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A bit stream is mapped to symbols during baseband modulation, each symbol (say for QPSK) represents two bits. Each symbol has a corresponding I & Q value, which in QPSK, are used as the amplitude for a cosine and sine carrier wave, which are then summed for the QPSK passband signal. Where does the baud rate come into play? If it is the rate at which the bit stream is processed and modulating the carrier, how do I account for baud rate if I just have a bit vector converted into a complex vector of symbols, in say, Matlab?

When I have a receiver digitizing the passband signal I receive into I/Q samples, do these values correspond to the original I/Q values representing the symbol? If the sample rate 4x the baud rate, I should receive 4:1 received samples:transmitted symbols? I would then down sample into 1:1 Rx:Tx symbols, then map the symbols back to bits using the Tx constellation (assuming no rotation)?

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  • $\begingroup$ I think reading this book will help you a lot, since it answers very clearly precisely the sort of questions you have. There is a free pdf towards the end of the page: sethares.engr.wisc.edu/telebreak.html $\endgroup$ – MBaz Mar 4 '15 at 15:28
  • $\begingroup$ Thanks for the suggestion on that book, it looks to be a great companion to Sklar's Digital Communication book, and directly applicable to what I am doing. $\endgroup$ – bruno617 Mar 4 '15 at 17:20
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The baud rate/symbol rate is the inverse of the time distance $T_\mathrm{s}$ of two consecutive symbols. It determines the bandwidth of the modulated signal and thus comes into play when simulating frequency dependent components like filters, a wireless channel, a digital-to-analog converter, and so on. If in a simulation system the carrier modulation is simulated in the passband the ratio of baud rate and carrier frequency has to be taken into account. Imagine your carrier is modelled by $\sin(2\pi (f_\mathrm{c}/f_\mathrm{s}) n)$ with a period of $N=f_\mathrm{s}/f_\mathrm{c}$, where $f_\mathrm{c}$ is the carrier frequency and $f_\mathrm{s}$ is the sample rate of the simulation system. In the simulation system, one symbol then has to have the length $N_\mathrm{s}=T_\mathrm{s}f_\mathrm{s}$ or $N_\mathrm{s}=R_\mathrm{s}f_\mathrm{s}$, where $R_\mathrm{s}$ is the baud rate.

If however, you "just have a bit vector converted into a complex vector of symbols" without further upsampling, then the baud rate does not have to be taken into account.

Usually, communication systems are simulated in the equivalent baseband, i.e. the generally relatively fast oscillating carrier is not modelled by a sine wave. Instead, the signal is thought of being shifted to the carrier frequency and only the frequency band of interest that contains the modulated signal is taken into account. Here, the baud rate again determines the width of this band.

If the receiver applies oversampling, i.e. if the sampling rate $f_\mathrm{A}>R_\mathrm{s}$, then the received signal has to be downsampled to reconstruct the transmitted symbols. Oversampling is usually applied for clock recovery or a fractionally spaced equalizer. In a simulation system you could assume perfect sample time and $f_\mathrm{A}=R_\mathrm{s}$. But this depends on what should be simulated, of course.

Update

Generally, in a receiver applying oversampling, the received samples corresponding to a symbol are not identical due to various impairments: noise, inter-symbol-interference, impulse shape, etc. To reconstruct the transmitted symbols, several steps are necessary including clock recovery, synchronization, channel estimation and equalization, and resampling.

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  • $\begingroup$ Thanks, that clears up the baud rate question. However, in regard to the sampling of the received signal, if a receiver is providing I/Q data, if I have 1 cycle corresponding to 1 symbol, which is sampled at 4x the symbol rate, would I get 4 identical I/Q points? Does this represent the I/Q of the constellation point used to create the transmitted signal? $\endgroup$ – bruno617 Mar 4 '15 at 15:29
  • $\begingroup$ @jp0933 No, see my updated answer. $\endgroup$ – Deve Mar 4 '15 at 16:16

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