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# Calorimeter Optimization

## Introduction

The simulations simulate a calorimeter with 20 layers of sensors (0-19) in the z direction.

Given 15 layers alone, one must decide which of the 20 layers to include.

Therefore, a study is being made to optimize the calorimeter given that constraint.

First, a longitudinal profile of the energy deposited was made, which you can see under the page of the same name.

As seen in these plots, the peak of the shower is located in the first half of the calorimeter.

However, this peak is shifted as the positron energy increases, and we'd want to measure it for all energies.

Since the same layout has to be constructed for all energies, this imposes a limitation on the possible subsets of layers we can choose.

Two different subsets were examined:

• A continuous set of the first 15 layers.
• The first 10 layers (every X0) and then every other layer (every 2X0), denoted the 10+5 subset.

Note that no new simulations were conducted for the study, but the "missing" layers were simply ignored.

The studies were based on 7 different positron energies (denoted E0) - 2, 4, 6, 8, 10, 12, and 14 GeV.

A total of 35,000 events were examined, 5000 for each E0.

## Goals

There are two goals for an ideal subset of layers:

• For the total energy deposited to be as close as possible to the positron energy, denoted E_0.
• For the energy resolution to be as small as possible (explained in detail below).

Both these measurements are based on the total energy deposited in the calorimeter.

Therefore, throughout the studies the general flow was as follows:

• For each E0:
• For each event:
• Sum up the total energy deposited in the calorimeter (in all existing pads).
• Calculate the mean value and standard deviation (σ).
• Keep events within 3σ of the mean value only.
• Denote the number of the remaining events by N.
• Plot a histogram of the energy deposited and set the number of bins to √N (which turned out to be 70 for all E0).
• Fit a gaussian to the center of that histogram, i.e. the range of the mean value ± 2σ.
• Extract μ, σ of the gaussian and continue according to the desired goal.

Now let us look at the results for a total of 20 layers to understand what is our current best.

### Total energy for 20 layers

As described above, μ was calculated for each positron energy E0.

To get the positron energy from that value, the following relation was considered:

E0 = R · μ

That is to say, given the total energy deposited, one can get the positron energy by multiplying it by some R.

The various R values received are:

2 GeV4 GeV6 GeV8 GeV10 GeV12 GeV14 GeVAverage
85.585.084.985.085.185.285.585.2

Hence, from now on, to compare the total energy to the positron energy, the former was multiplied by 85.2.

We shall call this our new μ.

The result for the 20 layers can be seen visually below:

The red line represents the y=x curve, and it can be seen that all 7 points sit nicely on that line.

### Energy resolution for 20 layers

Again the values of μ and σ were calculated for each energy.

The histograms and gaussian fits are shown below:

And the Χ2 values, divided by the number of degrees of freedom, are as follows:

2 GeV4 GeV6 GeV8 GeV10 GeV12 GeV14 GeV
1.51.30.881.00.720.971.3

Then, the ratio σ/μ was plotted as a function of E0 and fitted to the following formula:

$//$

The energy resolution, in percentage, is 100 times the variable a.

The fit result for all 20 layers can be seen below:

Hence our resolution for all layers is 19.4%.

### Missing energy for 15 layers

One can see that to properly consider a given subset, a reconstruction of the missing energy is to be applied.

For each of the subsets, the percentage of the missing energy is presented below, averaging over 5000 events per E0.

For the 15 continuous layers, the calculation does not make up for the missing layers but simply ignores them.

For the 10+5 method, we multiply the energy in a layer after a missing one by factor 2.

Clearly, one can, and should, do better.

Even though the first 15 layers contain most of the energy, it would be preferable to add the average missing energy to the total.

As for the 10+5 layers, a factor of 2 to the energy in the last 5 layers gives solid results, but that factor could be tuned.

Hence, throughout the studies shown below, the total energy deposited in the calorimeter was fixed in the following ways:

• For the 15 continuous layers - the average fraction of energy missing, based on the plot above, was added to the total energy deposited.
• For the 10+5 layers - the energy deposited in the last 5 layers (these are 11, 13, 15, 17, 19) was multiplied by a constant factor. Several values between 1 and 2.5 were taken as the constant, and the results were compared to one another.

The new evaluations of the total energy deposited were then multiplied by 85.2 as explained previously, and the mean value μ was calculated based on these final values.

## Results

### 15 continuous layers

As seen above, when discarding the last 5 layers of the calorimeter, we lose about 3-7% of the total energy deposited, depending on E0.

Therefore, when summing up the energy in the first 15 layers, it was assumed to be only a fraction X of the total: Edep = X Etot

To get a better assumption of the true total energy, the given deposition of each event was divided by the proper X of that E0.

Since in reality we don't know E0, the same measurements were done while using the average value of X - 0.94.

Luckily, the quality of the results was identical, and therefore X was set to 0.94 for all energies.

#### Gaussian fits

And the Χ2 values, divided by the number of degrees of freedom, are:

2 GeV4 GeV6 GeV8 GeV10 GeV12 GeV14 GeV
1.50.61.01.10.811.01.2

#### Energy resolution

As can be seen below, the energy resolution is even better than before - 18.7%.

This result is slightly too good, and will therefore be further explored and checked.

#### The similarity of μ and E0

Again, the red line represents the y=x line.

Clearly, the points sit nicely on the desired curve, similar to the results of the 20 layers.

It seems that the energy estimate we got based on 15 layers alone, is quite well.

### 10+5 layers

As mentioned earlier, the energy in the 5 layers which are located every 2X0 was multiplied by some constant to estimate the total energy in 20 layers.

That constant was scanned for values between 1 and 2.5.

Below are the results for 1.25, 1.75, and 2.25.

As you will see, the lower the factor the better the resolution, however the further we are from E0.

Again, these results will also be further investigated to make sure no mistakes were done in the process.

#### Gaussian fits

And the Χ2 values, divided by the number of degrees of freedom, are:

2 GeV4 GeV6 GeV8 GeV10 GeV12 GeV14 GeV
1.30.831.00.91.10.921.0

#### Energy resolution

The energy resolution for a coefficient of 1.25 is 20%.

#### The similarity of μ and E0

As expected, the points deviate from the y=x line.

This is because, in the last 10 layers, the energy deposited in layer i+1 is lower than that of layer i.

Hence, multiplying the i+1 energy by 1.25 is an underestimate of the total energy in the calorimeter.

#### Gaussian fits

And the Χ2 values, divided by the number of degrees of freedom, are:

2 GeV4 GeV6 GeV8 GeV10 GeV12 GeV14 GeV
2.00.981.11.10.921.10.92

#### Energy resolution

The energy resolution for a coefficient of 1.75 is 20.5%.

#### The similarity of μ and E0

And again the points deviate from the y=x line since this is still an underestimate of the total energy.

#### Gaussian fits

And the Χ2 values, divided by the number of degrees of freedom, are:

2 GeV4 GeV6 GeV8 GeV10 GeV12 GeV14 GeV
1.61.20.720.981.51.01.0

#### Energy resolution

The energy resolution for a coefficient of 2.25 is 21.3%.

#### The similarity of μ and E0

Now, the points sit nicely on the y=x, but the resolution is a bit lesser.

Results for the other coefficients were excluded as they present a similar picture to those above.

As mentioned, the lower the coefficient the better the resolution (the best value was 19% for a coefficient of 0.75) but the further we are from the true positron energy.

## Conclusion

So far, the option of 15 continuous layers seems best.

However, since these results are even better than for the full calorimeter, they are to be double-checked.

The same goes for the 10+5 results, which show the best resolution for a coefficient so low.

To be continued...

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