Minimizing the average achievable distortion using multi-layer coding approach in two-hop networks

Abstract

Minimizing the average achievable distortion (AAD) of a Gaussian source at the destination of a two-hop block fading relay channel is studied in this paper. The communication is carried out through the use of a Decode and Forward (DF) relay with no source-destination direct links. The associated receivers of both hops are assumed to be aware of the corresponding channel state information (CSI), while the transmitters are unaware of their corresponding CSI. The current paper explores the effectiveness of incorporating the successive refinement source coding together with multi-layer channel coding in minimizing the AAD. In this regard, the closed form and optimal power allocation policy across code layers of the second hop is derived, and using a proper curve fitting approach, a close-to-optimal power allocation policy associated with the first hop is devised. It is numerically shown that the DF strategy closely follows the Amplify and Forward (AF) relaying, while there is a sizable gap between the AAD of multi-layer coding and single-layer source coding.

Appendix 1

1.1 1.1 Proof of Lemma 1

The last constraint of Eq. 18, results in Eq. 19. Now, one can formulate the Lagrangian function of Eq. 18 as follows:

$$ \begin{array}{@{}rcl@{}} \mathcal{L}(T)={\int}_{0}^{\infty}\left( f(x)\mathcal{G}\left( \frac{1}{T(x)^{b}}\right)+\lambda \frac{T(x)}{x^{2}}-T^{\prime}(x)\psi(x)\right)dx, \end{array} $$

(45)

where λ and ψ(.) are, respectively, the slack multiplier and the slack arbitrary positive function that ensure meeting the power constraint, as well as monotonically non-decreasing characteristics of $T^{\prime }(.)$. Defining

$$ \begin{array}{@{}rcl@{}} D(x)=\frac{1}{T(x)^{b}}, \end{array} $$

(46)

and applying the variational method in [29] to Eq. 45 gives the following:

$$ \begin{array}{@{}rcl@{}} -b\mathcal{G}_{D}\left( D(x)\right)\frac{f(x)}{T(x)^{b+1}}+\frac{\lambda}{x^{2}}+\psi^{\prime}(x)=0, \end{array} $$

(47)

where $\mathcal {G}_{D}(D(x))=\frac {\partial }{\partial D(x)}\mathcal {G}(D(x))$, and $\psi ^{\prime }(x)=\frac {d}{dx}\psi (x)$. Taking the slackness condition corresponding to the positivity of $T^{\prime }(.)$ into account, for positive power allocation to the code layers, the second constraint of Eq. 18 is met with inequality, so ψ(x) = 0, and Eq. 47 leads to:

$$ \begin{array}{@{}rcl@{}} T(x)=\left( \frac{bx^{2}f(x)\mathcal{G}_{D}\left( D(x)\right)}{\lambda}\right)^{\frac{1}{b+1}}. \end{array} $$

(48)

As the above equation is derived for the case of having $T^{\prime }(.)>0$, bringing both sides of Eq. 48 to the power of (b + 1) and taking derivation w.r.t. x, gives the requirements of having the aforementioned condition met, as follows:

$$ \begin{array}{@{}rcl@{}} (b+1)T(x)^{b}T^{\prime}(x) = \frac{b}{\lambda}\left[\mathcal{G}_{D}\left( D(x)\right)\frac{d}{dx}\left( x^{2}f(x)\right)\right.\\ \left.+x^{2}f(x)\frac{d}{dx}\mathcal{G}_{D}\left( D(x)\right)\right] \end{array} $$

(49)

Noting that $\mathcal {G}(.)$ is an increasing convex function, $\mathcal {G}_{D}\left (D(x)\right )$ has positive values. Going further through the left-most term of Eq. 49, x²f(x) is a non-negative value, as we are dealing with a non-negative f(.) function, so, the term $\frac {d}{dx}\mathcal {G}_{D}\left (D(x)\right )$ should be discussed. Incorporating the derivative chain rule leads to:

$$ \begin{array}{@{}rcl@{}} \frac{d}{dx}\mathcal{G}_{D}\left( D(x)\right)=D^{\prime}(x) \mathcal{G}_{DD}\left( D(x)\right), \end{array} $$

