On 2-D Direction-of-Arrival Estimation Performance for Rank Reduction Estimator in Presence of Unexpected Modeling Errors

Abstract

The rank reduction estimator (RARE) is one kind of autocalibration method used in the presence of sensor errors. It demonstrates high accuracy of direction-of-arrival (DOA) estimation in the absence of multidimensional search or iteration. However, its estimation performance is affected by “unexpected modeling errors.” There is a lack of research regarding the performance of 2-D RARE estimation, although 2-D RARE estimation is extensively employed in applications. This paper presents a theoretical derivation for the closed-form expression of the mean square error (MSE) of 2-D RARE estimation under the influence of small unexpected modeling errors in the first order analysis. First, three definitions of 2-D joint direction-finding success are introduced, in order to establish the criterion for estimate performance. Then corresponding theoretical formulas for three probabilities of direction-finding success are given with the circularly Gaussian assumption of unexpected modeling errors, and their relations are discussed. Finally, the results of simulations utilizing our analysis method are demonstrated, verifying the effectiveness of the MSE expression and the formulas for probabilities of success. Therefore, our first order approximation provides a good prediction of the necessary calibration accuracy in the presence of unexpected modeling errors in order to help RARE meet an expected performance specification.

Appendices

Appendix A

In this appendix, we prove (21). Before the proof, we will introduce Lemma A.1 and Theorem A.1.

Lemma A.1

A sequence composed of 1,2,…,K can be sorted from 1 to K through several exchanges between two integers, and the number of exchange times has the same parity as the inverse number of the original sequence.

Theorem A.1

Introduce the exchange operator J transforming a sequence composed of 1,2,…,K to the sorted sequence from 1 to K. When the sorted sequence is transformed by J, its inverse number has the same parity as that of the original sequence.

Proof

The number of exchange times of J has the same parity as the inverse number of the original sequence according to Lemma A.1. Besides, τ(1,2,…,K), the inverse number of the sorted sequence equals zero. One exchange leads to the change of the parity of the inverse number. Thus the inverse number of the sorted sequence after transformation by J will have the same parity as the number of exchange times of J. Based on these equivalence relations, the conclusion in Theorem A.1 is proven.

Proof of (21)

We only prove the first equation here; it is easy to extend the proof to the second and third ones.

First, we introduce $\tilde{\mathbf{T}}_{k}(\theta_{m},\varphi_{m}) \in \mathbf{C}^{M \times K}$, k=1,2,…,K as T(θ _m ,φ _m ) recomposed with the kth column replaced by $\mathbf{t}_{k}^{(\theta \theta )}(\theta_{m},\varphi_{m})$, its second partial derivative with respect to azimuth at the true DOAs (θ _m ,φ _m ). The rank defect of both $\tilde{\mathbf{T}}_{k}^{H}(\theta_{m},\varphi_{m})\boldsymbol{\pi}\mathbf{ T}(\theta_{m},\varphi_{m})$ and its transposed matrix follows from the subspace orthogonality principle, which is shown as

(46)

Equation (46) provides an important proof for the latter discussion.

Present $w_{ij}(\theta,\varphi,\boldsymbol{\varepsilon} ) = [ W(\theta,\varphi,\boldsymbol{\varepsilon} ) ]_{ij} = \mathbf{t}_{i}^{\mathbf{H}}(\theta,\varphi )\hat{\boldsymbol{\pi}}\mathbf{ t}_{j}(\theta,\varphi )$, and expand D _θθ (θ _m ,φ _m ,0) defined in (16):

