SLAC-TN-09-002
B A B AR Note # 327
October 1996

Noise in a Calorimeter
Readout System
Using Periodic Sampling

Walter R. Innes
SLAC National Accelerator Laboratory

Fourier transform analysis of the calorimeter noise problem gives quantitative
results on a) the time-height correlation, b) the effect of background on
optimal shaping and on the ENC, c) sampling frequency requirements, and d) the
relation between sampling frequency and the required quantization error.

##  1 Introduction

Noise in calorimeter readout electronics has been treated in some detail
[Radeka, Haller, Dow]. However, since each author builds the details of their
particular situation into their analysis, it is worthwhile to review the
subject in light of our own experiment. Most previous studies assume that a
single measurement will be made at a well determined time. Furthermore they
assume that filtering preceding the digitization is limited to relatively
simple analog devices. Since we have chosen to solve our lack of knowledge of
the event time by recording periodic samples, applying digital signal
processing techniques is natural. This also raises questions about the
required sampling rate and the quantization error that don’t occur in the
usual scheme.

Electronics noise and filters are usually analyzed using Fourier transforms
[Ambrozny, Humphreys, Papoulis]. In the frequency domain, filtering,
differentiation, integration, and time shifting are all represented by
multiplications. Stochastic noise is most easily represented by its power
spectral density (which is the Fourier transform of the more complicated auto-
correlation function in the time domain). Many questions such as the signal to
noise ratio for particular parameters and filter design can be handled
entirely in the frequency domain.

##  2 The Input Circuit Model

Figure 1: Simplified photodiode-preamp circuit

Figure 2: Model for the front end of the calorimeter readout. $i_{s}$ is the
current due to the signal and is a function of time. $I_{d}$ is the spectral
current noise generator corresponding to the shot noise of the photodiode.
$I_{b}$ is the equivalent current noise caused by background photons from lost
particles. $I_{f}$ is the FET input current noise, and $V_{n}$ is the FET
input voltage noise. $C_{s}$ is the capacitance of the photodiode and
$C_{iss}$ is the common source input capacitance of the FET.

Figure 1 shows a simplified version of the calorimeter input circuitry. Figure
2 shows a noise equivalent representation of the same circuit. The right hand
portion represents the input FET. This is one of many possible equivalent
representations [vanderZiel, page 29][Ambrozny, page 137]. This particular one
has a relatively simple connection between the physical processes which
generate noise and the elements of the representation. The spectral densities
of the noise generators will correspond to the those given in manufacturers
specification sheets.

###  2.1 The photo-diode

$i_{s}$ is a current generator corresponding to the signal generated by the
interactions in the calorimeter. It will be represented by the function:

$i_{s}(t)=\frac{1}{\tau_{s}}e^{-t/\tau_{s}}H(t).$ (1)

$\tau_{s}$ is the decay time of the CsI(Tl) scintillation and $H(t)$ is the
Heaviside unit step function. $\tau_{s}$ is 0.94$\,\mu\mbox{s}$ for CsI(Tl).
The Fourier transform of $i(t)$ is

$I_{s}(\omega)=\int_{-\infty}^{\infty}{i(t)e^{-j\omega
t}\,dt}=\frac{1}{1+j\omega\tau_{s}}\ .$ (2)

The spectral power density of the signal is:

$I_{s}^{*}(\omega)I_{s}(\omega)=\frac{1}{1+\omega^{2}\tau_{s}^{2}}.$ (3)

For characteristic times represented as $\tau_{x}$, I will use the notation
that the associated radial velocity is $\omega_{x}\equiv 1/\tau_{x}$, the
associated frequency is $f_{x}\equiv\omega_{x}/(2\pi)$, and the associated
period is $T_{x}\equiv 1/f_{x}$. Using this notation we can write:

$I_{s}(\omega)=\frac{-j\omega_{s}}{\omega-j\omega_{s}}\ \ {\rm and}\ \
I_{s}^{*}(\omega)I_{s}(\omega)=\frac{\omega_{s}^{2}}{\omega^{2}+\omega_{s}^{2}}.$
(4)

$I_{d}$ is the current generator corresponding to the shot noise caused by the
leakage current in the photodiode. This has a “white” (independent of
$\omega$) spectrum with the value $I_{d}^{2}=2eI_{leak}$ (units of current
squared per bandwidth). The order of magnitude for $I_{leak}$ is nanoamperes.
We’ll assume our diode has a leakage of 4 nA. Since we average two diodes we
can use the equivalent leakage of 2 nA for a single diode giving
$I_{d}=25\mbox{\,fA}/\rm\sqrt{\mbox{\,Hz}}$.

$C_{s}$ is the capacitance of the back-biased photodiode. A typical value is
80 pF, which we’ll use in our estimates.

A photodiode with these properties is the 1N2744.

As much as possible, the discussion will be restricted to the electronic
regime. However, some effects depend on the efficiency of the conversion of
calorimeter shower energy into charge. When this number is required, I will
use the value of 3000 photo-electrons per MeV in each diode.

###  2.2 The FET

$C_{iss}$ is the FET common source input capacitance. It can range from a few
picofarads to tens of picofarads depending on the choice of FET. We will
return to this choice later. In the circuit it functions in parallel with
$C_{s}$. Not shown in the figure since they are much less than $C_{s}$ are the
feedback capacitance $C_{f}$ and the calibration injection capacitance
$C_{cal}$. I will call the sum $C_{s}+C_{iss}+C_{f}+C_{cal}$ simply $C$.

$I_{f}$ is a noise generator corresponding the input current noise of the FET.
It is primarily due to the gate leakage current, although for very large gate
areas the effect the thermal noise of the real part of the input impedance
should be checked. The shot noise acts just like the diode shot noise and the
two leakage currents may be added. In practice the FET gate leakage current is
of the order of picoamperes and is negligible compared to the diode leakage
current. Let $I_{n}^{2}=I_{d}^{2}+I_{f}^{2}\approx I_{d}^{2}$.

$V_{n}$ is a voltage noise generator representing the input voltage noise of
the FET. This is primarily due the thermal noise in the FET channel. Many
authors put a formula, which shows the dependence of this noise on the
temperature and the forward transconductance, into their equations
[vanderZiel, page 75]. I prefer to keep $V_{n}$ as an explicit term. The value
of $V_{n}$ is usually given on specification sheets and is also easy to
measure. For junction FETs, $V_{n}$ is approximately independent of frequency
over the range of interest to us. Typical values are 1 to 2
nV/$\rm\sqrt{\mbox{\,Hz}}$.

In order to be more specific I’ll take as an example the SANYO 2SK932 JFET.
$C_{iss}$ is 20 pF and $V_{n}$ is $\approx
0.7\mbox{\,nV}/\rm\sqrt{\mbox{\,Hz}}$. Because we average the outputs two
diode-FET-preamp combinations on each crystal, we divide by $\sqrt{2}$ to get
0.5 nV/$\rm\sqrt{\mbox{\,Hz}}$. This is a good FET but not optimally matched
to our problem. We’ll return to this issue after we have developed the signal
to noise formulas.

