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1 7 his is the support document for the proofs of emmas and heorems in Paper Optimal Design Of inear Space Codes For Indoor MIMO Visible ight Communications With M Detection submitted to IEEE Photonics Journal A Main emmas In order to prove heorems and, we need to establish the following six lemmas for the investigation of some important properties on the objective function and its feasible domain emma : et θ and θ be defined by 8, where 0 < and 0 ψ hen, we have 0 θ θ Moreover, θ θ if and only if the condition number of the channel satisfies the following inequality: + + Proof : tanθ θ tanθ tanθ +tanθ tanθ sinψ > 0, θ [ /, 0], θ [0, /] and 0 ψ, we attain 0 θ θ Hence, requiring θ θ is equivalent to requiring tanθ θ sinψ, which, in turn, is also equivalent to the following inequality: ρ ρ 0, where ρ is the condition number of the channel H herefore, we have + + It is not difficult to examine + + his completes the proof of emma In α, β-plane, whether or not there is any intersection between ines α β and the square Ω as shown in Fig 4 plays a crucial role in obtaining a closed-from solution to Problem emma provides us with a necessary and sufficient condition to check when there exists an intersection he following emmas - 6 will lead us how to find the optimal solution under this condition emma : For ψ [0, ], let three angles θ, φ and τ be defined, respectively, by θ arctan tan 4 ψ 4 sin θ θ φ arctan 5 + cos θ θ sin θ θ τ arctan, 6 + cos θ θ where θ and θ are given in 8 hen, we have the following two statements: When 0 ψ < 4, we have θ θ and φ 6 if θ θ When 4 ψ, we have θ + θ and τ 6 if θ θ Proof : he whole proof captures the following two parts Proof of Statement : Since tanθ θ tan θ tan θ + tan β tan θ cot ψ tan 4 ψ + cot ψ tan 7 4 ψ C θ, θ H θ, θ + D θ, θ β G θ 0, θ B θ, θ F θ, θ α A θ E θ +, θ, θ Fig 4 Feasible domain Ω Ω Ω Ω in terms of α and β has intersections with ines β α and θ θ [, ], in this case, θ θ is equivalent to θ θ <, which, together with 7, is equivalent to saying ρ tan 4 ψ + + cot ψ tan 4 ψ ρ cot ψ 0, 8 where ρ Hence, we have + cot ψ tan ρ max, 4 ψ + D tan 4 ψ, 9 where the discriminant is D + cot ψ tan 4 ψ + 4 cot ψ tan 4 ψ > 0 Now, we need to prove that + cot ψ tan 4 ψ + D tan < 0 4 ψ o do that, we notice that + cot ψ tan 4 ψ + D tan 4 ψ cot ψ + cot ψ tan 4 ψ + D cot ψ + cot ψ tan a 4 ψ, since D + cot ψ tan 4 ψ In addition, we further note that cot ψ + cot ψ tan cot ψ + cot ψ cot ψ ψ + cot ψ cot < ψ + cot ψ + cot ψ + cot ψ cot ψ + < b

2 8 Now, combining 0 and, we claim that 0 is indeed true his, together with 9 gives us ρ, which implies θ θ In addition, on one hand, we note that θ θ [0, ] and cotx is monotonically decreasing in [0, ] On the other hand, we also notice that tanφ sin θ θ +cos θ θ t t + g t, where t cotθ θ, and that g t t t + Hence, g t is monotonically increasing when t >, which is equivalent to < θ θ <, and monotonically decreasing when t <, ie, < θ θ < However, at any rate, we always have tanφ g his, along with φ [, ] infers φ 6 So far, we have completed the proof of Statement Proof of Statement : When 4 ψ, θ 0 and as a result, θ θ, [, ] herefore, inequality θ + θ is equivalent to θ θ, which, in turn, is also equivalent to tanθ θ Since tanθ θ tan θ tan θ +tan θ tan θ only if ρ tan 4 ψ tan ψ tan ψ tan tan 4 ψ, θ θ 4 ψ if and tan ψ tan 4 ψ ρ + tan ψ 0, where ρ Now, let ψ ψ hen, 0 ψ 4 and becomes the exact same as 8 except for the fact that ψ is replaced by ψ Hence, following the same way, we can prove θ + θ Furthermore, we notice that θ θ [0, ] and