# Presentation: 297682

## BPM Signal Processing:              A Cartoon View

- Rob Kutschke
- Oct 13, 2003

## Outline

- Reminder about notation.
- Reminders about Fourier Transforms  (FT).
- Bunch shape.
- Response of resonant filter to a single pulse.
- Response of resonant filter to multiple pulses, batch mode and bunch mode.
- A proposal about how to analyze the output of the resonant filter.

## Reminder about Notation

- F(t) = sin( 2f0 t )  = sin( t )
- Frequency: f0
- Angular frequency: 0 = 2f0
- Period: T0 = 1/f0 = 2/0
- When someone says “frequency” they usually mean f0 but sometimes they mean 0  !

## FT of a Gaussian is Gaussian

- Bunch shape

- Power spectrum of a single bunch passing a fixed point.

## FT of Finite Wave Train

## FT of Finite Wave Train

- Define N = number of oscillations in the time domain ( = 10 here ).
- First zero near max is at:  () =  ± 1/ N        ( = in this example)
- Equivalent statement:  zero occurs when the number of oscillations in the interval differs by ±1.

## Bunch Shapes

- Shape of a single bunch is nominally a Gaussian.
- Never really is.
- Nominal scaling of bunch length is 1/sqrt(E).
- In real life bunches do get shorter with energy but slower than this.

- Coalesced bunches:
-    4 ns at 150 GeV   ( injection energy )
-    2 ns at 980 GeV
- Uncoalesced bunches
- About 1/9 as many particles/bunch.
-    2 ns at 150 GeV   ( injection energy )
-    1 ns at 980 GeV

## Coalesced vs Uncoalesced

- Coalescing: take n bunches, separated by 19 ns and put them on top of each other.
- Typically n=9.
- Nominal behaviour:
- Number of particles in the coalesced bunch is n times the number in each uncoalesced bunch.
- Time width increases as sqrt(n)
- This ideal case is usually not achieved and bunch width is a little wider.

##  Is 200 MHz is fast enough to digitize pulse with no shaping?

-  Is 200 MHz is fast enough to digitize pulse with no shaping?
-  A and B plates must be digitized at same time: time difference implies a position error.

## Three Fill Patterns of Interest

-  Third pattern is just a single bunch in the machine.
-  In batch mode there is only ever one batch in the  machine.  Number of bunches in a batch may change.
-  Present bunch mode is 3 trains of 12 bunches. The new BPM should allow other choices.

## Required Measurements

- First turn:
- Only needed for a single bunch in the machine.
- Report turn by turn positions for first turn.
- Turn by turn:
- Report turn by turn position.
- Only needed for a single bunch or single batch in the machine.
- Closed orbit:
- Means to average over many turns to average out the betatron oscillations.
- Needed for all fill patterns.

## Cartoon of Our Options

- Superfast digi, no shaping ( 2 GHz or more ).
- Very expensive.
- Shape the pulse ( Digitize once or many ).
- Need to work hard at timing?  Especially on first turn?
- Tradeoff: long pulse good for precision and accuracy but bad for separation of protons and anti-protons.
- How to deal with batch mode when bunches are 19 ns apart?
- Ring resonant filter and transfer position info to the frequency domain.
- Coherent addition of bunch signals is natural.
- Give up single bunch resolution.

## Existing RF Module

- Band Pass Filter

- Low Pass Filter

- B

- Position  Circuit

- Intensity  Circuit

- Look at waveform here ....

- ... and here

- Various ideas of the new system have either a band-pass + low-pass filter or just a low-pass fitler at their front end.

- See Beams-doc-815, page 8.  (a talk by Bob Webber ).

## Cartoon shows the 53.1 MHz component only.  The true signal will be complicated by other frequencies inside the band-pass.

- Why 53.1 MHz? How many rings do we want/need?     Answers later.

##  Bunch pattern in batch mode. 

-  Bunch pattern in batch mode.
-  Bunches are uncoalesced.

-  Cartoon response of the resonant filter, to the above batch of bunches.
-  Existing BPM electronics designed to use this signal.

##  Bunch pattern in “bunch mode”, 396 ns between bunches in a train.

-  Bunch pattern in “bunch mode”, 396 ns between bunches in a train.
-  Bunches are coalesced.

-  Cartoon response of the resonant filter, to the above batch of bunches.

## Comment on 53.1 MHz

- The 53.1 MHz component of the filter is kicked in phase by each passing bunch.
- Also true for harmonics of 53.1 MHz.
- All other frequencies will be excited out of phase by each bunch.
- Wave form at these frequencies is more complex and either we lose information or we need fancier signal processing.
- Following slides will talk about extracting the information at 53.1 MHz.

## How to process the above waveform to get a position?

- A and B signals differ in amplitude.
- For now, consider just protons or anti-protons in the machine.
- Also assume no significant reflections.