Figure 3: The amplitude spectra of the various noise sources expressed as
input current densities. For the background, nominal conditions are assumed
and no energy cutoff is used. Note that the signal has the same spectrum as
the background noise.

###  2.3 Background

$I_{b}$ is a current noise generator representing the sea of photons generated
by particles lost from the beam. This noise resembles shot noise in that near
impulsive elements arrive at random times. It differs in that the pulse shape
is not a delta function but has the shape of the signal and there is a
distribution of pulse areas instead of just the electron charge.111In radar
theory, noise with these properties is called “clutter.” If we return to the
derivation of the shot noise formula [Ambrozny, page 81], we see that the
first difference can be treated by giving this noise the power spectrum of the
signal (the Fourier transform of each pulse is the same as that of the signal
except for phase and magnitude). The second is handled by noting that the shot
noise formula is really a pulse area squared times a rate:
$I_{d}^{2}=2e^{2}f_{leak}$. $e$ is the size of the pulse and $f_{leak}$ is the
mean rate of their occurrence. Therefore we can get the low frequency limit of
the noise spectral density of the background by doing the integral:

$I_{b}^{2}=\int_{0}^{\infty}{2q^{2}B(q)\,dq}$ (5)

where $B$ is the rate of background events per pulse area interval per time
interval. The value of this integral can be inferred from the statistics of
histograms such as those in B A B AR Technical Design Report Figure 12-9
[BaBar]. Here I use more recent calculations of the background [Marsiske].
Assuming $3000\rm\,pe/\mbox{\,Me\kern-0.80002ptV}/diode$, the result for an
average crystal is $I_{b}=460\mbox{\,fA}/\rm\sqrt{\mbox{\,Hz}}$ at nominal
background.

For convenience I’ll use the symbol $I_{b}$ to represent this low frequency
limit and put the frequency dependence into the formulas explicitly. Since
$I_{n}$ and $V_{n}$ are approximately independent of frequency, I can then
treat $I_{b}$, $I_{n}$, and $V_{n}$ as constants.

Note that individual crystals can differ from the average by factors as large
as $4$. In particular, compared an average barrel crystal, an average endcap
crystal has $5/7$th the hit rate, twice the average energy deposit, and an
$I_{b}$ twice as large.

Figure 3 summarizes all the noise sources.

####  2.3.1 Limitations

The treatment of background as noise is a good description if the background
signal after all processing contains significant contributions from many
background events. If this is not the case, the data may fall into two
classes, those showers little affected by background, and those with a large
background contribution. The former will have a distribution of measured
values consistent with electronic noise, while the latter may have a wider,
highly skewed distribution. While the background as noise treatment may give a
near optimal procedure for minimizing the rms error, it will widen the
distribution for that class events unaffected by background. Depending on the
physics objective, treating the background affected events as an inefficiency
rather than events with errors may give the better result.

How well does our case satisfy this many hit condition? In the background
calculation in the preceding paragraph, an average crystal had $0.11\,\rm
hits/crystal/\,\mu\mbox{s}$. Assuming a window of $2\,\mu\mbox{s}$ and a sum
of 25 crystals, this implies 5.5 background hits contributing to the
measurement during nominal conditions. This is enough to give reasonably
Guassian behavior. However, much of $I_{b}$ is generated by the high energy
tail of the background distribution. If we count down from high energies until
get a mean number of hits of at least 2, the cut would be in the vicinity of
$1\mbox{\,Me\kern-0.80002ptV}$. At 10 times nominal this cut would be close to
$3\mbox{\,Me\kern-0.80002ptV}$.

Another consideration is that an approach which tries to separate a background
tail from other noise is hardly likely to succeed if the magnitude of the
background contribution is of the same size as other errors. The size of the
intrinsic error depends on the size of the signal. For a
$100\mbox{\,Me\kern-0.80002ptV}$ shower the expected error is $\approx
2\mbox{\,Me\kern-0.80002ptV}$. Assuming that background contributions must be
at least $3\mbox{\,Me\kern-0.80002ptV}$ before they can be separated from
noise seems conservative.

If only background hits with energies below $3\mbox{\,Me\kern-0.80002ptV}$ are
included, $I_{b}$ falls to $160\mbox{\,fA}/\rm\sqrt{\mbox{\,Hz}}$. The
background rate falls to $0.10\,\rm hits/crystal/\,\mu\mbox{s}$.

##  3 Theorems

###  3.1 $S/N$

The Radar Handbook [Skolnik, page 4-11] reviews the theory of finding a known
signal in noise from the starting point of doing a least square fit to a
finite set of measurements. Under fairly general conditions this is equivalent
to the maximum likelihood fit and is the best that can be done. The treatment
proceeds by taking the continuous infinite time limit which turns the usual
matrix equations into integral equations. These are then solved by applying
Fourier transforms. For a signal $i(t,a_{1},...,a_{n})$ with Fourier transform
$I(\omega,a_{1},...,a_{n})$ the elements of the inverse of the error matrix
for $a_{i}$ are given by:

$[{\bf
V}^{-1}]_{ik}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\left(\left(\frac{\partial
I(\omega)}{\partial a_{i}}\frac{\partial I^{*}(\omega)}{\partial
a_{k}}\right)/N(\omega)\right)d\omega},$ (6)

where $N$ is the double sided noise power spectral density. This matrix is
known as the information matrix and also as the signal to noise ratio ($S/N$)
matrix.

In our case the signal has two parameters, amplitude and time offset:

$i(t)=ai_{0}(t-t_{0}).$ (7)

This has the Fourier transform

$I(\omega)=aI_{0}(\omega)e^{-j\omega t_{0}}.$ (8)

If we define $I_{0}$ such that the actual value of $a$ is 1 we find

$\displaystyle\frac{\partial I}{\partial a}$ $\displaystyle=$ $\displaystyle
I(\omega)\mbox{,\rm\hskip 72.26999pt and}$ (9) $\displaystyle\frac{\partial
I}{\partial t_{0}}$ $\displaystyle=$ $\displaystyle-j\omega I(\omega).$ (10)

Thus the matrix of integrands in the $S/N$ formula is

$\left[\begin{array}[]{cc}~{}~{}~{}1{}{}{}&~{}~{}j\omega{}{}\\\
-j\omega&\omega^{2}\end{array}\right]\frac{|I(\omega)|^{2}}{N(\omega)}$ (11)

Because both $|I(\omega)|^{2}$ and $N(\omega)$ are even in $\omega$, the
integral of the off-diagonal elements is 0. Since the matrix is diagonal,
inverting it to get the error matrix ${\bf V}$ is trivial. If $i(t)$ is
normalized to unit area, the $aa$ element of ${\bf V}$ is the square of the
equivalent noise charge (ENC) and the $t_{0}t_{0}$ element is the square of
product of the pulse height and the time error (ENTC). Since there is no
correlation between these errors, a priori knowledge of the time does not
improve the determination of the amplitude. The final formulas for the signal
to noise are:

