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We currently have a radio that is transmitting at the following RF params:

  • Modulation: OQPSK
  • Data rate: 500Kbps
  • Uplink coding: BCH

I am interested in understanding what is the associated uplink bandwidth. From my understanding, the formula is R = 2B * log2(M) where R is data rate in bps, B is bandwidth in Hz, and M is the symbol level of modulation.

Therefore for OQPSK (4 symbols), the formula would be 500,000 bps = 2B * log2(4), and B = 125 KHz.

However, the answer i received from our radio vendor was that for a roll-off factor of 0.5, the bandwidth is 750 KHz.

I have attempted to understand the meaning of roll-off factor in the context of bandwidth and datarates, however, i have not been able to understand it. Can someone explain or provide links to good reading material regarding how the roll-off of 0.5 is accounted for in the equation above, and why the 750 KHz is so different than my 125 KHz calculation

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The rolloff factor (sometimes denoted $\beta$) is a measure of the excess bandwdith of a filter (i.e. pulse-shaping filter), normalized in terms of the Nyquist bandwidth of $1/2T$. Hence, if the excess bandwidth is $\Delta f$, then the rolloff factor is $$\beta=\frac{\Delta f}{\frac{1}{2T}}=2T\Delta f\,.$$

Ideally, we would like $\beta=0$ rolloff to be as spectrum-efficient as possible. However, the resulting pulse (rectangular, brick-wall) is ideal and non-realizable, and we must pay a price to be able to implement the communication system.

The baseband bandwidth is then the Nyquist bandwith plus excess bandwidth, which in terms of rolloff $\beta$ and data rate $R=\log_2(M)/T$ can be written as $$B=\frac{R}{2\log_2(M)}(\beta+1)=\frac{500\cdot10^3}{2\log_2(4)}(0.5+1)=375\cdot10^3$$ and the RF bandwidth is double this value, i.e. 750kHz.

Note that here I assume (as is convention) that the reported data rate of 500kbps is the gross data rate, i.e., containing both information and redundant bits. The original question did not state this explicitly, but the assumption makes sense as the transmit bandwidth needs to support the coded message in full.

However, in occasion the convention is not followed, and the reported data rate is just coded (information) data rate. This would require to consider the coding rate of the code into account to obtain $R$. Luckily this is not very common, but I add this comment for the sake of completeness.

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  • $\begingroup$ Thanks for the response. So the BCH is BCH(64,56). How does that affect the bandwidth? And for the downlink, it is ReedSolo(255, 223) + Conv Code (1/2). I assumed that would modify the downlink bandwidth too? $\endgroup$
    – ITried
    Sep 7 at 23:29
  • $\begingroup$ I edited the answer to further explain this. $\endgroup$
    – cjferes
    Sep 8 at 15:17
  • $\begingroup$ Hi @ITried, if the edited answer satisfies you, would you mind accepting it? $\endgroup$
    – cjferes
    Sep 14 at 18:18
  • $\begingroup$ sorry for delay. But yes, the 500 kbps includes both information and redundant bits. So with the BCH on the uplink and RS+CC on the downlink, the bandwidth is still 750 khz for both? $\endgroup$
    – ITried
    Sep 15 at 22:26
  • $\begingroup$ yes, the code would change the information rate but not raw data rate, and bandwidth is the same $\endgroup$
    – cjferes
    Sep 17 at 9:32

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