First-order sigma-delta quantization for fusion frames

for any vector v∈Rⁿ, there exits a vector u∈Nθ, such that the angle between v and u is less thanθ, and the cardinality ofNθ is at most(1+_sin(θ)² )ⁿ.

Proof. Letε =sin(θ), then by Theorem 4.2.2 there exists anε−net Nε with cardinality N (R^N,θ)at most(1+_sin(θ² ₎)ⁿ. Then let

Nθ ={u|u=−−→

x−0,x∈Nε}.

By the definition ofε−net, we know for any vectorv∈Rⁿand anyu∈Nθ, we have

|u−v| ≤sin(θ),

which means the angle between betweenvanduat mostθ.

After we have the concept of nets, we can give the first definition of quantizer Q_n as following:

Theorem 4.2.4. Let{(W_n,w_n)}_n∈I be a fusion frame for HandN^π₃,n={e_i,n}^N_i=1^(Rn,π3) be unit vectors set in W_n constructed in Corollary 4.2.3. The algorithm is defined by (4.2.1).

For any x∈W_n, we define:

Q_n(x) =e_i,n, i=min{k∈N (Rⁿ,π

3):hx/|x|,e_k,ni ≥ hx/|x|,e_j,ni f or all j6=k}

and

Q_n(x) =e_1,n, i f x=0.

Then ifkxk ≤δ <¹₂, for all n, we have:

ku_nk ≤C=max{δ²−δ+1 1−2δ ,1}

Proof. Let a_k =kπ_Wk(u_k−1) +x_kk, b_k=kq_kk, c_k=ku_kk, θ_k is the angle between vector

π_Wk(u_k−1) +x_kand vectorq_k. Then by the definition ofN^π₃,n, we knowθ_k≤^π₃.

We prove the theorem by induction. Whenn=1, sinceu₀=0 andθ1< ^π₃, then by the law of cosines, we have:

c²₁=a²₁+b²₁−2 cos(θ₁)a₁b₁≤a²₁+b²₁−a₁b₁.

Since|a₁| ≤ |δ+0|=δ < ¹₂ andb₁=1. Thus:

a²₁+b²₁−a₁b₁=a²₁+1−a₁= (a₁−1 2)²+3

4 ≤1 4+3

4=1 then we get

|u₁| ≤C=max{δ²−δ+1 1−2δ ,1}

so whenn=1 the result is true.

Suppose for all n≤k−1, the result is true, then when n=k, by Law of cosines, we have:

c²_k=a²_k+b²_k−2 cos(θ_k)a_kb_k. Sinceθ_k≤^π₃ andb_k=1, so we have:

c²_k=a²_k+1−2 cos(θ_k)a_k≤a²_k+1−2 cos(π

3)a_k≤a²_k+1−a_k.

Consider about the function:f(x) =x²+1−x, since f⁰(x) =2x−1 and f⁰⁰(x) =2>0, then f(x) has minimum at x= ¹₂. By the inductive assumption, 0<a_k ≤C+δ, so for x∈[0,C+δ], f(x)has its maximum value at f(0)or f(C+δ). Now, we claim

f(x)≤C=max{δ²−δ+1

1−2δ ,1},f or x∈(0,C+δ). (4.2.2) For f(0), we have f(0) =1.

For f(C+δ), if we can prove:

C²≥(C+δ)²+1−(C+δ) (4.2.3)

then we know our claim (4.2.2) is true. Since the inequality (4.2.3) is equivalent to:

C²+δ²+2Cδ+1−(C+δ)≤C²

which is equivalent to:

2C(δ−1

2)≤ −δ²+δ−1 we know(δ−¹₂)<0, then the inequality (4.2.3) becomes:

C≥ δ²−δ+1 1−2δ , which follows from the definition of C:

C=max{δ²−δ+1 1−2δ ,1}.

Thus we haveC≥ |u_k|. Hence by induction, we prove the Theorem.

By Corollary 4.2.3, we know

|R_Qn|=N (R^Mn,π

3)≤(1+ 2

√3 2

)^Mn ≈3.3^Mn.

Thus for each subspace we need log₂(3.3^Mn)1.7M_n bits to quantize the signal. Actu- ally, we can show that to keep |u_n| have uniform bound, we just need log₂(M_n+1) bits.

Moreover, it is the best we can get. For proving this, we need several lemmas.

