Re: [Discuss-gnuradio] Question on number of input items per work call

[Michael Dickens]

Hi Anil - In the case where your block gets 4095 items, it can't process all of those items. Your block performs seven 512-length FFTs, consumes 7x512 == 3584 items, and generates 3584 items, leaving 511 unprocessed. The code that calls your block's "::work" method will handle making sure that the input and output streams are updated correctly, and that those unprocessed 511 items are the next in the input data stream.

When doing streaming, you can, to some degree, control the amount of input and output items via special functions / methods such as "set_output_multiple". You can also set the input / output types to be a vector of a specific length, e.g., that of your FFT == 512. Using either "set_output_multiple" or a vector will guarantee that the number of input and/or output items meets the criteria you require for this specific block (for it to be a multiple of the FFT size, 512 items in this case).

Hope this helps! - MLD

ps> I'm pretty sure there are presentations on the internet freely available that cover the concepts of how GR works under the hood -- probably in basic detail, not getting bogged down by too many cases / specifics. I'd encourage you to see what you can find in that regard; GRCon and people's blogs are always good places to learn about how GR works.

On Wed, Nov 15, 2017, at 09:56 PM, Anil Kumar Yerrapragada wrote:

Hi all

I am experimenting with my own python block. I started by simply interrogating each variable to see their types and shapes.

I connected a vector source that just generates increasing numbers (0, 1, 2, 3, 4, 5 and so on upto a big number).

I noticed that the length of (input_items[0]) seems to oscillate between 4096 and 4095 in successive work calls. I also noticed that in each work call, a NEW set of samples appear. (attached file)

Here are my questions:

If I were just doing something really simple like squaring each element, in each work call I'd just square every element in the vector input_items[0]. No problem.

Suppose I need to do an FFT. (I know it can be done with stream to vector followed by FFT) But suppose I wanted to make it work with streams.

In each work call, I’d divide the input_items[0] vector into chunks of size = FFT size and take the FFT of each chunk. Again no problem. Since 4096 will divide perfectly by a power of 2 FFT size.

But what about the case when I have 4095? If I do FFT = 512, I will get 7 chunks of 512 and I will be left over with 511. Would I have to save those 511 elements and use them in the next work call when input_items[0] gets new elements?