$\left[\frac{S}{N}\right]_{aa}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{|I(\omega)|^{2}}{N(\omega)}\,d\omega},\\\
$ (12)

and

$\left[\frac{S}{N}\right]_{tt}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{|I(\omega)|^{2}}{N(\omega)}\,\omega^{2}d\omega}$
(13)

###  3.2 Matched filter

This result for the error in the amplitude is the same that is given in many
texts as the best that can be achieved using an optimal matched filter
[Humphreys, page 65][Papoulis, page 135]. Assume there exists a filter such
that the output peaks at $t=0$. The square of that peak value is given by:

$o^{2}(0)=\left|\frac{1}{2\pi}\int_{-\infty}^{\infty}{I(\omega)F(\omega)\,d\omega}\right|^{2},$
(14)

where $F(\omega)$ is the transfer function of the filter. The mean square
noise after filtering is the same at all times and is given by

$n_{o}^{2}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{N(\omega)|F(\omega)|^{2}\,d\omega}.$
(15)

The signal to noise is the ratio of $o^{2}(0)$ to $n_{o}^{2}$. Multiplying the
integrand of the numerator by $\sqrt{N(\omega)}/\sqrt{N(\omega)}$ and using
Schwarz’s inequality proves that the signal to noise is less than or equal to
the $S/N$ derived from the least square fit method (equation 12). Furthermore,
inspection of the $S/N$ equation before this manipulation shows that the
equality can be achieved using a filter with a transfer function of

$F(\omega)\propto\frac{I^{*}(\omega)}{N(\omega)}\ .$ (16)

From this I conclude that the best possible result can be achieved with a
filter technique and that the required filter is readily calculated given
knowledge of the shape of the signal and the spectrum of the noise.

The impulse response to the optimal filter is

$f(t)\propto\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{I^{*}(\omega)}{N(\omega)}\,e^{j\omega
t}d\omega}\ ,$ (17)

and the shape of a noiseless signal after filtering (the convolution of the
signal with impulse response) is

$o(t)\propto\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{|I(\omega)|^{2}}{N(\omega)}\,e^{j\omega
t}d\omega}\ .$ (18)

A useful normalization for the filter can be achieved by dividing by the
signal to noise ratio (equation 12). This yields $o(0)=1$. In the following
discussion, $F$, $f$, and $o$ are so normalized.

##  4 Application to our problem

We have now collected all the necessary input information and the tools we
need to design an optimal filter and calculate the signal to noise. Before
tackling the full problem lets explore some special cases so that we can get
an understanding of the effect of each input, can compare our results with
previous work, and gain confidence in the method. Some of the formulas are
followed by a bracketed reference to the source of the evaluation of the
previous integral. Actually most of the integrals are straightforward (but
sometimes tedious) to do using contour integration. This is a consequence of
the fact that the signals, backgrounds, and analog filters all have
exponential shapes in the time domain. More general signal shapes are less
tractable.

###  4.1 No background ($I_{b}=0$) and $\tau_{s}=0$

This is the usual treatment when the “ballistic deficit” is ignored.
$i_{s}(t)$ is a delta function, $I_{s}(\omega)=1$, and
$2N(\omega)=I_{n}^{2}+V_{n}^{2}\omega^{2}C^{2}$.

Before starting lets evaluate some useful constants related to the inputs so
that we can use them to evaluate the results as we go: $\tau_{n}\equiv
V_{n}C/I_{n}=2.0\,\mu\mbox{s}$, $\omega_{n}\equiv
1/\tau_{n}=0.50\rm\,M\,radians/s$, and $f_{n}=80\mbox{\,kHz}$.

$\displaystyle\left[\frac{S}{N}\right]_{aa}$ $\displaystyle=$
$\displaystyle\frac{\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{1}{\omega_{n}^{2}+\omega^{2}}\,d\omega}$
(19) $\displaystyle=$ $\displaystyle\frac{\omega_{n}}{I_{n}^{2}}$ (20)
$\displaystyle ENC^{2}$ $\displaystyle=$ $\displaystyle
I_{n}^{2}\tau_{n}=I_{n}V_{n}C$ (21)
$\displaystyle\left[\frac{S}{N}\right]_{tt}$ $\displaystyle=$
$\displaystyle\frac{\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{\omega^{2}}{\omega_{n}^{2}+\omega^{2}}\,d\omega}$
(22) $\displaystyle=$ $\displaystyle\infty$ (23) $\displaystyle ENTC^{2}$
$\displaystyle=$ $\displaystyle 0$ (24) $\displaystyle F(\omega)$
$\displaystyle=$ $\displaystyle\frac{2\omega_{n}}{\omega_{n}^{2}+\omega^{2}}$
(25) $\displaystyle f(t)$ $\displaystyle=$ $\displaystyle e^{-|t|/\tau_{n}}$
(26) $\displaystyle o(t)$ $\displaystyle=$ $\displaystyle e^{-|t|/\tau_{n}}$
(27)

Substituting our sample values, we find that $ENC=223\,e$, and the corner
frequency of the optimal filter is 80 kHz.

###  4.2 No background ($I_{b}=0$) and $\tau_{s}>0$

This is the case usually treated, ballistic deficit included.

