Todai Entrance Exam: Subject 2018 – Problem 1

Problem link

(1)

    \[ e(t) = \begin{Bmatrix} E &, 0 < t \leq T \\  0 &, \text{otherwise} \\ \end{matrix} \]

    \[ L\{e(t)\} = \int_{0}^{\infty} e^{-st}e(t)dt = 0 + \int_{0+}^{T} e^{-st}e(t)dt =  \int_{0+}^{T} e^{-st}Edt = - E\frac{e^{-st}}{s} \Big |^{T}_{0+} = \frac{E(1 - e^{-sT})}{s} \]

(2) Under the superpower of this Youtube Lecture, this WikiBooks Article, this Table of Laplace Transforms, and Khan Academy Lectures:

    \[ e(t) = e_{R}(t) + e_{C}(t) \]


    \[ \Leftrightarrow L\{ e(t) \} = L\{ e_{R}(t) \} + L\{ e_{C}(t) \} \text{ (linearity property of Laplace Transform)} \]


    \[ \Leftrightarrow \frac{E(1 - e^{-sT})}{s} = RI(s) + \frac{I(s)}{Cs} \]


    \[ \Leftrightarrow I(s) = \frac{E(1 - e^{-sT})}{s\left ( R + \frac{1}{Cs} \right )} = \[ \frac{E}{R} \cdot  \frac{1 - e^{-sT}}{s + \frac{1}{RC}} \]


    \[ \Leftrightarrow i(t) = L^{-1} \left \{ I(s) \right \} = \frac{E}{R} \cdot \left ( L^{-1} \left \{ \frac{1}{s + \frac{1}{RC}} \right \} - L^{-1} \left \{ e^{-sT}\frac{1}{s + \frac{1}{RC}} \right \} \right ) \]


    \[ \Leftrightarrow i(t) = \frac{E}{R}\left ( e^{-\frac{t}{RC}} - \mu_{T}(t)e^{\frac{T - t}{RC}} \right ) \]

As the Heaviside step function \mu has multiple definitions (see my unanswered question on Math StackExchange), so I will define it based on the definition of e(t).

(a) i(t) = \frac{E}{R}e^{-\frac{t}{RC}}.

(b) i(t) = \frac{E}{R}\left ( e^{\frac{T - t}{RC}} - e^{-\frac{t}{RC}} \right ).

In case (b), i(t) < 0 indicates that the direction of electrons in the circuit is reversed. By derivation and limit, it is easy to show that the current decreases over time and approaches to 0. I guess when the power supply is gone, the negative charges in the capacitor will move backwards to the positive charges, the voltage of the circuit will then decreases, hence the decreasing current (based on visualization from Khan Academy).

(3)

v(T) = e(t) - Ri(t) = E(1 - e^{-\frac{T}{RC}})

v(2T) = Ri(t) = E(e^{-\frac{T}{RC}} - e^{-\frac{2T}{RC}})

v(3T) = v(2T)e^{-\frac{T}{RC}} + v(T) = E(1 - e^{-\frac{T}{RC}} + e^{-\frac{2T}{RC}} - e^{-\frac{3T}{RC}}) (superposition?)

v(4T) = v(3T)e^{-\frac{T}{RC}} = E(e^{-\frac{T}{RC}} - e^{-\frac{2T}{RC}} + e^{-\frac{3T}{RC}} - e^{-\frac{4T}{RC}})

(4)

(a) i(t) = \frac{E - v(2nT)}{R}.

(b) i(t) = -\frac{v((2n + 1)T)}{R}.

(5)

    \[ v(2nT) = v((2n - 1)T)e^{-\frac{T}{RC}} = (v(2(n-1)T)e^{-\frac{T}{RC}} + v(T))e^{-\frac{T}{RC}} = v(2(n-1)T)e^{-\frac{2T}{RC}} + v(T)e^{-\frac{T}{RC}} \]

(6)

v(2nT) + \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} = \left (v(2(n-1)T) + \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} \right )e^{-\frac{2T}{RC}} = \left (v(2T) + \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} \right )e^{-\frac{2T}{RC}(n-1)}
v(2nT) = \left (v(2T) + \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} \right )e^{-\frac{2T}{RC}(n-1)} - \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1}

    \[ \lim_{n \rightarrow \infty} v(2nT) = - \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} \]

(7) As done in the previous question.

    \[ v(2nT) = \left (v(2T) + \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} \right )e^{-\frac{2T}{RC}(n-1)} - \frac{v(T)e^{-\frac{T}{RC}}}{e^{-\frac{2T}{RC}} - 1} \]


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