I would like to show at high level that, in principle, the spectral efficiency (SE) of frequency division duplex (FDD) equals the SE of time division duplex (TDD) and both of them are half the SE of full duplex (FD).

Recall that in FD, the downlink (DL) and uplink (UL) transmissions utilize the whole spectrum for the whole time. On the other hand, in FDD these transmissions utilize half of the spectrum for the whole time and in TDD they utilize the whole spectrum half of the time.

Let us assume that an FD system transmits $b$ bits in total (i.e., DL + UL) in $T_{\text{FD}} = T$ seconds. Hence, the aggregated data rate is $R_{\text{FD}} = b/T$ bits/s. The bandwidth utilized for the DL and UL transmissions is $B_{\text{FD}} = B$ Hz. Therefore, the SE is $\text{SE}_{\text{FD}} = R_{\text{FD}} / B_{\text{FD}} = b/BT$ bits/s/Hz.

An equivalent FDD system would also need $T_{\text{FDD}} = T$ seconds to send $b$ bits and, therefore, would have an aggregated data rate of $R_{\text{FDD}} = b/T$ bits/s. To this end, though, it would utilize a spectrum of $B_{\text{FDD}} = 2B$ Hz. Thus, the SE of this FDD system would be $\text{SE}_{FDD} = R_{\text{FDD}} / B_{\text{FDD}} = b/2BT = \text{SE}_{FD}/2$ bits/s/Hz.

Similarly, an equivalent TDD system would need $T_{\text{TDD}} = 2T$ seconds to send $b$ bits and, therefore, would have an aggregated data rate of $R_{\text{FDD}} = b/2T$ bits/s. For the DL and UL transmissions it would utilize a spectrum of $B_{\text{TDD}} = B$ Hz. Hence, the SE of this TDD system would be $\text{SE}_{TDD} = R_{\text{TDD}} / B_{\text{TDD}} = b/2BT = \text{SE}_{FD}/2$ bits/s/Hz.

We see that, in principle, FDD and TDD have the same SE, which is half the SE of FD.

Is my reasoning correct or some of my assumptions is wrong? Does indeed TDD achieves an aggregated data rate that is half that of FDD (but, of course, the same SE as FDD since it uses also half of the bandwidth)?

  • $\begingroup$ Is this on topic? Please see engineering.stackexchange.com/help/on-topic $\endgroup$ – Solar Mike Aug 8 '19 at 6:32
  • $\begingroup$ Normally full duplex needs a back channel that's as wide as the forward channel. This either means a separate cable or fiber, or a separate frequency. If you can manage antenna separation well enough to achieve communication on the same band, that implies that your antennas are directional enough that the back channel is effectively a different channel. So, basically, for full duplex you first double the spectrum available, and then you use it. $\endgroup$ – TimWescott Aug 8 '19 at 15:07

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