Lemma 4.2.5. InR^d, there exists a set of vectors{e_i}^d+1_i=1 such that:

he_i,e_ii=1, he_i,e_ji=c=cosθ, f or any i6= j,

d+1 i=1

∑

e_i=0

Moreover,θ =arccos(−¹_d).

X Y

X Z

Y

Figure 4.1: Example inR²andR³

Proof. Consider the subspace

A={(x₁,x₂, . . . ,x_d+1)|

d+1 i=1

∑

x_i=0} ∈R^d+1.

There exists a 1-to-1 invertible mapping fromAtoR^d which preserves the Euclidean met- ric. Hence,Ais isomorphic toR^d. Then let

V_i= (0,0, . . . ,1, . . . ,0) f or i=1,2, . . . ,d+1

where the i-th coordinate is 1 and the rest are all 0, and let

V₀= ( 1 d+1, 1

d+1, . . . , 1 d+1).

Letu_i=V_i−V₀fori=1,2, . . . ,d+1, then we know{u_i}is a subset inA.

For anyi6= j,

hu_i,u_ji=hV_i−V₀,V_j−V₀i=hV_i,V_ji − hV_i,V₀i − hV₀,V_ji+hV₀,V₀i=− 1

d+1 =C₁, hereC₁is a constant.

For alli=1,2, . . . ,d+1,

hu_i,u_ii= (1− 1

d+1)²+ d

(d+1)² = d

d+1 =C₂, hereC₂is a constant.

Also,

d+1

∑

i=1

u_i=

d+1

∑

i=1

V_i−(d+1)V₀= (1,1, . . . ,1)−(1,1, . . . ,1) =0.

Lete_i= _|u^ui

i|= ^√ui

C₂.

Firstly,{e_i}is also a subset inA, and for alli=1,2, . . . ,d+1, we know|e_i|=1.

Secondly, for anyi6= j,

he_i,e_ji=h u_i

|u_i|, u_j

|u_j|i=C₁ C₂ =C.

Lastly,

d+1 i=1

∑

e_i=

d+1 i=1

∑

u_i

√C₂ =0∗p C₂=0

Hence, {e_i} exists has the designed properties. Additionally, since ∑^d+1_i=1 e_i=0, then for anye_j,∑^d+1_i=1he_i,e_ji=0 which is equivalent to:

dc+1=0

.

Hencec=−_d¹, which impliesθ =arccos(−¹_d).

Lemma 4.2.6. Let {e_i}^d+1_i=1 ∈R^d be the vectors set in Lemma 4.2.5. For any unit vector

u∈R^d, we define the vector function Q(u)as following:

Q(u) =e_i, i f hx,e_ii ≥ hx,e_ji f or all i6= j

Letθ =arccoshu,Q(u)i, thenθ ≤π−arccos(−¹_d).

Proof. Let u= (x₁,x₂, . . . ,x_d+1) be a unit vector inA, here A is the same as in Lemma 4.2.5. As we know, there exists a 1-to-1 invertible mapping fromAtoR^d which preserves the Euclidean metric, thus we can consider u as a unit vector inR^d. Then we have:

d+1 i=1

∑

x_i=0;

d+1 i=1

∑

x²_i =1 (4.2.4)

Let{e_i}be the same as in Lemma 4.2.5. Fori=1,2, . . . ,d+1, let

hu,e_ii= (− 1 d+1

d+1 j=1

∑

x_j+x_i)/C₂=x_i/C₂.

Then by definition ofQ(u), we have

hu,Q(u)i=max{x_i}/C₂.

Sinceuis a unit vector, then

θ =arccoshu,Q(u)i ≤π−arccos(−1 d), if and only if,

hu,Q(u)i=max{x_i}/C₂≥ 1 d. Hence, to proveθ ≤π−arccos(−¹_d)is equivalent to prove

min max{x_i}=C₂ d =

s 1 d(d+1),

here{x_i}satisfy equation (4.2.4).

Without loss of generality, we supposex₁≥x₂≥. . .≥x_d+1. First step, ifx₁6=x₂, then we let

x₁⁰ = x₁+x₂

2 , x⁰₂= x₁+x₂

2 , x⁰_i=x_i, f or i=3,4, . . . ,d+1.

Firstly we know

d+1

∑

i=1

x⁰_i=

d+1

∑

i=1

x_i=0.

