1. Chapter 5 How the spectrometer works Magnet Probe Transmitter SynthesizerReceiver ADC Pulse programmer Computer 5.1 The magnet: 1. Strength: 14.1 T for 60-0 MHz and 22 T for 900 MHz Limitation: wire strength, triple points, material (Nb-T) 2. Homogeneity: 0.1 Hz at 900 MHz  0.1/9x10 8 ~ 10 -10 cm -3  shimming (30 shim sets, spherical harmonic function) 3. Stability: Drift rate 2 Hz/hr in out 800 MHz system Deuterium lock: Phase locked loop 4. Helium consumption. Synthesizer Receiver Pulse programmer Transmitter Magnet Probe Deuterium Lock System: Phase sensitive detection: Cos  1 txCos  2 t = Sin(  1 -  2 )t + Sin(  1 +  2 )t = Sin(  1 -  2 ) ~  1 -  2 Output a negative current which is proportional to the difference in phase, i.e. I   1 -  2 to the magnet to compensate for the drift 2 H Lock system

2. The magnet

3. The Probe Tuning (To tune the probe to a desired frequency): Matching (To maximize the power delivered to the coil): Make R L = 50  Transmitter RLRL Quality factor (Q):   Sensitivity: Johnson noise (Thermal noise) : V rms = (4kTR  ) 1/2

4. The Spectrometer: Power level (Decibels): dB = 10log(P out /P in ) = 20xlog(V out /V in )  “+” increase power, “-” decrease power  10 dB change the power by a factor of 10  3 db change power of 2, 6 dB by a factor of 2x2 = 4, 9 dB by a factor of 2x2x2 = 8  20 dB change voltage by a factor of 10 but power by a factor of 100  6 dB change voltage by a factor of 2 but power by a factor of 4 Transmitter: The part of spectrometer that delivers radio frequency power to the probe (High power, 100W ). Cross-diode (Diplexer): Permit only high power to pass (Block high power noise). XMTR Probe RCVR I V Diode I-V curve

5. Power level and pulse width: How many dB do you have to use for increasing the pulse width by a factor of 2 (Assuming a linear amplifier is use, class C amplifier) ?. Pulse width  i  (power)1/2 Increase pulse width by a factor of 2 need to decrease power by 4.  dB = 10 log 4 ~ - 6 dB In general: Power ratio in dB = 20log 10 (  initial /  new ) If the current 90 o pulse is  i = 10 us how do we adjust attenuator in order to get a 90 o pulse of  new = 8 us ? Ans: dB = 20 log (10/8) = 1.9 dB  One needs to reduce the attenuator by 1.9 dB, i.e. if the initial attenuation was set at 12 dB then for getting a 8 us 90 o pulse the new attenuation should be set at 10.1 dB. Phase: X-pulse (0 o phase-shifted, 90 X, a cosine wave) Y-pulse (90 o phase-shifted, 90 Y, a sine wave)

6. Receiver: The part of spectrometer that detect and Amplifies signal (Low power, uV)  Need to amplify by a factor of 10 6  120 dB amplification  The first stage of amplification is the most important  Preamplifier (Preamp) determines the receiver S/N ratio.  Broad band GaAs amplifier is used (Noise figure ~ 1.04dB) Digitizing the signal (Analog to Digital Converter, ADC): A device which convert analog signal voltage into digital numbers. Factors to be considered in choosing a ADC: 1. Resolution (How many “bits“): A n-bit ADC divide the full analog voltage into 2 N divisions. A 10-bit ADC convert the 1.0 V signal into 2 10 = 1024 division  Minimum signal that can be detected = 1/1024 ~ 1 mV.  Signal below 1 mV will be treated as noise.  Set receiver gain as high as possible without saturation. 2. Speed (Sampling rate):  Nyquist theorem: One need at least two points per cycle to correctly represent a sinusoidal wave.  Sampling rate must be at least twice the spectral width to be covered.  Dwell time  1/f max. For example to observe a signal which resonates at 1 kHz one needs to digitize at 2 kHz rate or DW = 1/2000 = 0.5 ms. But since in quadrature detection one can see both +f max and –f max one is able to observe 2f max range. 1.0 V - 1.0 V 1.0 V - 1.0 V