$\displaystyle I_{s}(\omega)$ $\displaystyle=$
$\displaystyle\frac{-j\omega_{s}}{\omega-j\omega_{s}}$ (28) $\displaystyle
2N(\omega)$ $\displaystyle=$
$\displaystyle(I_{n}^{2}+V_{n}^{2}\omega^{2}C^{2})=I_{n}^{2}(\omega^{2}+\omega_{n}^{2})/\omega_{n}^{2}$
(29) $\displaystyle\left[\frac{S}{N}\right]_{aa}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{1}{(\omega^{2}+\omega_{s}^{2})(\omega^{2}+\omega_{n}^{2})}\,d\omega}$
(30) $\displaystyle=$
$\displaystyle\frac{\omega_{s}\omega_{n}}{I_{n}^{2}(\omega_{s}+\omega_{n})}\mbox{\hskip
72.26999pt \rm[Dwight~{}856.31]}$ (31) $\displaystyle ENC^{2}$
$\displaystyle=$ $\displaystyle
I_{n}^{2}(\tau_{n}+\tau_{s})=I_{n}V_{n}C+I_{n}^{2}\tau_{s}$ (32)
$\displaystyle\left[\frac{S}{N}\right]_{tt}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{\omega^{2}}{(\omega^{2}+\omega_{s}^{2})(\omega^{2}+\omega_{n}^{2})}\,d\omega}$
(33) $\displaystyle ENTC^{2}$ $\displaystyle=$ $\displaystyle
I_{n}^{2}\tau_{n}\tau_{s}(\tau_{n}+\tau_{s})=ENC^{2}\tau_{n}\tau_{s}\mbox{\hskip
36.135pt \rm[Maple]}$ (34) $\displaystyle F(\omega)$ $\displaystyle=$
$\displaystyle\frac{2j\omega_{n}(\omega_{n}+\omega_{s})}{(\omega+j\omega_{s})(\omega^{2}+\omega_{n}^{2})}$
(35) $\displaystyle f(t)$ $\displaystyle=$
$\displaystyle\frac{\omega_{n}(\omega_{n}+\omega_{s})}{\pi}\int_{-\infty}^{\infty}{\frac{(\omega_{s}+j\omega)}{(\omega^{2}+\omega_{s}^{2})(\omega^{2}+\omega_{n}^{2})}e^{j\omega
t}d\omega}$ (36) $\displaystyle=$
$\displaystyle\frac{\omega_{n}(\omega_{n}+\omega_{s})}{(\omega_{n}^{2}-\omega_{s}^{2})}\left(e^{-|t|/\tau_{s}}-\frac{\omega_{s}}{\omega_{n}}e^{-|t|/\tau_{n}}-\rm
signum(t)\left(e^{-|t|/\tau_{s}}-e^{-|t|/\tau_{n}}\right)\right)$ (37)
$\displaystyle=$
$\displaystyle\left(e^{-|t|/\tau_{n}}+H(-t)\frac{2\tau_{s}}{\tau_{n}-\tau_{s}}\left(e^{-|t|/\tau_{n}}-e^{-|t|/\tau_{s}}\right)\right)$
(38) $\displaystyle o(t)$ $\displaystyle=$
$\displaystyle\frac{\omega_{n}\omega_{s}(\omega_{n}+\omega_{s})}{\pi}\int_{-\infty}^{\infty}{\frac{e^{j\omega
t}}{(\omega^{2}+\omega_{s}^{2})(\omega^{2}+\omega_{n}^{2})}\,d\omega}$ (39)
$\displaystyle=$
$\displaystyle\frac{1}{(\tau_{n}-\tau_{s})}\left(\tau_{n}e^{-|t|/\tau_{n}}-\tau_{s}e^{-|t|/\tau_{s}}\right)$
(40)

Substituting our sample values, we find that $ENC=239\,e$,
$ENTC=327\,\mu\mbox{s}\,e$ (1.1 ns for a 100 MeV deposit), and the corner
frequency of the optimal filter is still 80 kHz. The ENC corresponds to and
equivalent noise energy (ENE) of 80 keV.

###  4.3 $I_{b}>0$, $V_{n}=0$, and $I_{n}=0$

This special case of no electronic noise is explored to give us confidence
that the filter technique does separate the desired signal from the background
noise.

$\displaystyle I_{s}(\omega)$ $\displaystyle=$
$\displaystyle\frac{-j\omega_{s}}{\omega-j\omega_{s}}$ (41) $\displaystyle
2N(\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}I_{b}^{2}}{\omega_{s}^{2}+\omega^{2}}$ (42)
$\displaystyle\frac{|I_{s}|^{2}}{N}$ $\displaystyle=$
$\displaystyle\frac{2}{I_{b}^{2}}$ (43)
$\displaystyle\left[\frac{S}{N}\right]_{aa}$ $\displaystyle=$
$\displaystyle\infty\ ,\ ENC^{2}=0$ (44)
$\displaystyle\left[\frac{S}{N}\right]_{tt}$ $\displaystyle=$
$\displaystyle\infty\ ,\ ENCT^{2}=0$ (45) $\displaystyle f(t)$
$\displaystyle\propto$
$\displaystyle\frac{1}{\pi}\int_{-\infty}^{\infty}{(\omega_{s}-j\omega)e^{j\omega
t}d\omega}$ (46) $\displaystyle\propto$ $\displaystyle
2\left(\delta(t)-\tau_{s}\delta^{\prime}(t)\right)$ (47) $\displaystyle o(t)$
$\displaystyle\propto$
$\displaystyle\frac{1}{\pi}\int_{-\infty}^{\infty}{e^{j\omega
t}d\omega}\,\,\propto\,\,2\delta(t)$ (48)

Because the signal and (background) noise have the same spectrum, the signal
to noise ratio is independent of frequency. This results in infinite signal to
noise integrals and no error on the pulse height and time determinations. In
the absence of other noise, the matched filter turns both the signal and
background events into delta functions which are always separable.

###  4.4 The general case

Now we will treat our general case, with all noise and width processes active.
The signal is as for the previous two cases. The noise is the sum of the noise
in those cases.

