## 3 Cross-Layer Design for Finite Blocklength Coding
### Achieving the Hard Delay Constraint with Finite Average Power  
We use \(L(q_{1})\) to denote the number of symbol blocks corresponding to \(q_{1}\) data packets. Subsequently, we have \(L(q_{1})=a_{2}[t]\) under conditions where \(q_{1}[t]=q_{1}\) and \(q_{2}[t]=0\). Let \(k\) and \(l\) represent the number of time slots and the number of symbol blocks that have been successfully transmitted after a finite blocklength coding process, respectively. Then, we have  
\[k =t-\max\{\hat{t}\mid\hat{t}<t,a_{2}[\hat{t}]>0\}. \tag{14}\] \[l =a_{2}[t-k]-q_{2}[t]. \tag{15}\]Considering that the transmission in a deep fading channel requires infinite power, we should suspend symbol transmission during deep fading to save power, particularly when the remaining number of symbol blocks is less than the number of the remaining time slots. To satisfy the hard delay constraint, it is necessary to transmit the symbol block when the remaining number of symbol blocks is equal to the remaining number of time slots, regardless of the channel quality. As such, the service rate of the symbol queue is given by  
\[s_{2}[t]=\left\{ \begin{array}{cl}\mathds{1}\left\{g[t]\geq g_{th}(\zeta,k,l) \right\},g_{th}(\zeta,k,l)>0,&\text{if }L(q_{1})-l<K-k+1,\\ 1,&\text{if }L(q_{1})-l\geq K-k+1.\end{array} \right. \tag{16}\]  
In each time slot, the threshold of the truncated channel inversion is intrinsically related to three parameters: \(\zeta\), defined as the number of data packets concurrently encoded; \(k\); and \(l\). For a more concise expression, we abbreviate \(g_{th}(\zeta,k,l)\) to \(g_{th}\). Let \(P=g[t]P[t]\) denote the fixed receiver power. The average power for \(g[t]\geq g_{th},g_{th}\in[0,\infty)\) in a single time slot is defined as  
\[\bar{P}(g_{th})=\int_{g_{th}}^{\infty}\frac{P}{x}f(x)dx. \tag{17}\]  
The average power consumption of the system is given by  
\[P_{ave}=\frac{1}{T}\sum_{t=1}^{T}P[t]. \tag{18}\]  
Based on Equations (4) and (17), the average power for \(g_{th}\in[0,\infty)\) in a single time slot is defined as  
\[\bar{P}(g_{th})=\int_{g_{th}}^{\infty}\frac{P}{x}f(x)dx=\int_{g_{th}}^{\infty} \frac{PMe^{-x}(1-e^{-x})^{M-1}}{x}dx. \tag{19}\]  
We assume \(\sigma^{2}=1/2\) in this equation without a loss of generality. To facilitate the computation of Equation (19), we propose the following two lemmas. **Lemma 1**. : _The integral expression \(\int_{y}^{\infty}e^{-cx}\ln xdx\) satisfies_  
\[\int_{y}^{\infty}e^{-cx}\ln xdx=-\frac{1}{c}\big{(}\gamma(cy)+e^{-y}\ln c \big{)}<\infty,\forall c>0,\forall y\geq 0, \tag{20}\]  
_where we have_  
\[\gamma(cy)=-\int_{cy}^{\infty}e^{-z}\ln zdz. \tag{21}\]  
**Proof. ** Firstly, we consider the following integral expression:  
\[\int_{y}^{\infty}e^{-cx}\ln xdx =\frac{1}{c}\int_{cy}^{\infty}e^{-z}(\ln z-\ln c)dz \tag{22}\] \[=\frac{1}{c}\int_{cy}^{\infty}e^{-z}\ln zdz-\frac{1}{c}\int_{cy} ^{\infty}e^{-z}\ln cdz\] \[=-\frac{1}{c}\big{(}\gamma(cy)+e^{-cy}\ln c\big{)}.\]  
Secondly, we prove the convergence of Equation (22) by considering two cases, i.e., \(cy\geq 1\) and \(cy<1\). For the case when \(cy\geq 1\),  
\[\bigg{|}\int_{cy}^{\infty}e^{-z}\ln zdz\bigg{|}\leq\int_{cy}^{\infty}e^{-z}(z -1)dz=e^{-cy}, \tag{23}\]which proves there is convergence. When \(cy<1\), we have  
\[ \begin{split}\left|\int_{cy}^{\infty}e^{-z}\ln zdz\right|& \leq\left|\int_{cy}^{1}e^{-z}\ln zdz\right|+\left|\int_{1}^{\infty}e^{-z} \ln zdz\right|\\ &\leq\int_{cy}^{1}-\ln zdz+\int_{1}^{\infty}e^{-z}(z-1)dz\\ &=1-cy+cy\ln cy+e^{-1}.\end{split}  \tag{24}\]  
Therefore, the lemma is proved. **Lemma 2**. : _The following equality holds, i.e.,_  
\[ \begin{split}&\sum_{i=0}^{M-1}\binom{M-1}{i}\frac{(-1)^{i+1}}{i+1 }\phi(i+1)\\ =&(M-1)\sum_{i=0}^{M-2}\binom{M-2}{i}\frac{(-1)^{i+1 }}{i+2}\phi(i+2)+\sum_{i=0}^{M-1}\binom{M-1}{i}(-1)^{i+1}\phi(i+1).\end{split}  \tag{25}\]  
_Here, \(\phi(\cdot)\) is an arbitrary function of \(i\)._  
Proof. : In this proof, we consider the following expression, which is equivalent to Equation (25):  
\[