Also sincex²₁+x²₂≥x₁⁰²+x⁰²₂, we get

d+1 i=1

∑

x⁰²_i ≤

d+1 i=1

∑

x²_i =1.

Let

c=

d+1 i=1

∑

x⁰²_i and let x¹_i =x⁰_i/c f or i=1,2, . . . ,d+1.

Then for{x¹_i}, we know∑^d+1_i=1 x¹_i = ¹_c∑^d+1_i=1 x¹_i =0 and∑^d+1_i=1 x¹_i²=1. Sincec≤1, also we still havex¹₁≥x¹₂≥. . .≥x¹_d+1, then we getx¹_i ≥x_ifor alli=2,3, . . . ,d+1, Hencex¹₁≤x₁. Thus max{x_i} ≥max{x¹_i}.

Ifx₁=x₂, we just letx¹_i =x_ifor alli=1,2, . . . ,d+1. Thus we have

max{x_i} ≥max{x¹_i}.

Second step, by same method we can get {x²_i}^d+1_i=1 satisfy that equation(4.2.4), x₁²=x²₂= x²₃≥x²₄≥. . .≥x²_d+1and max{x¹_i} ≥max{x²_i}.

We can keep doing the same thing for d−1 times until we get {x^d−1_i }^d+1_i=1 satisfy that equation(4.2.4),x^d−1₁ =x^d−1₂ =. . .=x^d−1_d ≥x²_d+1 and max{x_i^d−1} ≤max{x_i^j}for all j= 1,2, . . . ,d−2. Since∑^d+1_i=1 x_i^d−1=0, then we can not do the next step. Thus min max{x_i}= max{x^d−1_i }.

Leta=x^d−1₁ andb=x^d−1_d+1, by equation (4.2.4), we have:

da²+b²=1, ad+b=0

By solving the equations above, we geta=q

1

d(d+1) which is we want.

After all the preparation, now we give the definition of the new quantizerQ_nas follow- ing and prove the stability of the scheme.

Theorem 4.2.7. Let {(W_n,w_n)}_n∈I be a fusion frame for Hwhere dim(W_n) =M_n and let {e_i,n}^M_i=1ⁿ⁺¹ be the set of unit vectors in W_n which we constructed in Lemma 4.2.5. The algorithm is defined as (4.2.1). For any x∈W_n, we define:

Q_n(x) =e_i,n, i=min{k∈ {1,2, ...,M_n+1}:hx/kxk,e_k,ni ≥ hx/kxk,e_j,ni f or all j6=k}

and

Q_n(x) =e_1,n, i f x=0

Let d=max{dim(W_n)},θ =π−arccos(−_d¹). Ifkxk ≤δ <cos(θ) =¹_d. Then for all n, we have:

ku_nk ≤C=max{δ²−2 cos(θ)δ+1 2(cos(θ)−δ) ,1}

Proof. Let a_k =kπ_Wk(u_k−1) +x_kk, b_k =kq_kk ,c_k =ku_kk, θ_k is the angle between vector π_Wk(u_k−1) +x_k and vectorq_k. Then by Lemma 4.2.6, we know θ_k≤θ =π−arccos(−_d¹).

Sinced≥2, thusθ ∈[^π₃,^π₂)

We prove the theorem by induction, whenn=1, sinceu₀=0 andθ1< ^π₂,then by the law of cosines,we have:

c²₁=a²₁+b²₁−2 cos(θ₁)a₁b₁≤a²₁+b²₁−2 cos(θ)a₁b₁

Since|a₁| ≤ |δ+0|=δ <cos(θ)andb₁=1, then

a²₁+b²₁−2 cos(θ)a₁b₁=a²₁+1−2 cos(θ)a₁= (a₁−cos(θ))²+1−cos²(θ)

which implies,

a²₁+b²₁−2 cos(θ)a₁b₁≤cos²(θ) +1−cos²(θ) =1

which means:

|u₁| ≤C=max{δ²−2 cos(θ)δ+1 2(cos(θ)−δ) ,1}

so whenn=1 the result is true.

