Hi, I have a question related to bit error rate and noise bandwidth. In a digital communications link that utilizes a matched filter design, a key factor in the filter design is the choice of an excess bandwidth parameter. This term affects whether or not the filter may be implementable, and also affects the occupied bandwidth of the signal. Larger excess bandwidth means more occupied bandwidth. The occupied bandwidth B is usually given in the form: B = (1+alpha)*Rs where alpha is said to be the excess bandwidth parameter and Rs is the symbol rate. alpha is unitless and both B and Rs have units of Hz. My question is, in computing the noise bandwidth to determine the expected signal to noise ratio, does one use the value Rs or the value B? In simulations with BPSK signals using a root-raised cosine matched filter design with a flat channel and AWGN I have matched the BER vs. Eb/N0 response using a variety of values for alpha and assuming Rs for the noise bandwidth. The simulation is largely insensitive to the choice of alpha. The results match the theoretical error curve nicely. This leads me to think the correct answer is to use Rs, regardless of alpha (and corresponding B). I am puzzled as to why this is the case since increasing alpha (and B) does allow more noise power into the detector. The only answer I can come up with (and I'm not satisfied hence the post) is that the downsampling of the received signal to one ideal sample (at max eye opening) per symbol must perform a filtering function that removes extra noise. Are my answer and justification correct? Is there a good published source I can use for verification/validation? Thanks for your help! ~Bando The actual bandwidth
BER and Noise Bandwidth
Started by ●June 17, 2010
Reply by ●June 17, 20102010-06-17
bando wrote:> Hi, > > I have a question related to bit error rate and noise bandwidth. In a > digital communications link that utilizes a matched filter design, a key > factor in the filter design is the choice of an excess bandwidth parameter.Wrong> This term affects whether or not the filter may be implementable, and also > affects the occupied bandwidth of the signal. Larger excess bandwidth > means more occupied bandwidth. The occupied bandwidth B is usually given > in the form: > > B = (1+alpha)*Rs > > where alpha is said to be the excess bandwidth parameter and Rs is the > symbol rate. alpha is unitless and both B and Rs have units of Hz. > > My question is, in computing the noise bandwidth to determine the expected > signal to noise ratio, does one use the value Rs or the value B?Neither. The noise is integrated from 0 to inf through the matched filter. You can convert this into the equvalent flat noise bandwidth.> In simulations with BPSK signals using a root-raised cosine matched filter > design with a flat channel and AWGN I have matched the BER vs. Eb/N0 > response using a variety of values for alpha and assuming Rs for the noise > bandwidth. The simulation is largely insensitive to the choice of alpha. > The results match the theoretical error curve nicely. > > This leads me to think the correct answer is to use Rs, regardless of alpha > (and corresponding B).No.> I am puzzled as to why this is the case since > increasing alpha (and B) does allow more noise power into the detector.No.> The only answer I can come up with (and I'm not satisfied hence the post) > is that the downsampling of the received signal to one ideal sample (at max > eye opening) per symbol must perform a filtering function that removes > extra noise.Matched filter = OPTIMAL filter. Does this ring a bell in your head ?> Are my answer and justification correct?No way.> Is there a good published source > I can use for verification/validation?A textbook on digital commucation such as Sklar or Proakis ? VLV
Reply by ●June 17, 20102010-06-17
VLV, Your post was not very helpful. Could you be more specific? Saying no, and wrong does not help me to learn. ~Bando
Reply by ●June 18, 20102010-06-18
On 6/17/2010 10:30 PM, bando wrote:> VLV, > > Your post was not very helpful. Could you be more specific? Saying no, > and wrong does not help me to learn.Be grateful for what you get. Dispelling misconceptions is a valuable service. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 18, 20102010-06-18
On Jun 17, 10:40�pm, Jerry Avins <j...@ieee.org> wrote:> > Be grateful for what you get. Dispelling misconceptions is a valuable > service. >When the pulse shape (and corresponding matched filter) is a root-raised-cosine signal, the noise-equivalent bandwidth (or equivalent flat noise bandwidth referred to by VLV) is Rs regardless of the value of alpha, 0 <= alpha <= 1. Increasing the value of alpha does increase B, but the additional noise let in by this increase in bandwidth is offset *exactly* by the decrease in the noise in the central portion of the band, that is, the noise-equivalent bandwidth is the same (Rs) regardless of the value of alpha. From this, the OP jumped to the more general (and incorrect) assumption that the noise equivalent bandwidth is always Rs, regardless of the pulse shape, and to this statement, VLV quite correctly and briefly responded No. Perhaps my more verbose statement will help dispel one or more of the OP's misconceptions, but I can only second VLV's suggestion that the OP consult a good book on digital communications, and VLV also suggested two good books to look at. In other words, the OP's assertion that VLV's post was not very helpful is also incorrect. VLV gave good (and free) advice. --Dilip Sarwate
