Sum of Exponential Model for Fitting Data ^†

Abstract

As an approach to feature estimation, exponential fitting has attracted research interests in mathematical modeling. Semantic networks are used for numerous applications in computers, physics, and biology. However, such applications may have fitting troubles with various mathematical tools. Therefore, we present a novel method of fitting 2n data points of a signal to a sum of n exponential functions. The experiments proved that the proposed method operated well for linear and nonlinear functions, as its algorithm was straightforward, practical, and easy to determine. At the same time, the computational intricacy was considerably low, which has specific worth in use.

1. Introduction

Relevant information with the linear combinations of real and complex exponentials is pervasive in science and engineering applications. Given that Gaspard Riche de Prony developed an approach [1] to resolve the problem for equally spaced samples, numerous advancements, and applications have been proposed. We surveyed the most effective ones to explain their applications and experiences and to allow their application in various fields. A linear combination of exponentials was used in regular differential equations to explain the different physical processes. After being modeled by the remedy of a formula, a combination of exponentials provided valuable information such as decay rates or product residential or commercial properties in a physical system. Likewise, the exponential fitting had an excellent approximation on the compact of the domain with Fourier transformation in complex exponentials [2,3,4,5,6].

The purpose of this study was to present a method of fitting real signal data sampled at a period T in a set of 2n data points $\left\{x(0), x(T), x(2T), \cdots , x([2n-1]T)\right\}$. The data points to the s curve were composed of n exponential functions with unknown weights and exponents. Mathematically, this involved the solution of the following Equations (1)–(5).

$$x(kT)={\displaystyle \sum _{i=1}^n{{\displaystyle c}}_i{{\displaystyle e}}^{{{\displaystyle p}}_i(kT)}}$$

(1)

For the unknown C_i and P_i in the complex conjugate pairs (P_i is an imaginary number), Equation (1) represents a sum of sinusoids. This curve fitting can have many applications. For example, if x(t) represents the impulse response of a linear time-invariant system, and the Laplace transform of Equation (1) yields the transfer function of an n^th-order model of the system.

2. Curve Fitting Method

We let ${{\displaystyle \varphi }}_i$ denote ${{\displaystyle e}}^{{{\displaystyle p}}_i(kT)}$ and ${{\displaystyle x}}_i$ denote x(kT). Then, Equation (1) could be rewritten as:

$$\begin{array}{llll}{{\displaystyle c}}_1& +{{\displaystyle c}}_2& +\cdots +{{\displaystyle c}}_n& ={{\displaystyle x}}_0\\ {{\displaystyle c}}_1{{\displaystyle \varphi }}_1& +{{\displaystyle c}}_2{{\displaystyle \varphi }}_2& +\cdots +{{\displaystyle c}}_n{{\displaystyle \varphi }}_n& ={{\displaystyle x}}_1\\ {{\displaystyle c}}_1{{\displaystyle \varphi }}_1^2& +{{\displaystyle c}}_2{{\displaystyle \varphi }}_2^2& +\cdots +{{\displaystyle c}}_n{{\displaystyle \varphi }}_n^2& ={{\displaystyle x}}_2\\ & & ⋮& \\ {{\displaystyle c}}_1{{\displaystyle \varphi }}_1^{2n-1}& +{{\displaystyle c}}_2{{\displaystyle \varphi }}_2^{2n-1}& +\cdots +{{\displaystyle c}}_n{{\displaystyle \varphi }}_n^{2n-1}& ={{\displaystyle x}}_{2n-1}\end{array}$$

(2)

Equation (2) is explained by the following theorem.

Theorem 1.

