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Chapter 5: Q21E (page 267)

(M) Find formulas for the current \({i_L}\) and the voltage \({v_C}\) for the circuit in Example 3, assuming that \({R_1} = 1\)ohm, \({R_2} = .125{\mathop{\rm ohm}\nolimits} \), \(C = .2\)farads, \(L = .125\)henry, and the initial current is 0 amp, and the initial voltage is 15 volts.

Short Answer

Expert verified

The formula for the current \({i_{_L}}\) is \(\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}}\\{{{v'}_C}}\end{aligned}} \right) = x\left( t \right) = 3\left( {\begin{aligned}{ {20}{c}}{2\cos 6t - 6\sin 6t}\\{5\cos 6t}\end{aligned}} \right){e^{ - 3t}} - \left( {\begin{aligned}{ {20}{c}}{2\sin 6t - 6\cos t}\\{5\sin 6t}\end{aligned}} \right){e^{ - 3t}} = \left( {\begin{aligned}{ {20}{c}}{ - 20\sin 6t}\\{15\cos 6t - 5\sin 6t}\end{aligned}} \right){e^{ - 3t}}\).

Step by step solution

01

Determine the eigenvalues and eigenvector of the matrix

The formula in example 1 is shown below:

\(\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}}\\{{{v'}_C}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - \frac{{{R_2}}}{L}}&{\frac{{ - 1}}{L}}\\{\frac{1}{C}}&{ - \frac{1}{{\left( {{R_1}C} \right)}}}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{{i_L}}\\{{v_C}}\end{aligned}} \right)\)

Substitute \({R_1} = 1,{R_2} = .125,C = .2,\) and \(L = .125\) into the above formula to obtain the matrix \(A\) as shown below:

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 1}&{ - 8}\\5&{ - 5}\end{aligned}} \right)\)

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

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( { - 1\,\,\, - 8;\,5\,\,\,\, - 5} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{ - 0.3000}& + &{0.6000i}\\{ - 0.3000}& - &{0.6000i}\end{aligned}} \right)\)

Thus, the eigenvalues of \(A\) are \({\lambda _1} = - 3 + 6i\) and \({\lambda _2} = - 3 - 6{\mathop{\rm i}\nolimits} \).

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

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

\(\left( {\begin{aligned}{ {20}{c}}{0.2000}& + &{0.6000i}\\{}&{5.0000}&{}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}{2 + 6{\mathop{\rm i}\nolimits} }\\5\end{aligned}} \right)\).

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

Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{2 - 6{\mathop{\rm i}\nolimits} }\\5\end{aligned}} \right)\).

02

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

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{2 + 6i}\\5\end{aligned}} \right){e^{\left( { - 3 + 6i} \right)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{2 - 6i}\\5\end{aligned}} \right){e^{\left( { - 3 - 6i} \right)t}}\). 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:

\(\begin{aligned}{c}{\mathop{\rm v}\nolimits} {e^{\left( { - 3 + 6i} \right)t}} = \left( {\begin{aligned}{ {20}{c}}{2 + 6i}\\5\end{aligned}} \right){e^{ - 3t}}\left( {\cos 6t + i\sin 6t} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{2\cos 6t - 6\sin 6t}\\{5\cos 6t}\end{aligned}} \right){e^{ - 3t}} + i\left( {\begin{aligned}{ {20}{c}}{2\sin 6t - 6\cos t}\\{5\sin 6t}\end{aligned}} \right){e^{ - 3t}}\end{aligned}\)

Therefore, the real general solution is of the form \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{2\cos 6t - 6\sin 6t}\\{5\cos 6t}\end{aligned}} \right){e^{ - 3t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{2\sin 6t - 6\cos t}\\{5\sin 6t}\end{aligned}} \right){e^{ - 3t}}\), with \({c_1}\) and \({c_2}\) are real numbers.

