Computing the PEP

3.4 Approximate non-coherent ML detection

3.4.3 Computing the PEP

3.2. Let φbe vector of phases of components of Lr. If ^2πm−π_q ≤φ_i < ^2πm+π_q ,1≤i≤ T, m is an integer, then ˆφ_i = ^2πm_q .

3.3. If Obj≤Tr(XX^∗φˆφˆ^∗) then Obj = Tr(XX^∗φˆφˆ^∗) and ¯φ= ˆφ.

3.4. count = count + 1.

4. If count≤10 go to 3.

5. Let ˆφ= ¯φ, ˆs= ^e√^jφˆ T.

6. Solve (3.44). Let ¯sbe the solution.

[Remark: We used 10 rounding iterations in the description of the SDPLS algorithm.

We would like to emphasize that there is no particular reason which would explain what is the optimal number of these iterations. We used 10 and obtained decent performance.]

Roughly speaking, the main idea of the SDPLS algorithm is to improve on SDP by doing an additional limited search over the codewords which have the squared inner product with ˆ

sgreater than α. With this improvement we will be able to provide sound proofs regarding the coding loss of the SDP relaxation in the following section.

algorithm and refer to it as theGenie. Its solution ˆs₁ is such that

if |ˆs^∗s_t|²< α ˆs₁ = ˆs

if |ˆs^∗s_t|²≥α ˆs₁ = ¯s (3.46)

where ˆs is as found in (3.38). The probability of error for the Geniealgorithm is given by

P_e^g=

q^T

X

i=1,i6=t

P_g(error|s_tis sent)P(s_tis sent) (3.47)

where Pg(error|s_tis sent) denotes the probability that an error occurred if the codeword s_t was sent and Genie algorithm was applied. Clearly, our SDPLS algorithm will have smaller probability of error than the Genie. Namely, the Genie and SDPLS differ in the case when|ˆs^∗s_t|< α. TheGeniekeeps ˆsas a solution which is incorrect since|ˆs^∗s_t|< α <1 implies ˆs 6= s_t. Since in the only case when they differ Genie certainly makes a mistake, it can not have smaller probability of error than SDPLS. Therefore if we prove that Genie attains a certain probability of error, then SDPLS does so too. Hence, we concentrate on bounding the probability of error of the Genie, i.e., on bounding P_g(error|s_tis sent). The bound obtained this way will also be a bound on the probability of error of the SDPLS. To this end, note that

P_g(error|s_tis sent) =P(ˆs₁ 6=s_t) =P(∃i: ˆs₁ =s_i6=s_t)≤ X

s_i6=st

P(ˆs₁ =s_i 6=s_t)

≤ X

|s^∗_ist|²<α

P(ˆs₁ =s_i6=s_t) + X

|s^∗_ist|²≥α

P(ˆs₁ =s_i6=s_t). (3.48)

Let us considerP(ˆs₁ =s_i6=s_t,|s^∗_is_t|² < α) in more detail. (For brevity of notation, in the following expressions we omit that everything is conditioned on s_t being transmitted, and that|s^∗_is_t|² < α.) So,

P(ˆs₁=s_i 6=s_t) =P(ˆs₁ =s_i6=s_t|ˆs₁ = ˆs)P(ˆs₁= ˆs) +P(ˆs₁ =s_i 6=s_t,ˆs₁6= ˆs). (3.49)

Let us define function C as C(s) = TrXX^∗ss^∗. Furthermore, let E denote the event that (ˆs₁ =s_i 6=s_t,ˆs₁ 6= ˆs). Clearly,E implies that C(s_i) =C(ˆs₁)≥C(ˆs)≥αC(s_ML)≥αC(s_t), which further means that C(s_i) ≥ αC(s_t). Using this, we obtain P(ˆs₁ = s_i 6= s_t,ˆs₁ 6= ˆs) ≤ P(C(si) ≥ αC(st)). Also, following similar argument, it is not difficult to see that P(ˆs₁ =s_i 6=s_t|ˆs₁ = ˆs)P(ˆs₁= ˆs))≤P(C(s_i)≥αC(s_t)). Replacing the obtained inequalities in (3.49) we have

P(ˆs₁ =s_i 6=s_t,|s^∗_is_t|² < α)≤2P(C(s_i)≥αC(s_t)). (3.50)

Now, let us consider P(ˆs₁=s_i 6=s_t,|s^∗_is_t|² ≥α). It is not that difficult to see that

P(ˆs₁ =s_i 6=s_t,|s^∗_is_t|² ≥α) ≤P(C(s_i)≥C(s_t)). (3.51)

In order to precisely establish (3.51) we need the following implication

(ˆs₁ =s_i6=s_t,|s^∗_is_t| ≥α) =⇒ |ˆs^∗s_t| ≥α.

We will show that the previous implication holds using a contradiction argument. Assume that it is not correct. Then it means that (ˆs₁ =s_i6=s_t,|s^∗_is_t| ≥α) is true and |ˆs^∗s_t| ≥α is not true. If|ˆs^∗s_t| ≥αis not true then|ˆs^∗s_t|< αis true. Then ˆs₁= ˆsis true as well. Further we have s_i = ˆs₁ = ˆs and |s^∗_is_t| =|ˆs^∗s_t| < α. This is a contradiction since it was assumed that (ˆs₁ =s_i6=s_t,|s^∗_is_t| ≥α) is true and hence|s^∗_is_t| ≥α. Therefore the implication

(ˆs₁ =s_i6=s_t,|s^∗_is_t| ≥α) =⇒ |ˆs^∗s_t| ≥α.

is indeed true.

