Question

Consider the \(R L\) network shown in the figure. Assume that the loop currents \(I_{1}\) and \(I_{2}\) are zero until a voltage source \(V_{S}(t)\), having the polarity shown, is turned on at time \(t=0 .\) Applying Kirchhoff's voltage law to each loop, we obtain the equations $$ \begin{aligned} -V_{S}(t)+L_{1} \frac{d I_{1}}{d t}+R_{1} I_{1}+R_{3}\left(I_{1}-I_{2}\right) &=0 \\ R_{3}\left(I_{2}-I_{1}\right)+R_{2} I_{2}+L_{2} \frac{d I_{2}}{d t} &=0 \end{aligned} $$ (a) Formulate the initial value problem for the loop currents, \(\left[\begin{array}{l}I_{1}(t) \\ I_{2}(t)\end{array}\right]\), assuming that $$ L_{1}=L_{2}=0.5 H, \quad R_{1}=R_{2}=1 k \Omega, \quad \text { and } \quad R_{3}=2 k \Omega . $$ (b) Determine a fundamental matrix for the associated linear homogeneous system. (c) Use the method of variation of parameters to solve the initial value problem for the case where \(V_{S}(t)=1\) for \(t>0\).

Short answer

(b) What are the eigenvectors corresponding to these eigenvalues? (c) Using the eigenvalues and eigenvectors found earlier, write the fundamental matrix X(t) used in solving the given initial value problem. (d) Write the final solution for the particular solution of the loop currents given the voltage source function of V_S(t) = 1.

Step-by-step solution

Write the system of equations in matrix form.

We can rewrite the given system of equations as: $$\frac{d}{dt}\begin{bmatrix} I_{1} \\ I_{2} \end{bmatrix} = \begin{bmatrix} -2k - 1k & 2k \\ 2k & -2k - 1k \end{bmatrix}\begin{bmatrix} I_{1} \\ I_{2} \end{bmatrix} + \begin{bmatrix} -1 \\ 0 \end{bmatrix} V_S(t).$$ Now we'll plug in the given values for \(L_1, L_2, R_1, R_2,\) and \(R_3\): $$\frac{d}{dt}\begin{bmatrix} I_{1} \\ I_{2} \end{bmatrix} = \frac{1}{0.5}\begin{bmatrix} -3 & 2 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} I_{1} \\ I_{2} \end{bmatrix} + \begin{bmatrix} -1 \\ 0 \end{bmatrix} V_S(t).$$ So, the initial value problem for the loop currents is: $$\frac{d}{dt}\boldsymbol{I}(t) = 2A \boldsymbol{I}(t) + \boldsymbol{b}V_S(t),$$ where $$A = \begin{bmatrix} -3 & 2 \\ 2 & -3 \end{bmatrix}, \quad \boldsymbol{b} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}, \quad \text{and} \quad \boldsymbol{I}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ (b) Fundamental matrix for the associated linear homogeneous system

Calculate the eigenvalues and eigenvectors of the matrix A.

First, we find the eigenvalues of the matrix \(A\) by solving the characteristic equation: $$\text{det}(A-\lambda I) = \begin{vmatrix} -3-\lambda & 2 \\ 2 & -3-\lambda \end{vmatrix}=(-3-\lambda)^2-4(2)=\lambda^2+6\lambda+1=0.$$ The eigenvalues of \(A\) are: $$\lambda_1=-3+2\sqrt{2}, \quad \lambda_2=-3-2\sqrt{2}.$$ Now we find the eigenvectors corresponding to these eigenvalues: For \(\lambda_1\): $$(A-\lambda_1 I)\boldsymbol{x}=\begin{bmatrix} -3-(-3+2\sqrt{2}) & 2 \\ 2 & -3-(-3+2\sqrt{2})\end{bmatrix}\boldsymbol{x}=\begin{bmatrix} 2\sqrt{2} & 2 \\ 2 & 2\sqrt{2} \end{bmatrix}\boldsymbol{x}=\boldsymbol{0}.$$ A possible eigenvector is \(\boldsymbol{x}_1=\begin{bmatrix} 1 \\ \sqrt{2} \end{bmatrix}.\) For \(\lambda_2\): $$(A-\lambda_2 I)\boldsymbol{x}=\begin{bmatrix} -3-(-3-2\sqrt{2}) & 2 \\ 2 & -3-(-3-2\sqrt{2})\end{bmatrix}\boldsymbol{x}=\begin{bmatrix} -2\sqrt{2} & 2 \\ 2 & -2\sqrt{2} \end{bmatrix}\boldsymbol{x}=\boldsymbol{0}.$$ A possible eigenvector is \(\boldsymbol{x}_2=\begin{bmatrix} 1 \\ -\sqrt{2} \end{bmatrix}.\)

Determine the fundamental matrix.

Now that we have the eigenvalues and eigenvectors of \(A\), we can write the fundamental matrix \(X(t)\) of the associated linear homogeneous system as: $$X(t) = \begin{bmatrix} e^{(-3+2\sqrt{2})t} & e^{(-3-2\sqrt{2})t} \\ \sqrt{2}e^{(-3+2\sqrt{2})t} & -\sqrt{2}e^{(-3-2\sqrt{2})t} \end{bmatrix}.$$ (c) Solving the initial value problem using variation of parameters

Calculate the inverse of the fundamental matrix.

