On time-triggered and event-based control of integrator systems over a shared communication system

Abstract

When analyzing networked control systems, where the control loop is closed over a communication system, it is crucial to take the communication system into account. Hence, time-triggered and event-based control of an integrator system with noise over a shared communication system is analyzed. Thereby, analytical models of different communication systems are used and the analysis is focused on the effect of the communication system on the performance, as well as the interaction between control and communication. For time-triggered control, where the event times are known in advance, a deterministic communication protocol can be used. Hence, time-triggered control with the two most well-known deterministic communication protocols, time division multiple access (TDMA) and frequency division multiple access (FDMA), is analyzed. For event-based control, where the events appear at random times, a contention-based communication protocol should be used. Hence, event-based control is analyzed with different contention-based communication protocols: pure ALOHA, slotted ALOHA, a queueing system, and Erlang’s loss model. It turns out that time-triggered control with either TDMA or FDMA outperforms event-based control with pure or slotted ALOHA. However, event-based control with a properly designed queueing system gives an even better performance. Thus, we conclude that it is crucial to take the details of the communication system into account.

Appendices

Appendix A: Remaining proofs of Sect. 3

In [1], the expected cost and interevent time for controlling system (2) with an impulsive input is studied for $\sigma _{i} = 1$ and an ideal communication. In [33], this result is extended to the case of a communication system with random packet loss. To prove Theorems 4, 7, and 8, we have to extend these results to arbitrary $\sigma _{i}$ and packet delay.

For simplicity of notation, we drop the subindex which indicates the agent.

1.1 Expected time between two successful transmissions

Lemma 35

Suppose, the expected time between two transmission attempts is $T$ and the packet loss probability is $p$. Then the expected time between two successful transmissions $T_{S}$ is

$$\begin{aligned} {{\mathrm{E}}}[ T_{S} ] = \frac{T}{1-p}. \end{aligned}$$

Proof

$$\begin{aligned} {{\mathrm{E}}}[ T_{S} ] = \sum _{k=1}^{\infty } (1-p) p^{k-1} k T = T \frac{(1-p)}{p} \sum _{k=0}^{\infty } k p^{k} = \frac{T}{1-p} \end{aligned}$$

$\square$

1.2 Additional cost due to delay

Time-triggered and event-based control with packet loss has been studied in [33]. In the considered setup packets are also delayed, so this effect must be taken into account. To show how delayed events affect the cost, we use $\tilde{x}$ to denote the state of the system with delayed events, whereas $x$ remains the state of the system without delayed events. Figure 10 shows an example of this case. Suppose an event is generated at time $t_{k}$. If the corresponding impulse is applied immediately, then the state will be reset to the origin. Due to the delay $d_{k}$, the impulse is applied at time $t_{k} + d_{k}$. Since the process continues in the meantime, the impulse will not reset the state to the origin. Thus, the cost of the system with delayed events will be different from the cost of the system without delayed events as stated in the following lemma.

Fig. 10

The effect of delayed events

Lemma 36

Suppose input (3) is used to control system (2). Then, the cost of the closed loop system with delayed events is $J + \sigma ^{2} d$, where $J$ is the cost of the system without delay and $d$ the expected delay.

Proof

Since the interevent times are iid, it suffices to integrate between the two successful packet transmissions $t_k$ and $t_{k+1}$. Thus, we get for the cost of the process with delayed events:

$$\begin{aligned} \tilde{J} = \frac{{{\mathrm{E}}}\left[ \int _{t_k}^{t_{k+1}} \tilde{x}(t)^2\,\mathrm{d}t\right] }{{{\mathrm{E}}}[t_{k+1}-t_{k}]}. \end{aligned}$$

(68)

Obviously, we get for the denominator ${{\mathrm{E}}}[t_{k+1} - t_{k}] = {{\mathrm{E}}}[T_{S}]$. Thus, we continue as follows:

$$\begin{aligned} \tilde{J}&= \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k}^{t_{k+1}} \tilde{x}(t)^2\,\mathrm{d}t\right] \\&= \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k}^{t_k+d_{k}} \tilde{x}(t)^2\,\mathrm{d}t\right] + \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k+d_{k}}^{t_{k+1}} \tilde{x}(t)^2\,\mathrm{d}t\right] \\&= \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k}^{t_k+d_{k}} \left( x(t)+x(t_k)\right) ^2 \,\mathrm{d}t\right] + \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k+d_{k}}^{t_{k+1}} x(t)^2\,\mathrm{d}t\right] \\&= \underbrace{ \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k}^{t_{k+1}} x(t)^2\,\mathrm{d}t\right] }_{J} + \underbrace{ \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k}^{t_k+d_{k}} x(t_k)^2\,\mathrm{d}t\right] }_{J_{d}}\\&\quad + \underbrace{\frac{2}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}\left[ \int _{t_k}^{t_k+d_{k}} x(t) x(t_k)\,\mathrm{d}t\right] }_{= J^*}. \end{aligned}$$

