Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 5: Q19E (page 267)

(M) Find formulas for the voltages \({v_1}\) and \({v_2}\)(as functions of time t) for the circuit in Example 1, assuming that \({R_1} = \frac{1}{5}\)ohm, \({R_1} = \frac{1}{3}{\mathop{\rm ohm}\nolimits} \), \({C_1} = 4\)farads, \({C_2} = 3\)farads, and the initial charge on each capacitor is 4 volts.

Short Answer

Expert verified

The formula for the voltages is \(\left( {\begin{aligned}{ {20}{c}}{{v_1}\left( t \right)}\\{{v_2}\left( t \right)}\end{aligned}} \right) = x\left( t \right) = \frac{5}{3}\left( {\begin{aligned}{ {20}{c}}1\\3\end{aligned}} \right){e^{ - .5t}} - \frac{2}{3}\left( {\begin{aligned}{ {20}{c}}{ - 2}\\3\end{aligned}} \right){e^{ - 2.5t}}\).

Step by step solution

01

Determine the eigenvalues and eigenvector of the matrix

The formula in example 1 is shown below:

\(\left( {\begin{aligned}{ {20}{c}}{{x_1}^\prime \left( t \right)}\\{{x_2}^\prime \left( t \right)}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - \left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right){C_1}}&{\frac{1}{{\left( {{R_2}{C_1}} \right)}}}\\{\frac{1}{{\left( {{R_2}{C_2}} \right)}}}&{ - \frac{1}{{\left( {{R_2}{C_2}} \right)}}}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\end{aligned}} \right)\)

Substitute \({R_1} = \frac{1}{5},{R_2} = \frac{1}{3},{C_1} = 4,\) and \({C_2} = 3\) into the above formula to obtain the matrix \(A\) as shown below:

\(\begin{aligned}{c}A = \left( {\begin{aligned}{ {20}{c}}{ - \left( {\frac{1}{{\frac{1}{5}}} + \frac{1}{{\frac{1}{3}}}} \right)}&{\frac{1}{{\frac{1}{3} \cdot 4}}}\\{\frac{1}{{\frac{1}{3} \cdot 3}}}&{ - \frac{1}{{\frac{1}{3} \cdot 3}}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{\frac{3}{4}}\\1&{ - 1}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( { - 2\,\,\,\,\frac{3}{4};\,\,1\,\,\,\,1} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{ - 0.5000}\\{ - 2.5000}\end{aligned}} \right)\)

Thus, the eigenvalues of \(A\) are \({\lambda _1} = - .5\) and \({\lambda _2} = - 2.5\).

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

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

\(\left( {\begin{aligned}{ {20}{c}}{1.0000}\\{2.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}1\\2\end{aligned}} \right)\).

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

\(\left( {\begin{aligned}{ {20}{c}}{ - 3.0000}\\{2.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{ - 3}\\2\end{aligned}} \right)\).

02

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

Therefore, the general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}1\\2\end{aligned}} \right){e^{ - .5t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 3}\\2\end{aligned}} \right){e^{ - 2.5t}}\).

03

Determine the formula for the voltages

The initial charge of each capacitor is 4 volts. Then, the initial condition is \(x\left( 0 \right) = \left( {\begin{aligned}{ {20}{c}}4\\4\end{aligned}} \right)\). It follows that \(\left( {\begin{aligned}{ {20}{c}}1&{ - 3}\\2&2\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{{c_1}}\\{{c_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}4\\4\end{aligned}} \right)\).

Use the MATLAB code to find \({c_1}\) and \({c_2}\) by solving the system of the linear equation as shown below:

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( {1\,\,\, - 3;\,\,2\,\,\,\,2} \right)\\ > > {\mathop{\rm B}\nolimits} = \left( {4\,\,\,\,4} \right)\\ > > {\mathop{\rm x}\nolimits} = {\mathop{\rm linsolve}\nolimits} \left( {{\mathop{\rm A}\nolimits} ,{\mathop{\rm B}\nolimits} } \right)\end{aligned}\)

\(\left( {\begin{aligned}{ {20}{c}}{2.5000}\\{ - 0.5000}\end{aligned}} \right)\). From \({\mathop{\rm x}\nolimits} \) we obtain \({c_1} = \frac{5}{2},{c_2} = \frac{{ - 1}}{2}\).

Thus, the formula for the voltages is \(\left( {\begin{aligned}{ {20}{c}}{{v_1}\left( t \right)}\\{{v_2}\left( t \right)}\end{aligned}} \right) = x\left( t \right) = \frac{5}{3}\left( {\begin{aligned}{ {20}{c}}1\\3\end{aligned}} \right){e^{ - .5t}} - \frac{2}{3}\left( {\begin{aligned}{ {20}{c}}{ - 2}\\3\end{aligned}} \right){e^{ - 2.5t}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude. (Hint: What would be true if \(I - A\) were not invertible?)

Consider an invertible n × n matrix A such that the zero state is a stable equilibrium of the dynamical system x→(t+1)=ATx→(t)What can you say about the stability of the systems.

x→(t+1)=ATx→(t)

Let\(D = \left\{ {{{\bf{d}}_1},{{\bf{d}}_2}} \right\}\) and \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{d}}_1}} \right) = 2{{\bf{b}}_1} - 3{{\bf{b}}_2}\), \(T\left( {{{\bf{d}}_2}} \right) = - 4{{\bf{b}}_1} + 5{{\bf{b}}_2}\)

Find the matrix for \(T\) relative to \(D\), and\(B\).

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

21. Use mathematical induction to prove that for \(n \ge {\bf{2}}\),\(\begin{aligned}{c}det\left( {{C_p} - \lambda I} \right) = {\left( { - {\bf{1}}} \right)^n}\left( {{a_{\bf{0}}} + {a_{\bf{1}}}\lambda + ... + {a_{n - {\bf{1}}}}{\lambda ^{n - {\bf{1}}}} + {\lambda ^n}} \right)\\ = {\left( { - {\bf{1}}} \right)^n}p\left( \lambda \right)\end{aligned}\)

(Hint: Expanding by cofactors down the first column, show that \(det\left( {{C_p} - \lambda I} \right)\) has the form \(\left( { - \lambda B} \right) + {\left( { - {\bf{1}}} \right)^n}{a_{\bf{0}}}\) where \(B\) is a certain polynomial (by the induction assumption).)

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free