**Spice Model**

Model of 8 GeV Style BPM includes 50 Ohm Terminating Resistor and 500
ft. of 3/8 '' Heliax cable . The cable is terminated in 50 Ohms. The BPM
model includes the plate-plate coupling capacitance and the plate
capacitance. The current source is shunted with 1 MOhm resistor for
modeling purposes.

![](media/image1.wmf)

![](media/image2.png){width="4.999305555555556in" height="3.8125in"}

Cable length is 600 ft RG-8/U attn = 1.3 dB/100 ft @ 50 MHz. Beam
Intensity is 3.94e12/batch. The signal at the BPM is 4Vpp /0.407 = 9.83
Vpp. For 1e13 p/batch the signal at the detector would be \~ 25 Vpp. If
the termination is 25 Ohms the amplitude would be 12.5 Vpp.

A 12.5 Vpp signal sent through 3/8 Heliax (atten=.848 dB/100 ft) 767 ft
of cable = 5.9 Vpp and for 175 ft = 10.53 Vpp. At receiver end.

![](media/image3.wmf)![](media/image4.wmf)

The 4 Vpp signal of Q846A was connected to a 53 MHz Bandpass filter and
terminated in 50 Ohms. The signal level changed from 4 Vpp to 1.8 Vpp a
--6.93 dB of filter insertion loss.

A 12.5 Vpp signal sent through 3/8 Heliax (atten=.848 dB/100 ft) 767 ft
of cable = 5.9 Vpp and for 175 ft = 10.53 Vpp. At receiver end. These
signals connected directly to a 53 MHz bandpass filter would have 2.66
Vpp and 4.73 Vpp of 53 MHz component at the input of the digital down
converter board.

![](media/image5.wmf)![](media/image6.wmf)

System Requirements:

  ---------------------------- ------------------------------------------
  NuMI spill (pbar operation)  5 batches x 84 bunches = 8.2 sec

  NuMI spill (no pbar          6 batches x 84 bunches = 9.8 sec
  operation)                   

                               

  Maximum intensity            4 x 10^13^ ppp (protons/spill)

  Total beam power             404 kW at maximum intensity
  ---------------------------- ------------------------------------------

Worst case design 1e13 particles per batch. plus factor of 2 for
off-center beam.

**Time Domain Calculations:**

Ib(t) = $\frac{\text{nq}}{\sigma\sqrt{\begin{matrix}
2 & \pi
\end{matrix}}}e^{- \frac{1}{2}\left( \frac{t}{\sigma} \right)^{2}}$
n=1^e^13 particles/80 bunches q=1.602^e^-19 C/particle =2^e^-9sec

Ib(0) = 3.994Amps.

The output of the BPM is dIb/dt times the length of the detector L/2c
that is 5.25 in (0.13335 m).

$\frac{l}{\begin{matrix}
2 & c
\end{matrix}}$= 222.25e-12 sec

$\frac{\text{dI}}{\text{dt}}$ =
$\frac{\text{nqt}}{\sigma^{3}\sqrt{\begin{matrix}
2 & \pi
\end{matrix}}}e^{- \frac{1}{2}\left( \frac{t}{\sigma} \right)^{2}}\frac{l}{2c}$

![](media/image7.png){width="3.872916666666667in"
height="2.904861111111111in"}

plot of dI(t)/dt using the following matlab program

function numi_i()

sigmat=2.0e-9;

t=8e-10:8e-10:100e-9;

l=0.13335; c=3e8;

n=1.0e13/(80); q=1.602e-19;

%detector length

bpm=l/(2\*c);

\% derivative of bunch

\%

dib1=((n\*q\*(10e-9 -
t))/(sigmat\^3\*sqrt(2\*pi))).\*exp(-0.5\*((10e-9 - t)/sigmat).\^2);

dib2=((n\*q\*(28.87e-9 -
t))/(sigmat\^3\*sqrt(2\*pi))).\*exp(-0.5\*((28.87e-9 - t)/sigmat).\^2);

%di_bpm1=dib1+dib2;

di_bpm1=dib1\*bpm+dib2\*bpm;

plot(t,di_bpm1);figure;grid on;

dIb(0)/dt = 0.52 Amps peak-peak or 0.26 Amps peak.

The load resistor is 25 Ohms. The voltage generated is

dIb(0)/dt \* R = 0.52 Amps peak-peak \* 25 = 13 Volts peak-peak.

Cable attenuation is .848 dB/100 ft. The shortest cable length is 175 ft
(1.48 dB attn) and the longest is 767 ft (6.5 dB attn)

dIb(0)/dt \* Ro // Rl \* Cable attn = 13 Volts peak-peak \* 0.8433=
10.96 Volts peak-peak and 13V p-p \* 0.473 = 6.15 Volts peak-peak.

A safety margin of 2 should be added for cases where beam is off-center
near one of the plates, doubling the total voltage seen by the receiver.

Vin = 21.92 Vp-p and 12.3 Vp-p for 1e13 particles per batch.

The signals at the coupled output are attenuated by 11 dB. This makes
the signal levels at the input of the op-amp

Vopamp = 6.17 Vp-p and 3.4 Vp-p.

**Frequency Domain calculations** :

In a batch the average current determined by the number of protons,
their charge and the batch period

Iav = $\frac{\text{nq}}{\tau}$

Iav= 1^e^13 \* 1.602e-19/ 1.6e-6 = 1.001 Amps

I(f) =
$I_{\text{av}}e^{- \frac{1}{2}{(\text{ωσ})}^{2}}\frac{\omega L}{2c}$

The output of the BPM is the derivative of I(f)

$$\frac{\text{dI}}{\text{df}} = \text{Fourier}\left\{ \frac{\text{di}}{\text{dt}} \right\}$$

$\frac{\text{dI}}{d\omega}$=$\frac{\begin{matrix}
n & q & \pi & f
\end{matrix}}{\begin{matrix}
2 & \sqrt{2} & \sigma & 
\end{matrix}}\frac{l}{c}e^{\begin{matrix}
 - & 2 & \begin{matrix}
\pi^{2} & \sigma^{2} & f^{2}
\end{matrix}
\end{matrix}}$

**Calculated values for dI/df case:**

For 53.1 MHz and 106.2MHz

I(f) 53.1 = 261.9e-3\* 0.895 = 234.48e-3 Amps; I(f)106.2 = 524.58e-3 \*
0.411 = 215.6e-3 Amps.

Consider a cable with its 25 Ω termination, the load voltage is

V(f) 53.1 = 5.86 Volts peak and V(f) 106.2 = 5.39 Volts peak