7. What happens if the resonance fall outside the range ? Ex: We digitize At 1 kHz but the signal resonate at 1.2 kHz ?  Fold over (Aliasing): If a peak occurs at f max + F then it will resonate at –f max + F. Thus, in this case f max = 1 kHz and F = 200 Hz, thus it will resonate at -1,000 + 200 = -800 Hz Mixing down to a low frequency (Mixer): Quadrature detection: Detect both X- and y-components of signal in order to Differentiate “+” and “-” frequencies. “+” “-”

8. Quadrature detection: Mixing of two RF signals, Cos  o t and Cos  rx t, where  o is the Larmor frequency and  rx is the reference frequency, we obtain: ACos  o t x Cos  rx t = ½A[Cos(  o +  rx )t + Cos(  o -  rx )t] -------- (1) Similarly, mixing of Cos  o t with -Sin  rx t we obtain: ACos  o t x (-Sin  rx t) = ½A[-Sin(  o +  rx )t + Sin(  o -  rx )t] ----- (2) After low pass filter only the low frequency component is detected. Thus, we see only Cos(  o -  rx ) for eq. 1 and Sin(  o -  rx ) for eq. 2.  By shifting the receiver phase by 90 o we can detect either X- or Y-component of the signal. If we then recombine eq. 1 and 2 we obtain: Signal = ½A[Cos (  o +  rx )t + Sin (  o -  rx )t ] = ½ A exp[-i (  o +  rx )t]  We can differentiate whether (  o -  rx ) is “+” or “-”.  It increases sensitivity by a factor of (2) 1/2 or 1.414. Dwell time (1/digitization rate) and spectral width: For obtaining a spectral width, f sw (or from - ½f sw to ½f sw ) one need to digitize at the same frequency with a dwell time  = 1/f sw.

9. Cos  o t Cos  o tCos  t Cos  o tSin  t Cos(  o -  )t Sin(  o -  )t Cos(  o -  )t+iSin (  o -  )t FT I. II. III. IV. I.Single channel detection. II.Quadrature detection but FT. III. Qua detection after low pass filter (Separate FT) IV. Quadrature detection and combined FT.

10. Bloch equation and Chemical Exchange: In the absence of relaxation: According to Bloch, the effect of relaxation can be approximated as exponential as follow: Or In the rotating frame (rotating at  wrt the Z-axis): We have: precession perturbation Relaxation

11. Solution to the Bloch equation: Under steady state condition: dM x /dt =dM y /dt = dM z /dt = 0 we can solve the equations and the following results: MyMy MxMx (Absorption) (Dispersion) For a small H 1 field, i.e.  2 H 1 2 T 1 T 2 <<1 we have: (Lorentzian) Two-site Chemical Exchange: A  B For a spin exchange in two magnetically different environments A and B having chemical shift  A /2  and  B /2  and life time  A and  B. Bloch equations become: 1/  A 1/  B

12. Similar equations for site B. Thus, we need to solve 6 simultaneous equations. Under steady state condition, i.e. dM i /dt=0 for all six magnetizations for the following conditions: I. Slow exchange (  A,  B >> 1/(  A -  B ): Where P A is the probability of finding the spin in state A, thus and T 2A ’ is the effective transverse relaxation time determined from the lineshape and is related to the relaxation time in the absence of exchange by 1/T 2A ’ = 1/T 2A + 1/  A. A similar relationship for the spin in site B having a peak at  B. II. Fast Exchange (  A,  B << 1/(  A -  B ): A single peak at = P A A + P B B will be observed and

13. III. Intermediate Exchange (  A,  B  1/(  A -  B ): : If (a) P A = P B = ½, (b)  =  A  B /(  A +  B ) =  A /2 =  B /2 and 1/T 2A = 1/T 2B  0 then