## Single pulse, coming out of the front end filters.  It contains:
The time structure of the beam.
Modified by the response of the pickup, cables and front end filters.
Call this shape:
Can also say the same things in frequency domain.
Call that shape: 

- The time structure of the beam.
- Modified by the response of the pickup, cables and front end filters.
- Call this shape:
- Can also say the same things in frequency domain.
- Call that shape:

- Time to Frequency Domain

- Details of this shape will not be important.

## Cartoons of Frequency Response

- Frequency content of a single gaussian bunch.

- Frequency response of pickup (Beams-doc-766, page 10 ).            Line matches scale in top plot.

- Frequency response of:     - band pass filter            - low pass filter

- Is the product of  these shapes.

## Comments on Previous Slide

- Really should also draw the phase information for these plots.
- The full function will be slowly varying in some neighbourhood about 53.1 MHz.
- Not sure what the phase does in this region or if it is important?

## Single Bunch, Multiple Turns

-  Time dependence of signal after filter of a single bunch which never changes shape and which follows a closed orbit past the same point N times:

## Same Thing, Frequency Domain

-  Fourier Transform of a single pulse:

-  FT of the full waveform:

## Repeat last equation on previous page.

- Change variables, u = t - tn

- Reorder sum and integral.

- FT of the single pulse shape comes out of the sum!

- Insert:

- Explore this shape for tn=nT0.  For large N it will be a comb.

##          Toy Model of:

- The model with f0 = 1 Hz
- No signal for a long time.
- N copies of one pulse, each separated by T0
- No signal for a long time.
- No edge effects of the FT window being too close to the first or last pulse.
- Try different values of N.

##     Fourier Transforms of                          for different N

-     Fourier Transforms of                          for different N

- These functions multiply the FT of a single pulse.

## Add Betatron Oscillations

-  Sinusoid dependence of An may be an approximation?
-  This has the approximation that betatron phase is constant over the time of the bunch ( 2-4 ns).

-  A(t) = signal on the A plate.
-  B(t) = Signal on B plate will have opposite phase.

## Repeat last equation on previous page.

- Change variables, u = t - tn

- Reorder sum and integral.

- FT of the single pulse shape cancels out.

- Insert:

- Do the math with A0  An

- Now look at this shape for betatron oscillations.

## Betatron Oscillations

- We have seen the first term already.  Look at the second term.

- Extend the previous toy model.  f0=1 Hz, f10 Hz,  = 0

## Fourier Transforms of                                           for different N

- Sidebands appear at  = 0 ±  .

## Add Synchrotron Oscillations

-  Approximation that the synchroton phase is constant across the bunch, ( 2-4 ns ).
-  Won’t go any farther with this for now.  Also will create a sideband structure.

## Two identical bunches, separated in time by 

- Still under construction.

## Two Different Bunches

## First Turn Mode

- Needs to work only for one bunch in the machine.
- For each of A and B signals:
- Digitize at  200 MHz.
- Compute FFT of digitized time series.
- Integration time can be as long as a full turn if the filter output after one turn is still above the noise and least count.
- Long integration time = narrow band.
- Output of FFT is the power in a frequency band centered at 53.1 MHz. This is A or B.
- Position = (A-B)/(A+B)

## Question

- I think that a long integration time in the FFT implies more precise values of A and B, which implies more precise position?   Is this right?
- At the same time, it changes the meaning of A and B since the bandwidth decreases.

## Turn by Turn Mode

- Only required to work for single bunch or single batch in the machine.
- Requirements say that it is OK to average over bunches:
- How many?
- Report this average on each turn.

## Turn by Turn mode

- Works as for first turn but the integration time window of the FFT may be changed.
- Might integrate over signal from one bunch, from several, ... up to all 36 bunches.
- Integration over several bunches produces some sort of average position.
- Question: if there are no bunch to bunch diseases, does accuracy improve if integration time is longer?

## Closed Orbit

- Must work for all fill patterns.
- As before but integration time of FFT is many turns.
- This averages out the betatron oscillations.
- Questions:
- If we integrate a long time the bandwidth narrows - can we lose important information located in the sidebands?  Maybe A becomes a histogram covering a range of frequencies, not just a single number.  Then do peak finding on it?

## What About Anti-Protons?

- For now, consider proton signal corrupting the anti-proton measurement.
- When the anti-proton intensity problems are solved, we will also need to worry about the antiprotons corrupting the proton measurement.

## What About Anti-Protons?

- Now we get a waveform from each end of each plate.  Processes each the same way as described before.
- In the limit of perfect directionality, we directly measure: AP, APbar
- Real numbers, the output of each FFT.
- For imperfect directionality, the observed numbers are contaminated with each other.

## Imperfect Directionality

- Define two feedthrough coefficients: 1, 2.
- Also need to know relative phase, ,  between proton and pbar waveforms.
- Depends on state of cogging.
- These differ from one BPM to the next.
- The instrument measures:

## Imperfect Directionality

- Measure P and Pbar, to get two equations and two unknowns.
- Solve for AP, APbar.
- Need to know 1, 2, and for each BPM and each cogging.
- Similarly to extract BP, BPbar.

