SVT Implications for IST

This document attempts to find an understanding between what is seen in numbers from reconstruction by Yuri (occupancies and signal-to-noise plots) and from simulations of the IST by Howard (analysis).

Gene Van Buren - BNL
19 Jan 2007

IST detector elements

Pad sizes are 0.192 cm x 0.12 cm = 0.0231 cm^2.
Strip sizes are 0.0064 cm x 3.84 cm = 0.0246 cm^2.
Assume an approximate area of 0.024 cm^2 for either element.
Assume an efficiency of 0.8 for a hit actually being reconstructed in the silicon (this is NOT the hit matching effiency, and it is not needed to correct the probabilities for it because occupancy already includes this efficiency).

The detector element size is not actually the area in which tracking will look for matching hits, but as an exercise, let's calculate a few quantities:

(A)
Occupancy (hits/cm^2)		 (B)
Probability of
at least one hit
in an element		 (C)
Probability of
at least two hits
in an element		 (D)
Probability of
at least two hits
in an element
given that
we have >=1 hit		 (E)
Average number
of hits
in an element
given that
we have >=1 hit
(Signal+Noise)		 (F)
Average number
of hits
in an element
given that
we have >=1 hit
weight by hits
(Signal+Noise
seen by tracks)		 (G)
Signal/Noise		 (H)
Signal/Noise
with
inefficiency
0.030000 0.000720 0.000000 0.000360 1.000360 1.000720 1388.889 3.985652
0.100000 0.002397 0.000003 0.001200 1.001200 1.002400 416.66667 3.952569
0.300000 0.007174 0.000026 0.003596 1.003604 1.007200 138.88889 3.861004
1.000000 0.023714 0.000283 0.011952 1.012048 1.024000 41.66667 3.571429
3.000000 0.069469 0.002471 0.035568 1.036432 1.072000 13.88889 2.941176
10.00000 0.213372 0.024581 0.115205 1.124795 1.240000 4.166667 1.818182

These were calculated using Poissonian distributions as follows: 

    mu = occ*area;
(A) res[0] = occ;
(B) res[1] = 1-TMath::PoissonI(0,mu);
(C) res[2] = 1-TMath::PoissonI(0,mu)-TMath::Poisson(1,mu);
(D) res[3] = (1-TMath::PoissonI(0,mu)-TMath::Poisson(1,mu))/(1-TMath::PoissonI(0,mu));
    //sum=0;for(i=1;i<100;i++) sum+= i*TMath::PoissonI(i,mu);  //definition of mu
(E) res[4] = mu / (1-TMath::PoissonI(0,mu));
    sum2=0;for(i=1;i<100;i++) sum2+= i*i*TMath::PoissonI(i,mu);
    //sum=0;for(i=1;i<100;i++) sum+= i*TMath::PoissonI(i,mu);  //definition of mu
(F) res[5] = sum2 / mu;
(G) res[6] = 1.0/(res[5]-1.0);
(H) res[7] = eff/(res[5]-eff);
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The signal+noise must be weighted by the hits as there is a correlation between the two: multiple tracks in one area will each see the elevated occupancy due to the fact that they are in the same area. However, it is worthwhile to note here two mistakes I have made for which I have not yet determined exactly how to correct:
1. I have weighted by hits, while what I really want is to weight by tracks (which should be something like hits/eff, but I'm not sure exactly where in the math this goes.)
2. I should have calculated signal+noise for the condition that I expect a hit, not that I have a hit. Again, this is because I know that I have a track and therefore expect a hit and have looked at the detector, but I do not necessarily have that hit in the detector.
Both mistakes make my signal-to-noise slightly worse than it should be.

NOTE: Before anyone gets too confused, I've figured out that my math for no inefficiency is actually exactly the same as Howard's formula. The hit-matching efficiency essentially comes out to be 1/(1+mu), where mu is the expected number of hits <n> in an area defined by the ellipse of area 2pi sigma_x sigma_y. My math got there a little more complicated in that I determined signal+noise = <n^2>/<n> , so signal/(signal+noise) = 1/(signal+noise) = <n>/<n^2> = mu / (mu + mu^2) = 1 / (1+mu).

Anyhow, we do learn that a moderate inefficiency for hit reconstruction takes a big hit on the signal-to-noise ratio (the hit-matching efficiency = signal/signal+noise cannot exceded the hit reconstruction efficiency). This means that any simulation of the IST performance must include the hit reconstruction efficiency if it is not very close to 1.0.

SVT detector elements

Let's repeat the exercise for Yuri's effective SVT elements by assuming:

area = 3 cm x 0.15 cm = 0.45 cm^2, where 0.15 cm is approximately the search window in r-phi.
efficiency = 0.7.