Note that $J^* = 0$ because $x(t)$ and $x(t_{k})$ are independent and ${{\mathrm{E}}}[x(t)] = 0$. Thus, it remains to check $J_{d} = \frac{1}{{{\mathrm{E}}}[T_{S}]} \int _{t_k}^{t_k+d_{k}} {{\mathrm{E}}}[x(t_k)^2]\,\mathrm{d}t$. Since the delay $d_{k}$ is independent from $x(t_{k})$ and ${{\mathrm{E}}}[d_{k}] = d$, we get for the additional cost due to delay $J_{d} = \frac{1}{{{\mathrm{E}}}[T_{S}]} {{\mathrm{E}}}[x(t_k)^2]d$. Since the process was reset to the origin at the previous successful transmission $t_{k-1}$, we have ${{\mathrm{E}}}[x(t_{k})^{2}] = \sigma ^{2} {{\mathrm{E}}}[T_{S}]$ and consequently $J_{d} = \sigma ^{2 }d$. $\square$

1.3 Proof of Theorem 4

To prove Theorem 4, we first extend the cost of time-triggered control as given in [1] to an arbitrary noise intensity $\sigma$. Based on this result, we then derive the cost of time-triggered control with packet loss.

Lemma 37

Suppose, system (2) is controlled by an impulsive time-triggered control scheme with sampling time $T_{\mathrm{TT }}$ and an ideal communication without loss and delay. Then the cost is

$$\begin{aligned} J_{\mathrm{TT }} = \sigma ^{2} \frac{T_{\mathrm{TT }}}{2}. \end{aligned}$$

(69)

Proof

$$\begin{aligned} J_{\mathrm{TT }} = \frac{1}{T_{\mathrm{TT }}} {{\mathrm{E}}}\left[ \int _{0}^{T_{\mathrm{TT }}} x^{2}(t)\,{\hbox {d}}t\right] = \frac{1}{T_{\mathrm{TT }}} \int _{0}^{T_{\mathrm{TT }}} {{\mathrm{E}}}[x^{2}(t)]\,{\hbox {d}}t = \frac{1}{T_{\mathrm{TT }}} \int _{0}^{T_{\mathrm{TT }}} \sigma ^{2}t\,{\hbox {d}}t=\sigma ^{2} \frac{T_{\mathrm{TT }}}{2}. \end{aligned}$$

$\square$

Lemma 38

Suppose, system (2) is controlled by an impulsive time-triggered control scheme with sampling time $T_{\mathrm{TT }}$ and a packet loss probability $p$. Then the cost is

$$\begin{aligned} J_{\mathrm{TT }} = \sigma ^{2} \left( \frac{T_{\mathrm{TT }}}{2} + \frac{T_{\mathrm{TT }} p}{(1 - p)} \right) . \end{aligned}$$

Proof

To prove Lemma 38, we use $J_{m}$ to denote the cost for the case that the sampling time is extended to $m T_{\mathrm{TT }}$. From Lemma 37, we have

$$\begin{aligned} J_{m} = \frac{{{\mathrm{E}}}\left[ \int _{0}^{m T_{\mathrm{TT }}} x^{2}(t)\,{\hbox {d}}t \right] }{m T_{\mathrm{TT }}} = \sigma ^{2} \frac{m T_{\mathrm{TT }}}{2} \end{aligned}$$

and thus,

$$\begin{aligned} {{\mathrm{E}}}\left[ \int _{0}^{m T_{\mathrm{TT }}} x^{2}(t)\,{\hbox {d}}t \right] = J_{m} m T_{\mathrm{TT }} = \sigma ^{2} \frac{m^{2} T_{\mathrm{TT }}^{2}}{2}. \end{aligned}$$

Now, we consider the term ${{\mathrm{E}}}[ \int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t]$, where $T_{S}$ is the expected time between two successful packet transmissions.

