*3.2. ESPRIT Algorithm*

The signal *g*(*mi*) can be sampled as [8]

$$m\_i = k\_i \left[ T\_{02} + \frac{T\_{01} - T\_{02}}{T\_{01}} t \right], 0 \le t \le T\_{01} \tag{23}$$

The value of *T*01 can be selected from 100~200, *T*02 usually set between 1~3, and in this paper, *T*01 = 200, *T*02 = 2, *t* is an integer. According to the relationship *m*<sup>2</sup> = *m*2*i* + *k*2*i* , the first sampling in *m*-plane is *ki*31 + *T*202. Therefore, the approximation of *g*(*mi*) starts from *m* = *ki*31 + *T*202, and the parameter *p* in formula (20) should be set not less than *ki*31 + *T*202toinsuretheintegrationaccuracy.

Then the sampling sequence can be expressed as

$$\log(m\_i) = y(t) \approx \sum\_{l=1}^{pb} A\_l(l) \exp[B\_l(l)t] \tag{24}$$

The relation of *<sup>A</sup>*(*l*), *<sup>B</sup>*(*l*), and *a* (*l*), *b* (*l*) can be obtained from Equations (23) and (24).

$$\sum\_{l=1}^{pb} a\ (l) \exp[b\ (l)m\_l] = \sum\_{l=1}^{pb} A(l) \exp[B(l)t] \tag{25}$$

Then, the unknown coefficients *a* (*l*), *b* (*l*) in (24) are obtained.

$$b\ (l) = \frac{B\ (l)T\_{01}}{k\_i(T\_{01} - T\_{02})} \tag{26}$$

$$a\left(l\right) = A\left(l\right) \exp\left(b\left(l\right) \cdot k\_{i} \cdot T\_{02}\right) \tag{27}$$

For the sampling sequence *y*(0), *y*(1), ··· , *y*(*<sup>N</sup>* − <sup>1</sup>), a data matrix can be constructed:

$$\mathbf{Y} = \begin{bmatrix} y(0) & y(1) & \cdots & y(L) \\ y(1) & y(2) & \cdots & y(L+1) \\ \vdots & \vdots & \ddots & \vdots \\ y(N-L-1) & y(N-L) & \cdots & y(N-1) \end{bmatrix}\_{(N-L)\times(L+1)} \tag{28}$$

where *N* is the sample number, *L* is called the pencil parameter, and its value should be between *N*/3 and *N*/2 [22].The data matrix **Y** can be decomposed by SVD,

$$\mathbf{Y} = \mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^{H} = \begin{bmatrix} \mathbf{U}\_{\text{s}} & \mathbf{U}\_{\text{n}} \end{bmatrix} \begin{bmatrix} \boldsymbol{\Sigma}\_{\text{s}} & \mathbf{0} \\ \mathbf{0} & \boldsymbol{\Sigma}\_{\text{n}} \end{bmatrix} \begin{bmatrix} \mathbf{V}\_{\text{s}}^{H} \\ \mathbf{V}\_{\text{n}}^{H} \end{bmatrix} \tag{29}$$

where **U** is (*N* − *L*) × (*N* − *L*) orthogonal matrix, and **V** is (*L* + 1) × (*L* + 1) orthogonal matrix, **Σ** is (*N* − *L*) × (*L* + 1) diagonal matrix with main diagonal element *σl*, which is the singular value of matrix **Y**. For signals without noise, **Y** has *pb* non-zero singular values *σl* (*l*= 1,2,..., *pb*), and *pb* represents the highest order of the exponential signal of formula (24). If the signal contains noise, mode number *pb* can be recorded by setting a minimum threshold for *σl*.

Take out the first *pb* dominant right singular vectors in **V***s* matrix to form (*L* + 1) × *pb* matrix **V***p<sup>b</sup> s* .The last line of **V***p<sup>b</sup> s* is deleted to obtain *L* × *M* matrix **V**1; the first line of **V***p<sup>b</sup> s* is deleted to ge<sup>t</sup> *L* × *M* matrix **V**2. Construct a matrix **Ψ** [23]

$$\mathbf{Y} = \left(\mathbf{V}\_1^H \mathbf{V}\_1\right)^{-1} \mathbf{V}\_1^H \mathbf{V}\_2 \tag{30}$$

Find the eigenvalues *λl* of the matrix **Ψ**,

$$B\left(l\right) = \log(\lambda\_{\,\,l}), \, l = 1, 2, \dots, \,\_r pb \tag{31}$$

For *N* sampled signals,

$$\mathbf{A} = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \lambda\_1 & \lambda\_2 & \cdots & \lambda\_{pb} \\ \vdots & \vdots & \vdots & \vdots \\ \lambda\_1^{N-1} & \lambda\_2^{N-1} & \cdots & \lambda\_{pb}^{N-1} \end{pmatrix}, \mathbf{Y} = \begin{pmatrix} y(0) \\ y(1) \\ \vdots \\ y(N-1) \end{pmatrix}, \mathbf{A} = \begin{bmatrix} A(1) & A(2) & \cdots & A(pb) \end{bmatrix} \tag{32}$$
 
$$\text{and}$$

$$\mathbf{Y} = \lambda \mathbf{A} \tag{33}$$

According to the least square method, we can obtain

$$\mathbf{A} = \left(\lambda^H \lambda\right)^{-1} \lambda^H \mathbf{Y} \tag{34}$$

So far, the coefficients *A* (*l*) and *B* (*l*) are obtained. According to (22), (26), and (27), the second part of Equation (20) is solved, and the integrand of the first part is gained, and then DE rules are applied to compute the finite integral.