(50)

where $\mathcal {G}_{DD}(D(x))=\frac {\partial ^{2}}{\partial D^{2}}\mathcal {G}(D(x))$. Considering the definition of D(.) in Eq. 46, the relation Eq. 50 converts to:

$$ \begin{array}{@{}rcl@{}} \frac{d}{dx}\mathcal{G}_{D}\left( D(x)\right)=\frac{-b}{T(x)^{b+1}}T^{\prime}(x)\mathcal{G}_{DD}\left( D(x)\right). \end{array} $$

(51)

Plugging the result of Eq. 51 into Eq. 49 results in Eq. 52:

$$ \begin{array}{@{}rcl@{}} \left[(b+1)T(x)^{b} + \frac{b^{2}}{\lambda T(x)^{b+1}}\mathcal{G}_{DD}\left( D(x)\right)\right]\\ T^{\prime}(x)=\frac{b}{\lambda}\mathcal{G}_{D}\left( D(x)\right)\frac{d}{dx}\left( x^{2}f(x)\right). \end{array} $$

(52)

Noting the convexity and the increasing properties of $\mathcal {G}(.)$, the positivity of $T^{\prime }(x)$ depends on:

$$ \begin{array}{@{}rcl@{}} \frac{d}{dx}\left( x^{2} f(x)\right)>0. \end{array} $$

(53)

In the sequel, the existence of at most one single interval of positive power allocation in any region like $x\in \mathcal {R}=[l,u]$ which satisfies the $\frac {d}{dx}\left (x^{2} f(x)\right )>0$ is discussed. To this end, let us assume the contradiction. For instance, assume there are two disjoint intervals of [x₁,x₂] and [x₃,x₄] (x₁ < x₂ < x₃ < x₄) in the region $\mathcal {R}$ where $T^{\prime }(x)=0$ for x ∈ (x₂,x₃), which leads to T(x₂) = T(x₃). Using Eq. 48, the latter equality leads to:

$$ \begin{array}{@{}rcl@{}} {{x}_{2}^{2}}f(x_{2})\mathcal{G}_{D}\left( D({x}_{2})\right) = {{x}_{3}^{2}}f(x_{3})\mathcal{G}_{D}\left( D(x_{3})\right), \end{array} $$

(54)

which contradicts the positivity of $\frac {d}{dx}(x^{2}f(x))$ in $\mathcal {R}$. Therefore, there should be at most one continues positive power allocation interval in the region satisfying Eq. 53.

1.2 1.2 Proof of the increasing property for $\mathcal {G}(.)$ function

Here, we demonstrate that $\mathcal {G}(.)$ is an increasing function. Inserting the derived auxiliary function of Eq. 16 into Eq. 8 we have:

$$ \begin{array}{@{}rcl@{}} \mathcal{G}\left( D_{r}(\alpha)\right)=\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)^{\frac{b}{b+1}}{\int}_{\beta_{1}}^{\beta_{2}}\frac{f_{r}(\beta)^{\frac{1}{b+1}}}{\beta^{\frac{2b}{b+1}}}d\beta \\+F_{r}(\beta_{1}) + \left( 1-F_{r}(\beta_{2})\right)D_{r}(\alpha). \end{array} $$

(55)

Re-writing the power constraint of Eq. 9 using Eq. 16, the following expression for the integral part of Eq. 55 is derived.

