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Matrix differentiation for capacity region of Gaussian multiple access channels under weighted total power constraint

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Abstract

Maximization the capacity region of Gaussian multiple access channels with vector inputs and vector outputs has been extensively studied in existing schemes. Although these schemes are proven effective in various real-life applications, they are inapplicable to deal with channels with matrix variables subjected to certain constraints. In this work, we present a new framework to estimate the capacity region of Gaussian multiple access channels with matrix inputs and outputs under weighted total power constraints. We propose an optimization model to address this issue and prove its concavity. By introducing an I-chain rule for matrix differentiation, the gradient of the objective function involving matrix variables can be obtained. An algorithm, named normalized projected gradient method (NPGM) is developed to find the global optimal solution for the proposed model. The convergence of NPGM is established by utilizing projection and normalization operators. Simulation results provide an interesting insight that NPGM can manage the existent situations within an unified framework, and provide a novel universal technical solution to optimize the capacity region under weighted total power constraints.

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Author information

Authors and Affiliations

  1. State Key Laboratory for Strength and Vibration of Mechanical Structures, Shaanxi Key Laboratory of Environment and Control for Flight Vehicle, School of Aerospace, Xi’an Jiaotong University, Xi’an, 710049, People’s Republic of China

    Zhao-Xu Yang & Hai-Jun Rong

  2. School of Electronic and Information Engineering, Xi’an Jiaotong University, Xi’an, 710049, People’s Republic of China

    Guang-She Zhao

  3. Center for Brain Inspired Computing Research, Department of Precision Instrument, Tsinghua University, Beijing, 100084, People’s Republic of China

    Guoqi Li

Authors
  1. Zhao-Xu Yang
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  2. Guang-She Zhao
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  3. Guoqi Li
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  4. Hai-Jun Rong
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Corresponding author

Correspondence to Guoqi Li.

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Funding Information

This work is funded in part by the National Basic Research Program of China (Grant No. 2015CB057406), Beijing Natural Science Foundation (4164086), National Natural Science Foundation of China (61603209), and Independent Research Plan of Tsinghua University (20151080467).

Appendix A: Proof of Lemma 7

Appendix A: Proof of Lemma 7

Proof

Denote X j, l k as the lk-th element of matrix X j ,we use I-chain rule to calculate the derivative of \(\frac {\partial F(X)}{\partial X_{j,lk}}\).According to Lemma 5, we have

$$\begin{array}{@{}rcl@{}} \frac{\partial F(X)}{\partial X_{j,lk}} &=&\frac{\partial\left( \frac{1}{2}\log\left|\mathcal{G}\right|-\frac{1}{2}\log|Z|\right)}{\partial X_{j,lk}} =\frac{1}{2}\frac{\partial\left( \log\left|\mathcal{G}\right|\right)}{\partial X_{j,lk}}\\ &=&\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1}\sum\limits^{\mathrm{m}}_{n=1}\frac{\partial\left( \log\left|\mathcal{G}\right|\right)} {\partial\left( \mathcal{G}\right)_{mn} }\frac{\partial\left( \mathcal{G}\right)_{mn}}{\partial X_{j,lk}}. \end{array} $$
(52)

Then, we need to drive \(\frac {\partial \left (\log \left |\mathcal {G}\right |\right )} {\partial \left (\mathcal {G}\right )_{mn} }\)and \(\frac {\partial \left (\mathcal {G}\right )_{mn}}{\partial X_{j,lk}}\).

