Sometimes it is nice to check and see whether a burst detected at some time in the past by some other instrument (like BATSE or HETE) would have produced a counts spectrum with a good signal-to-noise ratio if it had been detected by BAT. Here is a simple way to find out using XSPEC.
See How to Use XSPEC for Simple Tasks for instructions on setting up XSPEC.
You will need three things to find the spectral SNR of BAT for a particular GRB
In the example given here, I use the xspec spectral model reported for GRB 040924 at http://space.mit.edu/HETE/Bursts/GRB040924/, which is "cutoffpl".
XSPEC> model cutoffpl
Model: cutoffpl<1>
Input parameter value, delta, min, bot, top, and max values for ...
1 0.01 -3 -2 9 10
1:cutoffpl:PhoIndex>1.171
15 0.01 0.01 1 50 200
2:cutoffpl:HighECut>62.49
1 0.01 0 0 1E+24 1E+24
3:cutoffpl:norm>62.66
My inputs are printed in boldface.
XSPEC> fakeit none
For fake data, file # 1 needs response file: response.rsp
... and ancillary response file:
Use randomization in creating fake data? (y) n
Input optional fake file prefix (max 4 chars): temp
Fake data filename (tempresponse.fak) [/ to use default]: /
Exposure time, correction norm (1, 1): 150000
The last command specifies the exposure time. In the example we are using, the exposure time is actually 1.5 s, not 150000 s. However, xspec cannot produce fractional counts for its fake data, and BAT's mask-weighted counts can often be fractional. To avoid this problem, I multiplied the exposure time by 100000. Then, when I interpret the fake data, I will divide the counts by 100000.
Net count rate (cts/s) for file 1 5.679 +/- 6.1532E-03 using response (RMF) file... response.rsp Chi-Squared = 3.9088514E-02 using 80 PHA bins. Reduced chi-squared = 5.0764304E-04 for 77 degrees of freedom Null hypothesis probability = 1.00
At this point, you will want to exit from XSPEC:
XSPEC>exit Do you really want to exit? (y)y XSPEC: quit
and run IDL and use it to add up the counts
IDL> a=mrdfits("tempresponse.fak",1)
IDL> e=mrdfits("sample.pha",2)
IDL> print,total((a.counts)[3:78])/100000
6.98308
I added up bins 3 through 78, which correspond to energies from 14 keV through 194.9 keV.
So this means that if BAT had detected this burst on-axis (and if all of the detectors were enabled), it would have detected 6.98 counts (mask-weighted counts, that is).
Next we must determine the signal-to-noise ratio. This is given by
(see Mask-Weight Counts Errors):

If we assume a fairly high background (x=10), then the SNR for the entire energy range considered here comes out to be:
IDL> print,sqrt(total((a.counts)[3:78])/(100000*0.000112674*21))
54.3253
We can also find the SNR for each energy bin.
IDL> snr=sqrt((a.counts)/(100000*0.000112674*21))
IDL> e_min=e.e_min
IDL> e_max=e.e_max
IDL> all=transpose([[e_min],[e_max],[snr]])
IDL> print,all
0.00000 10.0000 13.7690
10.0000 12.0000 14.2822
12.0000 14.0000 15.9593
14.0000 16.0000 16.0748
16.0000 18.0000 15.6993
18.0000 20.0000 15.1241
20.0000 22.0000 14.4560
22.0000 24.0000 13.7455
24.0000 26.0000 12.5864
26.0000 28.0000 11.4598
28.0000 30.1000 10.8953
30.1000 32.1000 10.2406
32.1000 34.2000 9.81475
34.2000 36.3000 9.40959
36.3000 38.3000 8.85064
38.3000 40.4000 8.90467
40.4000 42.5000 8.84873
42.5000 44.6000 8.47873
44.6000 46.8000 8.59826
46.8000 48.9000 7.92266
48.9000 51.1000 7.80388
51.1000 53.2000 7.41225
53.2000 55.4000 7.36678
55.4000 57.6000 7.08397
57.6000 59.8000 6.78597
59.8000 62.0000 6.50650
62.0000 64.2000 6.19540
64.2000 66.4000 5.77014
66.4000 68.7000 5.62325
68.7000 70.9000 5.25695
70.9000 73.2000 5.03815
73.2000 75.4000 4.65172
75.4000 77.7000 4.49602
77.7000 80.0000 4.23163
80.0000 82.3000 3.98367
82.3000 84.6000 3.79013
84.6000 87.0000 3.64808
87.0000 89.3000 3.45654
89.3000 91.7000 3.26541
91.7000 94.0000 2.99610
94.0000 96.4000 2.95204
96.4000 98.8000 2.81050
98.8000 101.200 2.58163
101.200 103.600 2.42287
103.600 106.000 2.29936
106.000 108.400 2.19209
108.400 110.900 2.15711
110.900 113.300 1.92082
113.300 115.800 1.83415
115.800 118.200 1.68900
118.200 120.700 1.63561
120.700 123.200 1.54116
123.200 125.700 1.45075
125.700 128.300 1.39430
128.300 130.800 1.28384
130.800 133.300 1.21274
133.300 135.900 1.17018
135.900 138.400 1.07610
138.400 141.000 1.03608
141.000 143.600 0.970803
143.600 146.200 0.907812
146.200 148.800 0.829989
148.800 151.400 0.790878
151.400 154.100 0.771948
154.100 156.700 0.712146
156.700 159.400 0.691063
159.400 162.000 0.640272
162.000 164.700 0.609846
164.700 167.400 0.551626
167.400 170.100 0.524125
170.100 172.800 0.507742
172.800 175.500 0.482125
175.500 178.200 0.450401
178.200 181.000 0.416266
181.000 183.700 0.384603
183.700 186.500 0.384603
186.500 189.300 0.361959
189.300 192.100 0.325049
192.100 194.900 0.290732
194.900 6553.60 0.903144
The SNR is high for the low-energy bins (16 in the 14-16 keV bin), decreasing to about 3 in the 91.7-94 keV bin, and about 1 in the 138.4-141 keV bin. All in all, the BAT spectrum for this particular burst would have a good SNR, and the peak in the spectral model at 62.49 keV should be easy to detect.