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Hi all, Nyquist showed that minumum bandwidth, B required for baseband transmission of Rs symbols per sec without ISI is B >= Rs/2 while for bandpass transmission of Rs symbols per sec without is B >= Rs So does it mean that if we use BPSK to transmit 500Mbps at baseband (or passband) then the we require at least 250MHz (or 500MHz) of bandwidth? Equivalently, can we use 16-PSK to transmit 2Gbps (i.e. 4*500M symbol per sec) at baseband (or passband) which requires us to use same minumum bandwidth of 250MHz (or 500MHz) as in the case of BPSK? If the above statements are right then it followed from sampling theorem i.e. For Baseband signal (low pass) - The sampling rate must be greater than twice the highest frequency compenent in the baseband signal. For bandpass signal - The sampling rate must be greater than twice the signal bandwidth, Then, the sampling rate of the system also dependent on the symbol rate of the system. Re-use the example above, the samplig rate for the passband signal must be at least 2*500MHz=1GHz. This sampling rate of 1GHz is sufficient for BPSK signal (where symbol=bit) but how about the 16-PSK case. For 16-PSK system each bit is now 0.5ns which is lower than the sampling time 1ns. So can someone please let me know what are the problems? Many thanks, Regards, Suterr

s...@gmail.com wrote: > Hi all, > > Nyquist showed that minumum bandwidth, B required for baseband > transmission of Rs symbols per sec without ISI is B >= Rs/2 > > while for bandpass transmission of Rs symbols per sec without is > B >= Rs > > So does it mean that if we use BPSK to transmit 500Mbps at baseband (or > passband) then the we require at least 250MHz (or 500MHz) of > bandwidth? > > Equivalently, can we use 16-PSK to transmit 2Gbps (i.e. 4*500M symbol > per sec) at baseband (or passband) which requires us to use same > minumum bandwidth of 250MHz (or 500MHz) as in the case of BPSK? > > If the above statements are right then it followed from sampling > theorem i.e. > > For Baseband signal (low pass) - > The sampling rate must be greater than twice the highest frequency > compenent in the baseband signal. > > For bandpass signal - > The sampling rate must be greater than twice the signal bandwidth, > > Then, the sampling rate of the system also dependent on the symbol rate > of the system. Re-use the example above, the samplig rate for the > passband signal must be at least 2*500MHz=1GHz. > > This sampling rate of 1GHz is sufficient for BPSK signal (where > symbol=bit) but how about the 16-PSK case. For 16-PSK system each bit > is now 0.5ns which is lower than the sampling time 1ns. > So can someone please let me know what are the problems? > > Many thanks, > Regards, > Suterr > I'm not quite sure what your question is... The bit time is relatively independent of the symbol time. ie with 16-PSK you may have a bit time of 0.5nS into the encoder, but once encoded and placed into the constellation the entire symbol now has a 2nS symbol time. This is how the bandwidth is reduced for the same encoded bit rate. Thus the sampling only has to occur at 2x the bandwidth (or 500MSPS), however it must have a high enough signal to noise ratio to separate out the constellation points from each other (and the noise). In other words, by increasing the density of the encoded constellation you may have reduced the bandwidth requirements, and therefore reduced the sample rate requirement, but you have also increased the sampling accuracy (signal to noise ratio) required. This is all in sync with shannon's capacity ruling. For a given capacity if you increase the bandwidth you can decrease the required signal to noise ratio, or for a narrower bandwidth you must increase the signal to noise ratio. As for the baseband/passband sampling, the rate must be equivalent, as the same information is available in both situations. Hence you only need to sample the passband signal at twice the baseband bandwidth (just as in the baseband case). There is however a further requirement that the centre passband frequency must be an integer multiple of the sampling frequency. Alternatively (as will most often be done due to design constraints) the sampling frequency must be an integer divisor of the centre passband frequency. Hope that helps

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Hi Bevan,
> In other words, by increasing the density of the encoded constellation
> you may have reduced the bandwidth requirements, and therefore reduced
> the sample rate requirement, but you have also increased the sampling
> accuracy (signal to noise ratio) required.
Yes you are right and the equation related as follows
SNR(dB) = Eb/No(dB) + 10log10(k) + 10log10(Rs/Fs)
where k=log2(M) i.e. M is # of constellation. Rs and Fs are the symbol
rate and sampling rate, respectively.
So as the Fs decreases, the SNR will increase for a given Rb.
Similarly, if Rb decreases then the SNR will decrease as well. What
does it by decrease in SNR? It is the decrease in the required SNR
value to achiecve the same symbol error rate (SER) for the two
different rates? But I remember when I run a simulation for different
rates and plotted the SER vs. Eb/No, I obtained a strange result which
the higher data rate case is better than the lower data rate one.