$$\begin{aligned} {{\mathrm{E}}}\left[ \int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t\right]&= \sum _{m=1}^{\infty } (1-p) p^{m-1} {{\mathrm{E}}}\left[ \int _{0}^{m T_{\mathrm{TT }}} x^{2}(t)\,{\hbox {d}}t \right] \\&= \frac{\sigma ^{2} T_{\mathrm{TT }}^{2}}{2} \frac{1-p}{p} \sum _{m=0}^{\infty } m^{2} p^{m}= \frac{\sigma ^{2} T_{\mathrm{TT }}^{2}}{2} \frac{1+p}{(1-p)^{2}}. \end{aligned}$$

Thus, we get for the cost

$$\begin{aligned} J_{\mathrm{TT }} = \frac{{{\mathrm{E}}}\left[ \int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t \right] }{{{\mathrm{E}}}[T_{S}]} = \frac{\sigma ^{2} T_{\mathrm{TT }}}{2} \frac{1+p}{1-p} = \sigma ^{2} \left( \frac{T_{\mathrm{TT }}}{2} + \frac{T_{\mathrm{TT }} p}{(1 - p)} \right) . \end{aligned}$$

$\square$

Proof of Theorem 4

Theorem 4 follows from combining Lemmas 36 and 38. $\square$

1.4 Proof of Theorem 7

To prove Theorem 7, we use a result from [46].

Lemma 39

([46]) Suppose, the lower threshold is $\underline{\Delta }$ and the upper threshold is $\overline{\Delta }$ and system (2) is started at $x_{0}$; $\underline{\Delta } < x_{0} < \overline{\Delta }$. Then, the expected time to reach one of these thresholds is characterized as the unique solution of

$$\begin{aligned} \frac{1}{2} \sigma ^{2} \frac{\partial ^{2}}{\partial x^{2}} \Psi (x_{0}) = -1 \qquad \text {with }\Psi (\underline{\Delta }) = \Psi (\overline{\Delta }) = 0. \end{aligned}$$

(70)

Proof of Theorem 7

Proof of Theorem 7 For the considered event-based control, we have $\underline{\Delta } = - \Delta$, $\overline{\Delta } = + \Delta$. Thus, the solution of (70) is

$$\begin{aligned} \Psi (x_{0}) = \frac{1}{\sigma ^{2}} (\Delta ^{2} - x_{0}). \end{aligned}$$

Now, note that in the considered setup, the system is reset to the origin, i.e., $x_{0} = 0$. Thus, the expected interevent time is

$$\begin{aligned} T_{\mathrm{EB }} = \Psi (0) = \frac{\Delta ^{2}}{\sigma ^{2}}. \end{aligned}$$

$\square$

1.5 Proof of Theorem 8

Again, we extend the cost given in [1] to the case of an arbitrary noise intensity $\sigma$. Based on this result, we derive the cost of event-based control with packet loss similar to [33].

Lemma 40

Suppose, system (2) is controlled by an impulsive event-based control scheme with threshold increment $\Delta$ and an ideal communication without loss and delay. Then the cost is

$$\begin{aligned} J_{\mathrm{EB }} = \sigma ^{2} \frac{T_{\mathrm{EB }}}{6}. \end{aligned}$$

(71)

Proof

The distribution of the state follows from the steady state of the Kolmogorov forward equation

$$\begin{aligned} 0 = \frac{1}{2} \frac{\partial ^{2}}{\partial x^{2}} \sigma ^{2} f(x) \qquad \text {with }\int _{-\Delta }^{\Delta } f(x)\,{\hbox {d}}x = 1, f(-\Delta ) = f(\Delta ) = 0. \end{aligned}$$

This equation has the solution

$$\begin{aligned} f(x) = \frac{\Delta - |x|}{\Delta ^{2}}. \end{aligned}$$

Consequently, the variance is

$$\begin{aligned} J_{\mathrm{EB }} = \int _{-\Delta }^{\Delta } x^{2} f(x)\,{\hbox {d}}x= \frac{\Delta ^{2}}{6} = \sigma ^{2} \frac{T_{\mathrm{EB }}}{6}. \end{aligned}$$

$\square$

Lemma 41

Suppose, system (2) is controlled by an impulsive event-based control scheme with threshold increment $\Delta$ and a packet loss probability $p$. Then the cost is

$$\begin{aligned} J_{\mathrm{EB }} = \sigma ^{2} \left( \frac{T_{\mathrm{EB }}}{6} + \frac{T_{\mathrm{EB }} p}{(1 - p)}\right) . \end{aligned}$$

