# PSK Modulation-\>Demodulation **URL:** \ **Category:** GNU Radio\ **Created:** [June 16, 2015, 5:02pm UTC](https://www.ruby-forum.com/t/psk-modulation-demodulation/242335 "2015-06-16T17:02:26Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![Detlef\_R](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/d/58f4c7/32.png) [@Detlef\_R](https://www.ruby-forum.com/u/Detlef_R)\ **Post date:** [June 16, 2015, 5:02pm UTC](https://www.ruby-forum.com/t/psk-modulation-demodulation/242335/1 "2015-06-16T17:02:26Z")
Hello, I know a similar question has been asked before on this mailing list but I didn’t quite get a solution out of it. I am generating a random byte source, either ones or zeroes, modulating them with DBPSK and immediately demodulating them. The problem I am running into is that I get roughly 8 times the number of bytes out of the demodulation block than I am inputting into the modulation block. I believe its a packed vs. unpacked byte problem but every attempt at packing the output bytes yields an output that doesn’t match the input. Attached are the GRC file, and the output file streams. Does anyone have any suggestions? I appreciate all of your time, Logan Washbourne Electrical Engineering Graduate Student (Electromagnetics) ---
**Author:** ![WashbourneS\_Logan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ec9cab/32.png) [@WashbourneS\_Logan](https://www.ruby-forum.com/u/WashbourneS_Logan)\ **Post date:** [June 16, 2015, 5:15pm UTC](https://www.ruby-forum.com/t/psk-modulation-demodulation/242335/2 "2015-06-16T17:15:53Z")
On Tue, Jun 16, 2015 at 11:01 AM, Washbourne, Logan \< [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid)\> wrote: > Attached are the GRC file, and the output file streams. > > Does anyone have any suggestions? I appreciate all of your time, > > Logan Washbourne > Electrical Engineering Graduate Student > (Electromagnetics) The problem is that you don’t know which bit to put where in the byte. So just packing the output bits into 8 bits per byte doesn’t necessarily give you the right information. Say you transmit a byte with bits [a b c d e f g h]. After the delays introduced by the transmit and receiver filter and the channel, when you pack the results, you could get something like [x x x a b c d e] [f g h 0 0 0 0 0], for some unknown number of x’s. You need some logic that knows how to discover the start of your information and get rid of those x’s. The framer blocks do this with some assumed formatting and expectations. Johnathan and I have been working on a better version of this, but we haven’t had a chance to finish it off. Tom ---
**Author:** ![WashbourneS\_Logan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ec9cab/32.png) [@WashbourneS\_Logan](https://www.ruby-forum.com/u/WashbourneS_Logan)\ **Post date:** [June 16, 2015, 6:07pm UTC](https://www.ruby-forum.com/t/psk-modulation-demodulation/242335/3 "2015-06-16T18:07:18Z")
So would a good solution to this be the packet encoder and packet decoder blocks? Since they utilize a preamble and access code? I tried to use the simple framer block in conjunction with the unpacked to packed byte block but was unsuccessful in getting a matching output, I will continue to play around with that configuration but for now do you think the packet encoder route would be a good solution? Again, I really appreciate everyone’s willingness to help. Logan Washbourne Electrical Engineering Graduate Student (Electromagnetics) ---
**Author:** ![WashbourneS\_Logan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ec9cab/32.png) [@WashbourneS\_Logan](https://www.ruby-forum.com/u/WashbourneS_Logan)\ **Post date:** [June 17, 2015, 1:54am UTC](https://www.ruby-forum.com/t/psk-modulation-demodulation/242335/4 "2015-06-17T01:54:45Z")
On Tue, Jun 16, 2015 at 12:05 PM, Washbourne, Logan \< [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid)\> wrote: > Electrical Engineering Graduate Student > (Electromagnetics) Yep, those two blocks should handle this issue for you. It’s a simplistic approach to the problem, but it should work for just what you are talking about here. Tom