converting to spinev from simpson

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Vivien Yeh

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Sep 26, 2013, 10:26:24 PM9/26/13
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Hi,

I'm currently using simulation to study magnetisation transfer. I've been using SIMPSON but for a system of 6 spins (3x13C, 1x15N and 2x1H) my laptop simply could not compute, so I thought I would try to transfer to SPINEVOLUTION for bigger spin system. It's quite a simple pulse sequence, but in SPINEV the FID pattern appeared to be 2Dish. I feel like I'm missing something. Here is the SIMPSON pulse program for just 2 channels:

reset
for {set i 0} {$i < 3} {incr i} {
  set phase [expr $i*360.0/3.0]
  pulse $t90 $pwr $phase 0 0 
  pulse [expr 4.0*$t90] $pwr [expr $phase+180.0] 0 0 
  pulse [expr 3.0*$t90] $pwr $phase 0 0 
}
store 1
reset
  pulseid $th $par(rf) 0 $par(rf) 90 
  store 2
for {set j 0} {$j < $par(np)} {incr j} {
  reset
  prop 1 $j
  prop 2
  prop 1 $j
  acq
}

Here is my interpretation of it in SPINEVOLUTION:

CHN 1
timing(usec)      (2.5 10   7.5  2.5 10  7.5  2.5 10 7 .5)40D1B (0.5)1D1A (2.5 10  7.5 2.5 10  7.5 2.5 10  7.5)40D1B 
power(kHz)        100 100 100 100 100 100 100 100 100 500 100  100 100 100 100 100 100 100 100
phase(deg)        0 180  0    120 300 120 240 60   240 0 0 180 0   120 300 120 240 60  240
freq_offs(kHz)    0 0     0     0    0   0 0 0 0 0 0 0 0  0   0 0 0  0   0
CHN 2
timing(usec)      (60) (0.5) (60)
power(kHz)        0 500 0
phase(deg)        0 90 0
freq_offs(kHz)    0 0 0

From what I understand of it, the spinevolution pulse sequence should be
A
BAB
BBABB
BBBABBB
...
like stated in SIMPSON. but I this surely wouldn't give an oscillating FID.

The initial magnetisation is in channel one and we observe nuclei in channel two. If someone could have a look and tell me what I've been missing much would be appreciated. I've been using SPINEV for years but only for very minor applications such as echo, angles. This pulse should be similar to REDOR and somehow I just couldn't quite figure out what is wrong. I'm sure there is something very simple explanation...

Thanks for all your help.

Regards,
Vivien

Mikhail Veshtort

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Sep 27, 2013, 2:13:07 AM9/27/13
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Hi Vivien,

I don't remember the SIMPSON API any more, so I can't read your SIMPSON code very well, but if the sequence you want is indeed

A
BAB
BBABB
BBBABBB

then your timing line in spinev should look something like

(B)40 A (B)40

or (which is exactly the same)

(B)40D1 (A)1D0 (B)40D1


Best regards,
Mikhail


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Vivien Yeh

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Sep 27, 2013, 5:32:39 AM9/27/13
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Dear Mikhail,

Thank you for your prompt reply. I initially tried (B)40 A (B)40 but I wasn't entirely sure it was correct. However, if the pulse sequence is correct, I don't really know why I'm getting such signal:
0
0.002897578
-0.003400668
-0.02337859
0.02532581
0.0511903
-0.07571747
-0.06549128
0.1508096
0.04646867
-0.2335118
0.01617202
0.2991045
-0.1182818
-0.3239836
0.2417557
0.2940105
-0.3600909
-0.2093749
0.4467363
0.08433725
-0.4830073
0.05766372
0.4627716
-0.1913748
-0.3925788
0.2966484
0.287796
-0.3625481
-0.1667657
0.3876563
0.04546042
-0.37725
0.06548083
0.3391564
-0.1607597
-0.280297
0.2382541
0.2052731
-0.2962877
Since SIMPSON file produced (expectedly) a smooth curve which peak about 50% of total magnetisation. I thought perhaps it was because SPINEVOLUTION acquire at strange times (once a the first B pulse and again at the second B pulse?) when I want it to acquire only at the end of each complete cycle, but I'm not entirely sure how to correct that, if that is the problem at all.

Regards,
Vivien

Mikhail Veshtort

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Sep 27, 2013, 7:05:07 AM9/27/13
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Hi Vivien,

If you are getting strange-looking results, there must be some other problem with your input file, but it is apparently not what you are thinking. The pulse sequence specified as
(B)40 A (B)40
produces the following acquisition pattern:
A (ACQ)
BAB (ACQ)
BBABB (ACQ)
BBBABBB (ACQ)

which seems to be just what you want.

Best,
Mikhail

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