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My hardware is an analog device SDR+Xilinx FPGA (sdr = fmcomms4-ebz). The direct conversion transceiver (ad9364) has a tunable channel bandwidth of 200Khz - 56Mhz. I am employing BPSK modulation on a PRN code of 100K bits in length (repeats itself).My Tx Sampling rate is 6 Msps, thus the highest frequency present in my transmitted signal (barring harmonics) is 6Mhz.

My Questions:

1) According to Nyquist criterion the minimum bandwidth should be (Fs/2) where Fs is the sampling frequency. However, the main lobe of the BPSK spectrum is 2*Fs wide. Is my minimum Tx bandwidth 3Mhz or 6Mhz ?

2)On the Receiver side if I am oversampling by a factor of 2 (ie Fs_Rx = 2*Fs_Tx) should I adjust my Receiver's channel bandwidth to the sampling rate of the receiver or the transmitted signal? ie should my Rx channel bandwidth be equal to my transmitters channel bandwidth or double ?

3) Beyond the appearance of higher order harmonics and inefficient use of bandwidth, are their negative effects to using an arbitrarily large bandwidth for both my Tx and Rx nodes? (effects like ISI,RF interference)

EDIT 1:

Pulse shape currently using root raised cosine filter, however I am also testing unfiltered data i.e. rectangular pulses. How does this effect the bandwidth (i assume the root raised cosine would allow for lower bandwidth)

Inquiring about both (sorry I should have specified). My current understanding is that the required bandwidth at baseband is half that of passband. Is that correct?

Thanks !

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  • $\begingroup$ Please explain what is the pulse shape, and for your signals and bandwidth, whether you're talking about baseband or passband. $\endgroup$
    – MBaz
    Commented Apr 23, 2019 at 19:36
  • $\begingroup$ Pulse shape currently using root raised cosine filter (with a matched filter at Rx), however I am also testing unfiltered data i.e. rectangular pulses. How does this effect the bandwidth (i assume the root raised cosine would allow for lower bandwidth) Inquiring about both (sorry I should have specified). My current understanding is that the required bandwidth at baseband is half that of passband. Is that correct? $\endgroup$ Commented Apr 23, 2019 at 20:17

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It seems from your question that you are still lacking a few concepts and techniques that you'll have to master before succeding. I'll just give a few pointers:

  • Get a good book on SDR. I recommend "Software Receiver Design" by Johnson et al, and "Software Radio: A Modern Approach to Radio Engineering" by J. Reed.

  • The passband bandwidth is double the baseband bandwidth. However, your radio works with quadrature signals and complex sampling, which muddles that concept a little, since the baseband quadrature signal's spectrum is no longer symmetrical.

  • Do not use rectangular pulses in an RF application.

  • The actual bandwidth depends on the pulse shape (and in the case of root raised cosine, its rolloff factor). If your sampling rate is $f_s$, then your baseband bandwidth goes from $-f_s/2$ to $f_s/2$.

  • The receiver's sampling rate is independent of the transmitter's, because the receiver sees an analog signal. The only thing you need to worry about is meeting Nyquist and having enough samples for your synchronizers to work.

  • You may need to worry about radiating large amounts of energy on a wide band. You should restrict your transmissions to one of the ISM bands and in general comply with all the relevant regulations in your country.

  • Regarding transmit bandwidth: Let's say you want to transmit at baud (pulse) rate $R_p$, using (root) raised cosine pulses with rolloff factor $\beta$. Then, the baseband signal bandwidth is $$B = \frac{R_p(\beta+1)}{2}.$$ The transmitter sampling rate needs to be larger than $2B$, but for practical reasons it is usually a good idea to oversample at least by a factor of 2. Also, you'll need to select a frequency that is compatible with your hardware. As an example, if I wanted to transmit at $R_p = 100,000$ pulses per second with $\beta=0.75$, I'd calculate $B=87.5$ kHz, and then oversampling by 4 I'd end up with $f_s = 350$ kHz. I'd configure the hardware for the smallest supported sampling frequency that is larger than 350,000.

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  • $\begingroup$ Thanks for the reply! I'll look into the book. I've read through some DSP books already. I should have been more clear: the rectangular pulses are used as for demonstration purposes prior to applying the pulse shaping filter. I understand that the my Rx sample rate is independent of my Tx, and I am oversampling on Rx by a factor of 4 currently. I guess my question (which I really didn't state clearly) is: with filter should my Tx and Rx RF bandwidth follow this equation? excess_bw = B/(2T) where B=rolloff factor, T=symbol period RF_bw = excess_bw + nyquist_bw $\endgroup$ Commented Apr 24, 2019 at 17:15
  • $\begingroup$ @StefanOrosco I added a bit about that to my answer. $\endgroup$
    – MBaz
    Commented Apr 24, 2019 at 19:37

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