Fully Complex Valued Wavelet Network for Forecasting the Global Solar Irradiation

Abstract

Forecasting solar irradiation is very important to plane and size PV systems. In this paper, the fully complex valued wavelet network (FCWN) for forecasting the global solar irradiation is proposed. The complex valued gradient descent-learning algorithm is used to find the optimal complex-valued parameters of the network. An improved fully wavelet function is proposed and used as an activation function of the hidden neurons of the FCWN. The meteorological measured data of Tamanrasset city, Algeria (latitude: $22^{\circ }48$N; longitude: $05^{\circ }26$E) is used to validate the developed model. The hourly and the daily solar irradiations are forecasted using the multi input single output and the multi input multi output strategies. Several results are presented to test the feasibility and the performance of the FCWN for forecasting either daily or hourly solar irradiation. Results obtained throughout this paper show that the FCWN is a promising technique for forecasting daily and hourly solar irradiation.

Appendix

$f\left( z\right)$ proposed in this paper is a complex valued wavelet function. This one is a valid function for the complex valued network if it satisfied the five conditions cited before in the current paper and is should be admissible (Eqs. (2)–(5)) to be acceptable like wavelet function.

Proof

Let us take: $f\left( z\right) =u(x,y)+jv(x,y)$

It could prove that it is nonlinear for all $z\in C$, and the its real and imaginary parts are nonlinear and bounded for all $\hbox {x},y\in R$.

$f\left( z\right)$ is not constant, so it not entire.

$u\left( {x,y}\right)$ and $v\left( {x,y}\right)$ should be bounded, therefore $f\left( z\right)$ is bounded.

The partial derivatives $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial x}$, and $\frac{\partial v}{\partial y}$ exist, and they are given as follows:

Decomposing the wavelet function into real and imaginary parts, one can find:

$u(x,y)=\frac{\alpha _1 \beta _1 +\alpha _2 \beta _2 }{\beta _1^2 +\beta _2^2 }$ and $v(x,y)=\frac{\alpha _2 \beta _1 -\hbox {}\alpha _1 \beta _2 }{\beta _1^2 +\beta _2^2 }$

where

$$\begin{aligned} \alpha _1= & {} -\hbox {2}\left( {\hbox {e}^{\mathrm{2x}}+\hbox {e}^{\mathrm{-2x}}}\right) \cos (2y)+12 \end{aligned}$$

(28)

$$\begin{aligned} \alpha _2= & {} -\hbox {2}\left( {\hbox {e}^{\mathrm{2x}}-\hbox {e}^{\mathrm{-2x}}}\right) \sin (2y) \end{aligned}$$

(29)

$$\begin{aligned} \beta _1= & {} \left( {\hbox {e}^{\mathrm{3x}}+\hbox {e}^{\mathrm{-3x}}}\right) \cos (3y)+\left( {\hbox {e}^{\mathrm{x}}+\hbox {e}^{\mathrm{-x}}}\right) \cos (y) \end{aligned}$$

(30)

$$\begin{aligned} \beta _2= & {} \left( {\hbox {e}^{\mathrm{3x}}-\hbox {e}^{\mathrm{-3x}}}\right) \sin (3y)+\left( {\hbox {e}^{\mathrm{x}}-\hbox {e}^{\mathrm{-x}}}\right) \sin (y) \end{aligned}$$

(31)

$$\begin{aligned} \frac{\partial u(x,y)}{\partial y}= & {} \frac{\left( \beta _1^2 +\beta _2^2 \right) \left[ {\left( \beta _1 \frac{\partial \alpha _1 }{\partial y}+\beta _2 \frac{\partial \alpha _2 }{\partial y}\right) } \right] +\left( \alpha _1 \beta _2^2 \hbox {-}\alpha _1 \beta _1^2 \hbox {-2}\alpha _2 \beta _1 \beta _2 \right) \frac{\partial \beta _1 }{\partial y}}{} \nonumber \\&\frac{+\left( \alpha _2 \beta _1^2 -\alpha _2 \beta _2^2 -2\alpha _1 \beta _1 \beta _2 \right) \frac{\partial \beta _1 }{\partial y}}{\left( \beta _1^2 +\beta _2^2 \right) ^{2}} \end{aligned}$$

(32)

$$\begin{aligned} \frac{\partial v(x,y)}{\partial x}= & {} \frac{\left( \beta _1^2 +\beta _2^2 \right) \left[ {\left( \beta _1 \frac{\partial \alpha _2 }{\partial x}-\beta _2 \frac{\partial \alpha _1 }{\partial x}\right) } \right] +\left( \alpha _1 \beta _2^2 -\alpha _1 \beta _1^2 -2\alpha _2 \beta _1 \beta _2 \right) \frac{\partial \beta _2 }{\partial x}}{} \nonumber \\&\frac{-\left( \alpha _2 \beta _1^2 -\alpha _2 \beta _2^2 -2\alpha _1 \beta _1 \beta _2 \right) \frac{\partial \beta _1 }{\partial x}}{\left( \beta _1^2 +\beta _2^2 \right) ^{2}} \end{aligned}$$

(33)

One can remark from Eqs. (28)–(31) that:

$$\begin{aligned} \frac{\partial \alpha _1 }{\partial x}=\frac{\partial \alpha _2 }{\partial y}, \quad \frac{\partial \alpha _1 }{\partial y}=-\frac{\partial \alpha _2 }{\partial x}, \quad \frac{\partial \beta _1 }{\partial x}=\frac{\partial \beta _2 }{\partial y}\,\hbox {and}\,\frac{\partial \beta _1 }{\partial y}=-\frac{\partial \beta _2 }{\partial x} \end{aligned}$$

(34)

Using some algebra, we can find that: $\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y}$.

Using the same method, the relations of the two other partial derivatives $\frac{\partial u(x,y)}{\partial y}$ and $\frac{\partial v(x,y)}{\partial x}$ could be identified like: $\frac{\partial u(x,y)}{\partial y}=-\frac{\partial v(x,y)}{\partial x}$

They are continuous and bounded because they satisfy the two relations $\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}\ne \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}$ except if $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}=0$ and $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}=0$, because:

$$\begin{aligned} \frac{\partial u(x,y)}{\partial x}\frac{\partial v(x,y)}{\partial y}= -\frac{\partial u(x,y)}{\partial y} \frac{\partial v(x,y)}{\partial x}\ne \frac{\partial u(x,y)}{\partial y} \frac{\partial v(x,y)}{\partial x} \end{aligned}$$

(35)

So the function proposed in this paper could be used as fully activation function.

As proved before throughout this paper (Eqs. (2)– (5)) that it is an admissible function, therefore, the function could be used as wavelet activation function.