$\displaystyle I_{s}(\omega)$ $\displaystyle=$
$\displaystyle\frac{-j\omega_{s}}{\omega-j\omega_{s}}$ (49)
$\displaystyle|I_{s}(\omega)|^{2}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}}{\omega_{s}^{2}+\omega^{2}}$ (50)
$\displaystyle 2N(\omega)$ $\displaystyle=$
$\displaystyle\frac{I_{n}^{2}}{\omega_{n}^{2}}(\omega_{n}^{2}+\omega^{2})+\frac{\omega_{s}^{2}I_{b}^{2}}{\omega_{s}^{2}+\omega^{2}}$
(51) $\displaystyle\left[\frac{S}{N}\right]_{aa}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{1}{\omega^{4}+(\omega_{s}^{2}+\omega_{n}^{2})\omega^{2}+\omega_{n}^{2}\omega_{s}^{2}(1+I_{b}^{2}/I_{n}^{2})}\,d\omega}$
(52) $\displaystyle b^{2}$ $\displaystyle\equiv$ $\displaystyle
1+I_{b}^{2}/I_{n}^{2}\mbox{\hskip 36.135pt\rm and\hskip 36.135pt
}\omega_{b}\equiv\omega_{n}b$ (53)
$\displaystyle\left[\frac{S}{N}\right]_{aa}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}\omega_{b}}{(I_{n}^{2}+I_{b}^{2})\sqrt{2\omega_{b}\omega_{s}+(\omega_{n}^{2}+\omega_{s}^{2})}}\mbox{\hskip
72.26999pt \rm[Dwight~{}857.11]}$ (54) $\displaystyle ENC^{2}$
$\displaystyle=$
$\displaystyle(I_{n}^{2}+I_{b}^{2})\sqrt{\tau_{b}\tau_{s}}\sqrt{2+\tau_{b}/\tau_{s}+\tau_{s}/(b^{2}\tau_{b})}$
(55) $\displaystyle\left[\frac{S}{N}\right]_{tt}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{\omega^{2}}{\omega^{4}+(\omega_{s}^{2}+\omega_{n}^{2})\omega^{2}+\omega_{n}^{2}\omega_{s}^{2}(1+I_{b}^{2}/I_{n}^{2})}\,d\omega}$
(56) $\displaystyle ENTC^{2}$ $\displaystyle=$ $\displaystyle
I_{n}^{2}\tau_{n}\tau_{s}\sqrt{\tau_{n}^{2}+\tau_{s}^{2}+2b\tau_{n}\tau_{s})}=ENC^{2}\tau_{b}\tau_{s}\mbox{\hskip
25.29494pt\rm[Prudnikov 2.2.10 \\#4 page 313]}\hfil$ (57) $\displaystyle
F(\omega)$ $\displaystyle=$ $\displaystyle
ENC^{2}\,\,\frac{2\omega_{s}\omega_{n}^{2}(\omega_{s}+j\omega)}{I_{n}^{2}(\omega^{4}+(\omega_{s}^{2}+\omega_{n}^{2})\omega^{2}+\omega_{s}^{2}\omega_{n}^{2}b^{2})}$
(58) $\displaystyle f(t)$ $\displaystyle=$ $\displaystyle
ENC^{2}\,\,\frac{\omega_{s}\omega_{n}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{(\omega_{s}+j\omega)e^{j\omega
t}}{(\omega^{4}+(\omega_{s}^{2}+\omega_{n}^{2})\omega^{2}+\omega_{s}^{2}\omega_{n}^{2}b^{2})}d\omega}$
(59) $\displaystyle=$ $\displaystyle
ENC^{2}\,\,\frac{2\omega_{s}\omega_{n}^{2}}{\pi
I_{n}^{2}}\left(\int_{0}^{\infty}{\frac{\omega_{s}\cos(\omega
t)-\omega\sin(\omega
t)}{(\omega^{4}+(\omega_{s}^{2}+\omega_{n}^{2})\omega^{2}+\omega_{s}^{2}\omega_{n}^{2}b^{2})}d\omega}\right)$
(60) $\displaystyle\omega_{+}$ $\displaystyle\equiv$
$\displaystyle\sqrt{(\omega_{b}\omega_{s}+(\omega_{s}^{2}+\omega_{n}^{2})/2)/2}$
(61) $\displaystyle\omega_{-}$ $\displaystyle\equiv$
$\displaystyle\sqrt{(\omega_{b}\omega_{s}-(\omega_{s}^{2}+\omega_{n}^{2})/2)/2}$
(62) $\displaystyle f(t)$ $\displaystyle=$ $\displaystyle
ENC^{2}\,\,\frac{\omega_{n}\omega_{s}}{2I_{n}^{2}}e^{-\omega_{+}|t|}\left(\frac{1}{b\omega_{+}}\cos(\omega_{-}t)+\frac{1}{b\omega_{-}}\sin(\omega_{-}|t|)-\frac{\omega_{n}}{\omega_{+}\omega_{-}}\sin(\omega_{-}t)\right)$
[Prudnikov 2.5.10 #15 and#17 page 397] $\displaystyle=$ $\displaystyle
e^{-|t|/\tau_{+}}\left(\cos(t/\tau_{-})+\left(\frac{\tau_{-}}{\tau_{+}}\rm
signum(t)-\frac{\tau_{+}}{\tau_{b}}\right)\sin(t/\tau_{-})\right)$ (64)
$\displaystyle o(t)$ $\displaystyle=$ $\displaystyle
ENC^{2}\,\frac{\omega_{n}^{2}\omega_{s}^{2}}{\pi
I_{n}^{2}}\int_{-\infty}^{\infty}{\frac{e^{j\omega
t}}{(\omega^{4}+(\omega_{s}^{2}+\omega_{n}^{2})\omega^{2}+\omega_{s}^{2}\omega_{n}^{2}b^{2})}\,d\omega}$
(65) $\displaystyle=$ $\displaystyle
e^{-|t|/\tau_{+}}\left(\cos(t/\tau_{-})+\frac{\tau_{-}}{\tau_{+}}\sin(|t|/\tau_{-})\right)$
(66)

The filter output for a noiseless signal pulse is shown in Figure 4.

Figure 4: A noiseless signal after passing through the optimal filter. The
peak gives the estimate for the signal charge and time. In this case t=0 and
the charge is unity.

###  4.5 $I_{n}=0$, $\omega_{s}<<\omega_{b}$

The results in the previous section are a bit complex, but can be simplified
for our case if we note that $I_{b}$ is more than 10 times $I_{n}$ even at
nominal background levels. In addition, $\omega_{b}$ is more than five times
$\omega_{s}$. Setting $I_{n}=0$, using the approximation
$\omega_{b}>>\omega_{s}$, and defining
$\omega_{sb}\equiv\sqrt{\omega_{b}\omega_{s}/2}\approx\omega_{-}\approx\omega_{+}$,
we find:

$\displaystyle I_{s}(\omega)$ $\displaystyle=$
$\displaystyle\frac{-j\omega_{s}}{\omega-j\omega_{s}}$ (67)
$\displaystyle|I_{s}(\omega)|^{2}$ $\displaystyle=$
$\displaystyle\frac{\omega_{s}^{2}}{\omega^{2}+\omega_{s}^{2}}$ (68)
$\displaystyle 2N(\omega)$ $\displaystyle\approx$
$\displaystyle\frac{I_{b}^{2}(\omega^{4}+4\omega_{sb}^{4})}{\omega_{b}^{2}(\omega^{2}+\omega_{s}^{2})}$
(69) $\displaystyle|I_{s}(\omega)|^{2}/N(\omega)$ $\displaystyle\approx$
$\displaystyle\frac{8\omega_{sb}^{4}}{I_{b}^{2}(\omega^{4}+4\omega_{sb}^{4})}$
(70) $\displaystyle ENC^{2}$ $\displaystyle\approx$ $\displaystyle
I_{b}^{2}\tau_{sb}=\sqrt{2\tau_{s}I_{b}^{3}V_{n}C}$ (71) $\displaystyle
ENTC^{2}$ $\displaystyle\approx$ $\displaystyle
ENC^{2}\tau_{s}\tau_{b}=\sqrt{2I_{b}\tau_{s}^{3}V_{n}^{3}C^{3}}$ (72)
$\displaystyle F(\omega)$ $\displaystyle\approx$
$\displaystyle\frac{j\omega_{b}\omega_{sb}(\omega-j\omega_{s})}{(\omega^{4}+4\omega_{sb}^{4})}$
(73) $\displaystyle f(t)$ $\displaystyle\approx$ $\displaystyle
e^{-|t|/\tau_{sb}}\left(\cos(t/\tau_{sb})+\sin(|t|/\tau_{sb})-\sqrt{2\tau_{s}/\tau_{b}}\sin(t/\tau_{sb})\right)$
(74) $\displaystyle o(t)$ $\displaystyle\approx$ $\displaystyle
e^{-|t|/\tau_{sb}}\left(\cos(t/\tau_{sb})+\sin(|t|/\tau_{sb})\right)$ (75)

Figure 5: The optimal filter in the frequency domain.

Figure 6: The optimal filter in the time domain.

The optimal filter is shown in the frequency and time domains in Figures 5 and
6.

Now for some values: $\tau_{b}=V_{n}C/I_{b}=0.11\,\mu\mbox{s}$, the filter
corner $\sqrt{\tau_{s}\tau_{b}}=0.32\,\mu\mbox{s}$, $ENC=1930$ e, and
$ENTC=620$$\,\mu\mbox{s}$ e or 2.1 ns for a 100 MeV deposit. The frequency
corner of the optimum filter is 0.50 MHz.

