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Re: [Discuss-gnuradio] what does noutput mean in a FFT block whic fft_le
From: |
Marcus Müller |
Subject: |
Re: [Discuss-gnuradio] what does noutput mean in a FFT block whic fft_len=2048? |
Date: |
Mon, 10 Mar 2014 09:03:21 +0100 |
User-agent: |
Mozilla/5.0 (X11; Linux x86_64; rv:24.0) Gecko/20100101 Thunderbird/24.2.0 |
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Hi BZS,
you're mixing concepts here: the FFT len is your FFT size, but noutput
is the number of items you are allowed to produce (maximum).
Please re-read the writing an out-of-tree module wiki page to
understand the difference better :)
Greetings,
Marcus
On 10.03.2014 03:21, ??????? wrote:
> hi, i want deal the data after FFT in my own block,i copy a
> fft_len*sizeof(gr_complex) length of data from in to a buffer,then
> deal the data ,and copy the dealed data from buffer to the out, in
> a FFT blocks,if there is size_t
> block_size=output_signature()->sizeof_stream_item(0); when i use
> memcpy(out,in,block_size*noutput ); it works normally , but when i
> do it like this :memcpy(out,in, fft_len*sizeof(gr_complex) );,it
> does not work
>
>
> does the result of block_size*noutput have any relations with
> fft_len*sizeof(gr_complex) or some other suggestions thanks, BZS
>
>
>
> _______________________________________________ Discuss-gnuradio
> mailing list address@hidden
> https://lists.gnu.org/mailman/listinfo/discuss-gnuradio
>
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