The Worst-Case DFT Filter Bank Design with Subchannel Variations

Abstract

In this paper, we consider an optimal design of a DFT filter bank subject to subchannel variation constraints. The design problem is formulated as a minimax optimization problem. By exploiting the properties of this minimax optimization problem, we show that it is equivalent to a semi-infinite optimization problem in which the continuous inequality constraints are only with respect to frequency. Then, a computational scheme is developed to solve such a semi-infinite optimization problem. Simulation results show that, for a fixed distortion level, the aliasing level between different subbands is significantly reduced, in some cases up to 28 dB, when compared with that obtained by the bi-iterative optimization method without consideration of the subchannel variations.

Appendices

Appendix 1

$$\displaystyle{ \gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta }) = \frac{1} {D}\sum \limits _{m=0}^{M-1}\sum \limits _{ n=0}^{M-1}\left (1 +\delta _{ m}\right )\left (1 +\delta _{n}\right )\sum \limits _{d=1}^{D-1}\sum \limits _{ l=1}^{D-1}\boldsymbol{h}^{T}\varPhi _{ m,n,d,l}\left (\boldsymbol{g}\right )\boldsymbol{h,} }$$

(10.41)

where $\varPhi _{m,n,d,l}\left (\boldsymbol{g}\right )$ is a $L_{a} \times L_{a}$ matrix. The $\left (i,j\right )$-th element of $\varPhi _{m,n,d,l}\left (\boldsymbol{g}\right )$ is given by

$$\displaystyle\begin{array}{rcl} & & \left [\varPhi _{m,n,d,l}\left (\boldsymbol{g}\right )\right ]_{i,j} =\sum \limits _{ t=0}^{L_{s}-1}\sum \limits _{ s=0}^{L_{s}-1}\cos \left [\left (m - n\right )\left (i + t - 2\right ) \frac{2\pi } {M}\right. {}\\ & & \left.+\left (d\left (i - 1\right ) - l\left (j - 1\right )\right ) \frac{2\pi } {D}\right ]\delta \left (i + t - j - s\right )g\left (t\right )g\left (s\right ), {}\\ \end{array}$$

where $\delta \left (\cdot \right )$ is the delta function, i.e.,

$$\displaystyle{ \delta \left (t\right ) = \left \{\begin{array}{c} 1,\text{ if}\ t = 0,\\ 0, \text{ if} \ t\neq 0. \end{array} \right. }$$

Appendix 2

Proof.

Clearly, $\mathcal{U}$ is a convex set and $\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}$ are extreme points of $\mathcal{U}.$ It remains to show that $\mathcal{U} = co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ).$ For any $\boldsymbol{\delta =} \left [\delta _{0},\delta _{1},\cdots \,,\delta _{M-1}\right ]^{T} \in co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ),$ there exists $\lambda _{i} \geq 0,\ i = 1,\cdots \,,2^{M},$ such that $\boldsymbol{\delta }=\sum \limits _{ i=1}^{2^{M} }\lambda _{i}\boldsymbol{\bar{\delta }}_{i}$ and $\sum \limits _{i=1}^{2^{M} }\lambda _{i} = 1.$ Then, $\delta _{k} =\sum \limits _{ i=1}^{2^{M} }\lambda _{i}\bar{\delta }_{i,k},$ where $\bar{\delta }_{i,k}$ denotes the $k$ th element of $\boldsymbol{\bar{\delta }}_{i}.$ From (10.18), we have

$$\displaystyle{ \left \vert \delta _{k}\right \vert = \left \vert \sum \limits _{i=1}^{2^{M} }\lambda _{i}\bar{\delta }_{i,k}\right \vert \leq \sum \limits _{i=1}^{2^{M} }\lambda _{i}\left \vert \bar{\delta }_{i,k}\right \vert =\sum \limits _{ i=1}^{2^{M} }\lambda _{i}\varepsilon _{k} =\varepsilon _{k}. }$$

Thus, $\boldsymbol{\delta }\in \mathcal{U},$ and hence, $co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ) \subset \mathcal{U}$. On the other hand, let $\boldsymbol{\delta =} \left [\delta _{0},\delta _{1},\cdots \,,\delta _{M-1}\right ]^{T}$ with $\left \vert \delta _{i}\right \vert \leq \varepsilon _{i},\ i = 0,1,\cdots \,,M - 1.$ Since $\left \vert \delta _{0}\right \vert \leq \varepsilon _{0},$ there exists a $\lambda _{0},\ 0 \leq \lambda _{0} \leq 1,$ such that $\delta _{0} =\lambda _{0}\varepsilon _{0} -\left (1 -\lambda _{0}\right )\varepsilon _{0}.$ Since $co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right )$ is convex and

$$\displaystyle\begin{array}{rcl} & \left [\delta _{0},\varepsilon _{1},\cdots \,,\varepsilon _{M-1}\right ]^{T} =\lambda _{0}\left [\varepsilon _{0},\varepsilon _{1},\cdots \,,\varepsilon _{M-1}\right ]^{T}& {}\\ & +\left (1 -\lambda _{0}\right )\left [-\varepsilon _{0},\varepsilon _{1},\cdots \,,\varepsilon _{M-1}\right ]^{T}, & {}\\ \end{array}$$

we have

$$\displaystyle{ \left [\delta _{0},\varepsilon _{1},\cdots \,,\varepsilon _{M-1}\right ]^{T} \in co\left (\boldsymbol{\bar{\delta }}_{ 1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ). }$$

