Laplace transforms

Three storage elements, three equations

• Solving one differential equation by hand is fine. A circuit with three capacitors and inductors gives three coupled ones, and the hand method stops scaling.
• The Laplace transform turns every derivative into a multiplication by $s$. Whole systems of differential equations become algebra.

The transform and its key pair

• The Laplace transform 拉普拉斯变换 is $F(s) = \mathcal{L}\{f(t)\} = \displaystyle\int_{0}^{\infty} f(t)\,e^{-st}\,dt$.
• The key pair: $\mathcal{L}\{e^{at}\} = \dfrac{1}{s - a}$, because $\displaystyle\int_{0}^{\infty} e^{-(s-a)t}\,dt = \dfrac{1}{s-a}$. With $a = 0$ it gives $\mathcal{L}\{1\} = \dfrac{1}{s}$.

Derivatives become multiplication

• $\mathcal{L}\left\{\dfrac{dy}{dt}\right\} = sY - y(0)$: the initial condition enters the algebra directly.
• Applied twice: $\mathcal{L}\left\{\dfrac{d^{2}y}{dt^{2}}\right\} = s^{2}Y - s\,y(0) - y'(0)$.

Partial fractions split the answer

• After solving for $Y(s)$, partial fractions 部分分式 break it into simple pieces, each of which matches a table entry.
• For distinct factors, $\dfrac{10}{s(s+2)} = \dfrac{A}{s} + \dfrac{B}{s+2}$. Cover up each factor in turn: $A = \dfrac{10}{0 + 2} = 5$ and $B = \dfrac{10}{-2} = -5$.

A first-order equation, by transform

• $\dfrac{dy}{dt} + 2y = 10$, $y(0) = 1$ becomes $sY - 1 + 2Y = \dfrac{10}{s}$, so $Y = \dfrac{10}{s(s+2)} + \dfrac{1}{s+2} = \dfrac{5}{s} - \dfrac{4}{s+2}$.
• Inverting term by term: $y = 5 - 4e^{-2t}$, the same answer as the classical route. The initial condition was handled by bookkeeping, not at the end.

Poles predict the behaviour

• The poles 极点 of $Y(s)$ are the values of $s$ that make its denominator zero. A pole at 0 gives a constant; a real pole at $-a$ gives $e^{-at}$; a complex pair gives an oscillation.
• Above, poles at $s = 0$ and $s = -2$ say "a constant plus a decay" before any inversion. A zero (numerator root) shapes the mix but does not create a term.

Second order: the poles are the decays

• $\dfrac{d^{2}y}{dt^{2}} + 3\dfrac{dy}{dt} + 2y = 0$, $y(0) = 1$, $y'(0) = 0$ gives $(s^{2} + 3s + 2)Y = s + 3$, so $Y = \dfrac{s+3}{(s+1)(s+2)} = \dfrac{2}{s+1} - \dfrac{1}{s+2}$.
• Then $y = 2e^{-t} - e^{-2t}$. The pole at −1 decays more slowly and owns the tail: the dominant-pole idea later courses reuse.

Three common slips

Recap