Re: [Discuss-gnuradio] sample wise execution of blocks in grc flow graph?

[Henry Matt]

Hi John and Colby,

Thanks for the detailed and helpful replies about the sampling rates. Now, I have another issue which is unclear to me. I am printing out "noutput_items" for my custom block which come out to be either 1 or 2. As I mentioned before, the custom block is intended to do sample by sample calculation, i.e., for an input sample it calculates the output sample which is then transmitted by the a usrp to another usrp which sends a feedback to first usrp; that feedback, after some processing goes into the input of the custom block. So, you can imagine that  I am kind of running my algorithm iteratively. Now, question is that how can I force noutput_items=1? Apparently it seems to me if noutput_items is set to 2 then scheduler will wait for the arrival of second input sample to compute two output samples at the same time but what I want from scheduler is to compute a single output sample, no less and no more. So, if the input to the custom block is coming at a rate of 20 samples/sec, how frequent will work() or general_work be called? Also, if during a call to work(), noutput_items comes out to be "n" (more than 1) then it computes "n" output samples at the same time which I do not want. I want to generate 1 sample, send it, wait for the feedback, and then after arrival of feedback compute the new output sample from custom block and so on. How to do it? Any ideas?

Thanks a lot,

H.

   

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From: John Andrews <address@hidden>

On Wed, Jun 15, 2011 at 1:31 AM, Henry Matt <address@hidden> wrote:

Hi Colby,

So it means that it if one input comes at 20 samples/sec then it limits the output rate of my custom block to exactly 20 samples/sec? That is, the other input, gr_noise_source can provide samples at a rate faster than 20 samples/sec to the custom block input but the custom block produces output only when both inputs are available. Am I right?

 

Yes, the custom block will execute its general_work() function only when the right number of input items are available on all the input streams.

I am sorry for not being clear about the other part of question. The question is that if I further multiply the output of the custom block (which has a sample rate of 20 samples/sec) with another signal of sampling rate 2M samples/sec then what will be the sampling rate of the product signal?

Look at the general_work() function for the simplest of blocks such as gr_add_cc etc to understand the behaviour. Look at the following example. In this I am producing 1/4th the number of input items I am using. I am sending out every 4th input sample to the output, and effectively reducing the sampling rate by 4. So as long as the input is available I keep sending every 4th item to the output. If the input doesn't arrive then the executing thread waits until it has something to work with.

int
custom_block_b::general_work(int noutput_items,gr_vector_int &ninput_items,gr_vector_const_void_star &input_items,gr_vector_void_star &output_items)
{ 
  const unsigned char *in = (const unsigned char *)input_items[0];
 unsigned char *out = (unsigned char *)output_items[0];
 int j=0;
  for(int i=0;i<noutput_items;i++) {
      if(i%4==0){
         out[j] = in[i];
         j++;
    }

The output rate is determined by your work function. In the following code I am producing more output than what I am using at the input.

int
custom_block_b::general_work(int noutput_items,gr_vector_int &ninput_items,gr_vector_const_void_star &input_items,gr_vector_void_star &output_items)
{ 
  const unsigned char *in = (const unsigned char *)input_items[0];
 unsigned char *out = (unsigned char *)output_items[0];
 int j=0;
 int k=0;
  for(int i=0;i<noutput_items;i++) {
      if(in[i]==0x00){
        for (int j=0;j<4;j++){
           out[k]=in[i] ;
           k++;
        }
       else{
         for(int j=0;j<4;j++){
             out[k]=~in[i];
             k++;
         }
      }
      
   consume(noutput_items);
   return k;
}


As Colby has already mentioned, you must be careful about having mismatched rates.

I hope this helps.