Hi again, Follup from thread started 16 Feb 08...Thanks to everyone who responded. I didn't get a chance to respond as I was disecting the filters to see where the problem lay....(it took me a while, but it eventually turned out to be the phase of the signal reading out the data from the filter) Managed to get a decent output from the RRC filters on the TX side (see link). http://i248.photobucket.com/albums/gg170/stenasc/RRC_pulses2.jpg So now I see something,,,notice the small peak to the left of the cursor on the second and fourth traces. Should this not be the same as the peaks beside it? The next two links are what I see on the I and Q branches on the demod side. I have zoomed in on one of the pics. The RRC on both branches are exactly the same as on the I and Q branches of the modulator.Again, I ask the question...is this what is expected? If so, how on earth does one try to decode this into something sensible. The peaks are of varying amplitudes, some are inverted (see just left of the cursor on the second trace. http://i248.photobucket.com/albums/gg170/stenasc/RX_Data1.jpg http://i248.photobucket.com/albums/gg170/stenasc/RRC_pulses2.jpg I had constructed a PLL for the non-RRC version of the QPSK demodulator, and I could easily get it to recover timing by hanging it of the I or Q branches after the LPF following the mixer, but with this, it drifts all over the place and won't lock. Are there better methods like early-late, Gardner that would be more suitable for timing recovery? I would be grateful if someone could check the filter spec. to see if the coeffs are correct. Regards Bob Here is the detail of the filter fsym = 128 Beta = 0.25 Fsamp = 1024 No of Coefficients = 81 -0.0070229023149443 -0.0101567256377105 -0.0106995463613499 -0.0081513284932908 -0.0027454853524680 0.0045119843518652 0.0119844900085342 0.0177165731088947 0.0198637673884311 0.0171561105887407 0.0092932360980081 -0.0028279163188171 -0.0171265141152235 -0.0306220077689427 -0.0399632629130674 -0.0421207092441888 -0.0351145115424028 -0.0186283183298937 0.0056366964189598 0.0339772545072063 0.0611209771631681 0.0810472328405277 0.0880859678347950 0.0781178174747294 0.0496594184521266 0.0046158687756456 -0.0514788123965642 -0.1098366649262891 -0.1594084600837957 -0.1883864949522413 -0.1860109812352232 -0.1444040487260019 -0.0601297025887537 0.0647967923675360 0.2226524648114345 0.4005929560133712 0.5820384315208919 0.7486846121353156 0.8828574301801004 0.9698660159441028 1.0000000000000000 0.9698660159441028 0.8828574301801004 0.7486846121353156 0.5820384315208919 0.4005929560133712 0.2226524648114345 0.0647967923675360 -0.0601297025887537 -0.1444040487260019 -0.1860109812352232 -0.1883864949522413 -0.1594084600837957 -0.1098366649262891 -0.0514788123965642 0.0046158687756456 0.0496594184521266 0.0781178174747294 0.0880859678347950 0.0810472328405277 0.0611209771631681 0.0339772545072063 0.0056366964189598 -0.0186283183298937 -0.0351145115424028 -0.0421207092441888 -0.0399632629130674 -0.0306220077689427 -0.0171265141152235 -0.0028279163188171 0.0092932360980081 0.0171561105887407 0.0198637673884311 0.0177165731088947 0.0119844900085342 0.0045119843518652 -0.0027454853524680 -0.0081513284932908 -0.0106995463613499 -0.0101567256377105 -0.0070229023149443
Pulse Shaping using RRC filter (again)
Started by ●February 20, 2008
Reply by ●February 20, 20082008-02-20
> > So now I see something,,,notice the small peak to the left of the > cursor on the second and fourth traces. Should this not be the same as > the peaks beside it? >remember you are looking at the RRC filter..which is only 1/2 the picture pass this output though another RRC filter so that you have the full RC filter and you will see what the receiver sees... the Rx and Tx have matched filters and the NYquist filter is both of them in cascade, you are looking 1/2 way through the cascade.. Mark
Reply by ●February 20, 20082008-02-20
On 20 Feb, 20:41, Mark <makol...@yahoo.com> wrote:> > So now I see something,,,notice the small peak to the left of the > > cursor on the second and fourth traces. Should this not be the same as > > the peaks beside it? > > remember you are looking at the RRC filter..which is only 1/2 the > picture > > pass this output though another RRC filter so that you have the full > RC filter and you will see what the receiver sees... > > the Rx and Tx have matched filters and the NYquist filter is both of > them in cascade, > > you are looking 1/2 way through the cascade.. > > MarkHi Mark, The pulses you see on the first link is what happens when I pass the data through the RRC on the TX link. On the demod side links 2 and 3 are when the data from the modulator is passed through idential RRCs. So, in effect I already have the RC response at the receiver. But is it correct...that's what I would like to know? Bob