For our worst case of times $10\times$nominal background, $I_{b}$ goes up
$\times 3.1$, $\tau_{b}$ goes down $\times 3.1$, and $\sqrt{\tau_{s}\tau_{b}}$
goes down $\times 1.78$ to 0.17$\,\mu\mbox{s}$. $ENC=4460\,e$,
$ENTC=760\,\mu\mbox{s}\,e$, and the corner frequency goes to
$0.90\mbox{\,MHz}$. If the $3\mbox{\,Me\kern-0.80002ptV}$ cut were used to
calculate $I_{b}$, $I_{b}$ would be lower by a factor of $2.4$ and the values
for $\times 10$ would be about the same as the uncut nominal values.

##  5 Picking the best FET

We see that the $ENC$ depends on the FET properties only in the combination
$V_{n}$($C_{iss}$+$C_{s}$). The smaller this value the better. For a given FET
design, these parameters can be varied by changing the gate area $A$ (or
equivalently connecting FETs in parallel). $V_{n}$ decreases as $1/\sqrt{A}$
while $C_{iss}$ is proportional to $A$. $V_{n}$$C$ is a minimum with respect
to $A$ when $C_{iss}=C_{s}$. This implies that our sample FET should be scaled
to 4 times its original gate area. Making this change would decrease
$V_{n}$$C$ by a factor of 1.25. Since $ENC$ goes as the fourth root of
$V_{n}$$C$, $ENC$ would decrease only 5%. On the other hand, $ENTC$ would
decrease by 16%.

##  6 Sampling Requirements

So far we have assumed that the signals are ideally and continuously measured.
In fact they will be sampled with limited precision. What restrictions on the
sampling rate, length, and accuracy will prevent the loss of information?

###  6.1 Sampling frequency

The sampling frequency must be high enough to capture all frequencies with
useful $S/N$. It must also be high enough to avoid misrepresenting noise at
higher frequencies as noise in the signal region (aliasing). On the other hand
the sampling frequency should be no higher than necessary.

What is the highest useful frequency? We note that $S/N$ is approximately
constant up to $\omega=\omega_{sb}$ and then decreases as $1/\omega^{4}$. If
we throw away all information above $\omega=\omega_{c}$ the fraction of
information lost is approximately $(3/4)(\omega_{sb}/\omega_{c})^{3}$. For a
20% loss of information this implies that $\omega_{c}\geq 1.5\omega_{sb}$, and
that $f_{c}>0.6\mbox{\,MHz}$. The fractional loss in time signal noise is
somewhat larger, being given approximately by $(3/4)(\omega_{sb}/\omega_{c})$.
Thus our choice based on the amplitude error will lose us half of our time
information and increase our time error by 25%. This is probably acceptable.
The exact formula for the fractional loss of amplitude information is

$1-\frac{1}{\pi}\left(\frac{1}{2}\log\left(\frac{\omega_{c}^{2}+2\omega_{sb}\omega_{c}+2\omega_{sb}^{2}}{\omega_{c}^{2}-2\omega_{sb}\omega_{c}+2\omega_{sb}^{2}}\right)+\tan^{-1}\left(\frac{2\omega_{c}\omega_{sb}}{(2\omega_{sb}^{2}-\omega_{c}^{2})}\right)\right)$
(76)

Assume that we have one stage of near ideal integration before the digitizer
(the charge sensitive preamp) followed by a near ideal differentiator. This
pair gives a band pass filter with a pass band from a low frequency to
$f_{c}$. The fall in the stop band is $1/\omega$. Further assume that there is
an additional two pole low pass filter with a corner at $f_{c}$ (drop off like
$(\omega_{c}/\omega)^{2}$ for $\omega>\omega_{c}$) between the preamp and the
digitizer. If we sample at frequency $f_{d}$, noise at frequencies above
$f_{d}/2$ will appear at $f_{obs}=f_{d}-f$. We are interested in studying the
contribution to $N$ from $V_{n}^{2}$ which drops off as $(\omega_{c}/w)^{4}$
after all the analog filters. We have sufficient protection against aliasing
if $V_{n}^{2}$ at $f_{d}/2-f_{c}\ll V_{n}^{2}$ at $f_{c}$. This is satisfied
if $(f_{c}/(f_{d}/2-f_{c}))^{4}<4$. This implies $f_{d}>3.4f_{c}$, which is
$3.4(1.5)/(2\pi\sqrt{\tau_{s}\tau_{b}})=0.8/\sqrt{\tau_{s}\tau_{b}}$. This is
2.5 MHz at nominal background and 4.7 MHz at $\times 10$ nominal. 3.7 MHz is
adequate adequate for nominal backgrounds and marginal for $\times 10$.

###  6.2 Sample length

The number of samples required is related to the lowest frequency of interest.
So far we have assumed that all integrals go from zero frequency. Pile up
makes this impractical as does the fact that we can’t use all past history and
wait forever to get an answer.

Lets look at the pile up problem first. Allowing for variations from crystal
to crystal, and warm diodes, we should plan for at least 20 nA of total
current in a diode. If the offset from this current is to stay within 10% of
full scale, the product of this current and the integration time must be less
than 10% of the charge of a 14 GeV shower, i.e., $<0.5\rm\,pC$. This limits
the integration time to 200$\,\mu\mbox{s}$. This sets the minimum value for
the frequency corner of the first integrator.

This gives 740 samples as the maximum useful number at 3.7 MHz. This is a
large number. What is the smallest number of samples we can use without
significant loss of information? For the amplitude parameter, the $S/N$ is
flat up to the corner frequency and then drops rapidly. For our lowest noise
case ($I_{b}=0$ during source calibration for example), we have a corner
frequency of 80 kHz. If we permit a 20% loss of information, we can cut off
the integral at the low frequency side at 13 kHz. This is less than three
times the previous integration limit leaving us little choice and many
samples. The problem here is that the sample rate is much higher than needed
for this case. Once filtering has been performed, the sample set could be
decimated, but most of the work is done by then. In the nominal background
case the corner frequency is 400 kHz, and the integral could be cut off at 80
kHz for a sample length of 12.5$\,\mu\mbox{s}$ or 46 samples. This would be
the normal operating mode. If the sampling frequency were optimized for this
nominal background level, the number of samples would be less than 25.

###  6.3 Sampling summary

We can summarize the conclusions of the last two sections by noting that the
number samples that must be dealt with is proportional to the ratio of the
highest and lowest frequencies of used. The proportionality constant lies
somewhere between 2.5 and 5 depending of the details of the system and how
“highest” is defined. If the signal to noise ratio is approximately constant
from low frequencies up to the highest frequency of interest then the fraction
of the available information retained (the efficiency) of the system is given
by $1-f_{lowest}/f_{highest}$. An interesting but not necessarily relevant
observation: the maximum information per sample is obtained when $f_{lowest}$
is one half of $f_{highest}$ and the efficiency is 50%.