Suppose for alln≤k−1, the result is true, then whenn=k, by the Law of cosines, we have:

c²_k=a²_k+b²_k−2 cos(θ_k)a_kb_k. Sinceθ_k≤θ < ^π₂ andb_k=1, then we get:

|c_k|²=a²_k+1−2 cos(θ_k)a_k≤a²_k+1−2 cos(θ)a_k

Consider the function: f(x) =x²+1−2 cos(θ)x, since f⁰(x) =2x−2 cos(θ)and f⁰⁰(x) = cos(θ)> 0, then f(x) has its minimum at x=cos(θ). Since 0<a_k ≤C+δ, then for x∈[0,C+δ], f is maximal at f(0)or f(C+δ). Now we claim:

f(x)≤C², f or x∈[0,C+δ] (4.2.5)

For f(0), we have f(0) =1.

For f(C+δ), if we can prove:

C²≥(C+δ)²+1−2 cos(θ)(C+δ), (4.2.6)

then we know our claim (4.2.5) is true. The inequality (4.2.6) is equivalent to:

C²+δ²+2Cδ+1−2 cos(θ)(C+δ)≤C²,

which is equivalent to:

2C(δ−cos(θ))≤ −δ²+2 cos(θ)δ−1.

Sinceδ <cos(θ), then(δ−cos(θ))<0, hence the inequality (4.2.6) is equivalent to:

C≥δ²−2 cos(θ)δ+1 2(cos(θ)−δ)

which holds by the definition of C. Hence by induction, we have proven the Theorem.

Remark 4.2.8. Suppose for every subspace W_n, the cardinality|R(Q_n)|is less then M_n+1, for example, R(Q_n) =M_n. Then no matter how we define the quantize Q_n, the situation θ_k ≥π/2 will happen, then cos(θ_k)<0, so for any δ >0 we have 2(δ−cos(θ_k))>0.

Hence we can only get

C≤ δ²−2 cos(θ_k)δ+1 2(cos(θ_k)−δ)

which is always false since the right part of the inequality is negative, so we cannot get a positive uniform bound C.

Actually, if we suppose all θ_k =π/2, by the law of cosine we can get that |u_k+1|²=

|π_Wk+1(u_k) +x_k+1|²+1 for all k, and we can easily get |u_k+1| ≥ |u_k|+σ, where σ is a positive constant. Hence {|u_n|} is increasing with n, which means {|u_n|} does not have uniform bound. Hence|R(Q_n)|=M_n+1is the best we can hope for Theorem 4.2.7.

However, in practice, quantizers are never perfect. In [DD03], the authors assume the quantizerq(x) =sign(x+ρ), whereρis unknown noise except for the specification|ρ|<τ.

Then the authors define the non-idealQas following:

Q(x) =sign(x), f orkxk ≥τ;

or

Q(x)≤1, f orkxk ≤τ.

and proved the 1-bitΣ∆quantization scheme is still stable.

Here, we will show the robustness of the quantizersQ_nin Theorem 4.2.7. Observe the definition of the quantizer Q_n in Theorem 4.2.7, not same as the quantizer Q above, Q_n mapping the vectorx∈R^dto alphabet vectors not depend on thekxkbut on thehx/|x|,e_k,ni which is the angle between x and each alphabet vector e_i,n. Now, define the non-ideal quantizerQ_nwith a noiseρ as following:

Q_n(x) =e_i,n, i=min{k∈ {1,2, ...,M_n+1}:hx/|x|,e_k,ni − hx/|x|,e_j,ni ≥ρ f or all j6=k}, (4.2.7) Q_n(x) =e_i,n, i={k∈ {1,2, ...,M_n+1}: 0<hx/|x|,e_k,ni − hx/|x|,e_j,ni ≤ρ f or all j6=k}.

(4.2.8) In equation (4.2.8), we just letibe any vector which satisfies the condition:

0<hx/|x|,e_k,ni − hx/|x|,e_j,ni ≤ρ f or all j6=k.

Then we have the proposition for our new quantizerQ_nas:

Proposition 4.2.9. Let{(W_n,w_n)}_n∈I be fusion frame forH and {ei,n}^M_i=1ⁿ⁺¹ be the set of unit vectors in W_nwhich we constructed in Lemma 4.2.5. The algorithm is defined as 4.2.1.

For any x∈W_n, we define Q_n by equation (4.2.7) and (4.2.8). Let d=max{dim(W_n)}, θ =π−arccos(−¹_d). Ifkxk ≤δ <cos(θ+arccos(ρ))andarccos(ρ)<arccos(−¹_d)−^π₂.

Then for all n, we have:

ku_nk ≤max{δ²−2 cos(θ+arccos(ρ))δ+1 2(cos(θ+arccos(ρ))−δ) ,1}.

