Re: channel question for programming Olimex EEG-SMT

Hi Again Daniel,

I forgot to mention, have you tried the Olimex EEG forum?

https://www.olimex.com/forum/index.php?board=25.0

Cheers

Paddy

 

From: Daniel Baker [mailto:[email protected]] 
Sent: 01 February 2013 09:23
To: [email protected]
Subject: Re: [Openeeg-list] channel question for programming Olimex EEG-SMT

 

Hi Arnold,

That's very interesting.  I wonder if my device has a loose connection that
causes crosstalk between the channels. That would explain the extra
activity. I guess it's not necessarily a problem, if the data I get out of
the first two channels are OK, perhaps I can ignore the other ones.

 - Daniel


On 31/01/2013 19:04, nocom wrote:

Hi Daniel,

I am trying to understand my recently received EEG-SMT as well. My results
are somewhat different than yours. From the six data channels numbers 0 and
1 yield data, while 2, 3 and 5 yield 6 and channel 4 holds one or other high
constant (something like 265436) which changes each time when you start to
sample. I assumed that ch- is a reference for ch+. The Olimex is able to
output to six channels (look at the schema's) but comes with only 2 to keep
costs down. 

I am somewhat alarmed that our two devices differ that much, maybe one of
them is somewhat defective?

Arnold

On 31-Jan-13 17:13, Daniel Baker wrote:

Hi Paddy,

Thanks for your response. If this is the case, then for a two-channel
device, four of the outputs should be silent, but they are clearly not. I've
written a blog post with some example data, code, and details of exactly
what I've done:

http://bakerdh.wordpress.com/2013/01/31/a-first-look-at-the-olimex-eeg-smt/

I've tried unplugging some of the electrodes to see what effect this has on
the output. When I remove electrodes CH2+ and CH2- I get activity only on
the first two channels, and it looks a little odd (the second channel never
goes above 511 (out of 1023)). When I plug only those ones in and remove
CH1+ and CH1- then the first channel goes silent, but the other five are
active.

I guess if you're right and the first two channels are the differences
between the electrodes, then maybe the other four outputs are just
interference. Still, it's puzzling that they go silent when CH2± are
unplugged.

 - Daniel



On 31/01/2013 15:17, Paddy Duncan wrote:

Hi Daniel,
As I understood it, starting at byte 5, the 2-byte 10-bit numbers are the
data for each channel ie the difference between the 2 channel electrodes.
There is no data representing a single electrode, as the channels are
differential. The six outputs are for the six channels that the protocol
supports.
Hope this helps.
Paddy
 
Below is the relevant section of the P2 firmware file:
////////// Packet Format Version 2 ////////////
 
// 17-byte packets are transmitted from the ModularEEG at 256Hz,
// using 1 start bit, 8 data bits, 1 stop bit, no parity, 57600 bits per
second.
 
// Minimial transmission speed is 256Hz * sizeof(modeeg_packet) * 10 = 43520
bps.
 
struct modeeg_packet
{
     uint8_t        sync0;         // = 0xa5
     uint8_t        sync1;         // = 0x5a
     uint8_t        version;       // = 2
     uint8_t        count;         // packet counter. Increases by 1
each packet.
     uint16_t       data[6];       // 10-bit sample (= 0 - 1023) in big
endian (Motorola) format.
     uint8_t        switches;      // State of PD5 to PD2, in bits 3 to
0.
};
 
//////////////////////////////////////////////////////////////
 
-----Original Message-----
From: Daniel Baker [mailto:[email protected]] 
Sent: 30 January 2013 20:26
To: [email protected]
Subject: [Openeeg-list] channel question for programming Olimex EEG-SMT
 
Dear all,
 
Last week I ordered and received an Olimex EEG-SMT box. I have successfully
written code to read from the device using Matlab, directly over the virtual
serial port interface. The data packets arrive in blocks of 17 bytes, and I
have converted the six outputs (from bytes 5-16) specified in the packet
protocol v2 into ten bit numbers.
 
All six of the outputs produce a sensible looking trace, with mains
artefacts at 50Hz as expected. What I don't understand is how these traces
map onto the electrodes. There are four active electrodes and one ground
electrode on the device. I initially thought that the mapping must be:
 
Bytes 5&6: Channel 1+
Bytes 7&8: Channel 1-
Bytes 9&10: Channel 2+
Bytes 11&12: Channel 2-
Bytes 13&14: All of channel 1 (i.e. the difference between positive and
negative) Bytes 15&16: All of channel 2
 
Presumably all referenced to the ground. However subtracting the first two
traces does not produce the fifth trace (and so on). I'm at a loss to
understand why else a device with four electrodes and two notional channels
would produce six outputs.
 
All help much appreciated. I intend to make some code publicly available
once I'm happy with it.
 
Many thanks,
 
 - Daniel Baker
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