(72)

Proof

To prove Lemma 41, we use $T_{m}$ to denote the time between $m$ transmission attempts and $T_{S}$ the time between two successful transmissions. We start by considering the term ${{\mathrm{E}}}[\int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t ]$

$$\begin{aligned} {{\mathrm{E}}}\left[ \int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t \right]&= \sum _{m=1}^{\infty } (1-p) p^{m-1} {{\mathrm{E}}}\left[ \int _{0}^{T_{m}} x^{2}(t)\,{\hbox {d}}t\right] \\&=\frac{1-p}{p} \sum _{m=0}^{\infty } p^{m} \sum _{n=1}^{m} {{\mathrm{E}}}\left[ \int _{T_{n-1}}^{T_{n}} x^{2}(t)\,{\hbox {d}}t\right] . \end{aligned}$$

To proceed, we consider the term $\nu _{n} := {{\mathrm{E}}}[\int _{T_{n-1}}^{T_{n}} x^{2}(t)\,{\hbox {d}}t]$.

$$\begin{aligned} \nu _{n} := {{\mathrm{E}}}\left[ \int _{T_{n-1}}^{T_{n}} x^{2}(t)\,{\hbox {d}}t\right] = {{\mathrm{E}}}\left[ x^{2}(T_{n-1}) \int _{T_{n-1}}^{T_{n}}\,{\hbox {d}}t + \int _{T_{n-1}}^{T_{n}} (x(t) - x(T_{n-1}))^{2}\,{\hbox {d}}t\right] . \end{aligned}$$

Now, note that $x(T_{n})$ is always an integer multiple of $\Delta$ and similar to the random process $\sum _{m=1}^{n} \theta (m) \Delta$, where $\theta (m) \in \{-1, 1\}$ with and . Thus, $\nu _{n}$ becomes

$$\begin{aligned} \nu _{n} = {{\mathrm{E}}}\left[ \left( \sum _{m=1}^{n} \theta _{m} \Delta \right) ^{2}\right] {{\mathrm{E}}}[T_{\mathrm{EB }}] + {{\mathrm{E}}}\left[ \int _{0}^{T_{\mathrm{EB }}} x^{2}(t)\,{\hbox {d}}t |x(0) = 0\right] = (n-1) \frac{\Delta ^{4}}{\sigma ^{2}} + \frac{\Delta ^{4}}{6 \sigma ^{2}}. \end{aligned}$$

Now, we proceed with the term ${{\mathrm{E}}}[\int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t]$

$$\begin{aligned} {{\mathrm{E}}}\left[ \int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t\right]&= \frac{1-p}{p} \sum _{m=0}^{\infty } p^{m} \sum _{n=1}^{m} \nu _{n} \\&= \frac{1-p}{p} \sum _{m=0}^{\infty } p^{m} \sum _{n=1}^{m} (n-1) \frac{\Delta ^{4}}{\sigma ^{2}} + \frac{\Delta ^{4}}{6 \sigma ^{2}} \\&= \frac{1-p}{p} \frac{\Delta ^{4}}{\sigma ^{2}} \sum _{m=0}^{\infty } p^{m} \left( \frac{m(m-1)}{2} + \frac{m}{6} \right) \\&= \frac{\Delta ^{4}}{\sigma ^{2}} \left( \frac{p}{(1-p)^{2}} + \frac{1}{6 (1-p)} \right) . \end{aligned}$$

Finally, Lemma 41 follows from $J_{\mathrm{EB }, i} = \frac{{{\mathrm{E}}}[\int _{0}^{T_{S}} x^{2}(t)\,{\hbox {d}}t ]}{{{\mathrm{E}}}[T_{S}]}$. $\square$

Proof of Theorem 8

Theorem 8 follows from combining Lemmas 36 and 41. $\square$

1.6 Proof of Theorem 12

To show that the arrival process of event-based control converges to a Poisson process for $N \rightarrow \infty$, we use the Palm–Khintchine Theorem and the following two assumptions, definition and short discussion, from [17].

Assumption 42

For all $N$ sufficiently large,

$$\begin{aligned} \lambda _{1, N} + \cdots + \lambda _{N, N} = \lambda _{\Sigma } < \infty , \end{aligned}$$

(73)

where $\lambda _{j, N}$ is the sending rate of agent $j$ for the case that there are $N$ agents.

