Packet Level Wireless Channel Model for OFDM System Using SHLPNs

Abstract

In Chap. 2, we have introduced the SHLPNs and the compound marking technique for model aggregation. In this chapter, we adopt this technique to form a wireless channel model for OFDM systems in order to simplify the cross-layer performance analysis of modern wireless systems [1]. Compared with existing FSMC model whose state space grows exponentially with the number of OFDM subchannels, our proposed SHLPN model uses state aggregation technique to deal with this problem. Closed-form expressions to calculate the transition probabilities among the compound markings of the SHLPN model are provided. When applied to derive the performance measures for OFDM system in terms of the average throughput, average delay, and packet dropping probability, the SHLPN model can accurately capture the correlated time-varying nature of wireless channels. Simulation is performed to show that the numerical results offered by the proposed model are more accurate compared with other simplified channel models for avoiding state space complexity.

Appendix: Proof of Lemma 5.1

According to Theorem 5.1, the transition probability $p_{\hat{l},\hat{n}}$ from compound marking $\{k_{1},k_{2},\ldots,k_{L}\}$ with index $\hat{l}$ to compound marking $\{k_{1}^{{\prime}},k_{2}^{{\prime}},\ldots,k_{L}^{{\prime}}\}$ with index $\hat{n}$ equals the sum of the transition probabilities from any one of the individual markings $\vec{l} \in \mathcal{L}$ to all of the individual markings $\vec{n} \in \mathcal{N}$. Given an individual marking $\vec{l} \in \mathcal{L}$, we not only know that there are k _l subchannels in local channel state l for any $l \in \{ 1,\ldots,L\}$, but also the specific subset of subchannels in each local state l. In order to calculate $p_{\hat{l},\hat{n}}$, we need to enumerate all the events that lead to the number of subchannels in local channel state l transited to $k_{l}^{{\prime}}$ for any $l \in \{ 1,\ldots,L\}$.

As illustrated in Fig. 5.5, let a _l, (l+1) and a _(l+1), l be the number of subchannels that transit from local channel state l to (l + 1) and vice versa, respectively. With an individual marking $\vec{l} \in \mathcal{L}$, and assuming that the values of a _l, (l+1) and a _(l+1), l are known for all $l \in \{ 1,\ldots,L - 1\}$, the probability of this event can be derived as

$$\displaystyle{ \prod _{l=1}^{L}C_{ k_{l}}^{a_{l,(l-1)} }(p_{l,(l-1)})^{a_{l,(l-1)} }C_{(k_{l}-a_{l,(l-1)})}^{a_{l,(l+1)} }(p_{l,(l+1)})^{a_{l,(l+1)} }(p_{l,l})^{(k_{l}-a_{l,(l-1)}-a_{l,(l+1)})}, }$$

(5.11)

where we set $a_{1,0} = a_{0,1} = a_{L,(L+1)} = a_{(L+1),L} = 0$.

Fig. 5.5

Illustration of the state transition of compound markings from $\{k_{1},k_{2},\ldots,k_{L}\}$ to $\{k_{1}^{{\prime}},k_{2}^{{\prime}},\ldots,k_{L}^{{\prime}}\}$. k _l is the number of subchannels originally in local state l. a _l, (l+1) and a _(l+1), l are the number of subchannels that transit from local channel state l to (l + 1) and vice versa, respectively. After the transition, the number of subchannels in local state l becomes $k_{l}^{{\prime}}$

Therefore, in order to derive $p_{\hat{l},\hat{n}}$, we only need to find all the possible values of a _l, (l+1) and a _(l+1), l that result in k _l transited to $k_{l}^{{\prime}}$ for every $l \in \{ 1,\ldots,L - 1\}$, and calculate the sum probabilities of these events. For this purpose, the following equation needs to be true for any $l \in \{ 1,\ldots,L - 1\}$:

$$\displaystyle{ k_{l} - a_{l,(l-1)} - a_{l,(l+1)} + a_{(l-1),l} + a_{(l+1),l} = k_{l}^{{\prime}}. }$$

(5.12)

Adding the first l equations of (5.12) together, we can get (5.6) in Lemma 5.1, which establishes the relationship between a _l, (l+1) and a _(l+1), l . Now, we only need to find the upper and lower bounds of a _l, (l+1).

We first derive the upper bound of a _l, (l+1). Since the number of subchannels transited from state l to other states must not be larger than the number of subchannels originally in state l, and the number of subchannels transited to state l from other states must not be larger than the number of subchannels in state l after the transition, the following four inequalities related to a _l, (l+1) and a _(l+1), l must hold:

$$\displaystyle{ \left \{\begin{array}{ll} a_{l,(l+1)} + a_{l,(l-1)} \leq k_{l}, \\ a_{(l+1),l} + a_{(l+1),(l+2)} \leq k_{(l+1)}, \\ a_{(l+1),l} + a_{(l-1),l} \leq k_{l}^{{\prime}}, \\ a_{l,(l+1)} + a_{(l+2),(l+1)} \leq k_{(l+1)}^{{\prime}}. \end{array} \right. }$$

(5.13)

Since we need to enumerate all the possible values of a _l, (l+1) for every $l \in \{ 1,\ldots,L - 1\}$, we start from l = 1 and proceed to increasing values of l in sequence. Therefore, when we set the upper bounds for a _l, (l+1) and a _(l+1), l , we can assume that the values of a _l, (l−1) and a _(l−1), l are known, and ignore the values of $a_{(l+1),(l+2)}$ and $a_{(l+2),(l+1)}$ since the upper bounds of these two values will be set as a function of the values of a _l, (l+1) and a _(l+1), l . Combining with (5.6), we find that the first and the third inequalities are equivalent. Therefore, the above four inequalities becomes:

$$\displaystyle{ \left \{\begin{array}{ll} a_{l,(l+1)} \leq k_{l} - a_{l,(l-1)}, \\ a_{l,(l+1)} \leq \sum _{i=1}^{(l+1)}k_{i} -\sum _{i=1}^{(l)}k_{i}^{{\prime}}, \\ a_{l,(l+1)} \leq k_{(l+1)}^{{\prime}},\end{array} \right. }$$

(5.14)

which establishes the upper bound of a _l, (l+1) by taking the minimum value of the right hand sides of the above three inequalities and results in (5.5) in Lemma 5.1.

Next, we derive the lower bound of a _l, (l+1) from the fact that a _l, (l+1) ≥ 0 and a _(l+1), l  ≥ 0. Therefore, we have

$$\displaystyle{ \left \{\begin{array}{ll} a_{l,(l+1)} \geq 0, \\ a_{l,(l+1)} \geq \sum _{i=1}^{(l)}(k_{i} - k_{i}^{{\prime}}), \end{array} \right. }$$

(5.15)

which establishes the lower bound of a _l, (l+1) by taking the maximum value of the right hand sides of the above two inequalities and results in (5.4) in Lemma 5.1.

Now armed with the upper and lower bounds of a _l, (l+1), we can enumerate all the possible values of a _l, (l+1) for every $l \in \{ 1,\ldots,L - 1\}$ and calculate the sum probabilities of these events, which results in (5.3) in Lemma 5.1 and completes the proof.