Since $\left \vert \delta _{1}\right \vert \leq \varepsilon _{1},$ there exists a $\lambda _{1}$ such that $\delta _{1} =\lambda _{1}\varepsilon _{1} -\left (1 -\lambda _{1}\right )\varepsilon _{1}$ and 0 ≤ λ ₁ ≤ 1. Thus,

$$\displaystyle\begin{array}{rcl} & \left [\delta _{0},\delta _{1},\varepsilon _{2},\cdots \,,\varepsilon _{M-1}\right ]^{T} =\lambda _{1}\left [\delta _{0},\varepsilon _{1},\varepsilon _{2},\cdots \,,\varepsilon _{M-1}\right ]^{T} & {}\\ & +\left (1 -\lambda _{1}\right )\left [\delta _{0},-\varepsilon _{1},\varepsilon _{2},\cdots \,,\varepsilon _{M-1}\right ]^{T} \in co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ).& {}\\ \end{array}$$

Continuing this process, we can show that $\boldsymbol{\delta \!=\!} \left [\delta _{0},\delta _{1},\cdots \,,\delta _{M-1}\right ]^{T}\!\! \in \! co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ).$ Hence, $\mathcal{U}\subset co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ).$ Therefore, $\mathcal{U} = co\left (\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}\right ).$

Proof.

From (10.8), we see that $\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta })$ is in quadratic form with respect to $\boldsymbol{\delta }$. Furthermore, $\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta }) \geq 0$ for any $\boldsymbol{\delta.}$ Thus, we can write $\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta })$ in the form below.

$$\displaystyle{ \gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta }) =\boldsymbol{\delta } ^{T}Q\boldsymbol{\delta + q}^{T}\boldsymbol{\delta +} q, }$$

where $Q = \left (Q^{1/2}\right )^{T}Q^{1/2}$ is a semi-positive definite matrix, $\boldsymbol{q}$ and q are corresponding vector and constant. For any $\lambda,\ 0 \leq \lambda \leq 1$, $\boldsymbol{\delta }_{1}$ and $\boldsymbol{\delta }_{2},$ we have

$$\displaystyle\begin{array}{rcl} & & \lambda \gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta }_{1}) + \left (1-\lambda \right )\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta }_{2}) {}\\ & & -\gamma (\boldsymbol{h},\boldsymbol{g},\lambda \boldsymbol{\delta }_{1} + \left (1-\lambda \right )\boldsymbol{\delta }_{2}) {}\\ & =& \lambda \boldsymbol{\delta }_{1}^{T}Q\boldsymbol{\delta }_{ 1} + \left (1-\lambda \right )\boldsymbol{\delta }_{2}^{T}Q\boldsymbol{\delta }_{ 2} -\lambda ^{2}\boldsymbol{\delta }_{ 1}^{T}Q\boldsymbol{\delta }_{ 1} {}\\ & & -\left (1-\lambda \right )^{2}\boldsymbol{\delta }_{ 2}^{T}Q\boldsymbol{\delta }_{ 2} - 2\lambda \left (1-\lambda \right )\boldsymbol{\delta }_{1}^{T}Q\boldsymbol{\delta }_{ 2} {}\\ & =& \lambda \left (1-\lambda \right )\left (\boldsymbol{\delta }_{1}^{T}Q\boldsymbol{\delta }_{ 1} +\boldsymbol{\delta }_{ 2}^{T}Q\boldsymbol{\delta }_{ 2} - 2\boldsymbol{\delta }_{1}^{T}Q\boldsymbol{\delta }_{ 2}\right ) {}\\ & =& \lambda \left (1-\lambda \right )\left [\left (Q^{1/2}\boldsymbol{\delta }_{ 1}\right )^{T}Q^{1/2}\boldsymbol{\delta }_{ 1}\right. {}\\ & & \left.+\left (Q^{1/2}\boldsymbol{\delta }_{ 2}\right )^{T}Q^{1/2}\boldsymbol{\delta }_{ 2} - 2\left (Q^{1/2}\boldsymbol{\delta }_{ 1}\right )^{T}Q^{1/2}\boldsymbol{\delta }_{ 2}\right ] {}\\ & \geq & 0. {}\\ \end{array}$$