$$ \begin{array}{@{}rcl@{}} &&{\int}_{\beta_{1}}^{\beta_{2}}\left( \frac{\beta^{2}f_{r}(\beta)}{{{\beta}_{1}^{2}}f_{r}(\beta_{1})}\right)^{\frac{1}{b+1}}\frac{1}{\beta^{2}}d\beta+\frac{T_{r}(\beta_{2})}{\beta_{2}}-\frac{1}{\beta_{1}}=P_{r} \rightarrow \\ &&{\int}_{\beta_{1}}^{\beta_{2}}\frac{f_{r}(\beta)^{\frac{1}{b+1}}}{\beta^{\frac{2b}{b+1}}}d\beta = \left( P_{r}+\frac{1}{\beta_{1}}-\frac{D_{r}(\alpha)^{\frac{-1}{b}}}{\beta_{2}}\right)\\&&\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)^{\frac{1}{b+1}}. \end{array} $$

(56)

Noting that β₁ and β₂ in Eq. 55 are dependent to D_r(α), substituting Eq. 56 in Eq. 55, and taking partial derivative from Eq. 55, yields:

$$ \begin{array}{@{}rcl@{}} \frac{\partial\mathcal{G}\left( D_{r}(\alpha)\right)}{\partial D_{r}(\alpha)}\!&=&\!\left[\frac{D_{r}(\alpha)^{\frac{-1}{b}}}{{{\beta}_{2}^{2}}} \left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)-f_{r}(\beta_{2})D_{r}(\alpha)\right]\\&&\!\times\frac{\partial \beta_{2}}{\partial D_{r}(\alpha)} + \left[f_{r}(\beta_{1}) - \frac{{{\beta}_{1}^{2}}f_{r}(\beta_{1})}{{{\beta}_{1}^{2}}}\right]\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)} \\&&\!+\frac{D_{r}(\alpha)^{-\frac{b+1}{b}}}{b\beta_{2}}\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)+\frac{\partial \left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)}{\partial \beta_{1}}\\&&\!\times\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)}\left( P_{r}+\frac{1}{\beta_{1}}-\frac{D_{r}(\alpha)^{\frac{-1}{b}}}{\beta_{2}}\right)\\ &&\!+\left( 1-F_{r}(\beta_{2})\right). \end{array} $$

(57)

According to Eq. 17, the first term in Eq. 57 is 0. Simplifying the second component, it is equal to 0, too. Thus, Eq. 57 changes to:

$$ \begin{array}{@{}rcl@{}} \frac{\partial\mathcal{G}\left( D_{r}(\alpha)\right)}{\partial D_{r}(\alpha)}&=&\frac{D_{r}(\alpha)^{-\frac{b+1}{b}}}{b\beta_{2}}\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right) \\ &&+\frac{\partial \left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)}{\partial \beta_{1}}\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)}\\&&\times\left( P_{r}+\frac{1}{\beta_{1}}-\frac{D_{r}(\alpha)^{\frac{-1}{b}}}{\beta_{2}}\right)\\ &&+\left( 1-F_{r}(\beta_{2})\right) \end{array} $$

(58)

In order to characterize $\frac {\partial \beta _{1}}{\partial D_{r}(\alpha )}$, one can re-write the integral of Eq. 56 as ${\int \limits }_{\beta _{1}}^{\beta _{2}}\frac {1}{\beta ^{2}} \left (\beta ^{2}f_{r}(\beta )\right )^{\frac {1}{b+1}}d\beta$. Thus, taking derivative from Eq. 56 with respect to D_r(α) yields:

$$ \begin{array}{@{}rcl@{}} &&\frac{\left( {{\beta}_{2}^{2}}f_{r}(\beta_{2})\right)}{{\beta}_{2}^{2}}^{\frac{1}{b+1}}\!\!\frac{\partial \beta_{2}}{\partial D_{r}(\alpha)}-\frac{\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)}{{\beta}_{1}^{2}}^{\frac{1}{b+1}}\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)} \\ &&=\frac{1}{b+1}\frac{\partial\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)}{\partial \beta_{1}}\left( \!P_{r} + \frac{1}{\beta_{1}} - \frac{D_{r}(\alpha)^{\frac{-1}{b}}}{\beta_{2}}\!\right)\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)}\\ &&+\left( {\beta_{1}^{2}}f_{r}(\beta_{1})\right)\\ &&\times\left( \!\frac{-1}{b}\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)} + \frac{D_{r}(\alpha)^{\frac{-(b+1)}{b}}}{b\beta_{2}} + \frac{D_{r}(\alpha)^{\frac{-1}{b}}}{{\beta_{2}^{2}}}\frac{\partial \beta_{2}}{\partial D_{r}(\alpha)}\!\right) \end{array} $$