cotx is monotonically decreasing in [0, ] On the other hand, we also notice that tanτ sin θ θ +cos θ θ t t + g t, where t cotθ θ, and that g t t t + Hence, g t is monotonically decreasing when t >, which is equivalent to < θ θ <, and monotonically increasing when t <, ie, < θ θ < However, in either case, we can always obtain tanφ g his, along with τ [, ] implies τ 6 hus far, we have completed the proof of Statement and thus, emma emma : et function F α, β be defined by a hen, the minimum of F α, β in Ω Ω as shown in Fig is achieved at one of the six vertices, ie, min α,β Ω Ω F α, β minf C, F G, F H, F E, F F, F A Proof : aking partial derivatives on both sides of a, we have f f α sinα θ 0 and β sinβ θ 0 Hence, α θ k and β θ l, where both of k and l are integers Since α θ [ /, /], β θ [ /, /], and α β θ θ, we have k l 0 However, at this point, f α, β achieves its maximum value Since f α, β is continuous in the compact domain Ω Ω, it must have the minimum value, which is achieved on the boundary of Ω Ω In addition, since f α, β is symmetrical in its feasible domain Ω Ω with respect to the line of α β, we only need consider one of triangular domains, say Ω as shown in Figure 4 et us now consider the following three lines ine HC On this line, the objective function f α, β is reduced to f θ, β cosθ θ + cosβ θ, where θ + β θ Since β θ [, ] and cosβ θ is monotonically increasing for β < θ, and is monotonically decreasing for β > θ, in this case, min f α, β minf H, f C ine CG On this line, the objective function f α, β becomes f θ, β cosα θ + cosθ θ, where θ α θ In the same token, Since cosα θ is monotonically increasing for α < θ, and is monotonically decreasing for α > θ, in this case, min f α, β minf C, f G ine HG On this line, the objective function f α, β is simplified into f α, β cosθ θ + cosβ θ cos β 6 θ, where θ + β θ Similarly, since cosβ 6 θ is monotonically increasing for β < θ + /6, and is monotonically decreasing for β > θ + /6, in this situation, min f α, β minf H, f G Summarizing the above discussions, we conclude that f α, β achieves its minimum at one of the three vertices of the triangular Ω, ie, min f α, β minf H, f C, f G his completes the proof of emma emma 4: et function F α, β be defined by b hen, the minimum of F α, β in Ω as shown in Fig is achieved at one of the six vertices, ie, min α,β Ω F α, β minf B, F D, F G, F H, F E, F F Proof : aking partial derivatives on both sides of b yield f α f sinα β θ 0 and β sinα β θ sinβ θ 0, which gives us sinα β θ sinβ θ 0 Hence, α β θ k and β θ l, where both of k and l are integers Since α θ [ /, /], β θ [ /, /], and α β θ θ, we have k l 0 and thus, α β θ At this point, f α, β achieves its maximum value Since f α, β is continuous in the compact domain Ω, it must have the minimum value, which is achieved on the boundary of Ω he boundary consists of six lines: GB, BF, F E, ED, DH and HG as shown in Figure 4 Notice that on ines F E and HG, f α, β f α, β, which has been discussed in emma In addition, on ine GB, f α, θ cosα θ θ + cosθ θ and on ine ED, f α, θ cosα θ θ+ cosθ θ, which can also be dealt with by following the way similar to the proof of emma Hence, in the following we only need to consider the other two lines, ie, BF and DH ine BF On this line, the objective function f α, β is reduced to f θ, β cosθ β θ + cosβ θ + cos θ θ cosβ θ + sin θ θ sinβ θ + cos θ θ + sin θ θ cosβ τ θ, where θ β θ and τ arctan sin θ θ +cos θ θ Now, following the same argument as in the proof of emma, we can conclude that in this situation, min f α, β minf B, f F ine DH On this line, the objective function f α, β becomes f θ, β cosθ β θ + cosβ θ + cos θ θ + sin θ θ cosβ φ θ, where θ β θ + and φ arctan sin θ θ +cos θ θ Following the discussion similar to ine BF, we can claim

