Question

(M) The circuit in the figure is described by the equation

\(\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}}\\{{{v'}_C}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}0&{\frac{1}{L}}\\{ - \frac{1}{C}}&{\frac{1}{{\left( {RC} \right)}}}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{{i_L}}\\{{v_C}}\end{aligned}} \right)\)

Where \({i_L}\) is the current through the inductor L and \({v_C}\) is the voltage drop across the capacitor C. Find a formula for \({i_L}\) and \({v_C}\) when \(R = .5\)ohm, \(C = 2.5\)farads, \(L = .5\)henry, and the initial current is 0 amp, and the initial voltage is 12 volts.

Short answer

The formula for the current \({i_{_L}}\) and \({v_L}\) is \(\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}\left( t \right)}\\{{{v'}_C}\left( t \right)}\end{aligned}} \right) = x\left( t \right) = \left( {\begin{aligned}{ {20}{c}}{30\sin .8t}\\{12\cos .8t - 6\sin .8t}\end{aligned}} \right){e^{ - .4t}}\).

Step-by-step solution

Determine the eigenvalues and eigenvector of the matrix

Substitute \(R = .5,C = 2.5,L = .5\) into the given formula to obtain the matrix \(A\) as shown below:

\(A = \left( {\begin{aligned}{ {20}{c}}0&2\\{ - .4}&{ - .8}\end{aligned}} \right)\)

Use the MATLAB code to compute the eigenvalues of the matrix \(A\) as shown below:

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( {0\,\,\,2;\, - .4\,\,\,\, - .8} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{ - 0.4000}& + &{0.8000i}\\{ - 0.4000}& - &{0.8000i}\end{aligned}} \right)\)

Thus, the eigenvalues of \(A\) are \({\lambda _1} = - .4 + 8i\) and \({\lambda _2} = - .4 - 8i\).

Use the MATLAB code to compute the eigenvector of the matrix A as shown below:

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 1 \right) {\mathop{\rm eye}\nolimits} \left( 2 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{ - 0.1000}& - &{0.2000i}\\{}&{1.0000}&{}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}{ - 1 - 2{\mathop{\rm i}\nolimits} }\\1\end{aligned}} \right)\).

The second eigenvalue-eigenvector pair is formed by their conjugates.

Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{ - 1 + 2{\mathop{\rm i}\nolimits} }\\1\end{aligned}} \right)\).

Determine the general solution of \(x' = Ax\)

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 1 - 2{\mathop{\rm i}\nolimits} }\\1\end{aligned}} \right){e^{\left( { - .4 + .8i} \right)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 1 + 2{\mathop{\rm i}\nolimits} }\\1\end{aligned}} \right){e^{\left( { - .4 - .8i} \right)t}}\). Where \({c_1}\) and \({c_2}\) are arbitrary complex numbers.

Determine the real general solution

Rewrite the first eigenfunctions and take its real and imaginary parts to obtain as shown below:

\(\begin{aligned}{c}{\mathop{\rm v}\nolimits} {e^{\left( { - .4 + .8i} \right)t}} = \left( {\begin{aligned}{ {20}{c}}{ - 1 - 2i}\\1\end{aligned}} \right){e^{ - .4t}}\left( {\cos 8t + i\sin 8t} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - \cos .8t + 2\sin .8t}\\{\cos .8t}\end{aligned}} \right){e^{ - .4t}} + i\left( {\begin{aligned}{ {20}{c}}{ - \sin .8t - 2\cos .8t}\\{\sin .8t}\end{aligned}} \right){e^{ - .4t}}\end{aligned}\)

Therefore, the real general solution is of the form \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - \cos .8t + 2\sin .8t}\\{\cos .8t}\end{aligned}} \right){e^{ - .4t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - \sin .8t - 2\cos .8t}\\{\sin .8t}\end{aligned}} \right){e^{ - .4t}}\), with \({c_1}\) and \({c_2}\) are real numbers.

Determine the formula for the current and voltages

Compute \({c_1}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right) + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 2}\\0\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}0\\{12}\end{aligned}} \right)\) to satisfy the initial condition \(x\left( 0 \right) = \left( {\begin{aligned}{ {20}{c}}0\\{12}\end{aligned}} \right)\) to obtain as shown below:

\({c_1} = 12,{c_2} = - 6\)

Thus, the formula for the current \({i_{_L}}\) and \({v_L}\) is \(\begin{aligned}{l}\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}\left( t \right)}\\{{{v'}_C}\left( t \right)}\end{aligned}} \right) = x\left( t \right) = 12\left( {\begin{aligned}{ {20}{c}}{ - \cos .8t + 2\sin .8t}\\{\cos .8t}\end{aligned}} \right){e^{ - .4t}} - 6\left( {\begin{aligned}{ {20}{c}}{ - \sin .8t - 2\cos .8t}\\{\sin .8t}\end{aligned}} \right){e^{ - .4t}}\\ = \left( {\begin{aligned}{ {20}{c}}{30\sin .8t}\\{12\cos .8t - 6\sin .8t}\end{aligned}} \right){e^{ - .4t}}\end{aligned}\).