Using Lemma 5, it is obtained that

$$ \frac{\partial\left( \log\left|\mathcal{G}\right|\right)} {\partial\left( \mathcal{G}\right)_{mn} } =\sum\limits^{\mathrm{m}}_{f=1}\sum\limits^{\mathrm{m}}_{z=1} \frac{\partial\left( \log\left|\mathcal{G}\right|\right)}{\partial\left( \mathcal{G}\right)_{zf}} \frac{\partial\left( \mathcal{G}\right)_{zf}}{\partial\left( \mathcal{G}\right)_{mn}}. $$
(53)

For matrix A, a special case of differentiation (see [16]) is

$$ \frac{\partial \log\left|A\right|}{\partial A}=(A^{-1})^{T}. $$
(54)

Refer to Eq. 54, the left fraction term of Eq. 53becomes

$$ \frac{\partial\left( \log\left|\mathcal{G}\right|\right)}{\partial\left( \mathcal{G}\right)_{zf}} =\left( \left( \mathcal{G}\right)^{-1}\right)^{T}_{zf} =\left( \left( \mathcal{G}\right)^{-1}\right)_{fz}. $$
(55)

Using Lemma 6, the right fraction term of Eq. 53becomes

$$ \frac{\partial\left( \mathcal{G}\right)_{zf}}{\partial\left( \mathcal{G}\right)_{mn}} =\delta_{mz}\delta_{fn}. $$
(56)

By substituting Eqs. 55and 56into Eq. 53, we have

$$\begin{array}{@{}rcl@{}} \frac{\partial\left( \log\left|\mathcal{G}\right|\right)} {\partial\left( \mathcal{G}\right)_{mn} } &=&\sum\limits^{\mathrm{m}}_{f=1}\sum\limits^{\mathrm{m}}_{z=1} \left( \left( \mathcal{G}\right)^{-1}\right)_{fz} \delta_{mz}\delta_{fn} \\ &=&\sum\limits^{\mathrm{m}}_{f=1} \left( \left( \mathcal{G}\right)^{-1}\right)_{fm} \delta_{fn} = \left( \left( \mathcal{G}\right)^{-1}\right)_{nm}. \end{array} $$
(57)

Likewise, using Lemma 5, also we can obtain that

$$\begin{array}{@{}rcl@{}} &&\frac{\partial\left( \mathcal{G}\right)_{mn}}{\partial X_{j,lk}}\\ &=&\sum\limits^{\mathrm{m}}_{v=1}\sum\limits^{\mathrm{m}}_{c=1}\frac{\partial\left( \sum\limits^{K}_{i=1}H_{i}{X_{i}^{T}}X_{i}{H_{i}^{T}}+Z\right)_{mn}}{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{cv}} \\ &&\qquad\quad\frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{cv}}{\partial X_{j,lk}}\\ &=&\sum\limits^{\mathrm{m}}_{v=1}\sum\limits^{\mathrm{m}}_{c=1}\frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{mn}} {\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{cv}}\\ &&\qquad\quad\frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{cv}}{\partial X_{j,lk}}. \end{array} $$
(58)

Using Lemma 6, the left fraction term of Eq. 58becomes

$$ \frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{mn}} {\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}+Z\right)_{cv}}=\delta_{cm}\delta_{nv}. $$
(59)

By substituting Eq. 59into Eq. 58, we have

$$\begin{array}{@{}rcl@{}} &&\frac{\partial\left( \sum\limits^{K}_{i=1}H_{i}{X_{i}^{T}}X_{i}{H_{i}^{T}}+Z\right)_{mn}}{\partial X_{j,lk}}\\ &=&\sum\limits^{\mathrm{m}}_{v=1}\sum\limits^{\mathrm{m}}_{c=1} \left( \delta_{cm}\delta_{nv}\left( \frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}\right)_{cv}}{\partial X_{j,lk}}\right)\right). \end{array} $$
(60)

