Have you tried to unplug and re-plug the board a few times? According to our database, you should have three boards. Do all three show the same behavior or only this board? In case all three show this, it could be a hint of a software problem. If two boards are good and one is bad, this would be a hint of a hardware problem (broken board).

Stefan

The serial number is 2586. This board is about two years old, and it might be in use (but I do not know exactly).

The code in the macro is correct. The misconception lies in the definition what "sample 0" means. Please view the attached picture. This is simplified case with a DRS chip with only 8 cells (instead of 1024). There are two events (blue and red). In the first event, the chip is stopped at trigger cell (tc) 2, in the second case at 5. Since the readout starts with the trigger cell, the first readout sample in the first event belongs to cell #2, the next one to cell #3 and so on. In the second (red) case, the first sample belongs to cell #5, the second to cell #6 and so on. "Aligning cells 0" now means that the physical cell 0 (not the readout sample) is aligned for each channel. In the first event, the 7th readout sample will have the same time in all channels, in the second event the fourth readout cells will have the same time. This is because physical cell #0 is always at different places inside the readout array.

Stefan

for (ch=0 ; ch<5 ; ch++) {
         i = fread(hdr, sizeof(hdr), 1, f);
         if (i < 1)
            break;
         if (hdr[0] != 'C') {
            // event header found
            fseek(f, -4, SEEK_CUR);
            break;      
         }
         chn_index = hdr[3] - '0' - 1;
         fread(voltage, sizeof(short), 1024, f);
         
         for (i=0 ; i<1024 ; i++) {
            // convert data to volts
            waveform[chn_index][i] = (voltage[i] / 65536. - 0.5);
            
            // calculate time for this cell
            for (j=0,time[chn_index][i]=0 ; j<i ; j++)
              time[chn_index][i] += bin_width[chn_index][(j+eh.trigger_cell) % 1024];            
         }
      }
    
      // align cell #0 of all channels
      t1 = time[0][(1024-eh.trigger_cell) % 1024];
      for (ch=1 ; ch<4 ; ch++) {
         t2 = time[ch][(1024-eh.trigger_cell) % 1024];
         dt = t1 - t2;
         for (i=0 ; i<1024 ; i++)
            time[ch][i] += dt;
      }

Which serial number has the board? Has it been in use before or is it a new board?

Stefan

Hello, everybody!

I installed DRSosc and DRScl. Command line works normally (at least, it can "see" the board). But when I start the oscilloscope, I have an error: "PLLs did not lock on USB board #0, serial number #...". In Info section I can see the board type = 9 (and in the error message I have "USB board #0").

After that I have a warning: "Board on USB0 has invalid voltge calibration. Only raw data will be displayed". I tried to execute voltage calibration using DRSosc and DRScl, but it did not help.

Did anybody face such broblems? Does anybody know, how to fix them?

Thank you. Alexey.

Hello,

I was running through a particular binary file containing data taken on all 4 channels of the DRS4 and printing out the value of the first time sample for each channel (per event). While doing so, I noticed that some of these times were negative. For this dataset, channel 1 was chosen as the reference channel (which is the default setup in Stefan's DRS4 macro).  From my understanding, the calibration procedure delineated in the DRS4 manual and shown in the code below is supposed to sync the timing of each channel relative to sample 0. However, this does not appear to be the case for when I print out the time value of the first sample, I notice that only channel 1's 0th sample is set to 0. The first sample for the other channels is nonzero (and most often negative). 

Even more strange is when I test another 4-channel dataset with the same code, this issue does not appear. More specifically, the first time sample on each waveform on all channels is set to 0, as should be the case.

