(RLC circuit) The circuit in the figure consists of a resistor (R ohms), and inductor (Lhenrys), a capacitor (Cfarads) and an initial voltage source. Let \(b = \frac{R}{{{\bf{2}}L}}\), and suppose R, Land C have been selected so that b also equals \(\frac{{\bf{1}}}{{\sqrt {LC} }}\). (This is done, for instance, when the circuit is used in a voltmeter). Let \(v\left( t \right)\) be the voltage (in volts) at time t, measured across the capacitor. It can be shown that v is in the null space H of the linear transformation that maps \(v\left( t \right)\) into \(Lv''\left( t \right) + Rv'\left( t \right) + \left( {\frac{{\bf{1}}}{C}} \right)v\left( t \right)\), and H consists of all functions of the form \(v\left( t \right) = {e^{ - bt}}\left( {{c_{\bf{1}}} + {c_{\bf{2}}}t} \right)\). Find a basis for H.

Step by step solution

01

Find the set of vectors for V

Suppose the set \(\left\{ {{{\bf{b}}_1},...,{{\bf{b}}_n}} \right\}\) is a basis for the space \(f\left( V \right)\), and it consists linearly independentset.

Let,

\(\begin{array}{c}v\left( t \right) = {e^{ - bt}}\left( {{c_1} + {c_2}t} \right)\\ = {c_1}{e^{ - bt}} + {c_2}t{e^{ - bt}}.\end{array}\)

The function \(v\left( t \right)\) is a linear combination of functions \(\left\{ {{e^{ - bt}},t{e^{ - bt}}} \right\}\).

So, the set \(\left\{ {{e^{ - bt}},t{e^{ - bt}}} \right\}\) spans H.

02

Check the linear independency

The functions \({e^{ - bt}}\) and \(t{e^{ - bt}}\)are not scalar multiples of each other, so they are linearly independent.

So, the set \(\left\{ {{e^{ - bt}},t{e^{ - bt}}} \right\}\) forms a basis of H.

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!