Matrix user forums

Hiyah!

I seem to be the first person to try out your EB085 board judging by the lack of discussion in the forum. I would be grateful if you would direct me to the test file DSP_Input.hex mentioned in the board's document. Meanwhile I shall try to do my own test...
Richard

Hello,

Here are the files mentioned in the datasheet. Let me know if you have any problems.

Hello
Thanks for the code. The EB085 works remarkably well. The code has given me access to the type of manipulations necessary for DSP operations. I plan to use Flowcode 6 and the dsPIC33FJ128GP802 to build my sonar system based on the newly-arrived EB064, which is intended to process chirp signals, as used in radar and sonar, for greater range and precision in noisy conditions. This will supplant a working system I have built on an Arduino Due. However, despite its mighty ARM processor the Due's signal handling speed isn't up to the job.

I can't find any HELP reference to your use of variable names preceded by a dot, as in the macros you have written in the example files. But presumably this is just what one does, in macros?
Richard

randomrichard wrote: I can't find any HELP reference to your use of variable names preceded by a dot, as in the macros you have written in the example files. But presumably this is just what one does, in macros?

Any variable with a name preceded by a dot is a local variable or a parameter for that macro and is only accessible within the macro. Variable names without a dot are global and are accessible from any macro.

Anyone who understands it please explain!

In the above flowcode program "DSP_Input.fcf" posted by Ben, in the Read_ADC macro at the bottom there is a calculation performed which I do not understand. A variable with a period before it is a local variable, OK. Is there something unique about .Return or can any variable return a local variable value into a global variable when the macro ends? Also I don't understand the shifting sequence and the or commands. It seems that the .Return can equal several different things. Will someone please explain the sequence. Also, above the calculation , the the read ADC is used via the SPI three times in succession returning each in different variables ( .upper, .mid, .lower). Why is that done, I know that it all ties together but I can't wrap my head around it. Please enlighten me. All help will definitely be appreciated.

Larry

I am starting to get it. The process is combining the binary data by shifting (<<) and or (l) statements.
I still don't understand why shift 12 places, I would have thought 8 places. The digital signal processing book from the course explains it but I think the example in the book is wrong for the:
.Return = .upper<<12
Initially .upper = 10101100
and after the <<12 16bit = 1010000000000000 is this right or is it wrong. please explain.

Larry

Hello Larry,

Is there something unique about .Return or can any variable return a local variable value into a global variable when the macro ends?

The .Return variable is only available when you have specified a return variable for your macro. Double click on the macro start icon and you can edit things like the parameters, return and locals.

I still don't understand why shift 12 places,

The ADC on the EB085 returns a 16-bit value but spread out over three bytes with 4 unused bits at either side of the 16-bit data. If you look at page 17 of the ADC datasheet then it should help to clear things up.

www.farnell.com/datasheets/1511560.pdf

Therefore the return variable is the combination of the 16-bit value from the 3 bytes received from the ADC.

Hi Ben

That was extremely helpful. I have a much better understanding. I do have 1 loose end.
The explanation in the DSP training manual.
If I understand it correctly, then Initially should read .upper 00001010 and not (101001100) and .lower 10000000 and not (10000001) to account for the leading and trailing 4 zeros.

Please help me with this..

Larry

Hi Larry,

If I understand it correctly, then Initially should read .upper 00001010 and not (101001100) and .lower 10000000 and not (10000001) to account for the leading and trailing 4 zeros.

Yes you are right the values from the ADC will be 0's outside of the 16-bit data window. I think the values shown are simply examples to help show which parts of the data are collected rather then actual demonstration values from the ADC.

Hi Ben
Ok, good news. Then I do understand what is going on.

Thank you very much for your help.

Larry