Chapter 5 How the spectrometer works

1. Chapter 5 How the spectrometer works Magnet Probe 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  0.1/9x108 ~ 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. Transmitter Synthesizer Receiver Computer ADC Pulse programmer Magnet Probe Synthesizer Transmitter Deuterium Lock System: Phase sensitive detection: Cos1txCos2t = 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 Pulse programmer Receiver 2H 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 RL = 50 RL Transmitter  Quality factor (Q):  Sensitivity: Johnson noise (Thermal noise) : Vrms = (4kTR)1/2

4. The Spectrometer: 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). I Probe XMTR RCVR V Diode I-V curve • Power level (Decibels): • dB = 10log(Pout/Pin) = 20xlog(Vout/Vin) • “+” 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

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 = 20log10(initial/new) If the current 90o 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 90o pulse the new attenuation should be set at 10.1 dB. Phase: X-pulse (0o phase-shifted, 90X, a cosine wave) Y-pulse (90o phase-shifted, 90Y, a sine wave)

6. Receiver: The part of spectrometer that detect and • Amplifies signal (Low power, uV) •  Need to amplify by a factor of 106 •  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) 1.0 V - 1.0 V 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: Resolution (How many “bits“): A n-bit ADC divide the full analog voltage into 2N divisions. A 10-bit ADC convert the 1.0 V signal into 210 = 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. 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/fmax. 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 +fmax and –fmax one is able to observe 2fmax range. 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 fmax + F then it will resonate at –fmax + F. Thus, in this case fmax = 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ot and Cosrxt, where o is the Larmor frequency and rx is the reference frequency, we obtain: ACos ot x Cosrxt = ½A[Cos(o + rx)t + Cos(o - rx)t] -------- (1) Similarly, mixing of Cosot with -Sinrxtwe obtain: ACos ot x (-Sinrxt) = 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 90o 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, fsw (or from - ½fsw to ½fsw) one need to digitize at the same frequency with a dwell time  = 1/fsw.

9. FT I. Cosot CosotCost CosotSint II. Cos(o-)t Sin(o-)t III. FT IV. Cos(o-)t+iSin(o-)t • Single channel detection. • Quadrature detection but FT. • Qua detection after low pass filter (Separate FT) • 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: Relaxation perturbation precession

11. Solution to the Bloch equation: Under steady state condition: dMx/dt =dMy/dt = dMz/dt = 0 we can solve the equations and the following results: My (Absorption) Mx (Dispersion) For a small H1 field, i.e. 2H12T1T2 <<1 we have: (Lorentzian) 1/A 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/B

12. Similar equations for site B. Thus, we need to solve 6 simultaneous equations. Under steady state condition, i.e. dMi/dt=0 for all six magnetizations for the following conditions: I. Slow exchange (A,B >> 1/(A - B): Where PA is the probability of finding the spin in state A, thus and T2A’ is the effective transverse relaxation time determined from the lineshape and is related to the relaxation time in the absence of exchange by 1/T2A’ = 1/T2A + 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  = PAA + PBB will be observed and

13. III. Intermediate Exchange (A,B  1/(A - B): : If (a) PA = PB = ½, (b)  = A B/(A + B) = A/2 = B/2 and 1/T2A = 1/T2B  0 then