Digital Discovery Buffer size, Compression

[BlueSky]

 

On Digital Discovery, When using Waveforms SDK.   

Could let me know the answer as following questions?

1) Does Digital Discovery logic analyzer have the same buffer size Always in 8, 16, 32 bits?
   FDwfDigitalInSampleFormatSet() 8, 16, 32 bit
   I found out the buffer size using FDwfDigitalInBufferSizeInfo(), One channel buffer size is 64 Mbits(64 x 1024 x1024).
   Whenever I select 8 channel( not 32 channel), Is the buffer size is the same ?
  
2) How long is the real transferred data width? 8, 
   I saw the usage of dwf.FDwfDigitalInSampleSensibleSet from DigitalIn_Record_Compress.py
   # enable data compression by select used lines
   When I use dwf.FDwfDigitalInSampleSensibleSet(hdwf, c_int(0x11111110)),
   Whic is the right FDwfDigitalInSampleFormatSet(hdwf, c_int(32))
        Ordwf.FDwfDigitalInSampleFormatSet(hdwf, c_int(8)) 
  
3)  What does the compression mean?  
   

[attila]

Hi @BlueSky

1. The 64M info refers to the default 32bit setting. With 8bits you have 4x buffer, for 256M samples.

2. The compression is not supported by the Digital Discovery, only by EExplorer, Analog Discovery 1, 2

[BlueSky]

12 hours ago, attila said:

Hi @BlueSky

1. The 64M info refers to the default 32bit setting. With 8bits you have 4x buffer, for 256M samples.

2. The compression is not supported by the Digital Discovery, only by EExplorer, Analog Discovery 1, 2

 

Thank you for a good reply!

I used FDwfDigitalInSampleFormatSet() to set 8, 16, 32 bits channels,

But FDwfDigitalInBufferSizeInfo() answer that the buffer size always 64Mbits....

The sequence is that First call FDwfDigitalInSampleFormatSet() and then read FDwfDigitalInBufferSizeInfo().

Could you repair my ambiguous thoughts?

 

 

[attila]

Hi @BlueSky

The 64M (2^26) refers to the default 32bit sampling, 64M*32 = 256Mbits

[BlueSky]

Hi @attila

1) Maybe "64M*32 = 256Mbits" is not correct or Am I misunderstanding?

I know that Digital Discovery has the 2G DDR for Logic Analyzer sampling buffer.

64M(2^26 = 67108864) x 32 channels = 2,147,483,648(2G bits)

15 hours ago, attila said:

The 64M (2^26) refers to the default 32bit sampling, 64M*32 = 256Mbits

 

2) Does FDwfDigitalInBufferSizeInfo() always return the same value of 64M?

     DWFAPI BOOL FDwfDigitalInSampleFormatSet(HDWF hdwf, 8); <------- 8 bits

     FDwfDigitalInBufferSizeInfo(HDWF hdwf, int *pnSizeMax);   ===> I get 64M(2^26 = 67108864) from pnSizeMax. 

     --------------------------------------------

     DWFAPI BOOL FDwfDigitalInSampleFormatSet(HDWF hdwf, 32);   <--- 32 bits

     FDwfDigitalInBufferSizeInfo(HDWF hdwf, int *pnSizeMax);   ===> I get 64M(2^26 = 67108864) from pnSizeMax. 

 

[attila]

Hi @BlueSky

1. I wanted to say: 64M (2^26) refers to the default 32bit sampling, or 256MBytes for 8bit sampling, or 128MWords for 16bit sampling...
2. Yes.