Best Regards,
Suterr
```

s...@gmail.com wrote: > Hi Bevan, > >> In other words, by increasing the density of the encoded constellation >> you may have reduced the bandwidth requirements, and therefore reduced >> the sample rate requirement, but you have also increased the sampling >> accuracy (signal to noise ratio) required. > > Yes you are right and the equation related as follows > SNR(dB) = Eb/No(dB) + 10log10(k) + 10log10(Rs/Fs) umm... what? Why would you want to put the equation like that? And where did you get it from? The sampling rate and symbol rate are quite separate from the signal to noise ratio. I mean there is a coupling (in terms of over-sampling improving the signal to noise ratio a certain amount) but it's certainly not as linear as this equation depicts. The absolute basics are... Sampling rate *must* be greater than twice the symbol rate. This means that it must be greater than the bit rate Rb / log2(M) where M is the number of constellation points. As the number of constellation points (M) increases then the signal to noise ratio must improve, as for a given transmitted bit energy there will be less euclidean distance between each constellation point. Hence any noise present will cause a greater error rate, thus why TCM etc are very popular in modern wireless communications. They help to offset this error rate by introducing error detection and correction in the transmitted constellation, and thereby providing some 'encoding gain' to offset the stricter requirements on SNR. > where k=log2(M) i.e. M is # of constellation. Rs and Fs are the symbol > rate and sampling rate, respectively. > > So as the Fs decreases, the SNR will increase for a given Rb. > Similarly, if Rb decreases then the SNR will decrease as well. What > does it by decrease in SNR? It is the decrease in the required SNR > value to achiecve the same symbol error rate (SER) for the two > different rates? But I remember when I run a simulation for different > rates and plotted the SER vs. Eb/No, I obtained a strange result which > the higher data rate case is better than the lower data rate one. As Fs decreases, the SNR will be largely unchanged, until you begin to alias, in which case everything will go to crap. If Rb decreases, then the SNR will also stay the same, all that has happened is that things are now slower, the constellation spacing is still the same, because you haven't specified a different constellation mapping, just a slower bit rate, and hence a slower symbol rate. As the density of the constellation increases (i.e. we change the constellation mapping) then the SNR of the system must increase to allow us to distinguish between adjacent constellation points. I have no idea why you plotted SER vs Eb/No... do you not understand the correspondence between bits and symbols?

Hi Mark, > the BER/SER will not change if you increase the bit rate by increasing > the symobl rate IF the SNR stays the same. However to pass a higher > symbol rate requires more bandwidth so to keep the SNR the same you > will need more power in the Tx. I understand what you said. I mentioned above that I obtained results from two different symbol rate simulation and indeed it showed the same thing you explained here. But the SER vs EbNo is worse for the lower symbol rate, would you be able to explain this abit to me? Do you know the usage of the following equation SNR(dB) = Eb/No(dB) + 10log10(k) + 10log10(Rs/Fs) I think it is important for simulation setup but I don't want to use it wrongly. > If you increase the bit rate by keeping the symnbol rate the same but > instead increase the constellation points, then the BER/SER will degrde > beacuse the points are closer together so for a given noise power you > will have more errors. Again you can increase the Tx power to maintain > the BER/SER. Bottom line, raising the bit rate requires more Tx power > to obtain a given BER. FEC coding can reduce the BER at the expense > of some overhead, but in general (in nature) you don't get something > for nothing. Yes, I understand this one too.... Many thanks for your helps. Regards, Suterr

On 3 Nov 2005 07:41:58 -0800, s...@gmail.com wrote: >Do you know the usage of the following equation >SNR(dB) = Eb/No(dB) + 10log10(k) + 10log10(Rs/Fs) >I think it is important for simulation setup but I don't want to use it >wrongly. This equation makes some assumptions that may or may not be true. Here's what is always true: Eb = Ps - 10log(Rb) No = Pn - 10log(BWn) where Ps is the total signal power and Pn is the total noise power. Rb is the _information_ (i.e., payload) bit rate and BWn is the bandwidth of the noise. Note that the symbol rate and the sampling frequency don't enter into this anywhere. SNR is simply SNR = Ps/Pn = Ps - 10log(Rb) - Pn + 10log(BWn) So the equation you're using assumes no coding since it related Rs and Eb directly via k in order to account for the modulation type. That's fine if it's really the case. It also assumes that the noise bandwidth is equal to the Nyquist bandwidth which may or may not be the case depending on whether you do any filtering in your simulation. So there are some caveats with using the equation you've shown. It'll work fine if you manage those assumptions appropriately, but if the noise isn't flat and isn't of the BW you expect then the results will differ. If there is coding overheard the results will differ. Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org