Thus, $\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta })$ is a convex function with respect to $\boldsymbol{\delta }$. Suppose that $\boldsymbol{\delta }^{{\ast}} =\arg \max \limits _{\boldsymbol{\delta }\in \mathcal{U}}\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta })$ and $\bar{\gamma }=\max \left \{\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\bar{\delta }}_{1}),\cdots \,,\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\bar{\delta }}_{2^{M}})\right \}.$ From Theorem 10.1, we know that there exists $\lambda _{i} \geq 0,\ 0 \leq i \leq 2^{M},$ with $\sum \limits _{i=1}^{2^{M} }\lambda _{i} = 1,$ such that $\boldsymbol{\delta }^{{\ast}} =\sum \limits _{ i=1}^{2^{M} }\lambda _{i}\boldsymbol{\bar{\delta }}_{i}.$ Since $\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta })$ is a convex function with respect to $\boldsymbol{\delta }$, we have

$$\displaystyle\begin{array}{rcl} \gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta }^{{\ast}})& =& \gamma (\boldsymbol{h},\boldsymbol{g},\sum \limits _{ i=1}^{2^{M} }\lambda _{i}\boldsymbol{\bar{\delta }}_{i}) {}\\ & \leq & \sum \limits _{i=1}^{2^{M} }\lambda _{i}\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\bar{\delta }}_{i}) \leq \sum \limits _{i=1}^{2^{M} }\lambda _{i}\bar{\gamma } =\bar{\gamma }. {}\\ \end{array}$$

Thus, the maximum of $\gamma (\boldsymbol{h},\boldsymbol{g},\boldsymbol{\delta })$ in $\mathcal{U}$ is attained at one of $\boldsymbol{\bar{\delta }}_{1},\cdots \,,\boldsymbol{\bar{\delta }}_{2^{M}}.$

Proof of Theorem 10.3.

Since

$$\displaystyle\begin{array}{rcl} & & \boldsymbol{h}^{T}Re\left \{\varPsi (e^{j\omega },\boldsymbol{\delta })\right \}\boldsymbol{g} {}\\ & \boldsymbol{=}& \frac{1} {D}\sum _{m=0}^{M-1}(1 +\delta _{ m})\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ) {}\\ & =& \frac{1} {D}\sum _{m=0}^{M-1}\psi _{ r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ) + \frac{1} {D}\sum _{m=0}^{M-1}\delta _{ m}\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ). {}\\ \end{array}$$

Hence,

$$\displaystyle\begin{array}{rcl} & & \max _{\boldsymbol{\delta }\in \mathcal{U}}\boldsymbol{h}^{T}Re\left \{\varPsi (e^{j\omega },\boldsymbol{\delta })\right \}\boldsymbol{g} {}\\ & \boldsymbol{=}& \frac{1} {D}\sum _{m=0}^{M-1}\psi _{ r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ) +\max _{\boldsymbol{\delta }\in \mathcal{U}}\frac{1} {D}\sum _{m=0}^{M-1}\delta _{ m}\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ). {}\\ \end{array}$$

For any $\boldsymbol{\delta =} \left [\delta _{0},\delta _{1},\cdots \,,\delta _{M-1}\right ]^{T} \in \mathcal{U}$, we have

$$\displaystyle\begin{array}{rcl} & & \left \vert \frac{1} {D}\sum _{m=0}^{M-1}\delta _{ m}\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert {}\\ & \leq & \frac{1} {D}\sum _{m=0}^{M-1}\left \vert \delta _{ m}\right \vert \left \vert \psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert {}\\ & \leq & \frac{1} {D}\sum _{m=0}^{M-1}\varepsilon _{ m}\left \vert \psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert. {}\\ \end{array}$$

Hence,

$$\displaystyle{ \max _{\boldsymbol{\delta }\in \mathcal{U}}\frac{1} {D}\sum _{m=0}^{M-1}\delta _{ m}\psi _{r,m}\left (w,\boldsymbol{h},\boldsymbol{g}\right ) \leq \frac{1} {D}\sum _{m=0}^{M-1}\varepsilon _{ m}\left \vert \psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert. }$$

(10.42)

On the other hand, taking $\boldsymbol{\tilde{\delta }=} \left [\tilde{\delta }_{0},\tilde{\delta }_{1},\cdots \,,\tilde{\delta }_{M-1}\right ]^{T},$ where $\tilde{\delta }_{m} =\varepsilon _{m}\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )/\left \vert \psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert,$ yields

$$\displaystyle\begin{array}{rcl} & & \frac{1} {D}\sum _{m=0}^{M-1}\tilde{\delta }_{ m}\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ) {}\\ & =& \frac{1} {D}\sum _{m=0}^{M-1}\varepsilon _{ m}\left (\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right )^{2}/\left \vert \psi _{ r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert {}\\ & =& \frac{1} {D}\sum _{m=0}^{M-1}\varepsilon _{ m}\left \vert \psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert. {}\\ \end{array}$$

Combining (10.42) and (10.43), we obtain

$$\displaystyle{ \max _{\boldsymbol{\delta }\in \mathcal{U}}\frac{1} {D}\sum _{m=0}^{M-1}\delta _{ m}\psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right ) = \frac{1} {D}\sum _{m=0}^{M-1}\varepsilon _{ m}\left \vert \psi _{r,m}\left (\omega,\boldsymbol{h},\boldsymbol{g}\right )\right \vert. }$$

(10.43)

Thus, (10.20) is obtained. The validity of (10.21)–(10.23) can be established similarly. Thus, the proof is complete.