The nonlinear equation, such as Equation (2), possesses a unique solution {c_k, N_k} (k = 1, 2, …, n) if and only with the following n × n matrix, which is nonsingular.

$$A\equiv \left[\begin{array}{cccc}{{\displaystyle x}}_0& {{\displaystyle x}}_1& \cdots & {{\displaystyle x}}_{n-1}\\ {{\displaystyle x}}_1& {{\displaystyle x}}_2& \cdots & {{\displaystyle x}}_n\\ ⋮& ⋮& & ⋮\\ {{\displaystyle x}}_{n-1}& {{\displaystyle x}}_n& \cdots & {{\displaystyle x}}_{2n-2}\end{array}\right]$$

(3)

The solution for N_k is given by the n distinct roots of the n^th degree polynomial equation.

$$\det \left[\begin{array}{ccccc}1& {{\displaystyle x}}_0& {{\displaystyle x}}_1& \cdots & {{\displaystyle x}}_{n-1}\\ {{\displaystyle \varphi }}_k& {{\displaystyle x}}_1& {{\displaystyle x}}_2& \cdots & {{\displaystyle x}}_n\\ {{\displaystyle \varphi }}_k^2& {{\displaystyle x}}_2& {{\displaystyle x}}_3& \cdots & {{\displaystyle x}}_{n+1}\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_k^n& {{\displaystyle x}}_n& {{\displaystyle x}}_{n+1}& \cdots & {{\displaystyle x}}_{2n-1}\end{array}\right]=0$$

(4)

The solution for c_k can then be given by:

$$A={{\displaystyle \left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_k& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_k^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_k^{2n-1}& {{\displaystyle \varphi }}_2^{2n-1}& {{\displaystyle \varphi }}_3^{2n-1}& \cdots & {{\displaystyle \varphi }}_n^{2n-1}\end{array}\right]}}^{-1}$$

(5)

$$\left[\begin{array}{c}{{\displaystyle c}}_1\\ {{\displaystyle c}}_2\\ {{\displaystyle c}}_3\\ ⋮\\ {{\displaystyle c}}_n\end{array}\right]=A\left[\begin{array}{c}{{\displaystyle x}}_0\\ {{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ ⋮\\ {{\displaystyle x}}_{n-1}\end{array}\right]$$

(6)

3. Proof

3.1. Sufficiency Part

It could be supposed that A was nonsingular; the first n1 equation of Equation (2) could be arranged as:

$$B=\left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_k& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_k^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_k^{2n-1}& {{\displaystyle \varphi }}_2^{2n-1}& {{\displaystyle \varphi }}_3^{2n-1}& \cdots & {{\displaystyle \varphi }}_n^{2n-1}\end{array}\right]$$

(7)

$$B \left[\begin{array}{c}{{\displaystyle c}}_1\\ {{\displaystyle c}}_2\\ {{\displaystyle c}}_3\\ ⋮\\ {{\displaystyle c}}_n\end{array}\right]=\left[\begin{array}{c}{{\displaystyle x}}_0\\ {{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ ⋮\\ {{\displaystyle x}}_{n-1}\end{array}\right]$$

(8)

We let the columns of the left-most matrix in Equation (8) be denoted as:

$${{\displaystyle v}}_k\equiv {{\displaystyle \left[\begin{array}{ccccc}1& {{\displaystyle \varphi }}_k& {{\displaystyle \varphi }}_k^2& \cdots & {{\displaystyle \varphi }}_k^n\end{array}\right]}}^T,k=1, 2, \cdots , n$$

(9)

and let ${{\displaystyle x}}_0\equiv {{\displaystyle \left[\begin{array}{cccc}{{\displaystyle x}}_0& {{\displaystyle x}}_1& \cdots & {{\displaystyle x}}_n\end{array}\right]}}^T$. Then, Equation (8) showed that x₀ was a linear combination of {v₁, v₂, …, v_n}.

Next, if the set of n + 1 consecutive equations in Equation (2) was considered as the starting point with the second equation, they could be rearranged as:

$$\left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_1& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_1^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_1^n& {{\displaystyle \varphi }}_2^n& {{\displaystyle \varphi }}_3^n& \cdots & {{\displaystyle \varphi }}_n^n\end{array}\right] \left[\begin{array}{c}{{\displaystyle c}}_1{{\displaystyle \varphi }}_1\\ {{\displaystyle c}}_2{{\displaystyle \varphi }}_2\\ {{\displaystyle c}}_3{{\displaystyle \varphi }}_3\\ ⋮\\ {{\displaystyle c}}_n{{\displaystyle \varphi }}_n\end{array}\right]=\left[\begin{array}{c}{{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ {{\displaystyle x}}_3\\ ⋮\\ {{\displaystyle x}}_{n+1}\end{array}\right]$$

(10)

Equation (10) shows that x₁ ≡ [x₁, x₂, …, x_n+1]^T was a linear combination of {v₁, v₂, …, v_n}.