Substituting (3.50) and (3.51) in (3.48), we finally obtain

P_g(error|s_tis sent)≤ X

|s^∗_ist|²≤α

2P(C(s_i)≥αC(s_t)) + X

|s^∗_ist|²≥α

P(C(s_i)≥C(s_t)). (3.52)

Let

P_it||s^∗

ist|²<α = P(C(s_i)≥αC(s_t)|s_tis sent,|s^∗_is_t|² < α) P_it||s^∗

ist|²≥α = P(C(s_i)≥ C(s_t)|s_t is sent,|s^∗_is_t|² ≥α).

In the remainder of this section, we compute bounds onP_it||s^∗

ist|²<α and P_it||s^∗

ist|²≥α. P_it||s^∗

ist|²<α=P(Tr(X^∗s_i)(X^∗s_i)^∗≥αTr(X^∗s_t)(X^∗s_t)^∗|s_t is sent). (3.53) Since we assume that s_t was transmitted, it holds that X =√

ks_th+W where, as earlier, k=ρT. Replacing this value for X in (3.53), we obtain

P_it||s^∗

ist|²<α=P(Tr(



h W





∗

Q_n



h W



≥0|s_t is sent), (3.54)

where

Q_n =





√ks^∗_t

I



(s_is^∗_i −αs_ts^∗_t) √

kst I

=





√ks^∗_t

I





s_i s_t 

1 0 0 −α







s^∗_i s^∗_t



 √

kst I

=





√kψ^∗_it √ k s_i s_t







1 0 0 −α









√kψ_it s^∗_i

√k s^∗_t



,

and ψ_it =s^∗_is_t. Although it is possible to compute explicitly the probability in (3.54), we will find that it is sufficient to find its Chernoff bound. In particular,

P_it||s^∗

ist|²<α ≤min

µ Ee

µ(Tr( 2 66 64

h W

3 77 75

∗

Qn

2 66 64

h W

3 77 75))

= Z e

−Tr( 2 66 64

h W

3 77 75

∗

(I−µQn) 2 66 64

h W

3 77 75)

dhdW

π^N = 1

det(I −µQ_n)^N

where{µ|I−µQ_n≥0}. We first simplify the determinant in the denominator as

det(I−µQ_n) = det(I−µ



 kψ_itψ_it^∗ + 1 (k+ 1)ψ_it

−α(k+ 1)ψ^∗_it −α(k+ 1)1



).

After some further algebraic transformations we obtain

det(I−µQ_n) = (k+ 1)α(V^(it)−1)(−µ+ξ⁽¹⁾)(−µ+ξ⁽²⁾)

with

ξ⁽¹⁾ = V^(it)−α+¹⁻_k^α + q

(V^(it)−α+¹⁻_k^α)²+^4α(1−V_k^(it)2 ^)(k+1)

2α(V^(it)−1)^k+1_k ξ⁽²⁾ = V^(it)−α+¹⁻_k^α −

q

(V^(it)−α+¹⁻_k^α)²+^4α(1−V_k^(it)2 ^)(k+1)

2α(V^(it)−1)^k+1_k

and V^(it) = ψ_itψ^∗_it. As earlier, our results can be made precise so that they hold for any SNR. However, to make writing less tedious in the rest of this section we consider only the case of large SNR. Assumingµ= ¹₂, the previous results simplify to

P_it||s^∗

ist|²<α≤ 1 (k^{(α−V(it))2}

4(1−V^(it)))^N. (3.55)

To compute the bound on P(C(si) ≥ C(st)|s_tis sent,|s^∗_is_t|² ≥α) we will use a well-known result from the literature (see, e.g., [51])

P_it||s^∗

ist|²≥α ≤ 1

(k^(1−V₄^(it)))^N. (3.56) Now we can substitute the results from (3.55) and (3.56) in (3.52) and obtain

Pg(error|s_tis sent)≤ X

|s^∗_ist|²<α

2 1

(k^{(α−V(it))2}

4(1−V^(it)))^N + X

|s^∗_ist|²≥α

1

(k^(1−V₄^(it)))^N =B^pep(ρ). (3.57) Recall that in the case of the exact ML detection, which requires algorithms, none of which

are of polynomial complexity, we have for the same probability of error

P_{M L}(error|s_tis sent)≤ X

|s^∗_ist|²<α

1 (k^(1−V₄^(it)))^N

+ X

|s^∗_ist|²≥α

1 (k^(1−V₄^(it)))^N

=B^pep_{M L}(ρ). (3.58)

Clearly, comparing (3.57) and (3.58) it follows that the SDPLS algorithm based on the well- known SDP relaxation (slightly refined here for the purposes of the valid proof) has thesame diversity as the exact ML and the AR algorithm. Of course, since the SDPLS algorithm is only an approximation, the exact ML solution still has an advantage of (^1−V(it)

α−V^(it))² in the coding gain. However, as the analysis conducted in Section 3.2 hints (and the simulation result on Figure 3.6 confirms) the AR algorithm has a significantly bigger coding loss than the SDPLS algorithm analyzed in this section.

It should also be noted that a very similar result related to the diversity of the SDP- based algorithm in the context of coherent (channel known at the receiver) ML detection has recently been shown in [60].

We summarize the previous results in the following theorem.

Theorem 3.4. Consider the problem of non-coherent ML detection for a SIMO system described in (3.1) in high-SNR regime. Assume that the codeword s_t was transmitted. Then the probability that an error occurred if SDPLS algorithm was applied to solve (3.34) can be upper bounded in the following way

P(error|s_tis sent)≤ X

|s^∗_ist|²<α

2 1

(ρT^(α₄₍₁^−V(it))2

−V^(it)))^N

+ X

|s^∗_ist|²≥α

1 (ρT^(1−V₄^(it)))^N

.

Proof. Follows from the previous discussion.