## Questions

- The 2-species response also depends on the relative phases of the betatron and synchrotron oscillations of the p and pbar?  Is this a signifcant effect?  Is it stable?  Is it repeatable from store to store?
- Closed orbit is probably immune from this regardless of its size, but not turn by turn?

## Second Comment on 53.1 MHz

- Could this work with other frequencies ?
- Sure, but.
- Only at 53.1 MHz and its harmonics is the interference between bunches in phase for all fill patterns.
- At other frequencies, some fill patterns will excite destructive interference, which reduces the sensitivity of the measurement.
- Could do 53.1 MHz for batch mode and  53.1/21 MHz for bunch?

## Back to Short Pulse

- The alternative to the ringing filter is to stretch the pulse out a little and then sample it ( either single measurement or 200 MHz sample ).
- Tightest time window is set by proton antiproton separation.

## For each BPM, compute arrival time of proton bunch.
Compute position of all antiproton bunches at that time.
Plot position difference of P and Pbar.
Above figure:
For “Collision Point Cogging”, P1=A1 at F0.
Not sure of sign.
Only 6 instances with separation < 2.35 buckets ( 44.5 ns ).

- Compute position of all antiproton bunches at that time.
- Plot position difference of P and Pbar.
- Above figure:
- For “Collision Point Cogging”, P1=A1 at F0.
- Not sure of sign.
- Only 6 instances with separation < 2.35 buckets ( 44.5 ns ).

## P/Pbar Overlaps for Collision Point Cogging  

-  BPM     Proton AntiProton
- HPA0U      1        13
- HPA0U     25       25
- HPB0U       1       25
- VPA0U       1       13
- VPA0U     25       25
- VPB0U       1       25

## Protons and Anti-Protons

-  Define shapes of after pickup, cables and filters
-  f(t): proton signal from the proton end.
-  g(t): anti-proton signal on the proton end.

## Backup Slides

## Signal on proton end from proton bunch in a regular store.
Tiggered either by “intensity signal” or by a TeV marker???
Very little ringing.
FWHM of positive piece is about 4 ns.
Gaussian:   1.7 ns.

- Signal on proton end from proton bunch in a regular store.
- Tiggered either by “intensity signal” or by a TeV marker???
- Very little ringing.
- FWHM of positive piece is about 4 ns.
- Gaussian:   1.7 ns.

- A Plate

- B Plate

- Scope picture from Fritz D.
- Uses HPA17
- Signals on cables in house.
- No filters or attenuators

## Signal on anti-proton end induced by the proton bunch.
Tiggered same way as previous slide.
Unsure how to interpret timing wrt previous slide.
As before:
Very little ringing.
FWHM of first negative piece  4 ns.

- A Plate

- B Plate

- Scope picture from Fritz D.
- Uses HPA17
- Direct from cables in house.
- No filters or attenuators

- Signal on anti-proton end induced by the proton bunch.
- Tiggered same way as previous slide.
- Unsure how to interpret timing wrt previous slide.
- As before:
- Very little ringing.
- FWHM of first negative piece  4 ns.

## Slide 52

## Comments for Next Slide

- This is a scope picture showing the response of a BPM to a train of about 30 uncoalesced bunches.
- Top trace shows signal on cables
- Bottom 3 traces characterize BPM response.
- These are all in time wrt each other
- The top trace has an unknown time shift.

## Train of 30 bunches, 19 ns apart.  

- Train of 30 bunches, 19 ns apart.
- Different t0 than other traces

- Intensity (A+B)

- Gate

- Position Signal

## Notes on Previous Slide

- The bottom 3 traces have the same timing.
- The top trace should shift to the right by ??? ns.
- Rise time of intensity signal is about 200 ns.
- Gate internally generated by threshold on intensity signal. (circled glitch)
- Is gate width programmable????

- Position Signal:
- Overshoot at small time.
- Position signal stable after about 200 ns.
- Sample and hold on falling edge of gate (circled glitch)
- Why is overshoot at end not in the opposite direction from that at start?

## BPM Response to a Single Coalesced Bunch

- Intensity A+B

- Position

- Next bunch comes at this time  ( 396 ns ).
- Does this signal ring at longer times?

- Scope picture from Fritz D. ( HPA17 ).
- Before removal of PSD boards.  Now ringing in position signal is less.

## Comments on Previous Slide

- Intensity Signal
- Rises to peak in 80 ns
- Much faster than batch mode.
- Conclude: this signal is never fully developed.
- Is there ringing at larger times – if so, then one bunch talks to the next bunch.

- Position Signal:
- Overshoot has same width as for batch mode.
- Never really gets flat.
- Integration time of sample+hold  a few ns.
- When does sample and hold take place?  Earlier than for batch mode?

## Constraints

- Two Possible answers:
- As short as possible to separate protons and anti-protons in the time domain.
- It does not matter so long as the center frequency is a harmonic of the shortest possible bunch spacing.  It can even have significant power after the next bunch has already arrived.
- This talk will concentrate on explaining how this works.