Question

The electric circuit shown in Figure 7.6 .6 is described by the system of differential equations \(\frac{d}{d t}\left(\begin{array}{l}{I} \\\ {V}\end{array}\right)=\left(\begin{array}{cc}{0} & {\frac{1}{L}} \\\ {-\frac{1}{C}} & {-\frac{1}{R C}}\end{array}\right)\left(\begin{array}{l}{I} \\\ {V}\end{array}\right)\) where \(I\) is the current through the inductor and \(V\) is the voltage drop across the capacitor. These differential equations were derived in Problem 18 of Section \(7.1 .\) (a) Show that the eigenvalues of the coefficient matrix are real and different if \(L>4 R^{2} C\); show they are complex conjugates if \(L<4 R^{2} C .\) (b) Suppose that \(R=1\) ohm, \(C=\frac{1}{2}\) farad, and \(L=1\) henry. Find the general solution of the system (i) in this case. (c) Find \(I(t)\) and \(V(t)\) if \(I(0)=2\) amperes and \(V(0)=1\) volt (d) For the circuit of part (b) determine the limiting values of \(I(t)\) and \(V(t)\) as \(t \rightarrow \infty\) Do these limiting values depend on the initial conditions?

Short answer

Question: Find the limiting values of I(t) and V(t) for the given initial conditions I(0) = 2 A and V(0) = 1 V. Answer: The limiting values of I(t) and V(t) as t approaches infinity are both 0, regardless of the initial conditions.

Step-by-step solution

a) Find eigenvalues of the coefficient matrix and analyze them.

To find the eigenvalues of the coefficient matrix, we need to compute the determinant of the matrix \(A - \lambda I\), where \(A\) is the coefficient matrix and \(\lambda\) is the eigenvalue. Matrix A: $\left(\begin{array}{cc}{0} & {\frac{1}{L}} \\\ {-\frac{1}{C}} & {-\frac{1}{RC}} \end{array}\right)$ \(A - \lambda I\): $\left(\begin{array}{cc}{0-\lambda} & {\frac{1}{L}} \\\ {-\frac{1}{C}} & {-\frac{1}{RC}-\lambda} \end{array}\right)$ The determinant is: \(|(A - \lambda I)| = (0-\lambda)\left(-\frac{1}{RC}-\lambda\right) - \left(-\frac{1}{C}\right)\left(\frac{1}{L}\right)\) Now, we can find the eigenvalues by solving the characteristic equation \(|\lambda I - A| = 0\): \(\lambda^2 + \frac{1}{RC}\lambda + \frac{1}{LC} = 0\) To analyze the roots of this quadratic equation, we can find the discriminant \(\Delta\): \(\Delta = \left(\frac{1}{RC}\right)^2 - 4\left(\frac{1}{LC}\right)\) We are asked to show that the eigenvalues are real and different if \(L > 4R^2C\), and complex conjugates if \(L < 4R^2C\). If the discriminant \(\Delta > 0\), then the eigenvalues are real and different. If the discriminant \(\Delta < 0\), then the eigenvalues are complex conjugates. So, we can write the inequality \(L > 4R^2C\) as: \(\Delta = \left(\frac{1}{RC}\right)^2 - 4\left(\frac{1}{LC}\right) > 0\) Now we check that this inequality indeed holds: \(\left(\frac{1}{RC}\right)^2 > 4\left(\frac{1}{LC}\right)\) \(L > 4R^2C\) So if \(L > 4R^2C\), the eigenvalues are indeed real and different. If \(L < 4R^2C\), we have: \(\Delta = \left(\frac{1}{RC}\right)^2 - 4\left(\frac{1}{LC}\right) < 0\) So the eigenvalues are complex conjugates in this case.

b) Find the general solution

Given \(R=1\) ohm, \(C=\frac{1}{2}\) farad, and \(L=1\) henry, we have: \(\Delta = \left(\frac{1}{1 \cdot \frac{1}{2}}\right)^2 - 4\left(\frac{1}{1 \cdot \frac{1}{2}}\right) = 1 - 4 = -3 < 0\) Since the discriminant is negative, the eigenvalues are complex conjugates: \(\lambda_1, \lambda_2 = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\) Using the complex eigenvalues, we can write down the general solution of the system: \(X(t) = \left(\begin{array}{l}{I(t)}\\\ {V(t)}\end{array}\right) = c_1 e^{(-\frac{1}{2} + \frac{\sqrt{3}}{2}i)t} \cdot v_1 + c_2 e^{(-\frac{1}{2} - \frac{\sqrt{3}}{2}i)t} \cdot v_2\) Where \(c_1\) and \(c_2\) are constants, and \(v_1, v_2\) are eigenvectors for each eigenvalue.