Reply by ●February 21, 20082008-02-21
On Feb 20, 4:38�pm, Bob <sten...@yahoo.com> wrote:> On 20 Feb, 20:41, Mark <makol...@yahoo.com> wrote: > > > > > > > > So now I see something,,,notice the small peak to the left of the > > > cursor on the second and fourth traces. Should this not be the same as > > > the peaks beside it? > > > remember you are looking at the RRC filter..which is only 1/2 the > > picture > > > pass this output though another RRC filter so that you have the full > > RC filter and you will see what the receiver sees... > > > the Rx and Tx have matched filters and the NYquist filter is both of > > them in cascade, > > > you are looking 1/2 way through the cascade.. > > > Mark > > Hi Mark, > > The pulses you see on the first link is what happens when I pass the > data through the RRC on the TX link. On the demod side links 2 and 3 > are when the data from the modulator is passed through idential RRCs. > So, in effect I already have the RC response at the receiver. But is > it correct...that's what I would like to know? > > Bob- Hide quoted text - > > - Show quoted text -Hello Bob, The feeding of pulses into the RRC filter (via convolution) assumes your pulses become Kronecker deltas. Don't feed rectangular pulses, instead feed spikes into the RRC filter. Then after going through two of these in sucession, you will see a nice looking signal. You are currenly getting a bunch of ringing resulting from the rectangular pulse's response comingled (convolved) with the RRC's response. For simplifying the testing, just try an RC filter and look at its output. Two RRCs convolved is an RC. Try it! IHTH, Clay
Reply by ●February 21, 20082008-02-21
On 21 Feb, 08:14, c...@claysturner.com wrote:> On Feb 20, 4:38�pm, Bob <sten...@yahoo.com> wrote: > > > > > > > On 20 Feb, 20:41, Mark <makol...@yahoo.com> wrote: > > > > > So now I see something,,,notice the small peak to the left of the > > > > cursor on the second and fourth traces. Should this not be the same as > > > > the peaks beside it? > > > > remember you are looking at the RRC filter..which is only 1/2 the > > > picture > > > > pass this output though another RRC filter so that you have the full > > > RC filter and you will see what the receiver sees... > > > > the Rx and Tx have matched filters and the NYquist filter is both of > > > them in cascade, > > > > you are looking 1/2 way through the cascade.. > > > > Mark > > > Hi Mark, > > > The pulses you see on the first link is what happens when I pass the > > data through the RRC on the TX link. On the demod side links 2 and 3 > > are when the data from the modulator is passed through idential RRCs. > > So, in effect I already have the RC response at the receiver. But is > > it correct...that's what I would like to know? > > > Bob- Hide quoted text - > > > - Show quoted text - > > Hello Bob, > > The feeding of pulses into the RRC filter (via convolution) assumes > your pulses become Kronecker deltas. Don't feed rectangular pulses, > instead feed spikes into the RRC filter. Then after going through two > of these in sucession, you will see a nice looking signal. You are > currenly getting a bunch of ringing resulting from the rectangular > pulse's response comingled (convolved) with the RRC's response. For > simplifying the testing, just try an RC filter and look at its output. > Two RRCs convolved is an RC. Try it! > > IHTH, > > Clay- Hide quoted text - > > - Show quoted text -Hi Clay, I did already try it. I fed the output of one RRC into another and you're correct...I did get a nice RC pulse output. In the test case, this was done for spikes well seperated. However, for real test data, I don't have much choice as I have strings of 1's and in most cases very few 0's between them. In this case, what are my options? Regards Bob
Reply by ●February 21, 20082008-02-21
