Bjam
The movement of Fly B as seen by Fly A
Fly A (Rotating frame)
Precession in the laboratory frame
0
dM/dt=M^B dM/dt=M^(B-)
L.F.R.F. at freq.
If = 0 dM/dt=0
If 0
dM/dt=M^B1 =
If = 0+B1
dM/dt=M^(B0 +B1 -0)
Rotation!
dM/dt=M^(B +B1 -0)
dM/dt=M^(B1+ (0))
dM/dt=M^(B1+ ())
B1
0
Precession in the laboratory frame
0
Mxy from any nuclear spin not exactly on resonance, will also precess in the x’y’ plane at the difference frequency 0.
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 t t
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 B1, the signal rotates on the xy plane, with a precession ferquency 1-0
FOURIER TRANSFORMATIONS
F()=(0)
F()=A(sin)/ centered at 0
F()=T2/1+(2T2)2 -i 2(T2)2/1+(2T2)20
F()=T2/1+(2T2)2 -i 2(T2)2/1+(2T2)20
F(t)=exp(-t/T2)
F(t)=exp(-t/T2)exp(i2A)
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
Excitation pulses• A single resonance at Larmor frequency 1= excitation
frequency 0 (precession of the rotating frame)
Transmitter B1
A Pulse
B1 switched off
B1 on. A pulse!
The My magnetization is observed by the receiver coil
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
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
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