Having some confusion about digital pulse shaping for complex baseband (passband) signals. The complex baseband linear modulation equation is $$s(t)=\sum_{m=-\infty}^{\infty}\text{Re}\{a_m\}h(t-mT)+j\sum_{m=-\infty}^{\infty}\text{Im}\{a_m\}h(t-mT)\tag{2}$$ where $a_m$ are the symbols and $h$ is the impulse response of the pulse shaping filter. Consider for now this is a root raised cosine filter with QPSK symbols.

It is often stated in references that the output rate of the pulse shaping filter should be at least twice the input rate - in other words, the pulse shaping filter should be at least a 2x interpolator. For example the following equation holds for the required (positive) bandwidth:

$$BW = (1+a)\frac{R_b}{2\log_2(M)} = (1+a)\frac{R_s}{2}$$ where $R_b$ is the bit rate, $a$ is the excess bw, $R_s$ is the symbol rate, and $M$ is the constellation size. Thus to satisfy the nyquist criterion, we must sample with at least $f_s = (1+a)R_s$. I believe another way of stating this is that the two-sided bandwidth $(1+a)R_s$ is equal to the complex sampling rate.

So considering some values, take $a=0$ which gives the sinc pulse, the digital pulse shaping filter could theoretically operate at 1 (complex) sample per symbol, i.e. the input and output rates are both $R_s$ (disregarding all the other obvious drawbacks of using sinc pulses).

Another common value $a=0.35$, gives us a sampling rate of $1.35R_s$, so we would need at least a fractional interpolating filter (e.g. upsample by 27, downsample by 20) to avoid aliasing. So clearly it seems possible to achieve less than 2 complex samples per symbol. My question is, if this is correct, why do so many references and libraries mandate a minimum of 2 (complex) samples per symbol when the actual limit is 1 complex sample per symbol (e.g. GnuRadio constellation modulator is one). I am assuming this is just due to simplicity and 2 being an easy value to interpolate and cover the range of allowable RRC bandwidths.

A second related question, assume we are doing BPSK, e.g. with $\text{Im}\{a_m\} = 0$. It seems to me in that case if you're in an IQ system still, you'd actually need 2 complex samples per symbol in order to actually give 2 real samples per symbol to the BPSK I channel. Perhaps this is yet another reason for mandating the 2 complex samples per symbol minimum?

  • $\begingroup$ $a_m = 0$, what do you mean? $\endgroup$
    – Gilles
    Commented Jun 26, 2020 at 13:05
  • 2
    $\begingroup$ @DanBoschen, my first comment was on the OP's last paragraph, "e.g. with $a_m = 0$". Did the OP mean $\operatorname{Im}\left\{a_m\right\} = 0$ or $\alpha = 0$? That was my confusion. $\endgroup$
    – Gilles
    Commented Jun 26, 2020 at 13:40
  • $\begingroup$ @gilles good catch, yes that should be $\text{Im}\{a_m\} = 0$ $\endgroup$
    – user67081
    Commented Jun 26, 2020 at 23:20

1 Answer 1


Yes the OP is correct in that you can implement pulse shaping in less than 2 samples per symbol for exactly the reasons that was outlined. However importantly we must also keep in mind having excess bandwidth to simplify subsequent filtering required (such as after the DAC on the transmitter side). The Nyquist criteria is the sampling rate must be twice the highest bandwidth of the signal. Therefore slower roll-off signals can be done with less samples per symbol as the OP has outlined. However they will require a much larger time response, meaning more samples overall for each symbol pulse extending well beyond the symbol period, at less than 2 samples per symbol. The reason 2 samples per symbol is commonly stated is for the convenience of implementation given any synchronization between symbol rates and sample rates.

A BPSK system that is complex would be complex for purpose of carrier recovery only; once the received symbol is rotated to the real axis there would be no further need for the imaginary axis. Yet the decision for number of samples per symbol is the same whether the signal is complex or real, exactly as first outlined (must be greater than twice the bandwidth of the signal, and specifically it is the bandwidth of significance- where beyond we are not concerned with the aliasing of residual spectrum that exists beyond the bandwidth given by the sampling rate chosen.

To help see all considerations here are spectral plots for two variants of a root-raised cosine implementation implemented with a 100 Tap FIR. The first is done with 10 samples per symbol while the second is with 2 samples per symbol. Specifically notice the much greater rejection achieved close to the primary bandwidth due to the longer overall time response of the filter (so same complexity but with fewer samples per symbol the time duration of the impulse response is longer, thus closer approximating the ideal infinite long response). The advantage of the first one with more samples per symbol is in the ability to have more relaxed filtering in subsequent stages, such as required analog filtering after the DAC since the image that would be at the sampling rate is now that much further away from the desired passband (or subsequent digital interpolation stages).

enter image description here

100 tap RRC 2 samples per symbol

The sampling rate for the lower figure specifically is 2 samples per symbol and here specifically in this example it is 2000 Hz. We see from the plot, as long as we are not concerned about folding noise below -60 dB that we could theoretically decrease this sampling rate to approximately 700 Hz while still maintaining the passband characteristics of the pulse. However consider what would be required for a transmit filter after the DAC to filter out the image of the passband that is centered on the sampling rate (in addition to the natural Sinc filter the DAC provides; which is not nearly sufficient to meet out of band emission requirements!). One solution is to interpolate between the lower rate pulse sampled waveform and the DAC, but even the proper interpolation filter design would be challenged by not having any usable transition band!

This illustrates a fundamental challenge in decreasing the sampling rate to exactly twice the bandwidth (where bandwidth would be the significant bandwidth after which any folding/aliasing is far enough below that dictated by a performance requirement), which is more of a concern on the transmitter side dictated by the subsequent filtering required to meet out of band emissions: given the image frequency response of the passband that would be symmetrically around the sampling rate; if we sampled at exactly twice the significant bandwidth in the transmitter the required filtering would not be realizable given there is no transition band for the filter to reject the images after the DAC that would be positioned at integer multiples of the sampling rate.

On the receiver side, we only need sufficient excess bandwidth to handle frequency offsets due to Doppler and Tx/Rx synchronization, so there is much more of an opportunity/practicality to reduce the sampling rate below 2 samples/symbol. Two samples per symbol is still convenient for common timing recovery implementations such as the Gardner TED (Isn't Gardner's algorithm and Early-Late gate the same thing?), and allows for a simple extraction of the one sample per symbol decisions.

Related to receiver considerations are baud rate equalizers which operate at 1 sample per symbol, with performance implications related to exactly the points the OP makes, but point is they are able to still function. This is detailed further here The benefits of a fractionally spaced equalizer. This is also similar to how the M&M timing recovery algorithm operates at one sample per symbol, but also has inferior performance to the Gardner in the presence of carrier offsets.

  • $\begingroup$ Hi Dan, I am having trouble with a similar question and I see your name on a lot of good replies so wondering if you can assist. dsp.stackexchange.com/questions/71574/… $\endgroup$ Commented Jan 19, 2021 at 19:59
  • $\begingroup$ Hi @NatalieJohnson and good to see you again. It looks like User07981 has done substantial work to try and help you since you wrote that comment but please let me know if you are still confused. $\endgroup$ Commented Jan 20, 2021 at 3:16

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