Assumption 43

Given $\epsilon > 0$, for each $t > 0$ and $N$ sufficiently large,

$$\begin{aligned} F_{j, N}(t) \le \epsilon , \qquad j = 1, \cdots , N, \end{aligned}$$

(74)

where $F_{j, N}$ is the CDF of agent $j$ for the case that there are $N$ agents.

Definition 44

For each $N$ define

$$\begin{aligned} L_{0, N}(t) = L_{1, N}(t) + \cdots + L_{N, N}(t), \end{aligned}$$

(75)

where $L_{j, N}(t)$ is a stochastic process, which counts the number of events of agent $j$ that occur by time $t$ for the case that there are $N$ agents.

Assumption 43 asserts that as $N$ increases, the processes being combined have renewals very infrequently. Assumption 42 shows that $L_{0, N+1}(t)$ is not formed by adding another process to $L_{0, N}(t)$. As $N$ increases, the processes being combined are changed so that (at least for large $N$) the asymptotic rate at which renewals occur is a constant.

Theorem 45

(Palm–Khintchine) [17, Theorem 5.15] Under Assumptions 42 and 43, as $N \rightarrow \infty$, $\{L_{0, N}(t); t \ge 0\}$ approaches a Poisson process.

Proof of Theorem 12

Since Assumption 42 is part of Theorem 12, it remains to show that Assumption 43 holds. From the scaling property and the definition of the CDF, it follows that . Since $f(x |1)$ is continuous, if follows from the mean value theorem that for some $z \in (0,t)$. Moreover, since for $N \rightarrow \infty$, we see that for each $\epsilon > 0$ and $t > 0$ there exists an $N$ such that $F_{j, N}(t) \le \epsilon$, i.e., Assumption 43 holds. $\square$

Appendix B: Remaining proofs of Sect. 4

1.1 Proof of Theorem 13

For simplicity of notation, we drop the subindex indicating the agent and use it to indicate the scheme instead. Moreover, we use $\tilde{p}$ to denote the probability of an unsuccessful transmission and $\tilde{d}$ for the delay between a packet transmission and its reception.

When using Scheme 1, all packets that are not successfully transmitted are lost and the delay is only the delay between event generation and the reception of the corresponding packet. Thus, for Scheme 1, the additional cost due to loss and delay is

$$\begin{aligned} \frac{\tilde{p}_{1}}{1-\tilde{p}_{1}} T + \tilde{d}_{1}. \end{aligned}$$

(76)

When using Scheme 2, all packets that are not transmitted successfully are retransmitted at some later time. Thus, there is no packet loss but an additional delay due to the retransmissions.

$$\begin{aligned} d_{2}&= \tilde{d}_{2} + 0 T (1-\tilde{p}_{2}) + 1 T (1-\tilde{p}_{2}) \tilde{p}_{2} + 2 T (1-\tilde{p}_{2}) \tilde{p}_{2}^{2} + 3 T (1-\tilde{p}_{2}) \tilde{p}_{2}^{3} \\&\quad + \cdots =\tilde{d}_{2} + (1-\tilde{p}_{2}) \tilde{p}_{2} T \sum _{m=0}^{\infty } m \tilde{p}_{2}^{m-1} = \tilde{d}_{2} + \frac{\tilde{p}_{2}}{1-\tilde{p}_{2}} T. \end{aligned}$$

Thus, for Scheme 2, the additional cost due to delay becomes

$$\begin{aligned} \frac{\tilde{p}_{2}}{1 - \tilde{p}_{2}} T + \tilde{d}_{2}. \end{aligned}$$

(77)

Now, we can compare the additional cost due to loss and delay of Scheme 1 and Scheme 2, given by (76) and (77). Due to the retransmission of packets, the load when using Scheme 2 is larger than the load when using Scheme 1. Since we assumed that either the probability of an unsuccessful transmission or delay, or both, is strictly increasing with the network load, Scheme 1 gives a better performance than Scheme 2. $\square$

1.2 Proof of Lemma 17

We start by bounding the probability of self-interference, i.e., the term $(1-\int _{0}^{\rho _{i}} f_{i}(x |1)\,{\hbox {d}}t )^{2}$. Since $f_{i}(t |1)$ is a PDF, $0 \le 1-\int _{0}^{x} f_{i}(t |1)\,{\hbox {d}}t \le 1$ holds. From the assumption $f_{i}(t |1) \le 1$, we also have $1 - x \le 1 - \int _{0}^{x} f_{i}(t |1)\,{\hbox {d}}t$, i.e.,

$$\begin{aligned} \max \{0, 1-x\} \le 1 - \int _{0}^{x} f_{i}(t |1)\,{\hbox {d}}t \le 1. \end{aligned}$$