3 9 that in this case, min f α, β minf D, f H Summarizing the above discussions, we conclude that f α, β achieves its minimum at one of the six vertices of Ω, ie, min f α, β minf G, f B, f F, f E, f D, f H his completes the proof of emma 4 emma 5: et a, b and c be given three positive real numbers with c If a function fx is defined on a closed interval [, c] by fx x cx+ ax+b, then, fx achieves its maximum at x c, its minimum at x when b a and at x x 0 when b a, where x 0 a+bc b+ac Proof : It is not difficult to compute the first order derivative with respect to x such that f x b+ac x a+bc b+ac ax+b Now, it can be seen that there is only one possible root x 0 for f x 0 We need to know when x 0 belongs to an open interval, c Hence, this leads us to considering the following two situations: b > a In this case, it can be observed that x 0 < and thus, f x > 0, ie, fx is monotonically increasing herefore, min fx f and max fx fc b a In this case, we first know that x 0 On the other hand, judging whether x 0 c is equivalent to examining whether the following inequality holds: a + bc c, b + ac which is ac + bc a 0 here are two roots for ax + bx a 0 in terms of x, ie, x b b +8a 4a and x b+ b +8a 4a Since x < 4 and c, c indeed meets herefore, fx achieves its minimum at x x 0 and its maximum at one of the two ending points: x and x c Notice that f c a+b and fc ac+b Now, we need to prove that f fc, which is equivalent to proving that the following inequality is true: c a + b ac + b 4 his leads us to considering a function fx xax+ b for / x Since f x ax+b4ax a b, ax + b > 0 and 4ax a b a a b b > 0, f x < 0 and hence, fx is monotonically decreasing Since f/ a + b, fc f a + b, implying that c indeed satisfies 4 his completes the proof of emma 5 emma 6: For 0 ψ, let functions F α, β and F α, β be defined by hen, we have F C F A F B F C F D cos ψ + sin ψ + sin ψ + cos ψ sin ψ sin ψ + cos ψ sin ψ cos ψ + sin ψ sin ψ + cos ψ sin ψ cos ψ + sin ψ sin ψ sin ψ + cos ψ cos ψ + sin ψ Moreover, the following four statements are true: F B F D if and only if 0 ψ 4 F C F D if and only if 0 ψ < arctan 4 5 sin ψ 5 sin ψ F C F B if and only if ψ > arctan and 5 sin ψ 5 sin ψ 4 4 nder the condition of > + +, and F A F C if and only if 0 ψ 4 Proof : For presentation clarity, we discuss how to prove each statement separately Proof of Statement : F B F D if and only if sin ψ + cos ψ cos ψ + sin ψ his is equivalent to cos ψ 0, which is equivalent to 0 ψ 4, since > > 0 and 0 ψ Proof of Statement : F C F D if and only if sin ψ + cos ψ cos ψ + sin ψ his, in turn, is equivalent to 5 sin ψ ρ 5 sin ψ 4, where ρ herefore, 5 sin 4 5 sin ψ < 0 and ρ ψ 5 sin ψ his completes the proof of Statement Proof of Statement : It can be proved by following the way similar to the proof of Statement Proof of Statement 4: First, we note that F A F C F B F D sin ψ 5 sin ψ + 4 On the other hand, we also notice that inequality sin ψ > is equivalent to sin ψ+4 sin ψ, which is sin ψρ ρ