(59)

Re-ordering Eq. 59 leads to:

$$ \begin{array}{@{}rcl@{}} &&\left[\frac{({\beta}_{2}^{2}f_{r}(\beta_{2}))}{{\beta}_{2}^{2}}^{\frac{1}{b+1}}-\frac{D_{r}(\alpha)\frac{-1}{b}({\beta}_{1}^{2}f_{r}(\beta_{1}))}{{\beta}_{2}^{2}}^{\frac{1}{b+1}}\right] \frac{\partial\beta_{2}}{\partial{D_{r}}(\alpha)}\\ &&= \left[\frac{\left( {\beta}_1^2f_r(\beta_1)\right)}{{\beta}_1^2}^{\frac{1}{b+1}}\! - \frac{\left( {\beta}_1^2f_r(\beta_1)\right)}{{\beta}_1^2}^{\frac{1}{b+1}}\right. \\&&+ \frac{1}{b+1}\frac{\partial \left( {\beta}_1^2f_r(\beta_1)\right)}{\partial \beta_1}\left( \beta_1^2f_r(\beta_1)\right)^{\frac{-b}{b+1}}\\ && \times\left. \left( P_r+\frac{1}{\beta_1}-\frac{D_r(\alpha)^{\frac{-1}{b}}}{\beta_2}\right)\right]\times\frac{\partial \beta_1}{\partial D_r(\alpha)} \\&&+\frac{D_r(\alpha)^{\frac{-(b+1)}{b}}\left( \beta_1^2f_r(\beta_1)\right)^{\frac{1}{b+1}}}{b\beta_2} \end{array} $$

(60)

and simplifying Eq. 60 by the use of Eq. 17 gives:

$$ \begin{array}{@{}rcl@{}} &&\!\!\!\!\frac{\partial \beta_{1}}{\partial D_{r}(\alpha)}\\&& \!\! = -\frac{(b+1)D_{r}(\alpha)^{\frac{-(b+1)}{b}}\left( {\beta_{1}^{2}}f_{r}(\beta_{1})\right)^{\frac{1}{b+1}}}{b\beta_{2}\frac{\partial \left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)}{\partial \beta_{1}}\left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right)^{\frac{-b}{b+1}}\left( P_{r}+\frac{1}{\beta_{1}}-\frac{D_{r}(\alpha)^{\frac{-1}{b}}}{\beta_{2}}\right)}.\\ \end{array} $$

(61)

Applying Eq. 61 into Eq. 58 and simplifying the relation gives:

$$ \begin{array}{@{}rcl@{}} \frac{\partial \mathcal{G}\left( D_{r}(\alpha)\right)}{\partial D_{r}(\alpha)} &=&-\frac{D_{r}(\alpha)^{-(\frac{b+1}{b})}}{\beta_{2}} \left( {{\beta}_{1}^{2}}f_{r}(\beta_{1})\right) \\&&+ \left( 1-F_{r}(\beta_{2})\right), \end{array} $$

(62)

and finally using Eqs. 17, 62 simplifies to:

$$ \begin{array}{@{}rcl@{}} \frac{\partial \mathcal{G}\left( D_{r}(\alpha)\right)}{\partial D_{r}(\alpha)} =1 - \beta_{2}f_{r}(\beta_{2}) -F_{r}(\beta_{2}). \end{array} $$

(63)

To give an example, in the Rayleigh fading case, the necessary condition to have $\frac {\partial \mathcal {G}\left (D_{r}(\alpha )\right )}{\partial D_{r}(\alpha )}\geq 0$ is β₂ ≤ 1.