(78)

Since $(1-\rho _{i})^{2} = 1 - 2 \rho _{i} + \rho _{i}^{2} \le 1 - 2 \rho _{i}$, we can bound the probability of self-interference as follows

$$\begin{aligned} b_{p}(\rho _{i}) \le \left( 1-\int _{0}^{\rho _{i}} f_{i}(x |1)\,{\hbox {d}}t \right) ^{2} \le 1, \end{aligned}$$

(79)

with $b_{p}(\rho )$ as defined in (24).

Now, we continue with the probability of a collision with another user, i.e., the term $p_{co}(\rho _{j}) := 1 - \int _{0}^{2\rho _{j}} [1-\int _{0}^{x}f_j(t |1)\,{\hbox {d}}t]\,{\hbox {d}}x$. Since $p_{co}(\rho _{j})$ is a probability, we immediately have $0 \le p_{co}(\rho _{j}) \le 1$. Moreover, from (78) we get $1 - 2 \rho _{j} \le p_{co}(\rho _{j}) \le 1 - 2 \rho _{j} + 2 \rho _{j}^{2}$. Since $p_{co}(\rho _{j})$ is increasing with $\rho _{j}$ and the upper bound $1 - 2 \rho _{j} + 2 \rho _{j}^{2}$ has a minimum at , we use $a_{p}(\rho _{j})$ as upper bound for $p_{co}(\rho _{j})$ for . Therefore, we get

$$\begin{aligned} b_{p}(\rho _{j}) \le 1 - \int _{0}^{2 \rho _{j}}\left[ 1 - \int _{0}^{x} f_{j}(t |1)\,{\hbox {d}}t\right] \,{\hbox {d}}x \le a_{p}(\rho _{j}), \end{aligned}$$

(80)

with $a_{p}(\rho )$ and $b_{p}(\rho )$ as defined in (23) and (24), respectively.

Finally, (22) follows using (79) and (80) in (21). $\square$

1.3 Proof of Theorem 20

We start by proving the case without self-interference. Equation (29) follows from (20) due to the fact that the vulnerable period is $\tau$ for slotted ALOHA, instead of $2 \tau$ for pure ALOHA.

Next, we analyze the losses due to self-interference. Figure 11 shows an event generation within a slot and the corresponding times. The time interval between the event generation and the end of the slot is called  forward recurrence time $\tau _F$. Similarly, the time interval between the event generation and the begin of the slot is denoted backward recurrence time $\tau _B$. As already stated, a self-interference occurs if the same user generates a new event while its packet is still waiting for the start of the next slot, i.e., during $\tau _F$. Thus, the steady-state probability that a packet is not lost due to self-interference is

$$\begin{aligned} {{\mathrm{E}}}\left[ \left( 1 - \int _{0}^{\tau _F} f_i(t |\lambda _{i})\,{\hbox {d}}t \right) \right] = \frac{1}{\tau } \int _{0}^{\tau } \left( 1 - \int _{0}^{\tau -x} f_i(t |\lambda _{i})\,{\hbox {d}}t \right) \,{\hbox {d}}x. \end{aligned}$$

(81)

Finally, (28) follow from the fact that a packet of user $i$ is lost if it interferes with a packet of user $i$ or with a packet of any other user. $\square$

Fig. 11

Forward and backward recurrence time

1.4 Proof of Lemma 23

First, we look at the losses due to self-interference. Since the corresponding term is a probability and the assumption $f_{i}(x |1) \le 1$, we get

$$\begin{aligned} b_{s}(\rho ) \le \frac{1}{\rho _{i}} \int _{0}^{\rho _{i}} \left( 1 - \int _{0}^{\rho _{i}-x} f(t |1)\,{\hbox {d}}t \right) \,{\hbox {d}}x \le 1. \end{aligned}$$

(82)

By following the same arguments as in the proof of Lemma 17, we get

$$\begin{aligned} b_{s}(\rho _{j}) \le 1 - \int _{0}^{ \rho _{j}} \left[ 1 - \int _{0}^{x} f_{j}(t |1)\,{\hbox {d}}t\right] \,{\hbox {d}}x \le a_{s}(\rho _{j}). \end{aligned}$$

(83)

Finally, (31) follows using (82) and (83) in (30). $\square$