## 2 System Composition and Protocol Design
### SU Channel Access and Data Transmission  
In the feedback phase, there is no packet transmission on the forward link when PU waits for the feedback from AP, which wastes the spectrum resources. However, in overlay mode, SU can sense this spectrum gap and use it to transmit data. Thus, let SU send packets in the feedback phase to use the spectrum resources fully. Moreover, there is also no data transmission on the channel in the next round of the backoff phase, so packets can also be sent during this period. As shown in Figure 3, the BSR frame of SU M4 collides with the  
Figure 3: Channel access and data transmission process of the CR-OFDMA.  
BSR frames of PU C3 and C4. Thus, data from SU M4 are transmitted during the feedback phase and the next round of the backoff phase of the PU.  
Since the value of the backoff counter is chosen randomly, the backoff time is a random variable. Let \(t_{bo}\) represent the backoff time. In order to not affect the transmission of the next round of BSR frames, we let the SU only transmit packets within the minimum backoff time \(\min(t_{bo})=DIFS+slot\), and \(t_{f1}=\frac{K_{f}}{R}\) denotes the transmission delay of the PU, where \(K_{f}\) is the number of bits of a packet, \(R_{1}\) is the transmission speed, \(t_{f2}\) denotes the propagation delay, and \(t_{MBA}\) denotes the transmission delay of MBA frame. Therefore, the time PU waits for feedback is \(t_{w}=2t_{f2}+2 SIFS+t_{MBA}\) and the transmission time available by the SU is \(t_{s}=t_{w}+\min(t_{bo})=2t_{f2}+2 SIFS+t_{MBA}+DIFS+slot\). To make it clear, the main notations used in this paper are listed in Table 1.  
Since the data to be sent for critical and mild patients are health indicators such as heartbeat and blood pressure, we assume that the packet size to be sent for critical and mild patients is identical. Let \(a=\frac{t_{f1}}{t_{s}}\), as shown in Figure 4. If \(a\leq 1\), SU can complete the transmission before the next round of BSR frames is transmitted, there is no collision with the BSR frames.  
If \(a>1\), the packet of SU cannot be completely transmitted within \(t_{s}\), and the transmission at the end of the SU packet will collide with the next round of BSR frames, as shown in the shaded portion of Figure 5.  
In order to avoid the collision between the SU packet and the BSR frame, the size of the SU packet needs to be reduced in the case of \(a>1\), which enables SU to complete the packet transmission in \(t_{s}\). It is necessary to equate the packet into \(n_{s}\) small packets so that the small packets can finish the transmission in \(t_{s}\).  
\[n_{s}=Int\bigg{(}\frac{K_{f}}{R_{1}t_{s}}\bigg{)}+1. \tag{1}\]  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Symbol** & **Definition** \\ \hline \(t_{f1}\) & transmission delay of the PU \\ \(t_{f2}\) & the propagation delay \\ \(t_{w}\) & the time PU waits for feedback \\ \(t_{s}\) & the transmission time available by the SU \\ \(K_{f}\) & the number of bits in a packet \\ \(R_{1}\) & the transmission speed \\ \(t_{bo}\) & the backoff time \\ \(a\) & ratio of SU transmission delay to propagation delay \\ \(n_{s}\) & the number of small packets after the packet division of SU \\ \hline \hline \end{tabular}
\end{table}
Table 1: The list of the main symbols in the protocol design.  
Figure 4: Utilization of spectrum resources by SU (\(a\leq 1\)).  
After the SU transmission is completed, the AP sends an MBA frame to inform whether the data packet is successfully received. Since the MBA frame is transmitted on the backward link, it does not affect the next round of BSR frames transmitted on the forward link.