Reply by ●June 18, 20102010-06-18
OP here... let me refer to one instance where I was told by VLK that I was incorrect. Is alpha is not an important parameter? Alpha does dictate the occupied bandwidth in the channel. This is very important, though perhaps importance is a subjective valuation (adjacent channel interference, spectral efficiency etc). I can not see holding the opinion that alpha is important is totally wrong. I am fine with being wrong, in fact I expect it when I'm asking a question in a post. That's why I'm looking for advice, to beome more right. I apologize for my dismissal of the terse reply, I do appreciate even minimal free help. However, I was hoping for more. Specifically, I expected a more verbose explanation as to why I was wrong. Thanks everyone who has replied. I think from Dilip's last reply I have the answer I sought. Appreciatively, Bando
Reply by ●June 18, 20102010-06-18
OP again... I feel the need to clarify. I did not assume that varying pulse shape (to something other than RRC pulse shaping) would be immaterial to the value of equivalent noise bandwidth. I was, in the question posed, intending to refer only to RRC pulse shaping. Furthermore, I admit that I clearly did not understand the difference between equivalent noise bandwidth, and the actual noise bandwidth. In the receive filter, the actual bandwidth over which noise is received is larger than the equivalent noise bandwidth (except for alpha = 0). The total noise power received in that band however is only Rs times the noise power spectral density because of the shape of the filter. Hopefully I have now stated this correctly (though if I'm wrong I'd love to hear it!). The reason I wanted to know what noise bandwidth to use, and maybe I should have explained my reason for asking the question in first place, is that I am maintaining a link budget that accounts for many impairments not modeled in the digital comm simulation (atmospheric loss, gain variation etc.). In this link budget I need to compute the "ideal Eb/N0" that should used to feed the simulation. The question I had to ask myself was whether to use Rs or Rs(1+alpha) to compute N = kTB and obtain the noise power for computing C/N. Now I clearly understand the right answer was to use Rs (for noise equivalent bandwidth when dealing with RRC pusle shaped signals). I knew this had to be the case since otherwise increasing alpha to 1 would have reduced my ideal Eb/N0 by 3dB. From modeling this could not be so. Now I not only know Rs is the correct equivalent noise bandiwidth, but I know why. My boss had put together the link budget with the (1+alpha) factor included for computation of noise power. Fixing this problem improves our C/N by 20%! Again, thanks. ~Bando
Reply by ●July 6, 20102010-07-06
Following up on this thread, is a common confusion I noticed among cable operator. RRC is extensively used in DVB. When one measures the SNR after equalizer and the Nyquist filter, the is always the question on equivalent noise bandwidth. Often, these operator assumes SNR is exactly equal to CNR, as the equivalent noise bandwidth is equal to that of Rs in RRC. What I found is that for RRC, while the signal power is like a S-curve around Rs (after eq and filter), a white noise is not exposed to the RRC on the transmit side. Thus the equvalent noise bandwidth, while is very close to Rs, it is NOT equal to Rs. What I found is that the noise bandwidth is slightly larger than Rs. The SNR thus has a factor of 10*log(alpha/4) lower than that of CNR before the receiver. Any comment or disagreement are welcome.
Reply by ●July 6, 20102010-07-06
On Jul 6, 7:27�am, "cl7teckie" <paul@n_o_s_p_a_m.pixelmetrix.com> wrote:> Following up on this thread, is a common confusion I noticed among cable > operator. RRC is extensively used in DVB. When one measures the SNR after > equalizer and the Nyquist filter, the is always the question on equivalent > noise bandwidth. Often, these operator assumes SNR is exactly equal to CNR, > as the equivalent noise bandwidth is equal to that of Rs in RRC. > > What I found is that for RRC, while the signal power is like a S-curve > around Rs (after eq and filter), a white noise is not exposed to the RRC on > the transmit side. Thus the equvalent noise bandwidth, while is very close > to Rs, it is NOT equal to Rs. What I found is that the noise bandwidth is > slightly larger than Rs. The SNR thus has a factor of 10*log(alpha/4) lower > than that of CNR before the receiver. > > Any comment or disagreement are welcome.yes it is often assumed that noise BW of the receiver is equal to the symbol rate Note the -3 dB points of an RRC (square root raised cosine) Rx filter are equal to the symbol rate BW so as ALPHA approaches 1 the above assumption gets closer and closer to true... Mark
Reply by ●July 6, 20102010-07-06