Using Lemma 5, the formula in parentheses of Eq. 60 becomes,

$$\begin{array}{@{}rcl@{}} \!\!&&\frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}\right)_{cv}}{\partial X_{j,lk}}\\ \!\!\!&=&\frac{\partial\sum\limits^{\mathrm{m}}_{a=1}(H_{j}{X_{j}^{T}})_{ca}(X_{j}{H_{j}^{T}})_{av}} {\partial X_{j,lk}}\\ \!\!\!&=&\sum\limits^{\mathrm{m}}_{a=1}\frac{\partial(H_{j}{X_{j}^{T}})_{ca}(X_{j}{H_{j}^{T}})_{av}} {\partial X_{j,lk}} \end{array} $$
(61)
$$\begin{array}{@{}rcl@{}} \!\!&=&\sum\limits^{\mathrm{m}}_{a=1}\frac{(H_{j}{X_{j}^{T}})_{ca}\partial(X_{j}{H_{j}^{T}})_{av}+(X_{j}{H_{j}^{T}})_{av}\partial(H_{j}{X_{j}^{T}})_{ca}} {\partial X_{j,lk}}\\ \!\!&=&\!\sum\limits^{\mathrm{m}}_{a=1}\!\left( \!(H_{j}{X_{j}^{T}})_{ca}\frac{\partial(X_{j}{H_{j}^{T}})_{av}} {\partial X_{j,lk}} \,+\,(X_{j}{H_{j}^{T}})_{av}\frac{\partial(H_{j}{\!X_{j}^{T}})_{ca}} {\partial X_{j,lk}} \right). \end{array} $$

Using Lemmas 5 and 6, the left fraction term in parentheses of Eq. 61 becomes,

$$\begin{array}{@{}rcl@{}} &&\frac{\partial(X_{j}{H_{j}^{T}})_{av}} {\partial X_{j,lk}}\\ &=&\frac{\partial\sum\limits^{\mathrm{n}_{j}}_{p=1}X_{j,ap}H_{j,pv}^{T}} {\partial X_{j,lk}}\\ &=&\sum\limits^{\mathrm{n}_{j}}_{p=1}\left( \frac{\partial X_{j,ap}H_{j,vp}} {\partial X_{j,lk}}\right)\\ &=&\sum\limits^{\mathrm{n}_{j}}_{p=1}\left( H_{j,vp}\frac{\partial X_{j,ap}} {\partial X_{j,lk}}\right)\\ &=&\sum\limits^{\mathrm{n}_{j}}_{p=1}\left( H_{j,vp}\delta_{la}\delta_{pk}\right)\\ &=&H_{j,vk}\delta_{la}. \end{array} $$
(62)

Similarly, the right fraction term in parentheses of Eq. 61becomes,

$$ \frac{\partial(H_{j}{X_{j}^{T}})_{ca}} {\partial X_{j,lk}} =H_{j,ck}\delta_{la}. $$
(63)

By substituting Eqs. 62 and 63 into Eq. 61 and using Lemma 6, we can obtain

$$\begin{array}{@{}rcl@{}} &&\!\!\!\!\!\sum\limits^{\mathrm{m}}_{a=1}\left( (H_{j}{X_{j}^{T}})_{ca}\frac{\partial(X_{j}{H_{j}^{T}})_{av}} {\partial X_{j,lk}} +(X_{j}{H_{j}^{T}})_{av}\frac{\partial(H_{j}{X_{j}^{T}})_{ca}} {\partial X_{j,lk}} \right)\\ &=&\sum\limits^{\mathrm{m}}_{a=1}\left( (H_{j}{X_{j}^{T}})_{ca}H_{j,vk}\delta_{la} +(X_{j}{H_{j}^{T}})_{av}H_{j,ck}\delta_{la} \right)\\ &=&\left( (H_{j}{X_{j}^{T}})_{cl}H_{j,vk} +(X_{j}{H_{j}^{T}})_{lv}H_{j,ck} \right). \end{array} $$
(64)