(A)
Occupancy (hits/cm^2)		 (B)
Probability of
at least one hit
in an element		 (C)
Probability of
at least two hits
in an element		 (D)
Probability of
at least two hits
in an element
given that
we have >=1 hit		 (E)
Average number
of hits
in an element
given that
we have >=1 hit
(Signal+Noise)		 (F)
Average number
of hits
in an element
given that
we have >=1 hit
weight by hits
(Signal+Noise
seen by tracks)		 (G)
Signal/Noise		 (H)
Signal/Noise
with
inefficiency
0.030000 0.013409 0.000090 0.006735 1.006765 1.013500 74.07407 2.232855
0.100000 0.044003 0.000983 0.022331 1.022669 1.045000 22.22222 2.028986
0.300000 0.126284 0.008332 0.065982 1.069018 1.135000 7.407407 1.609195
1.000000 0.362372 0.075439 0.208182 1.241818 1.450000 2.222222 0.933333
3.000000 0.740760 0.390785 0.527547 1.822453 2.350000 0.740741 0.424242
10.00000 0.988891 0.938901 0.949448 4.550552 5.500000 0.222222 0.145833

Here we learn that we can reproduce ballpark numbers for Yuri's signal-to-noise ratios (about 2.0 for occupancies of about 0.03-0.10 for the 3 SVT barrels in minbias CuCu62) by assuming properties similar to what is used for the SVT in his study and a notable hit reconstruction inefficiency (it would only take a little tweeking of the hit reconstruction efficiency to get closer to his signal-to-noise results), so there is unlikely to be any significant problems with his analysis.

Simulated IST

In order to tie Yuri's analysis to Howard's, let's use the IST strips with a search area that approximates the resolution of the track-hit-matching which Howard uses (my only hope to match Howard's least chi-squares method to Yuri's search window analysis is to treat the resolutions as the approximate search window):

area = (0.03 cm x 2.25 sigma) x ((3.8/sqrt(12)) cm x 2.25 sigma) = 0.1666 cm^2
and let's assume a hit reconstruction efficiency of 0.8

(A)
Occupancy (hits/cm^2)		 (B)
Probability of
at least one hit
in an element		 (C)
Probability of
at least two hits
in an element		 (D)
Probability of
at least two hits
in an element
given that
we have >=1 hit		 (E)
Average number
of hits
in an element
given that
we have >=1 hit
(Signal+Noise)		 (F)
Average number
of hits
in an element
given that
we have >=1 hit
weight by hits
(Signal+Noise
seen by tracks)		 (G)
Signal/Noise		 (H)
Signal/Noise
with
inefficiency
0.030000 0.004986 0.000012 0.002497 1.002501 1.004998 200.0781 3.902476
0.100000 0.016522 0.000137 0.008307 1.008353 1.016660 60.02342 3.692419
0.300000 0.048752 0.001208 0.024782 1.025198 1.049980 20.00781 3.200250
1.000000 0.153463 0.012429 0.080989 1.085613 1.166602 6.002342 2.182205
2.000000 0.283375 0.044594 0.157367 1.175837 1.333203 3.001171 1.500366
3.000000 0.393351 0.090145 0.229172 1.270633 1.499805 2.000781 1.143176
10.00000 0.811002 0.496127 0.611746 2.054270 2.666016 0.600234 0.428721

Now, admittedly, I have chosen the search window width of 2.25 sigma in each direction to try to reproduce Howard's results, and I have achieved that if you look at the bold numbers for the signal-to-noise with no inefficiency: for occupancies of {0.3,1.0,2.0} hits/cm^2, I find signal-to-noise of {20,6,3}, which converts to hit-matching efficiencies of {95%,86%,75%}, agreeing well with Howard's graph for these occupancies. This implies that my formulas can do a rather good job of matching both Howard's and Yuri's analysis results.

Conclusions

Howard's results of signal-to-noise vs. occupancy are consistent with Yuri's results of signal-to-noise in minbias CuCu62 data vs. occupancy modulo a hit reconstruction efficiency. Therefore, it is important that any simulations like Howard's try to use a hit reconstruction efficiency if such efficiency is not very close to 1.0. Observe that an efficiency of 80% brings the signal-to-noise ratio down by a factor of two from 3 to 1.5 (or stated differently, the hit-matching efficiency goes from 75% to 60%) for the IST strips at an occupancy of 2.0 hits/cm^2. It is important to understand that an occupancy observed in real data already includes this efficiency.

It is also important that the pointing resolution of tracks to the IST be understood. In Yuri's analysis, the resolution is notably worse in r-phi than Howard assumes. In part this is because Yuri includes some tracks which do not have an SSD hit to further constrain the pointing. This probably needs to be incorporated into Howard's analysis.

As a concluding statement, I believe that the IST strips by themselves would be severely detrimental to tracking if the signal-to-noise gets anywhere near 1.0 (hit-matching efficiency of 50%). Using a search area which matches Howard's results, I find that this will happen if the occupancies reach 6.0 hits/cm^2 for perfect hit reconstruction, 4.75 hits/cm^2 for 90% efficiency, and 3.6 hits/cm^2 for 80% efficiency. I am unsure of what hit reconstruction efficiency should be expected out of the IST detectors, and I think that Howard's r-phi resolution is a bit optimistic, so I consider Howward's current numbers as an upper bound on the performance of the strips alone. I provide below a plot of hit-matching efficiency vs. occupancy which demonstrates that we are considering a technology which will operate (maximum occupancy of perhaps 1-2 hits/cm^2) near the knee of these curves. Perhaps this is where we want to be (we don't want to over-design by a factor of 10 spending more than we need), but we are close to where we may have only a factor of 2 margin.