##  7 Quantization Error

###  7.1 With no net integration or differentiation

With the same assumptions about filtering as in the previous section we can
estimate the requirements on the quantization error. We assume there is a low
pass filter such that there is no aliasing, no loss of information, and there
are offsetting integrators and differentiators such that there is no net
effect in the pass band. Under such conditions, quantization error may be
referred back to the input where it appears as another current noise source.

According to the sampling theorem [Papoulis, page 141], if the anti-aliasing
conditions are met, the original signal may be reconstructed from the samples
by the interpolation formula

$i_{s}(t)=\sum_{n=-\infty}^{\infty}i_{s}(nT_{d})\frac{\sin((\omega_{d}/2)(t-nT_{d}))}{(\omega_{d}/2)(t-nT_{d})}$
(77)

where $T_{d}$ is the sampling period, and $\omega_{d}$ is related to it in the
usual way. An error in a sample can be represented as a function proportional
to one element of this sum. The Fourier transform of such a function is flat
out to $f_{d}/2$. The amplitude of this noise pulse is drawn from a square
distribution whose width is given by the bit resolution (or the effective bit
resolution for a non-ideal ADC). The rms area of the pulse is then
$T_{d}i_{bit}/\sqrt{12}$. The rate of pulses with effectively random phase is
$f_{d}\equiv 1/T_{d}$, and the noise spectral density is flat with a value of
$I_{q}^{2}=i_{bit}^{2}/(12f_{d})$. This is to be added to the FET input
current noise. We should compare this to $I_{n}^{2}$ for very quiet
conditions, and to $I_{b}^{2}$ for nominal conditions.

First lets treat the quiet case for which $ENC^{2}$ goes as
$\sqrt{I_{n}^{2}}$. If we wish to restrict our error increase to less than 5%
due to this source we require that $I_{q}^{2}<I_{n}^{2}/5$. Therefore
$i_{bit}$ should be less than $\sqrt{2f_{d}}\,I_{n}$. For the example diode
and our sampling rate, this is $i_{bit}<68\mbox{\,pA}$. The current at the
signal peak is $q/\tau_{s}$ if there is no analog shaping. For a triple RC
filter with a time constant of 0.25$\,\mu\mbox{s}$, this is reduced by a
factor of 2. Assuming that 12 GeV somehow gets into one crystal, the maximum
peak current would be 3 $\mu$A. This implies a dynamic range requirement of
43,000 or 15.4 bits. Since the signal is not calibrated, we need to add a bit
for gain variations suggesting that 16.5 effective bits of dynamic range are
required during quiet conditions. Note that better light collection imposes a
greater dynamic range requirement.

During nominal conditions, $ENC^{2}$ goes as $\sqrt{I_{b}^{3}}$. The 5% error
increase condition implies that $I_{q}^{2}<I_{b}^{2}/7$, and that $i_{bit}$ be
equivalent to less than $\sqrt{1.7f_{d}}\,I_{b}$. For nominal background
conditions and our sampling rate this is $i_{bit}<710\mbox{\,pA}$. The dynamic
range requirement is 4,200 or 12 bits. With an extra bit for calibration
differences, this is 13 bits.

In practice, many crystals are summed to measure a shower. Both the
electronics and background noise increase as $\sqrt{m}$, where $m$ is the
number of crystals included. The dynamic range required decreases as
$\sqrt{m}$. For a 9 crystal sum and nominal background conditions, the number
of bits required is 10.5.

###  7.2 With one net integration

If we assume that instead of offsetting integrators and differentiators, we
have only one integrator and then the low pass filter, the quantization error
behaves as a voltage noise when referred to the input. The rms area of the
error pulse is $T_{d}v_{bit}/\sqrt{12}$, and the noise spectral density is
flat with a value of $V_{q}^{2}=v_{bit}^{2}/(12f_{d})$. This is to be added to
the FET input voltage noise $V_{n}^{2}$. Recall that in our case $ENC^{2}$
goes as $\sqrt{V_{n}^{2}}$ in the quiet case and as $\sqrt[4]{V_{n}^{2}}$ in
the nominal case. If we wish to restrict our error increase to less than 5%
due to this source we require that $V_{q}^{2}<V_{n}^{2}/10$ in the quiet case
and $V_{q}^{2}<V_{n}^{2}/5$ in the nominal case. This implies that $v_{bit}$
be equivalent to less than $\sqrt{1.2f_{d}}\,V_{n}$ and
$\sqrt{2.4f_{d}}\,V_{n}$, respectively. For the example FET and our sampling
rate this is $v_{bit}<1.05\mbox{\,$\mu$V}$ and $v_{bit}<1.5\mbox{\,$\mu$V}$,
which given the source capacitance corresponds to a charges of $<660$ and
$<940$ e, and to a shower energies of $<0.22$ and
$<0.31\mbox{\,Me\kern-0.80002ptV}$. Assuming that 12 GeV somehow gets into one
crystal, the dynamic ranges required are 55,000 and 42,000, or 15.7 bits for
quiet conditions and 15.4 bits for nominal conditions. Since the signal is not
calibrated, we need to add a bit for gain variations suggesting that 17
effective bits of dynamic range may be sufficient.

All of the above dynamic range calculations address the sufficient conditions
for which the quantization error will not contribute to the electronic (and
background) noise. They do not treat whether or not this dynamic range is
necessary given the inherent energy fluctuations of the shower process.

The considerations for multi-range digitizing have been addressed by Al Eisner
and Gunther Haller and remain unchanged. For larger pulses, sources of error
other than electronics dominate.

##  8 Filter Implementation and Interpolation

There is still the problem of signal extraction. Because it is non-causal, a
matched filter cannot be implemented as a simple analog filter. A procedure
for directly calculating the coefficients of a tapped delay line approximation
to the optimum matched filter is described by Papoulis [Papoulis, page 327].
The result is exactly parallel to the usual least square fit formulas. The
autocorrelation function, $n_{f}$, is the inverse Fourier transform of the
noise power spectral density after the analog filter. The elements of $\rm r$,
the error matrix for the samples, are given by $n_{f}(|l-k|T)$. The expected
values $\bf a$ for the samples is given by $i_{sf}((l-m)T+t)$, where $i_{sf}$
is the expected signal after analog filtering and $t$ is the time offset of
the signal from sample $m$. The coefficients of the tapped delay line filter
are given by $\rm r^{-1}\bf a$. The optimally filtered estimate for the signal
at time $t$ is given by $o(t)=\bf d\rm r^{-1}\bf a$, where $\bf d$ is the set
of samples. For the case of the triple RC analog filter, the coefficients each
have four terms with different $t$ dependencies. The estimate for $o(t)$ also
has four such terms and its derivative with respect to $t$ has three.
Calculating $o(t)$ takes four multiply and adds for each sample used. Finding
the root of the expression for the derivative with Newton’s method should not
take many iterations. Each iteration involves calculating one exponential and
approximately a dozen multiply and adds. The presence of multiple peaks in the
time window or the absence of any peaks would complicate this last step. A
possibility is to not try to find the peak, but to report the coefficients of
the four terms instead.