On Feb 21, 5:35�am, Bob <sten...@yahoo.com> wrote:> On 21 Feb, 08:14, c...@claysturner.com wrote: > > > > > > > On Feb 20, 4:38�pm, Bob <sten...@yahoo.com> wrote: > > > > On 20 Feb, 20:41, Mark <makol...@yahoo.com> wrote: > > > > > > So now I see something,,,notice the small peak to the left of the > > > > > cursor on the second and fourth traces. Should this not be the same as > > > > > the peaks beside it? > > > > > remember you are looking at the RRC filter..which is only 1/2 the > > > > picture > > > > > pass this output though another RRC filter so that you have the full > > > > RC filter and you will see what the receiver sees... > > > > > the Rx and Tx have matched filters and the NYquist filter is both of > > > > them in cascade, > > > > > you are looking 1/2 way through the cascade.. > > > > > Mark > > > > Hi Mark, > > > > The pulses you see on the first link is what happens when I pass the > > > data through the RRC on the TX link. On the demod side links 2 and 3 > > > are when the data from the modulator is passed through idential RRCs. > > > So, in effect I already have the RC response at the receiver. But is > > > it correct...that's what I would like to know? > > > > Bob- Hide quoted text - > > > > - Show quoted text - > > > Hello Bob, > > > The feeding of pulses into the RRC filter (via convolution) assumes > > your pulses become Kronecker deltas. Don't feed rectangular pulses, > > instead feed spikes into the RRC filter. Then after going through two > > of these in sucession, you will see a nice looking signal. You are > > currenly getting a bunch of ringing resulting from the rectangular > > pulse's response comingled (convolved) with the RRC's response. For > > simplifying the testing, just try an RC filter and look at its output. > > Two RRCs convolved is an RC. Try it! > > > IHTH, > > > Clay- Hide quoted text - > > > - Show quoted text - > > Hi Clay, > > I did already try it. I fed the output of one RRC into another and > you're correct...I did get a nice RC pulse output. > In the test case, this was done for spikes well seperated. However, > for real test data, I don't have much choice as I have strings of 1's > and in most cases very few 0's between them. In this case, what are my > options? > > Regards > Bob- Hide quoted text - > > - Show quoted text -Hello Bob, The overall RC response does two major things for you. First it limits your spectral splattering by effectively making your spectral shape fit inside of a rectangular window[1]. And second you have periodic zeroes so your ISI is kept down to a minimum. Look at your data's eye pattern and make sure the eye is open enough at the decision points to tell your ones from your zeroes. The data under a low S/N is going to look bad. IHTH, Clay [1] The rectangular shape ideally will have vertical sides made up as pieces of cosines. But a practical implementation will have other distortion as you will end up limiting your impulse repsonse's duration to something finite.
Reply by ●February 21, 20082008-02-21
On Thu, 21 Feb 2008 02:35:33 -0800 (PST), Bob <stenasc@yahoo.com> wrote:>On 21 Feb, 08:14, c...@claysturner.com wrote: >> On Feb 20, 4:38�pm, Bob <sten...@yahoo.com> wrote: >> >> >> >> >> >> > On 20 Feb, 20:41, Mark <makol...@yahoo.com> wrote: >> >> > > > So now I see something,,,notice the small peak to the left of the >> > > > cursor on the second and fourth traces. Should this not be the same as >> > > > the peaks beside it? >> >> > > remember you are looking at the RRC filter..which is only 1/2 the >> > > picture >> >> > > pass this output though another RRC filter so that you have the full >> > > RC filter and you will see what the receiver sees... >> >> > > the Rx and Tx have matched filters and the NYquist filter is both of >> > > them in cascade, >> >> > > you are looking 1/2 way through the cascade.. >> >> > > Mark >> >> > Hi Mark, >> >> > The pulses you see on the first link is what happens when I pass the >> > data through the RRC on the TX link. On the demod side links 2 and 3 >> > are when the data from the modulator is passed through idential RRCs. >> > So, in effect I already have the RC response at the receiver. But is >> > it correct...that's what I would like to know? >> >> > Bob- Hide quoted text - >> >> > - Show quoted text - >> >> Hello Bob, >> >> The feeding of pulses into the RRC filter (via convolution) assumes >> your pulses become Kronecker deltas. Don't feed rectangular pulses, >> instead feed spikes into the RRC filter. Then after going through two >> of these in sucession, you will see a nice looking signal. You are >> currenly getting a bunch of ringing resulting from the rectangular >> pulse's response comingled (convolved) with the RRC's response. For >> simplifying the testing, just try an RC filter and look at its output. >> Two RRCs convolved is an