(47)

in which $w_{ij}^{(\theta \theta )}(\theta_{m},\varphi_{m},\mathbf{0}) = \frac{\partial^{2}w_{ij}(\theta,\varphi,\boldsymbol{\varepsilon} )}{\partial \theta^{2}} \vert _{(\theta,\varphi,\boldsymbol{\varepsilon} ) = (\theta_{m},\varphi_{m},\mathbf{0})} = f(\mathbf{t}_{i}^{(\theta \theta )},\mathbf{t}_{j}) + 2f(\mathbf{t}_{i}^{(\theta )},\mathbf{t}_{j}^{(\theta )}) + f(\mathbf{t}_{i},\mathbf{t}_{j}^{(\theta \theta )})$ where f(•,•) is defined in (23). $\mathbf{t}_{i}^{(\theta )}$ is short for the first partial derivative of t _i (θ _m ,φ _m ) as to azimuth and $\mathbf{t}_{i}^{(\theta \theta )}$ is short for $\mathbf{t}_{i}^{(\theta \theta )}(\theta_{m},\varphi_{m})$.

Now we reduce the higher order terms in (47):

(48)

where $\tilde{w}_{ij}^{(\theta \theta )}(\theta_{m},\varphi_{m},\mathbf{0})$ denotes $2f(\mathbf{t}_{i}^{(\theta )},\mathbf{t}_{j}^{(\theta )}). Z_{1},Z_{2}$ can be left out because

(49)

Equation (49) is validated based on (46). Theorem A.1 and (46) yield

(50)

Discarding Z ₁,Z ₂, Eq. (48) becomes

(51)

Finally, substituting (51) back into (47), we have

(52)

where the notation is the same as in (21). Up to this point, the first equation in (21) is proven.

Appendix B

In this appendix, we prove (24). Based on the subspace orthogonality principle [20], the following equation can be established if ε _m =0, regardless of the value of other unexpected modeling errors ε _n (n≠m):

$$ \hat{\boldsymbol{\pi}}\mathbf{ T}(\theta_{m},\varphi_{m}) \boldsymbol{\rho} (\theta_{m},\varphi_{m},\boldsymbol{ \eta}_{m}) = \mathbf{0} $$

(53)

Now $\mathbf{W}(\theta_{m},\varphi_{m},\boldsymbol{\varepsilon} ) = \mathbf{T}(\theta_{m},\varphi_{m})^{H}\hat{\boldsymbol{\pi}}\mathbf{ T}(\theta_{m},\varphi_{m})$ is still the rank defect which is given by

$$ D(\theta_{m},\varphi_{m},\boldsymbol{\varepsilon} ) \vert _{\boldsymbol{\varepsilon}_{m} = \mathbf{0}} = \det \bigl[\mathbf{W}(\theta_{m}, \varphi_{m},\boldsymbol{\varepsilon} )\bigr] \big \vert _{\boldsymbol{\varepsilon}_{m} = \mathbf{0}} = 0 $$

(54)

Thus

$$ D_{\theta} (\theta_{m},\varphi_{m}, \boldsymbol{\varepsilon} ) \vert _{\boldsymbol{\varepsilon}_{m} = \mathbf{0}} = 0, \qquad D_{\varphi} ( \theta_{m},\varphi_{m},\boldsymbol{\varepsilon} ) \vert _{\boldsymbol{\varepsilon}_{m} = \mathbf{0}} = 0 $$

(55)

Because D _θ (θ,φ,ε) and D _φ (θ,φ,ε) are constant on (θ _m ,φ _m ,ε _m =0) according to (55), the gradient vectors for each unexpected modeling errors except ε _m are zeros on (θ _m ,φ _m ,ε _m =0):

$$ \frac{\partial D_{\theta} (\theta_{m},\varphi_{m},\boldsymbol{\varepsilon} ) \vert _{\boldsymbol{\varepsilon}_{m} = \mathbf{0}}}{\partial \boldsymbol{\varepsilon}_{n}^{(x)}} = \mathbf{0},\qquad \frac{\partial D_{\varphi} (\theta_{m},\varphi_{m},\boldsymbol{\varepsilon} ) \vert _{\boldsymbol{\varepsilon}_{m} = \mathbf{0}}}{\partial \boldsymbol{\varepsilon}_{n}^{(x)}} = \mathbf{0}, \quad x \in \{ r,i\} $$

(56)