Similarly, if we considered the set of n + 1 consecutive equations in Equation (2) starting with the third equation, we could see that x₂ ≡ [x₂, x₃, …, x_n+2]^T was a linear combination of {v₁, v₂, …, v_n}.

This continued until the last n + 1 of Equation (2) was taken, from which it was shown that x_n−1 ≡ [x_n−1, x_n, …x_2n+1]^T was a linear combination of {v₁, v₂, …, v_n}.

Equation (3) of the Theorem implies that the vectors {x₀, x₁, …x_n−1} are linear and independent of each other. Hence, they span an n-dimensional subspace in an (n + l) dimensional Euclidean space. This subspace must be the same as the one spanned by the vectors {v₁, v₂, …, v_n} since each x_i, i = 0, 1, …, n−1 is a linear combination of the set {v₁, v₂, …, v_n}. It follows that the vectors {v₁, v₂, …, v_n} are linearly independent and that, from Equation (7) of v_k (k = 0, 1, …, n), they must be distinct.

Moreover, each v_k is a linear combination of {x₀, x₁, …x_n+1}. This implies:

$${{\displaystyle v}}_k=\left[\begin{array}{c}1\\ {{\displaystyle \varphi }}_k\\ {{\displaystyle \varphi }}_k^2\\ ⋮\\ {{\displaystyle \varphi }}_k^n\end{array}\right]={{\displaystyle d}}_1\left[\begin{array}{c}{{\displaystyle x}}_0\\ {{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ ⋮\\ {{\displaystyle x}}_n\end{array}\right]+{{\displaystyle d}}_2\left[\begin{array}{c}{{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ {{\displaystyle x}}_3\\ ⋮\\ {{\displaystyle x}}_{n+1}\end{array}\right]+\cdots +{{\displaystyle d}}_n\left[\begin{array}{c}{{\displaystyle x}}_{n-1}\\ {{\displaystyle x}}_n\\ {{\displaystyle x}}_{n+1}\\ ⋮\\ {{\displaystyle x}}_{2n-1}\end{array}\right]$$

(11)

Equation (11) could be rearranged as an (n + 1) × (n + 1) equation system.

$$\left[\begin{array}{ccccc}1& {{\displaystyle x}}_0& {{\displaystyle x}}_1& \cdots & {{\displaystyle x}}_{n-1}\\ {{\displaystyle \varphi }}_k& {{\displaystyle x}}_1& {{\displaystyle x}}_2& \cdots & {{\displaystyle x}}_n\\ {{\displaystyle \varphi }}_k^2& {{\displaystyle x}}_2& {{\displaystyle x}}_3& \cdots & {{\displaystyle x}}_{n+1}\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_k^n& {{\displaystyle x}}_n& {{\displaystyle x}}_{n+1}& \cdots & {{\displaystyle x}}_{2n-1}\end{array}\right] \left[\begin{array}{c}-1\\ {{\displaystyle d}}_1\\ {{\displaystyle d}}_2\\ ⋮\\ {{\displaystyle d}}_n\end{array}\right]=0$$

(12)

Since the solution of Equation (12) was nontrivial, the determinant of the square matrix had to vanish, leading to Equation (4) which was an n^th-degree polynomial equation in N_k because the coefficient of the n^th power term of N_k could be seen from Equations (3) and (12) to be (−1)ⁿ det A, which was assumed to be nonzero. The n roots N_k of Equation (4) must be distinct because each N_k had to satisfy Equation (4) and be distinct. Having obtained the distinct values of N_k, k = 0, 1, …, n, c_k could be given by the first n equations (Equation (2)), which led to Equation (6) in the Theorem.

As for the uniqueness of the solution, since every solution {c_k, N_k} (k = 1, 2, …, n) had to satisfy Equation (4), according to the above arguments, Equation (4) produced exactly n distinct values for N_k, and the solution of Equation (2) was unique.

3.2. Necessity Part

Suppose Equation (2) has a unique solution {c_k, N_k} (k = 1, 2, …, n). This implies the following.

