The circuit in Figure 1 can be described by the differential equation [x'1(t) x'2(t)] = [-(1/R1 + 1/R2)/C1 1/(R2C1) 1/(R2C2) -1/(R2C2)][x1(t) x2(t)] where x1(t) and x2(t) are the voltages across the two capacitors at time t. Suppose resistor R1 is 1 ohm , R2 is 2 ohm, capacitor C1 is 1 farad , and

Question

The circuit in Figure 1 can be described by the differential equation

[latex] \left [ \begin{matrix} \acute{x}_{1}(t) \ \acute{x}_{2}(t) \end{matrix} ight ] = \left [ \begin{matrix} -(1/R_{1}+1/R_{2})/C_{1} & 1/(R_{2}C_{1}) \ 1/(R_{2}C_{2}) & -1/(R_{2}C_{2}) \end{matrix} ight ] \left [ \begin{matrix} x_{1}(t) \ x_{2}(t) \end{matrix} ight ] [/latex]

where [latex] x_{1}(t) [/latex] and [latex] x_{2}(t) [/latex] are the voltages across the two capacitors at time t . Suppose resistor [latex] R_{1} [/latex] is 1 ohm, [latex] R_{2} [/latex] is 2 ohms, capacitor [latex] C_{1} [/latex] is 1 farad, and [latex] C_{2} [/latex] is .5 farad, and suppose there is an initial charge of 5 volts on capacitor [latex] C_{1} [/latex] and 4 volts on capacitor [latex] C_{2} [/latex]. Find formulas for [latex] x_{1}(t) [/latex] and [latex] x_{2}(t) [/latex] that describe how the voltages change over time.

Accepted Answer

Let A denote the matrix displayed above, and let [latex] x(t) = \left [ \begin{matrix} x_{1}(t) \ x_{2}(t) \end{matrix} ight ] [/latex] . For the data given, [latex] A = \left [ \begin{matrix} -1.5 & .5 \ 1 & -1 \end{matrix} ight ] , ext{and} x(0) = \left [ \begin{matrix} 5 \ 4 \end{matrix} ight ] [/latex] . The eigenvalues of A are [latex] \lambda_{1} = -.5 ext{and} \lambda_{2} = -2 [/latex] , with corresponding eigenvectors

[latex] v_{1} = \left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] ext{and} v_{2} = \left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] [/latex]

The eigenfunctions [latex] x_{1}(t) = v_{1}e^{\lambda_{1}t} ext{and} x_{2}(t) = v_{2}e^{\lambda_{2}t} [/latex] both satisfy [latex] \acute{x} = Ax [/latex] , and so does any linear combination of [latex] x_{1} [/latex] and [latex] x_{2} [/latex]. Set

[latex] x(t) = c_{1}v_{1}e^{\lambda_{1}t} + c_{2}v_{2}e^{\lambda_{2}t} = c_{1}\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] e^{-.5t} + c_{2}\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] e^{-2t} [/latex]

and note that [latex] x(0) = c_{1}v_{1} + c_{2}v_{2} [/latex]. Since [latex] v_{1} [/latex] and [latex] v_{2} [/latex] are obviously linearly independent and hence span [latex] \mathbb{R} ^{2} [/latex], [latex] c_{1} [/latex] and [latex] c_{2} [/latex] can be found to make x(0) equal to [latex] x_{0} [/latex]. In fact, the equation

[latex] c_{1}\underset{\overset{\uparrow }{v_{1}} }{\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] } + c_{2}\underset{\overset{\uparrow }{v_{2}} }{\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] } = \underset{\overset{\uparrow }{x_{0}} }{\left [ \begin{matrix} 5 \ 4 \end{matrix} ight ] } [/latex]

leads easily to [latex] c_{1} = 3 ext{and} c_{2} = -2 [/latex]. Thus the desired solution of the differential equation [latex] \acute{x} = Ax [/latex] is

[latex] x(t) = 3\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ]e^{-.5t} - 2\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ]e^{-2t} [/latex]

or

[latex] \left [ \begin{matrix} x_{1}(t) \ x_{2}(t) \end{matrix} ight ] = \left [ \begin{matrix} 3e^{-.5t} + 2e^{-2t} \ 6e^{-.5t} - 2e^{-2t} \end{matrix} ight ] [/latex]

Figure 2 shows the graph, or trajectory, of x(t), for t ≥ 0, along with trajectories for some other initial points. The trajectories of the two eigenfunctions [latex] x_{1} [/latex] and [latex] x_{2} [/latex] lie in the eigenspaces of A.

The functions [latex] x_{1} [/latex] and [latex] x_{2} [/latex] both decay to zero as [latex] t ightarrow \infty [/latex], but the values of [latex] x_{2} [/latex] decay faster because its exponent is more negative. The entries in the corresponding eigenvector [latex] v_{2} [/latex] show that the voltages across the capacitors will decay to zero as rapidly as possible if the initial voltages are equal in magnitude but opposite in sign.

Question 5.7.1: The circuit in Figure 1 can be described by the differential...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions