Chapter 5 How the spectrometer works

MagnetProbe

Transmitter

Synthesizer Receiver 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 0.1/9x108 ~ 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.

Synthesizer

ReceiverPulse

programmer

TransmitterMagnetProbeDeuterium Lock System:

Phase sensitive detection:

Cos1txCos2t = 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 - 2to the magnet to compensate for the drift

2H Lock system

The magnet

The Probe

LCo

1

Tuning (To tune the probe to a desired frequency):

Matching (To maximize the power delivered to the coil):

Make RL = 50 TransmitterRL

Quality factor (Q):

oQ

2

2

)4(

)/(

RkT

VVQ

N

S Cso

Sensitivity:

Johnson noise (Thermal noise) : Vrms = (4kTR)1/2

The Spectrometer:

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

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

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)

Receiver: The part of spectrometer that detect and Amplifies signal (Low power, uV) Need to amplify by a factor of 106

120 dB amplificationThe 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 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. 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/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

1.0 V

- 1.0 V

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.

“+”

“-”

Quadrature detection: Mixing of two RF signals, Cosot and Cosrxt, where o is the Larmor frequency and rx is the reference frequency, we obtain:

ACos ot x Cosrxt = ½A[Cos(o + rx)t + Cos(o - rx)t] -------- (1)

Similarly, mixing of Cosot with -Sinrxt we obtain:

ACos ot x (-Sinrxt) = 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.

Cosot

CosotCost CosotSint

Cos(o-)t Sin(o-)t

Cos(o-)t+iSin(o-)t

FT

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.

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:

)( HMdt

Md

1T

MM

dt

dM ozz ;

2T

M

dt

dM xx ;2T

M

dt

dM yy

121

)()()()(

T

MMk

T

jMiMHMHM

dt

Md ozyxo

precession perturbation Relaxation

;1

1 T

MMMH

dt

dM ozy

z

;)(2

1 T

MMHM

dt

dM yzxo

y

2

)(T

MM

dt

dM xyo

x

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:

2121

2222

222

)(1

)(1

TTHT

TMM

o

ooz

2121

2222

21

)(1 TTHT

THMM

ooy

2121

2222

221

)(1

)(

TTHT

THMM

o

oox

My

Mx

(Absorption)

(Dispersion)

For a small H1 field, i.e. 2H12T1T2 <<1 we have:

222

21 )(1

)(

o

oy T

TMHM

222

221

)(1

)(

o

oox T

THMM(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

;1

1B

Bz

A

Az

A

Ao

AzA

y

Az MM

T

MMMH

dt

dM

;)(2

1B

By

A

Ay

A

AyA

zAxA

Ay MM

T

MMHM

dt

dM

;)(

2 B

Bx

A

Ax

A

AxA

yA

Ax MM

T

MM

dt

dM

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): 22'2

'2

1 )()(1 AoAA

AoA

Ay T

TMPHM

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.

BA

AAP

BA

BBP

II. Fast Exchange (A,B << 1/(A - B):

A single peak at = PAA + PBB will be observed and

22'2

'2

1 )()(1

BBAAA

AoAy PPT

TMPHM

B

B

A

A

T

P

T

P

T 222

1

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

2222

2

1

)()(})(21{

)(

4

1

BABA

BAoy MHM