04

Determine the formula for the current

Compute \({c_1}\left( {\begin{aligned}{ {20}{c}}2\\5\end{aligned}} \right) + {c_2}\left( {\begin{aligned}{ {20}{c}}6\\0\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}0\\{15}\end{aligned}} \right)\) to satisfy the initial condition \(x\left( 0 \right) = \left( {\begin{aligned}{ {20}{c}}0\\{15}\end{aligned}} \right)\) to obtain as shown below:

\({c_1} = 3,{c_2} = - 1\)

Thus, the formula for the current \({i_{_L}}\) is \(\left( {\begin{aligned}{ {20}{c}}{{{i'}_L}}\\{{{v'}_C}}\end{aligned}} \right) = x\left( t \right) = 3\left( {\begin{aligned}{ {20}{c}}{2\cos 6t - 6\sin 6t}\\{5\cos 6t}\end{aligned}} \right){e^{ - 3t}} - \left( {\begin{aligned}{ {20}{c}}{2\sin 6t - 6\cos t}\\{5\sin 6t}\end{aligned}} \right){e^{ - 3t}} = \left( {\begin{aligned}{ {20}{c}}{ - 20\sin 6t}\\{15\cos 6t - 5\sin 6t}\end{aligned}} \right){e^{ - 3t}}\).

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}{.4}&{ - .3}\\{.4}&{1.2}\end{aligned}} \right)\). Explain why \({A^k}\) approaches \(\left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\) as \(k \to \infty \).

19–23 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 \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)\)

Mark each statement as True or False. Justify each answer.

a. If \(A\) is invertible and 1 is an eigenvalue for \(A\), then \(1\) is also an eigenvalue of \({A^{ - 1}}\)

b. If \(A\) is row equivalent to the identity matrix \(I\), then \(A\) is diagonalizable.

c. If \(A\) contains a row or column of zeros, then 0 is an eigenvalue of \(A\)

d. Each eigenvalue of \(A\) is also an eigenvalue of \({A^2}\).

e. Each eigenvector of \(A\) is also an eigenvector of \({A^2}\)

f. Each eigenvector of an invertible matrix \(A\) is also an eigenvector of \({A^{ - 1}}\)

g. Eigenvalues must be nonzero scalars.

h. Eigenvectors must be nonzero vectors.

i. Two eigenvectors corresponding to the same eigenvalue are always linearly dependent.

j. Similar matrices always have exactly the same eigenvalues.

k. Similar matrices always have exactly the same eigenvectors.

I. The sum of two eigenvectors of a matrix \(A\) is also an eigenvector of \(A\).

m. The eigenvalues of an upper triangular matrix \(A\) are exactly the nonzero entries on the diagonal of \(A\).

n. The matrices \(A\) and \({A^T}\) have the same eigenvalues, counting multiplicities.

o. If a \(5 \times 5\) matrix \(A\) has fewer than 5 distinct eigenvalues, then \(A\) is not diagonalizable.

p. There exists a \(2 \times 2\) matrix that has no eigenvectors in \({A^2}\)

q. If \(A\) is diagonalizable, then the columns of \(A\) are linearly independent.

r. A nonzero vector cannot correspond to two different eigenvalues of \(A\).

s. A (square) matrix \(A\) is invertible if and only if there is a coordinate system in which the transformation \({\bf{x}} \mapsto A{\bf{x}}\) is represented by a diagonal matrix.

t. If each vector \({{\bf{e}}_j}\) in the standard basis for \({A^n}\) is an eigenvector of \(A\), then \(A\) is a diagonal matrix.

u. If \(A\) is similar to a diagonalizable matrix \(B\), then \(A\) is also diagonalizable.

v. If \(A\) and \(B\) are invertible \(n \times n\) matrices, then \(AB\)is similar to \ (BA\ )

w. An \(n \times n\) matrix with \(n\) linearly independent eigenvectors is invertible.

x. If \(A\) is an \(n \times n\) diagonalizable matrix, then each vector in \({A^n}\) can be written as a linear combination of eigenvectors of \(A\).

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.
See all solutions

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