My question is therefore whether or not the timing calibration varies from dataset to dataset. My initial thought was that this should not be the case, but I have two different data sets taken on the same set of channels which give different timing calibration results. Are there any circumstances under which this behavior can happen? 

for (ch=0 ; ch<5 ; ch++) {
         i = fread(hdr, sizeof(hdr), 1, f);
         if (i < 1)
            break;
         if (hdr[0] != 'C') {
            // event header found
            fseek(f, -4, SEEK_CUR);
            break;      
         }
         chn_index = hdr[3] - '0' - 1;
         fread(voltage, sizeof(short), 1024, f);
         
         for (i=0 ; i<1024 ; i++) {
            // convert data to volts
            waveform[chn_index][i] = (voltage[i] / 65536. - 0.5);
            
            // calculate time for this cell
            for (j=0,time[chn_index][i]=0 ; j<i ; j++)
              time[chn_index][i] += bin_width[chn_index][(j+eh.trigger_cell) % 1024];            
         }
      }
    
      // align cell #0 of all channels
      t1 = time[0][(1024-eh.trigger_cell) % 1024];
      for (ch=1 ; ch<4 ; ch++) {
         t2 = time[ch][(1024-eh.trigger_cell) % 1024];
         dt = t1 - t2;
         for (i=0 ; i<1024 ; i++)
            time[ch][i] += dt;
      }

Cell 700 is arbitrary. You can choose any cell to align the channels to each other. The only requirement is that it's always the same cell for each event. Historically, Daniel chose cell #700 more or less arbitrary, but later we found out that this works with any cell. So for the publication we went with cell #0 (and that's why we have t_ch,0 in the paper), but cell #700 was left in the code because of lazyness. Feel free to replace 700 with any other number and you should get the same result. In a newer version of the software I use

// align cell#0 of all channels
float t1 = time[0][(1024-tc) % 1024];
for (int ch=1 ; ch<8 ; ch++) {
   float t2 = time[ch][(1024-tc) % 1024];
   float dt = t1 - t2;
   for (int i=0 ; i<1024 ; i++)
      time[ch][i] += dt;
   }
 

which is also a bit simpler. So time[ch] contains already the integrated time array (like 0.2 ns, 0.4 ns, 0.6 ns if at 5 GSPS, not the delta_t values as in the DRS.cpp code). Since the readout starts with cell # tc, the cell time[channel][1024-tc] is the physical cell #0 of the chip. The code makes sure that cell #0 in all 8 channels has the same time value.

Best regards,
Stefan

Hello,

Did you start to write some VI to interface DRS4board with labview ?

i also have in mind to do that.I am surprised that nobody alraedy did it since there is no answer toyour question

gerard

Hello,

I have a question about the GetTime() method in DRS.cpp.  I understand how the DT values are applied for all channels, and I also understand from the evaluation board manual that the timing of each channel is synchronized at sample 0, so samples should really be aligned from channel-to-channel relative to sample 0.

However, DRS.cpp has the following snippet in DrsBoard::GetTime():

   if (channelIndex > 0) {
      // correct all channels to channel 0 (Daniel's method)
      iend = tc >= 700 ? 700+1024 : 700;
      for (i=tc,gt0=0 ; i<iend ; i++)
         gt0 += fCellDT[chipIndex][0][i % 1024];
      
      for (i=tc,gt=0 ; i<iend ; i++)
         gt += fCellDT[chipIndex][channelIndex][i % 1024];
      
      for (i=0 ; i<fChannelDepth ; i++)
         time[i] += (float)(gt0 - gt);
   }

I can see what this is calculating and applying such an offset, but I don't understand why things seem to be referenced to sample 700.  Is there a particular reason why sample 700 is chosen here?  This does not seem like a straightforward application of the attached instructions from the evaluation board user's manual.

Any insight would be much appreciated!

Thanks so much,

-Kurtis

You're right, fread() return the number of objects read, so indeed it should be one if successful.

for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < sizeof(eh))
         break;

 for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < 1)
         break;

Hi Stefan,

I have already read the paper. I was just unsure where the calibration code was located. Thank you so much for all your help.

Christian

Hello,

I am wondering why the code should be changed to i < sizeof(eh), since doesn't fread(&eh,sizeof(eh),1,f) return 1 in this scenario? I've confirmed with a cout statement that this is the case, so this break condition will therefore always trigger as sizeof(eh) is 32 bytes. 