(i) N_k must be distinct from each other; otherwise, non-unique combinations of c_k in Equation (2) exist and are fulfilled.

(ii) None of c_k vanishes; otherwise, the value of N_k associated with a vanishing c_k becomes non-unique.

The first n of Equation (2) gave:

$$C={{\displaystyle \left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_1& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_1^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_1^{n-1}& {{\displaystyle \varphi }}_2^{n-1}& {{\displaystyle \varphi }}_3^{n-1}& \cdots & {{\displaystyle \varphi }}_n^{n-1}\end{array}\right] }}^{-1}$$

(13)

$$\left[\begin{array}{c}{{\displaystyle c}}_1\\ {{\displaystyle c}}_2\\ {{\displaystyle c}}_3\\ ⋮\\ {{\displaystyle c}}_n\end{array}\right]=C\left[\begin{array}{c}{{\displaystyle x}}_0\\ {{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ ⋮\\ {{\displaystyle x}}_{n-1}\end{array}\right]$$

(14)

The next n of Equation (2), starting with the second equation, gave:

$$D={{\displaystyle \left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_1& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_1^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_1^{n-1}& {{\displaystyle \varphi }}_2^{n-1}& {{\displaystyle \varphi }}_3^{n-1}& \cdots & {{\displaystyle \varphi }}_n^{n-1}\end{array}\right] }}^{-1}$$

(15)

$$\left[\begin{array}{c}{{\displaystyle c}}_1{{\displaystyle \varphi }}_1\\ {{\displaystyle c}}_2{{\displaystyle \varphi }}_2\\ {{\displaystyle c}}_3{{\displaystyle \varphi }}_3\\ ⋮\\ {{\displaystyle c}}_n{{\displaystyle \varphi }}_n\end{array}\right]=D\left[\begin{array}{c}{{\displaystyle x}}_1\\ {{\displaystyle x}}_2\\ {{\displaystyle x}}_3\\ ⋮\\ {{\displaystyle x}}_n\end{array}\right]$$

(16)

This proceeded until the set of n in the consecutive Equation (2), starting with the n^th equation, was reached.

$$E={{\displaystyle \left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_1& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_1^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_1^{n-1}& {{\displaystyle \varphi }}_2^{n-1}& {{\displaystyle \varphi }}_3^{n-1}& \cdots & {{\displaystyle \varphi }}_n^{n-1}\end{array}\right] }}^{-1}$$

(17)

$$\left[\begin{array}{c}{{\displaystyle c}}_1{{\displaystyle \varphi }}_1^{n-1}\\ {{\displaystyle c}}_2{{\displaystyle \varphi }}_2^{n-1}\\ {{\displaystyle c}}_3{{\displaystyle \varphi }}_3^{n-1}\\ ⋮\\ {{\displaystyle c}}_n{{\displaystyle \varphi }}_n^{n-1}\end{array}\right]=E\left[\begin{array}{c}{{\displaystyle x}}_{n-1}\\ {{\displaystyle x}}_2\\ {{\displaystyle x}}_{n+1}\\ ⋮\\ {{\displaystyle x}}_{2n-2}\end{array}\right]$$

(18)

Combining Equations (14) to (18) yielded:

$$F={{\displaystyle \left[\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {{\displaystyle \varphi }}_1& {{\displaystyle \varphi }}_2& {{\displaystyle \varphi }}_3& \cdots & {{\displaystyle \varphi }}_n\\ {{\displaystyle \varphi }}_1^2& {{\displaystyle \varphi }}_2^2& {{\displaystyle \varphi }}_3^2& \cdots & {{\displaystyle \varphi }}_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {{\displaystyle \varphi }}_1^{n-1}& {{\displaystyle \varphi }}_2^{n-1}& {{\displaystyle \varphi }}_3^{n-1}& \cdots & {{\displaystyle \varphi }}_n^{n-1}\end{array}\right] }}^{-1}$$