Proof. Let a_k =kπ_Wk(u_k−1) +x_kk, b_k=kq_kk, c_k=ku_kk, θ_k is the angle between vector π_Wk(u_k−1) +x_k and vectorq_k. Notice that we assume arccos(ρ)<arccos(−¹_d)−^π₂. Then by Lemma 4.2.6 and the definition ofQ_n, we have

θ_k≤θ+arccos(ρ) =π−arccos(−1

d) +arccos(ρ)< π 2

thus we still haveθ+arccos(ρ)∈[^π₃,^π₂). Since every thing else is the same as Theorem4.2.7, following the same method, we have:

ku_nk ≤C=max{δ²−2 cos(θ+arccos(ρ))δ+1 2(cos(θ+arccos(ρ))−δ) ,1}.

In [DD03], the noise ρ do not have any limitation. Unlike as [DD03], in Proposition 4.2.9, we assume the noiseρ should satisfy arccos(ρ)<arccos(−¹_d)−^π₂ which is equiv- alent toρ <|cos(arccos(−_d¹)−^π₂)|. However, sinced=max{dim(W_n)}, in practice, the dimension of the subspaceW_nis always far less the the dimension of the whole spaceH, for example, whend=10, we only need arccos(ρ)< ₁₅^π which is a reasonable noise. Hence, Q_nis robust for a acceptable noise.

In Theorem 4.2.7, we know

|R_Qn|=M_n+1.

Thus for each subspace we need log₂(M_n+1)M_nbit to quantize the signal. Compared to the algorithm (4.0.4) we gave in the beginning of this section and the first algorithm (4.2.4) we gave before, it is a big improvement.

Now we give a example to show the error bound of the fisrt-order Sigma-Delta quanti- zation with fusion frame:

Example 4.2.10. Let{E(n/N)}^N_n=1⊂R³be defined as following:

E(n/N) = [cos(2πn/N),sin(2πn/N),0]^∗, 1≤n≤N, (4.2.9)

Let W_n be the two dimensional subspace ofR³ with normal vector E(n/N). By Theorem 1.3.5, we know there exists a constant w, such that {W_n,w} is a C−tight fusion frame.

Moreover, by Proposition 1.3.6, we have:

kS_W,wk_op=C= ∑^N_n=1w²_ndim(W_n)

dim(R³) = w²2N 3 Suppose f is a signal inR³that satisfieskfk ≤¹₂, then we know,

f =S_W,w⁻¹ S_W,wf =

N

∑

n=1

w²S⁻¹_W,wπ_Wn(f)

Now we use the fisrt-order Sigma-Delta quantization as in Theorem 4.2.7, then we have the reconstructed signal,

ef =

N

∑

n=1

w²S_W,w⁻¹ q_n

Thus, the error is equal to:

kf−efk=k

N

∑

n=1

w²S_W,w⁻¹ π_Wn(f)−

N

∑

n=1

w²S_W,w⁻¹ q_nk

=w²k

N

∑

n=1

S_W,w⁻¹ (π_Wn(f)−q_n)k

=w²k

N n=1

∑

S_W,w⁻¹ (u_n−π_Wn(u_n−1))k

≤w²kS_W,w⁻¹ k_opk

N n=1

∑

(u_n−π_Wn(u_n−1))k

=w²kS_W,w⁻¹ k_opk

N−1

∑

n=1

(u_n−πWn+1(u_n)) +u_Nk

By the definition of W_n, we have:

u_n−πWn+1(u_n) =u_nsin(2π N ), and by Theorem 4.2.7 we haveku_nk ≤C. Thus,

k

N−1 n=1

∑

(u_n−πW_n+1(u_n)) +u_Nk=k

N−1 n=1

∑

Csin(2π

N ) +Ck=kC(N−1)sin(2π N ) +Ck Since when N is large enough, we havesin(_2π^N)≈ ^2π_N. Then,

kC(N−1)sin( N

2π) +Ck ≈ kC(N−1)2π

N +Ck=O(1),

which meansk∑^N−1_n=1(u_n−πWn+1(u_n)) +u_Nkis equivalent to a constant C⁰. Hence,

w²kS_W,w⁻¹ k_opk

N−1

∑

n=1

(u_n−πWn+1(u_n)) +u_Nk=C⁰w² 3

w²2N =O(N⁻¹).

So the errorkf−efkis equal toO(N⁻¹).