I'm working on an echo implementation on the board adsp 21262. I've added a buffer to have more samples to work with. I end up then with 4 buffers of size 2048 each.

The problem is that I need to access at least 7000 samples back. (An echo can only be heard with a delay of 1/15th of a second, we have 2*48K~100k samples per second. So we need 100k/15~7k samples).

The problem is that 2048 samples per buffer is the max we can have because afterwards we're out of memory. I'm trying to access the ROM but I have no idea as to how to do that. I've spent a lot of time looking for this information finds nothing.

Thank you !

  • $\begingroup$ Do you think you could rephrase the question for clarity? The first paragraph makes it sound as if you already have acquired 4*2048=8192 samples. The second paragraph states that you need to go back 7000 samples (<8192) and the third states that the problem is that you can only acquire 1 buffer of 2048. Why are you trying to access the ROM? The ROM would be for things like static buffers (for example), you would not be able to write real-time data there (?). $\endgroup$ – A_A Dec 17 '18 at 10:52
  • $\begingroup$ Basically you can only make operations on 2 buffers, and since the buffers get $\endgroup$ – Courtney Dec 18 '18 at 13:53
  • $\begingroup$ We can only make operations on 2 buffers at a time (I think one of the buffers is used by the DAC and the other by the ADC do only 2 usable). So only 2 buffers of size 2048 each let's call them A and B. Output=A[i]+0.5*B[i]. B[i] and A[i] have a 2048 samples delay so I only have 2048 instead of 7000. $\endgroup$ – Courtney Dec 18 '18 at 13:57

Assuming you have this board, there are 512kByte of SRAM on the board. That should easily accommodate a 7k delay buffer.

| improve this answer | |
  • $\begingroup$ Thank you I'll look into that. I was looking for ROM answers while I should have been looking for SRAM. It should be easier now to solve my problem. $\endgroup$ – Courtney Dec 18 '18 at 14:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.