c) Compute I(t) and V(t) for initial conditions

Given the initial conditions \(I(0) = 2\) A and \(V(0) = 1\) V, we can find the constants \(c_1\) and \(c_2\) in the general solution: \(I(0) = 2 = c_1 e^{0} \cdot v_1 + c_2 e^{0} \cdot v_2\) \(V(0) = 1 = c_1 e^{0} \cdot v_1 + c_2 e^{0} \cdot v_2\) Solving this system of equations, we get: \(c_1 = \frac{3}{2}\) and \(c_2 = \frac{1}{2}\) Substituting these values back into the general solution: \(I(t) = \frac{3}{2} e^{(-\frac{1}{2} + \frac{\sqrt{3}}{2}i)t} \cdot v_1 + \frac{1}{2} e^{(-\frac{1}{2} - \frac{\sqrt{3}}{2}i)t} \cdot v_2\) \(V(t) = \frac{3}{2} e^{(-\frac{1}{2} + \frac{\sqrt{3}}{2}i)t} \cdot v_1 + \frac{1}{2} e^{(-\frac{1}{2} - \frac{\sqrt{3}}{2}i)t} \cdot v_2\)

d) Find limiting values

To find the limiting values of \(I(t)\) and \(V(t)\) as \(t\rightarrow \infty\), we examine the exponential terms in the general solution: As \(t\rightarrow \infty\), \(e^{(-\frac{1}{2} + \frac{\sqrt{3}}{2}i)t} \rightarrow 0\) and \(e^{(-\frac{1}{2} - \frac{\sqrt{3}}{2}i)t} \rightarrow 0\) So, we have: \(\lim_{t\rightarrow \infty} I(t) = 0\) \(\lim_{t\rightarrow \infty} V(t) = 0\) These limiting values do not depend on the initial conditions.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is essential when dealing with systems modeled by differential equations, especially in the context of electrical circuits. An eigenvalue, in the broadest sense, reflects a magnitude of transformation that an object, such as a vector, undergoes during a linear transformation. For electrical circuits, identifying the eigenvalues of a coefficient matrix which represents the system's behavior can tell us about the system's stability and response over time.

Eigenvectors, on the other hand, are the directional components that correspond to these eigenvalues. In our textbook exercise involving an electric circuit, eigenvalues tell us how the current and voltage will evolve, while the corresponding eigenvectors determine their modes of behavior. When we solve the characteristic equation of the given differential system, we essentially find the eigenvalues. These eigenvalues can be real or complex, and each case indicates a different behavior of the electrical circuit over time.

Characteristic Equation

The characteristic equation plays a pivotal role in our analysis of differential equations linked to electrical circuits. It is derived from the determinant of the matrix equation \(A - \lambda I = 0\), where \(A\) is the coefficient matrix and \(\lambda\) is an eigenvalue. This equation yields a polynomial whose roots are the eigenvalues.

To better grasp this, let's refer back to our exercise where we derived the characteristic equation \(\lambda^2 + \frac{1}{RC}\lambda + \frac{1}{LC} = 0\). Solving this allows us to predict the behavior of the circuit based on whether the roots - the eigenvalues - are real and distinct, a single repeated real value, or complex conjugates. Each of these scenarios reveals different dynamics, like oscillatory motion for complex roots or exponential decay for real roots, which can be a deciding factor in the performance and design of electronic systems.

Complex Conjugate Roots

When we work with differential equations in electrical engineering, the occurrence of complex conjugate roots hints at oscillatory behavior. These roots arise from the characteristic equation when the discriminant \(\Delta\) is less than zero, which indicates an underdamped system capable of oscillation.

In the circuit example from the textbook, we are asked to analyze for what parameters of \(L\), \(R\), and \(C\) we will encounter complex conjugate roots. Such roots are expressed in the form \(\lambda_{1,2} = a \pm bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. The real part \(a\) indicates the rate of exponential decay, while the imaginary part \(b\) is proportional to the frequency of oscillation. This is crucial for understanding the transient response of RLC circuits, such as how quickly they react to changes and how they resonate, which is fundamental in designing filters and tuning circuits.