By substituting Eq. 64 into Eq. 60 and using Lemma 6, we have

$$\begin{array}{@{}rcl@{}} &&\sum\limits^{\mathrm{m}}_{v=1}\sum\limits^{\mathrm{m}}_{c=1} \left( \delta_{cm}\delta_{nv}\left( \frac{\partial\left( H_{j}{X_{j}^{T}}X_{j}{H_{j}^{T}}\right)_{cv}}{\partial X_{j,lk}}\right)\right)\\ &=&\sum\limits^{\mathrm{m}}_{v=1}\sum\limits^{\mathrm{m}}_{c=1} \left( \delta_{cm}\delta_{nv}\left( (H_{j}{X_{j}^{T}})_{cl}H_{j,vk} +(X_{j}{H_{j}^{T}})_{lv}H_{j,ck} \right)\right)\\ &=&\sum\limits^{\mathrm{m}}_{v=1} \left( \delta_{nv}\left( (H_{j}{X_{j}^{T}})_{ml}H_{j,vk} +(X_{j}{H_{j}^{T}})_{lv}H_{j,mk} \right)\right)\\ &=&\left( (H_{j}{X_{j}^{T}})_{ml}H_{j,nk} +(X_{j}{H_{j}^{T}})_{ln}H_{j,mk} \right). \end{array} $$
(65)

By substituting Eqs. 57 and 65 into Eq. 52, we have

$$\begin{array}{@{}rcl@{}} &&\frac{\partial F(X)}{\partial X_{j,lk}}\\ &=&\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1}\sum\limits^{\mathrm{m}}_{n=1}\frac{\partial\left( \log\left|\mathcal{G}\right|\right)} {\partial\left( \mathcal{G}\right)_{mn} }\frac{\partial\left( \mathcal{G}\right)_{mn}}{\partial X_{j,lk}}\\ &=&\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1}\sum\limits^{\mathrm{m}}_{n=1} \left( \left( \mathcal{G}\right)^{-1}\right)_{nm}\\ &&\times\left( (H_{j}{X_{j}^{T}})_{ml}H_{j,nk}+(X_{j}{H_{j}^{T}})_{ln}H_{j,mk}\right)\\ &=&\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1}\sum\limits^{\mathrm{m}}_{n=1} \left( \left( \mathcal{G}\right)^{-1}\right)_{nm} \left( (H_{j}{X_{j}^{T}})_{ml}H_{j,nk}\right)\\ &&+\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1}\sum\limits^{\mathrm{m}}_{n=1} \left( \left( \mathcal{G}\right)^{-1}\right)_{nm} \left( (X_{j}{H_{j}^{T}})_{ln}H_{j,mk}\right). \end{array} $$
(66)

Using Lemmas 5 and 6, and switching the places of entities such that the derivation can be written as [⋅] l k , wehave

$$\begin{array}{@{}rcl@{}} \frac{\partial F(X)}{\partial X_{j,lk}} &=&\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1} \left( {H_{j}^{T}}\left( \mathcal{G}\right)^{-1}\right)_{km} (H_{j}{X_{j}^{T}})_{ml}\\ &&+\frac{1}{2}\sum\limits^{\mathrm{m}}_{m=1} \left( (X_{j}{H_{j}^{T}})\left( \mathcal{G}\right)^{-1}\right)_{lm} H_{j,mk}\\ &=&\frac{1}{2} \left( {H_{j}^{T}}\left( \mathcal{G}\right)^{-1}H_{j}{X_{j}^{T}}\right)_{kl}\\ &&+\frac{1}{2} \left( (X_{j}{H_{j}^{T}})\left( \mathcal{G}\right)^{-1}H_{j}\right)_{lk}\\ &=&\frac{1}{2} \left( X_{j}{H^{T}_{j}}\left( \left( \mathcal{G}\right)^{-1}\right)^{T}H_{j}\right)_{lk}\\ &&+\frac{1}{2} \left( X_{j}{H^{T}_{j}}\left( \mathcal{G}\right)^{-1}H_{j}\right)_{lk}. \end{array} $$
(67)

Thus Lemma 7 is proved. □

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Yang, ZX., Zhao, GS., Li, G. et al. Matrix differentiation for capacity region of Gaussian multiple access channels under weighted total power constraint. Ann. Telecommun. 72, 703–715 (2017). https://doi.org/10.1007/s12243-017-0610-7

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