It is proved that the operation order of derivation with respect to $\boldsymbol{\varepsilon}_{n}^{(x)},x \in \{ r,i \}$ and assignment of θ,φ can be changed, and we have

(57)

Therefore, $\mathbf{D}_{\theta \boldsymbol{\varepsilon}_{{n}}^{(x)}}(\theta_{m},\varphi_{m},\mathbf{0}) = \mathbf{0}$, $\mathbf{D}_{\varphi \boldsymbol{\varepsilon}_{{n}}^{(x)}}(\theta_{m},\varphi_{m},\mathbf{0}) = \mathbf{0}$, x∈{r,i} in terms of n≠m. The proof of (24) is completed.

Appendix C

In this appendix, we prove (26).

We choose

$$\mathbf{D}_{\boldsymbol{\theta} \boldsymbol{\varepsilon}_{n}^{(r)}} (\theta_{m},\varphi_{m},\mathbf{0}) = \sum_{(i_{1},i_{2}, \ldots, i_{K})} ( - 1)^{\tau (i_{1},i_{2}, \ldots,i_{K})} \sum_{k = 1}^{2K} \boldsymbol{\varGamma}_{n}^{(r)}\bigl[\boldsymbol{\varSigma}_{k}^{\theta} (i_{1},i_{2}, \ldots, i_{K})\bigr] $$

to prove; it is also easy to extend the proof to others.

(58)

where the notation is the same as in (26). The proof is finished.

Appendix D

In this appendix, we prove Proposition 2. To begin with, we give Lemma D.1, which is useful for the proof of the proposition.

Lemma D.1

[1]

Assume x∼χ ²(n), i.e., x is chi-square distributed with n degrees of freedom. Its characteristic function is thus given by M _x (t)=(1−2it)^−n/2.

Proof of Proposition 2

Introduce vector $\boldsymbol{\delta}_{0} = \mathbf{R}_{\boldsymbol{\delta}}^{ - 1/2}\boldsymbol{\delta}$; thus δ ₀ is Gaussian distributed with mean zero, and its variance matrix is the identity matrix. Then y is rewritten as

$$ y = \boldsymbol{\delta}_{0}^{T}\mathbf{R}_{\boldsymbol{\delta}}^{1/2T} \mathbf{QR}_{\boldsymbol{\delta}}^{1/2}\boldsymbol{\delta}_{0} = \sum_{n = 1}^{D} \lambda_{n}\bigl( \mathbf{u}_{n}^{T}\boldsymbol{\delta}_{0} \bigr)^{2} $$

(59)

where u _n indicates eigenvectors of $\mathbf{R}_{\boldsymbol{\delta}}^{1/2T}\mathbf{QR}_{\boldsymbol{\delta}}^{1/2}$. Each eigenvector is orthogonal to others, so we have $E [ (\mathbf{u}_{n}^{T}\boldsymbol{\delta}_{0})(\mathbf{u}_{m}^{T}\boldsymbol{\delta}_{0})^{T} ] = \mathbf{u}_{n}^{T}\mathbf{u}_{m} = 0$ when n≠m which implies that Gaussian random variable $\mathbf{u}_{n}^{T}\boldsymbol{\delta}_{0}$ is statistically independent of $\mathbf{u}_{m}^{T}\boldsymbol{\delta}_{0}$. Moreover, $(\mathbf{u}_{n}^{T}\boldsymbol{\delta}_{0})^{2}$ from n=1 to D (the rank of Q) are independently chi-square distributed with one degree of freedom. Using Lemma D.1, the characteristic function of $x_{n} = \lambda_{n}(\mathbf{u}_{n}^{T}\boldsymbol{\delta}_{0})^{2}$ can be expressed as $M_{x_{n}}(t) = (1 - 2it\lambda_{n})^{ - 1/2}$. If one random variable can be expressed as the sum of several independent variables, its characteristic function equals the product of those specified characteristic functions. Consequently, the characteristic function of y follows, as shown in (36).