4 0 sin ψ > 0, where ρ Hence, we have ρ > + + his is our assumption herefore, under this condition, equation 5 is equivalent to F B F D, which is Statement his completes the proof of Statement 4 and thus, emma 6 B Proofs of heorems Proof of heorem : Without loss of generality, we assume that d d and thus, we only need to consider the following three cases: d d d, max d d d d, max d d d d max d In addition, It is not difficult to observe that the optimal solution cannot be achieved when d 0 Hence, to facilitate our analysis, by utilizing the parameterization of the space coding matrix F proposed in Subsection III-B, Cases, and can be equivalently represented in terms of µ, α and βm respectively, as a µ cosα β, max d b µ max cosα β, cosα β, max d c µ cosα β, max d, where α, β Ω, ie, θ α, β θ Now, it is more clear to see that in order to solve these optimization problems, we need to consider when there is an intersection between the square Ω and ines: α β C θ, θ D θ, θ β, θ B θ 0, θ A θ α Fig 5 Feasible domain Ω in terms of α and β has no intersection with ines β α I No intersection By emma, we know that there is no intersection, as shown in Fig 5, between the square Ω and ines: α β if 0 θ θ <, which is equivalent to the fact that the condition number of the channel satisfies the following inequality: > + + nder this condition, the feasible domain of Case a is empty herefore, we only need to consider Cases b and c, each feasible set of which is the whole square Ω Case b: µ cosα β In this case, we need to maximize d sing 0, we have P max d max α,β max P µ P min α,β F α,β F α,β,µ min α,β min µ F α,β,µ, where we have utilized the fact that min µ F α, β, µ F α, β, cosα β, since µ cosα β and F α, β, µ is an increasing function of µ Now, following the discussion similar to the proof of emma 4, we can conclude that F α, β achieves its minimum at one of the four vertices of Ω, ie, min F α, β minf A, F B, F C, F D Case c: µ cosα β In this case, we need to maximize d Combining the definition of d and 0, we can represent d as d d µ µ cosα β + µ µ cosα β+ F α,β,µ 4P µ µ cosα β+,,ψ f α,β,µ 4P Hence, we have max d max α,β max µ d Now, applying emma 5 into function µ µ cosα β+ f α,β,µ yields max µ d 4P,,ψ f α,β, cosα,β P herefore, max d min F α,β 4P F α,β, cosα,β 4P F α,β, which is exactly the same case as Case b Now, summarizing the above discussions on Cases b and c, we attain that the optimum is reached P when d d and is given by min F, where α,β min F α, β minf A, F B, F C, F D Combining this with emma 6, we have min F α, β minf C, F D when 0 ψ 4 On the other hand, we know that in this case, f θ, β + cos θ θ + sin θ θ cosβ φ θ, where θ β θ From the definitions of θ and φ in emma, when 0 ψ 4, we can obtain tan φ tanθ θ + tan θ θ tanθ θ and as a consequence, θ θ + φ Similarly, we can prove θ θ + φ Hence, when 0 ψ 4 and θ + φ θ θ θ + φ, ie, θ + φ θ +θ, we have f θ, θ f θ, θ, ie, F D F C and thus, min F α, β F D Analogously, we can prove the case where 4 < ψ II Intersection By emma, there is an intersection, as shown in Fig 4, between the square Ω and ines: α β if θ θ, which is equivalent to the fact that the condition number of the channel satisfies + + nder this condition, different cases have different feasible sets So we need to deal with each individual optimization problem separately Case a: µ cosα β Note that this inequality implies that the feasible set with respect to design variables α and β must satisfy α β, as shown in Figure Our task in this case is to maximize d sing 0, P we have max d max α,β Ω Ω max µ F α,β,µ P P min α,β Ω Ω min µ F α,β,µ min α,β Ω Ω F, where α,β we have utilized the fact that min µ F α, β, µ F α, β,,