(19)

$$G=\left[\begin{array}{ccccc}{{\displaystyle x}}_0& {{\displaystyle x}}_1& {{\displaystyle x}}_2& \cdots & {{\displaystyle x}}_{n-1}\\ {{\displaystyle x}}_1& {{\displaystyle x}}_2& {{\displaystyle x}}_3& \cdots & {{\displaystyle x}}_n\\ {{\displaystyle x}}_2& {{\displaystyle x}}_3& {{\displaystyle x}}_4& \cdots & {{\displaystyle x}}_{n+1}\\ ⋮& ⋮& ⋮& \cdots & ⋮\\ {{\displaystyle x}}_{n-1}& {{\displaystyle x}}_n& {{\displaystyle x}}_{n+1}& \cdots & {{\displaystyle x}}_{2n-2}\end{array}\right]$$

(20)

$$\left[\begin{array}{cccc}{{\displaystyle c}}_1& {{\displaystyle c}}_1{{\displaystyle \varphi }}_1& \cdots & {{\displaystyle c}}_1{{\displaystyle \varphi }}_1^{n-1}\\ {{\displaystyle c}}_2& {{\displaystyle c}}_2{{\displaystyle \varphi }}_2& \cdots & {{\displaystyle c}}_2{{\displaystyle \varphi }}_2^{n-1}\\ {{\displaystyle c}}_3& {{\displaystyle c}}_3{{\displaystyle \varphi }}_3& \cdots & {{\displaystyle c}}_3{{\displaystyle \varphi }}_3^{n-1}\\ ⋮& ⋮& \cdots & ⋮\\ {{\displaystyle c}}_n& {{\displaystyle c}}_n{{\displaystyle \varphi }}_n& \cdots & {{\displaystyle c}}_n{{\displaystyle \varphi }}_n^{n-1}\end{array}\right]=FG$$

(21)

4. Examples

Consider the signal:

$$x(t)=2e^{-2t}-3e^t$$

(22)

Sampling this signal at a sampling period T = 1 yielded Equation (2) with c₁ = 2, c₂ = −3, N₁ = e⁻² = 0.13533283, N₂ = e¹ = 2.718281828, x₀ = −1, x₁ = −7.88417491, x₂ = −22.130537, and x₃ = −60.2516532. Let us reverse this process. After sampling four consecutive points of the signal x(t) at a uniform sampling period T = 1, we obtained the values of {x₀, x₁, x₂, x₃}, as indicated above, which could be solved for {c₁, c₂, N₁, N₂}. From Equation (2), we obtained:

$$c_1+c_2=-1$$

(23)

$$c_1{\varphi }_1+c_2{\varphi }_2=-7.88417491$$

(24)

$$c_1{\varphi }_1^2+c_2{\varphi }_2^2=-22.130537$$

(25)

$$c_1{\varphi }_1^3+c_2{\varphi }_2^3=-60.2516532$$

(26)

First, we could see that:

$$A=\left[\begin{array}{cc}{x}_0& x_1\\ x_1& x_2\end{array}\right]=\left[\begin{array}{cc}-1& -7.88417491\\ -7.88417491& -22.130537\end{array}\right]$$

(27)

was nonsingular. N_k was obtained as the root of Equation (4), which, in this case, led to:

$$\begin{array}{l}\det \left[\begin{array}{ccc}1& -1& -7.88417491\\ {\varphi }_k& -7.88417491& -22.130537\\ {\varphi }_k^2& -\mathrm{22.130.537}& -60.2516532\end{array}\right]\\ =-40.029677{\varphi }_k^2+114.2293714{\varphi }_k-14.7260955\\ =0\end{array}$$

(28)

This yielded N₁ = 0.135335283 and N₂ = 2.718281828. Substituting these values into Equation (6) provided:

$$\left[\begin{array}{cc}1& 1\\ 0.13533528& 2.71828182\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\left[\begin{array}{c}-1\\ -7.88417491\end{array}\right]$$

(29)

from which we obtained c₁ = 2 and c₂ = −3. From known N₁ and N₂, p₁ = −2 and p₂ = 1 were obtained according to Ф_i = e^piT.

5. Conclusions

The problem of fitting 2n data points to a curve consisting of n exponential functions was solved. The exponential functions were complex in general, with sinusoids being a special case. The curve-fitting problem was solved by a system of nonlinear algebraic equations. An example has been given to illustrate the procedure of this method.