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Chapter 5: Q22E (page 267)

(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

Expert verified

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

01

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

02

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.

03

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.

04

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 \(\begin{aligned}{l}\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}\left( t \right)}\\{{{v'}_C}\left( t \right)}\end{aligned}} \right) = x\left( t \right) = 12\left( {\begin{aligned}{ {20}{c}}{ - \cos .8t + 2\sin .8t}\\{\cos .8t}\end{aligned}} \right){e^{ - .4t}} - 6\left( {\begin{aligned}{ {20}{c}}{ - \sin .8t - 2\cos .8t}\\{\sin .8t}\end{aligned}} \right){e^{ - .4t}}\\ = \left( {\begin{aligned}{ {20}{c}}{30\sin .8t}\\{12\cos .8t - 6\sin .8t}\end{aligned}} \right){e^{ - .4t}}\end{aligned}\).

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Most popular questions from this chapter

Exercises 19鈥�23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right]\).

22. Let \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + {a_{\bf{2}}}{t^{\bf{2}}} + {t^{\bf{3}}}\), and let \(\lambda \) be a zero of \(p\).

  1. Write the companion matrix for \(p\).
  2. Explain why \({\lambda ^{\bf{3}}} = - {a_{\bf{0}}} - {a_{\bf{1}}}\lambda - {a_{\bf{2}}}{\lambda ^{\bf{2}}}\), and show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\).

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\).

  1. Reduce \(A\) to echelon form \(U\) with no row scaling, and use \(U\) in formula (1) (before Example 2) to compute \(\det A\). (If \(A\) happens to be singular, start over with a new random matrix.)
  2. Compute the eigenvalues of \(A\) and the product of these eigenvalues (as accurately as possible).
  3. List the matrix \(A\), and, to four decimal places, list the pivots in \(U\) and the eigenvalues of \(A\). Compute \(\det A\) with your matrix program, and compare it with the products you found in (a) and (b).

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)鈥檚 and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\)? If so, find the eigenvalue.

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)
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