Chapter 5 How the spectrometer works

1
Chapter 5 How the spectrometer works
Magnet Probe
Transmitter
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
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
½ACos(?o ?rx)t Cos(?o - ?rx)t --------
(1) Similarly, mixing of Cos?ot with -Sin?rxt we
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 ½ACos (?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
Cos?ot
I.
Cos?otCos?t
Cos?otSin?t
II.
Cos(?o-?)t
Sin(?o-?)t
III.
FT
IV.
Cos(?o-?)tiSin(?o-?)t

1. Single channel detection.
2. Quadrature detection but FT.
3. Qua detection after low pass filter (Separate
   FT)
4. 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 ltlt1 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/dt0 for all six
magnetizations for the following conditions I.
Slow exchange (?A,?B gtgt 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 ltlt 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
14
(No Transcript)
15
(No Transcript)