RC. Try it! >> >> IHTH, >> >> Clay- Hide quoted text - >> >> - Show quoted text - > >Hi Clay, > >I did already try it. I fed the output of one RRC into another and >you're correct...I did get a nice RC pulse output. >In the test case, this was done for spikes well seperated. However, >for real test data, I don't have much choice as I have strings of 1's >and in most cases very few 0's between them. In this case, what are my >options? > >Regards >BobI haven't looked at your latest plots, but I'm a little confused by your last statement. How many samples per symbol are there going into the transmit RRC filter? If the answer is one, then you don't have to worry about the rectangular symbol problem. If the answer is more than one, then only one of those samples per symbol should be non-zero if you want to avoid an additional sinx/x response in the filtering. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org
Reply by ●February 21, 20082008-02-21
On 21 Feb, 16:38, Eric Jacobsen <eric.jacob...@ieee.org> wrote:> On Thu, 21 Feb 2008 02:35:33 -0800 (PST), Bob <sten...@yahoo.com> > wrote: > > > > > > >On 21 Feb, 08:14, c...@claysturner.com wrote: > >> On Feb 20, 4:38�pm, Bob <sten...@yahoo.com> wrote: > > >> > On 20 Feb, 20:41, Mark <makol...@yahoo.com> wrote: > > >> > > > So now I see something,,,notice the small peak to the left of the > >> > > > cursor on the second and fourth traces. Should this not be the same as > >> > > > the peaks beside it? > > >> > > remember you are looking at the RRC filter..which is only 1/2 the > >> > > picture > > >> > > pass this output though another RRC filter so that you have the full > >> > > RC filter and you will see what the receiver sees... > > >> > > the Rx and Tx have matched filters and the NYquist filter is both of > >> > > them in cascade, > > >> > > you are looking 1/2 way through the cascade.. > > >> > > Mark > > >> > Hi Mark, > > >> > The pulses you see on the first link is what happens when I pass the > >> > data through the RRC on the TX link. On the demod side links 2 and 3 > >> > are when the data from the modulator is passed through idential RRCs. > >> > So, in effect I already have the RC response at the receiver. But is > >> > it correct...that's what I would like to know? > > >> > Bob- Hide quoted text - > > >> > - Show quoted text - > > >> Hello Bob, > > >> The feeding of pulses into the RRC filter (via convolution) assumes > >> your pulses become Kronecker deltas. Don't feed rectangular pulses, > >> instead feed spikes into the RRC filter. Then after going through two > >> of these in sucession, you will see a nice looking signal. You are > >> currenly getting a bunch of ringing resulting from the rectangular > >> pulse's response comingled (convolved) with the RRC's response. For > >> simplifying the testing, just try an RC filter and look at its output. > >> Two RRCs convolved is an RC. Try it! > > >> IHTH, > > >> Clay- Hide quoted text - > > >> - Show quoted text - > > >Hi Clay, > > >I did already try it. I fed the output of one RRC into another and > >you're correct...I did get a nice RC pulse output. > >In the test case, this was done for spikes well seperated. However, > >for real test data, I don't have much choice as I have strings of 1's > >and in most cases very few 0's between them. In this case, what are my > >options? > > >Regards > >Bob > > I haven't looked at your latest plots, but I'm a little confused by > your last statement. � How many samples per symbol are there going > into the transmit RRC filter? � If the answer is one, then you don't > have to worry about the rectangular symbol problem. � If the answer is > more than one, then only one of those samples per symbol should be > non-zero if you want to avoid an additional sinx/x response in the > filtering. > > Eric Jacobsen > Minister of Algorithms > Abineau Communicationshttp://www.ericjacobsen.org- Hide quoted text - > > - Show quoted text -Hi Eric, Hi Eric, One sample per symbol into each RRC filter. Bob
Reply by ●February 21, 20082008-02-21
Bob <stenasc@yahoo.com> writes:> [...] > Hi Eric, > > Hi Eric,Multipath? -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% <yates@ieee.org> % 'Waterfall', *Face The Music*, ELO http://www.digitalsignallabs.com
Reply by ●February 21, 20082008-02-21
On Thu, 21 Feb 2008 13:13:11 -0500, Randy Yates <yates@ieee.org> wrote:>Bob <stenasc@yahoo.com> writes: >> [...] >> Hi Eric, >> >> Hi Eric, > >Multipath?lol Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org