5 since µ cosα β and F α, β, µ is an increasing function of µ Case b: µ max cosα β, cosα β his inequality infers that the feasible set of α and β must meet the inequality: α β and θ α, β θ or α β and θ α, β θ, as shown in Figure Our goal in this case is to maximize d With 0, we obtain P max d max max max α,β Ω Ω µ F α, β, µ, P max max α,β Ω µ F α, β, µ max P min α,β Ω Ω min µ F α, β, µ, P min α,β Ω min µ F α, β, µ 6 For any fixed α and β, since F α, β, µ is an increasing function of µ, we have and min min F α, β, µ min F α, β α,β Ω µ α,β Ω min min α,β Ω Ω min α,β Ω Ω F F α, β, µ µ α, β, min F α, β α,β Ω Ω Combining 6 with 7 results in cosα β 7a 7b P max d min min α,β Ω Ω F α, β, min α,β Ω F α, β Case c: µ cosα β In this case, the feasible set in terms of α and β is Ω α, β : α β, θ α, β θ, as shown in Figure, and thus, we aim at maximizing d Combining the definition of d and 0, we can represent d as d d µ µ cosα β + µ µ cosα β+ F α,β,µ 4P µ µ cosα β+,,ψ f α,β,µ 4P Hence, we have max d max α,β max µ d Now, applying emma 5 into function µ µ cosα β+ f α,β,µ yields max µ d 4P,,ψ f α,β, cosα,β P min α,β Ω F α,β 4P F α,β, cosα,β 4P F α,β herefore, max d Now, summarizing the above discussions on Cases a, b and c, we conclude that the optimum is reached when d d and is given by P min min α,β Ω Ω F α, β, min α,β Ω F α, β On the other hand, by emmas and 4, we know that min F α, β F C, F H, F G, since F A F C, F H F E and F G F F, and that min F α, β F B, F D, F H, F G, F E, F F o further simplify this optimization problem, let us consider the case when 0 ψ 4 In this situation, we notice that f α, θ cosα θ + cosθ θ, where θ α θ by the definition of θ in emma and by emma, we know that θ θ and θ < θ As a result, we have f θ, θ f θ, θ and thus, F C F G In addition, we also note that f θ, β cosθ θ + cosβ θ, where θ + β θ Hence, if θ + θ, then, f θ, θ + f θ, θ and thus, we have F H F C If θ + < θ, then, by emma, we know φ + θ θ > In addition, we also know that f θ, β + cos θ θ + sin θ θ cosβ φ θ, where θ β θ + As a result, we obtain f θ, θ < f θ, θ + l + and thus, F D < F H herefore, in any case, the point H is not any optimal point Since F H F H and F F F G, we have minmin F α, β, min F α, β F C, F D Now, by emma 6, we can prove the result on heorem regarding the considered case Similarly, we can prove the case when 4 < ψ his completes the proof of heorem Proof of heorem : Since F VG, the nonnegative constraint on F is transfered to G such that g cos ψ g sin ψ 0 8a g sin ψ + g cos ψ 0 8b g cos ψ g sin ψ 0 8c g sin ψ + g cos ψ 0 8d From 8 we can attain g 0 and g 0 In addition, when 0 ψ 4, from 8b and 8d we have g g tan ψ 9a g g tan ψ 9b and when 4 < ψ, from 8a and 8c we obtain g g cot ψ 0a g g cot ψ 0b On the other hand, the nonnegative power constraint 4b in terms of F is transfered into that in terms of G such that cos ψ + sin ψg + g +cos ψ sin ψg + g P Now, combining 9 with results in the largest feasible domain in terms of design variables g and g, ie, g + g P cos ψ for g 0, g 0 when 0 ψ 4, while combining 0 with produces the largest feasible domain in terms of design variables g and g, ie, g + g P sin ψ for g 0, g 0 when 4 < ψ In addition, notice that in this case, from 5 we know that g d /, g d / and a g g d d Hence, by the definition of d, we obtain d d d and thus, the objective function in Problem is reduced to max mind, d d his optimization problem can be split into two sub-problems: max d, d d and

6 maxd d, d d subject to either constraint for 0 ψ 4, which is equivalent to d + d P cos ψ, or constraint for 4 < ψ, which is equivalent to d + d P sin ψ No mater which case occurs, the resulting optimization problem is linear and thus, the maximum is achieved when all the corresponding equalities hold, ie, when 0 ψ 4, we have d d, d + d P cos ψ and the equality in 9 holds, and when 4 < ψ, we have d d, d + d P cos ψ and the equality in 0 holds his completes the proof of heorem