Knowing the event time would reduce the time to get an estimate for $o(t)$ by
a factor of four, and there would no need to search for the maximum.

##  9 Less Than the Best

The previous sections deal largely with optimum solutions, albeit with some
practical limitations. This section will examine the loss of information if
other that optimal filtering is used. It may be less than optimal in that it
is not matched, e.g., an all analog filter, or in that the pass band is
optimized for another condition.

Figure 7: Raw input current signal and the signal after triple RC filters with
time constants of $0.25\,\mu\mbox{s}$ and $2.0\,\mu\mbox{s}$.

Figure 8: The peak signal current after a triple RC filter vs. the filter time
constant

Let us look at the case of an all analog filter. In this case all filtering is
done prior to digitizing and the only digital processing performed is
interpolation. I will take the B A B AR TDR design (described in subsection
6.1). This has a charge sensitive preamp followed by a CR and then two RC
filters. Together these give the equivalent of three RC filters, all with the
corner frequency $f_{c}$. This has the transfer function

$F(\omega)=\left(\frac{-j\omega_{c}}{\omega-j\omega_{c}}\right)^{3}.$ (78)

The signal is

$I(\omega)=\left(\frac{-j\omega_{s}}{\omega-j\omega_{s}}\right).$ (79)

Figure 9: The Equivalent Noise Energy (electronic only) of a single sample at
the peak after a triple RC filter versus the radial velocity of the filter
corner ($1/RC$).

Figure 10: The efficiency vs. 1/RC for nominal background conditions.

The time dependence of the output signal is given by the inverse Fourier
transform of their product. This integral can be done by contour integration.
If the contour is completed by a semi-circle at infinity around the upper half
plane (for $t>0$), the integral can be deduced from the residues at the poles
$j\omega_{c}$ and $j\omega_{s}$. The result is

$o(t)=\frac{\omega_{c}^{3}\omega_{s}}{(\omega_{s}-\omega_{c})^{3}}\left\\{\left(\frac{(\omega_{s}-\omega_{c})^{2}}{2}t^{2}-(\omega_{s}-\omega_{c})t+1\right)e^{-\omega_{c}t}-e^{-\omega_{s}t}\right\\}.$
(80)

This is shown for several time constants in figure 8. The peak value vs. time
constant is shown in figure 8. Since there are no poles in the lower half
plane, the integral is $0$ for $t<0$.

Figure 11: The efficiency vs. 1/RC for low background conditions.

Figure 12: The efficiency vs. 1/RC for $10\times$ nominal background
conditions.

The mean square noise is given by the inverse Fourier transform of the noise
evaluated at $t=0$, i.e., the autocorrelation for no time difference. The
noise before filtering is given by equation 51. The mean square noise is

$n=\frac{1}{\pi}\int_{0}^{\infty}{N(\omega)|F(\omega)|^{2}\,d\omega}.$ (81)

The contribution of the electronic noise is

$n_{elec}=\omega_{c}(V_{n}^{2}C^{2}\omega_{c}^{2}+3I_{n}^{2})/32.$ (82)

The peak of $\sqrt{n_{elec}/o^{2}(t)}$ gives the ENC for the electronic noise
after analog filtering. Appropriate scaling gives the ENE shown in Figure 10.
The background noise contribution is

$n_{back}=\frac{(8\omega_{c}^{2}+9\omega_{s}\omega_{c}+3\omega_{s}^{2})\omega_{c}\omega_{s}I_{b}^{2}}{32(\omega_{c}+\omega_{s})^{3}}$
(83)

$n$ is given by their sum. The signal to noise ratio for a single sample taken
at time $t$ is given by $o^{2}(t)/n$. If we divide this by the optimum signal
to noise given by equation 55 we get the “efficiency” for this single sample
versus the sample time. Figure 10 shows this function for the B A B AR
calorimeter during nominal operating conditions for several values of the
corner frequency. The highest efficiency achieved is 68% for a shaping time
constant of 0.2$\,\mu\mbox{s}$. For low background conditions (Figure 12) the
peak efficiency is 80% at a shaping time of 1.5$\,\mu\mbox{s}$. At ten times
nominal background, (Figure 12) the efficiency drops to 50% at a shaping time
of 0.12$\,\mu\mbox{s}$. The loss of efficiency for nominal and better
background conditions is not significant. The loss at during high background
conditions hits just when noise is the causing the most problems. Matched
filters utilize knowledge of the signal shape. In low background conditions,
the optimum shaping time is longer than the signal. The shape of the signal is
not seen and matched filters offer little advantage. The situation and
conclusions are reversed in high background conditions.

The cost of using an analog filter optimized for conditions other than those
encountered is more dramatic. If the corner frequency is set to the
0.2$\,\mu\mbox{s}$ which is optimal for nominal conditions, but the
backgrounds are actually at $10\times$ nominal levels, the efficiency is 44%.
If the corner frequency were optimal for low background conditions, the
efficiency at $10\times$ nominal background would be 12%.

##  10 Summary of Significant Findings

Since the important conclusions may have gotten lost in the derivations, I’ll
review them here.

  * •

Machine background can be treated as a current noise with the spectrum of the
signal.

  * •

The best possible amplitude and time measurements can be obtained with a
matched filter.

  * •

The errors in the amplitude and time are not correlated, therefore knowing the
time does not improve the ultimate precision of the amplitude determination,
although such knowledge may reduce the computation required.

  * •

The signal to noise methodology used here reproduces previous results.

  * •

The background current noise dominates the photodiode shot noise even at
nominal backgrounds and in the quietest part of the calorimeter.

  * •

For nominal background where $I_{b}\gg I_{n}$:

$\displaystyle ENC^{2}$ $\displaystyle\approx$ $\displaystyle
I_{b}^{2}\sqrt{2\tau_{s}\tau_{b}+\tau_{b}^{2}}=\sqrt{2\tau_{s}I_{b}^{3}V_{n}C+I_{b}^{2}V_{n}^{2}C^{2}}$
$\displaystyle ENTC^{2}$ $\displaystyle\approx$ $\displaystyle
ENC^{2}\tau_{s}\tau_{b}=\sqrt{2I_{b}\tau_{s}^{3}V_{n}^{3}C^{3}+\tau_{s}^{2}V_{n}^{4}C^{4}}$

  * •

The sampling frequency of 3.7 MHz is adequate at nominal backgraounds, but it
is marginal for $10\times$nominal background levels.

  * •

At this frequency, 64 samples will have to be processed under nominal
conditions.

  * •

The step of an ADC bit (on the most sensitive range) should be set so that
there are at least 13 bits to full scale (12 GeV) during nominal background.
For quiet conditions and with a differentiating stage in the analog filter,
16.5 bits may be required.

  * •

Simple analog filters can achieve near optimal results in low to moderate
background conditions if the shaping time is optimized for the actual
conditions.

##  11 Acknowledgements

I thank my BaBar colleagues for consultations and for providing the data used
in the examples. This work was supported by the U.S. Department of Energy
under contract number DE-AC02-76SF00515.

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