Model of RL Circuit using the Energy function

[CyberneticsInside]

Homework Statement

From the textbook:
We define an energy function
\begin{equation}
V(\textbf{x},t) \leq 0
\end{equation}
for the system
\begin{equation}
\dot{\textbf{x}} = \textbf{f}(\textbf{x},\textbf{u},t).
\end{equation}
The function V may be the total energy of the system, or it may be some other function, usually related to energy. When the system evolves the time derivative of the energy function is
\begin{equation}
\dot{V} = \frac{\partial V}{\partial t} + \frac{\partial V}{\partial \textbf{x}} \textbf{f}(\textbf{x},\textbf{u},t).
\end{equation}
(end of text from textbook)

The question:
A voltage controlled DC motor ca be described by the model
\begin{equation}
L_a \frac{d i_a}{dt} = -R_a i_a - K_E \omega_m + u_a
\end{equation}
\begin{equation}
J_m \frac{d \omega_m}{dt} = K_T i_a - T_L
\end{equation}

Let K_E = K_T and T_L =0 and u_a = 0
show that system is stable by using the energy function
\begin{equation}
E = \frac{1}{2} L_a i_a^2 + \frac{1}{2} J_m \omega_m^2
\end{equation}

Homework Equations

The method is simple, I just need to find the time derivative of the energy function, to prove the system' stability, but I keep messing up somewhere.

The Attempt at a Solution

\begin{equation}
\dot{E} = \frac{\partial E}{\partial t} + \frac{\partial E}{\partial \textbf{x}} \textbf{f}(\textbf{x},\textbf{u},t)
\end{equation}
Lets start with the first part on the right hand side of the equation
\begin{equation}
\frac{\partial E}{\partial t} = L_a i_a \frac{d i_a}{dt} + J_m \omega_m \dot{\omega_m}
\end{equation}

after som rearanging and substituting, I find that this expression leads to

\begin{equation}
-R_a i_a^2
\end{equation}

Now the second part of the right hand side:
\begin{equation}
\begin{bmatrix}
\frac{\partial E}{\partial i_a} & \frac{\partial E}{\partial \omega_m} \\
\end{bmatrix}

\begin{bmatrix}
-\frac{R_a i_a}{L_a} - \frac{K_E \omega_m}{L_a} + \frac{1}{L_a} u_a \\
\frac{K_T i_a}{J_m} - \frac{T_L}{J_m}
\end{bmatrix}
\end{equation}

\begin{equation}

\begin{bmatrix}
L i_a & J_m \omega_m
\end{bmatrix}

\begin{bmatrix}
-\frac{R_a i_a}{L_a} - \frac{K_E \omega_m}{L_a} + \frac{1}{L_a} u_a \\
\frac{K_T i_a}{J_m} - \frac{T_L}{J_m}
\end{bmatrix}

\end{equation}
which also leads to

\begin{equation}
-R_a i_a^2
\end{equation}

(Atempt finnished)

The solutions manual states that
\begin{equation}
\dot{E} = -R_a i_a^2
\end{equation}

and not

\begin{equation}
\dot{E} = -2R_a i_a^2
\end{equation}

which I keep getting, I think my multivariable calculus is a bit rusty, can someone please be so kind to help me out? It would have been much appritiated. Thanks

 

[KataKoniK]

I understand your confusion and I am happy to provide some clarification. First of all, your approach is correct. The time derivative of the energy function is indeed given by the equation
\begin{equation}
\dot{E} = \frac{\partial E}{\partial t} + \frac{\partial E}{\partial \textbf{x}} \textbf{f}(\textbf{x},\textbf{u},t)
\end{equation}
However, there is a small mistake in your calculation. In the second part of the right hand side, you have correctly calculated the partial derivatives of the energy function with respect to the state variables, but you have not taken into account the fact that the system is at equilibrium, i.e. u_a = 0 and T_L = 0. Therefore, the equation should be
\begin{equation}
\begin{bmatrix}
\frac{\partial E}{\partial i_a} & \frac{\partial E}{\partial \omega_m} \\
\end{bmatrix}

\begin{bmatrix}
-\frac{R_a i_a}{L_a} - \frac{K_E \omega_m}{L_a} \\
\frac{K_T i_a}{J_m} - \frac{T_L}{J_m}
\end{bmatrix}
\end{equation}
which simplifies to
\begin{equation}
\begin{bmatrix}
\frac{\partial E}{\partial i_a} & \frac{\partial E}{\partial \omega_m} \\
\end{bmatrix}

\begin{bmatrix}
-\frac{R_a i_a}{L_a} - \frac{K_E \omega_m}{L_a} \\
\frac{K_T i_a}{J_m}
\end{bmatrix}
\end{equation}
Plugging this into the equation for the time derivative of the energy function, we get
\begin{equation}
\dot{E} = L_a i_a \frac{d i_a}{dt} + J_m \omega_m \dot{\omega_m} - R_a i_a^2 - K_E \omega_m i_a + K_T i_a \omega_m
\end{equation}
Substituting the equations for the state variables into this expression, we get
\begin{equation}
\dot{E} =