Either way, I believe I figured out my problem. In my revised version of your code, I had two nested loops, the outer one being a loop over the channels and the inner one being a loop over the events. However, I really should have been doing the reverse considering the binary structure of the file.  Otherwise, the end of the file will be reached for only a single iteration of the channel loop if I choose to loop through all the events in the data file.

Once I modified the code to have the outer loop be over all the events and the inner one be over all the channels, I no longer suffered from breaks in the loops. 

Many thanks for your assistance. 

Abhishek 

for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < sizeof(eh))
         break;

 for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < 1)
         break;

Hi,

fread() returns the number of bytes read and zero (I believe) if there is an end of file. So this break statement is a simple end-of-file test. There might be other erros such as hard disk failures, but these are extremely rare. 

If course the file should not end in the middle of an event header. If it does, it means the file is corrupted and truncated, and we should not continue to read that file, that's why there is the break. The internal file is just a series of bytes, it does not know about the event header, so there will be no "error" if we have for example a missing event header but a voltage array. To be correct, the code should actually read

for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < sizeof(eh))
         break;

Hope this helps,

Stefan

 for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < 1)
         break;

Hello,

I recently modified the binary to ROOT convertor written by Stefan (https://midas.psi.ch/elogs/DRS4+Forum/361) so it can decode data taken with any channel or set of channels  on the DRS4. In the process of testing this modifed version for data taken on all 4 channels, I encountered problems with decoding some of the event data. More specifically, upon hitting a certain event in some channel, the histograms for that channel would no longer be filled and the histograms for subsequent channels would not be filled with any event at all. 

After considerable bug hunting, I discovered the source of this problem was due to the break statement in the following code extract from the ROOT to binary macro:

 for (n=0 ; n<5 ; n++) {
      // read event header
      i = fread(&eh, sizeof(eh), 1, f);
      if (i < 1)
         break;

For some events apparently, the event header fails to be read properly (fread line returns 0 in this case). Moreover, when I used the feof and ferror functions on a particular file I was testing, the feof function returned a value of 1.

So my questions deal with two scenarios.

Firstly, in the event of an fread error, is a break statement is necessary? Is it not possible to skip the voltage data for those events whose event header fails to be read properly? Or is it the case that when some "corrupted" event header is encountered, all waveform data subsequent to that event is likewise corrupted? If the former is the case, is it advisable to replace the break condition with an fseek line that advances the position indicator of the stream by an additional 2052*n_channels + 32 bytes (in accordance with the binary file specifications of page 25 on the DRS4 manual) so that the next set of voltage data can be read? 

Secondly, in the case of an end of file error, does there exist any possible solution? Or is such an error an indication of a faulty drs4 channel or corrupted binary file? 

Any help with the aforementioned issues would be greatly appreciated.

Abhishek

Best is to read this paper: https://arxiv.org/abs/1405.4975

The source code for that is in DRS.cpp in the DRS software distribution in the function DRSBoard::CalibrateTiming()

Stefan

Thank you Stefan, that was just what I needed.

Also, I have another question, if I am allowed to ask on this forum. I am trying to study how the time calibration of the DRS is done. Can you point me to the script in which this is done?

Thank you,

Christian

The web page from where you downloaded the software contains a sentence "requires libusb-1.0 package". Please install it. This package brings the "usb.h" header file.

Stefan

Hello everybody,

I am completely new to this, so please bear with me.

I am trying to install the applications on my laptop. I downloaded and untar-ed the drivers and applications for Linux as described in the evaluation board manual. However, when I do the make, I get the following error:

drs-4.0.0$ make
g++ -g -O2 -Wall -Wuninitialized -fno-strict-aliasing -Iinclude -DOS_LINUX -DHAVE_LIBUSB -DUSE_DRS_MUTEX -c src/musbstd.c
In file included from src/musbstd.c:14:0:
include/musbstd.h:17:17: fatal error: usb.h: No such file or directory
 #include <usb.h>
 

Can anybody help me please?

Thanks.