The inverse matrix \(X^{-1}(t)\) is given by: $$X^{-1}(t)=\frac{1}{2}\begin{bmatrix} -\sqrt{2}e^{(3+2\sqrt{2})t} & e^{(3+2\sqrt{2})t} \\ \sqrt{2}e^{(3-2\sqrt{2})t} & e^{(3-2\sqrt{2})t} \end{bmatrix}.$$

Find the particular solution using the method of variation of parameters.

We are given that \(V_S(t)=1\) for \(t>0\). The particular solution is given by: $$\boldsymbol{I}_p(t) = X(t)\int X^{-1}(t)\boldsymbol{b}V_s(t)dt$$ Because \(V_S(t)=1\) for \(t>0\), we have: $$\boldsymbol{I}_p(t) = X(t)\int X^{-1}(t)\boldsymbol{b}dt = X(t)\int \left[ \frac{1}{2}\begin{bmatrix} -\sqrt{2}e^{(3+2\sqrt{2})t} & e^{(3+2\sqrt{2})t} \\ \sqrt{2}e^{(3-2\sqrt{2})t} & e^{(3-2\sqrt{2})t} \end{bmatrix}\right] \begin{bmatrix} -1 \\ 0 \end{bmatrix} dt$$ This gives: $$\boldsymbol{I}_p(t) = X(t)\begin{bmatrix} \frac{1}{2(3+2\sqrt{2})}\left(1-e^{(3+2\sqrt{2})t}\right) \\ -\frac{1}{2(3-2\sqrt{2})}\left(1-e^{(3-2\sqrt{2})t}\right) \end{bmatrix} = \begin{bmatrix} e^{(-3+2\sqrt{2})t} & e^{(-3-2\sqrt{2})t} \\ \sqrt{2}e^{(-3+2\sqrt{2})t} & -\sqrt{2}e^{(-3-2\sqrt{2})t} \end{bmatrix} \begin{bmatrix} \frac{1}{2(3+2\sqrt{2})}\left(1-e^{(3+2\sqrt{2})t}\right) \\ -\frac{1}{2(3-2\sqrt{2})}\left(1-e^{(3-2\sqrt{2})t}\right) \end{bmatrix}$$ This is the particular solution for the loop currents given the voltage source function of \(V_S(t) = 1\).

Key concepts

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) is a fundamental principle used in the analysis of electrical circuits. It states that the total sum of voltages around any closed loop in a circuit must equal zero. When applying KVL, we consider both the source voltages and the voltage drops across components such as resistors and inductors.

In the context of differential equations and the circuit described in our exercise, KVL is used to set up a system of equations that models the behavior of loop currents over time. For each loop in the given R L network, KVL provides us with an equation that incorporates the rate of change of current (due to inductors), resistance values, and any external voltages. The result is a set of coupled first-order linear ordinary differential equations (ODEs) that describe the dynamic response of the circuit.

Initial Value Problem

The initial value problem (IVP) in the context of differential equations refers to finding a function (or functions) that satisfies a given ODE, subject to initial conditions at a certain starting point (hence 'initial' value). In electrical circuits with inductors and capacitors, the IVP typically represents the currents and voltages at the moment a source is switched on or some external influence is applied.

For our exercise, the initial value problem is formulated based on the coupled first-order linear ODEs derived from KVL and the initial conditions given, which are the currents in the loops that are zero at time t=0. These initial values are crucial for predicting the future behavior of the circuit currents I₁ and I₂ as the circuit evolves over time from the moment the voltage source is turned on.

Fundamental Matrix

A fundamental matrix is a concept from linear algebra and systems of linear differential equations. It plays a vital role in solving systems of linear homogeneous ODEs. The fundamental matrix is built from the solutions to the homogeneous equation. Specifically, it is an assortment of independent solutions to the system, placed into a matrix format.

In solving linear differential equation systems, particularly for our electrical circuit problem, the fundamental matrix expresses how various initial conditions would evolve over time. Given the elements of the fundamental matrix typically involve exponential functions of time, they describe how initial perturbations to the system's state—such as the loop currents I₁ and I₂—will exponentially decay or grow with time. This matrix becomes key for constructing the general solution to the differential equations and is also used when employing methods such as variation of parameters.

Variation of Parameters

Variation of parameters is an analytical method for obtaining the particular solution of non-homogeneous differential equations. The method involves finding functions that adjust the coefficients of the general solution of the associated homogeneous problem to 'fit' the non-homogeneous one.

Using this method to solve our exercise involves two main steps: we first determine the fundamental matrix associated with the homogeneous part of the system. Then, the variation of parameters technique adapts this matrix to accommodate the non-homogeneous part, introduced by the voltage source V_S(t). This approach yields a particular solution that, combined with the general homogeneous solution, accounts for the initial conditions and provides a complete description of the current behavior over time. It’s a powerful technique for circuits as it allows for the inclusion of real-world sources and inputs, which are often time-dependent.