Laplace II: Steps, Impulses & Convolution

Why the factor appears: $\mathcal L\{g(t-a)u(t-a)\} = \int_a^\infty g(t-a)e^{-st}\,dt$, because the step erases everything before $t = a$. Substituting $\tau = t - a$ splits $e^{-st}$ into $e^{-as}e^{-s\tau}$; the constant $e^{-as}$ comes out of the integral and what remains is $G(s)$. So the rule applies only once a piece is written in the delayed variable $t - a$ — that bookkeeping is the whole skill. The force on from $t = 3$ to $t = 7$ is $u(t-3) - u(t-7)$, with transform $\dfrac{e^{-3s} - e^{-7s}}{s}$.

Why the transform is $e^{-as}$: replace the blow by a rectangle of height $\tfrac1\varepsilon$ on $[a, a+\varepsilon]$. Its transform is $\dfrac{e^{-as} - e^{-(a+\varepsilon)s}}{\varepsilon s} = e^{-as}\cdot\dfrac{1 - e^{-\varepsilon s}}{\varepsilon s}$, and the second factor tends to $1$ as $\varepsilon \to 0$. What $\delta$ does to a solution follows by integrating across the instant: $y' = \delta(t-1)$, $y(0) = 0$ gives the step $y = u(t-1)$ — flat, a jump of exactly $1$, flat again. In a second-order equation the same integration makes $y'$ jump while $y$ stays continuous (next sections). The response of a system at rest to $\delta(t)$ is its impulse response $h(t)$.

One integral formula expresses every solution — the superposition of infinitely many scaled, delayed hammer blows.

The method works because the equation is linear and its coefficients do not change with time: the response to a delayed input is the delayed response. So the computation splits into one undelayed problem per switching time, and the factors $e^{-as}$ only record when each response begins. There is no case analysis and no matching of $y$ and $y'$ at the switch — continuity there comes out of the formula automatically.

Step 1 — invert the undelayed factor. Partial fractions give $\dfrac{3}{s(s+1)} = \dfrac3s - \dfrac{3}{s+1}$, the transform of $r(t) = 3(1 - e^{-t})$: the response to a step switched on at time $0$.

Step 2 — the factor $e^{-2s}$ starts that response at $t = 2$: $y(t) = 3\left(1 - e^{-(t-2)}\right)u(t-2)$.

Step 3 — read it: nothing happens before $t = 2$ (no forcing, zero start); afterwards $y$ climbs toward the new equilibrium $3$, closing $63\%$ of the gap in one time unit: $y(3) = 3(1 - e^{-1}) \approx 1.90$.

Step 4 — check the switch: $y(2) = 0$ from both sides, so $y$ is continuous, as a first-order equation with a bounded input requires; only $y'$ jumps, from $0$ to $3$.

Step 1 — transform the equation: $(s^2 + 4)Y = \dfrac{1 - e^{-s}}{s^2}$, so $Y = (1 - e^{-s})R(s)$ with $R(s) = \dfrac{1}{s^2(s^2+4)}$.

Step 2 — invert $R$ once, by partial fractions in $s^2$: $\dfrac{1}{s^2(s^2+4)} = \dfrac14\left(\dfrac{1}{s^2} - \dfrac{1}{s^2+4}\right)$, so $r(t) = \dfrac t4 - \dfrac{\sin 2t}{8}$, the response to the unending ramp.

Step 3 — the $e^{-s}$ term is the same response delayed by $1$: $y(t) = r(t) - r(t-1)\,u(t-1)$.

Step 4 — simplify for $t \ge 1$ with $\sin A - \sin B = 2\cos\tfrac{A+B}{2}\sin\tfrac{A-B}{2}$: $y = \dfrac14 - \dfrac{\sin 1}{4}\cos(2t - 1)$. The constant $\tfrac14$ is where the spring balances the constant force ($4y = 1$); the ramp stopped while the mass was still moving, so it keeps oscillating about $\tfrac14$ with amplitude $\tfrac{\sin 1}{4} \approx 0.210$.

Integrate the equation across a short interval $[t_0 - \varepsilon, t_0 + \varepsilon]$. The $\delta$ term contributes exactly $k$; the terms $by'$ and $cy$ contribute almost nothing, because $y$ and $y'$ stay bounded on an interval of length $2\varepsilon$. So $y'(t_0^+) - y'(t_0^-) = k$. A jump in $y'$ moves $y$ by nothing in zero time, so $y$ is continuous. This is what a hammer does to a mass: it changes the momentum at once, not the position.

Step 1 — the first term is the free motion $\cos t$, what the spring would do untouched.

Step 2 — the second is the impulse response $h(t) = \sin t$, started at $t = \pi$: $\sin(t - \pi)u(t-\pi) = -\sin t\;u(t-\pi)$.

Step 3 — so $y = \cos t$ for $t < \pi$ and $y = \cos t - \sin t$ for $t \ge \pi$.

Step 4 — check the jump rule: at $t = \pi$ the mass is at $y = -1$ with velocity $0$; just after the blow the velocity is $-\sin\pi - \cos\pi = 1$, a jump of exactly $1$, and the position is still $-1$.

Step 5 — the amplitude grows from $1$ to $\sqrt{1^2 + 1^2} = \sqrt2$: the blow added energy at a turning point.

Step 1 — an impulse changes only the velocity, by $k$. Rest at equilibrium needs position $0$ (already true at $t = \pi$) and velocity $0$, so $-1 + k = 0$: $k = 1$.

Step 2 — confirm by transform: $(s^2 + 1)Y - 1 = e^{-\pi s}$, so $y = \sin t + \sin(t - \pi)u(t - \pi) = \sin t\,\big(1 - u(t-\pi)\big)$: one arch of the sine, then $y \equiv 0$.

Step 3 — timing matters. The same blow at $t = \tfrac\pi2$, where the mass is at $y = 1$ and momentarily still, cannot stop it: it changes the velocity from $0$ to $1$, giving $y = \sin t - \cos t$ afterwards, an oscillation of amplitude $\sqrt2$. A blow can cancel velocity, never displacement.

Step 1 — $\dfrac{2}{s^2 + 4}$ is the transform of $\sin 2t$, and replacing $s$ by $s + 1$ multiplies the time function by $e^{-t}$; so $h(t) = \tfrac12 e^{-t}\sin 2t$.

Step 2 — linearity scales it by $3$ and the delay rule starts it at $t = 2$: $y = \tfrac32 e^{-(t-2)}\sin\big(2(t-2)\big)\,u(t-2)$.

Step 3 — check against the jump rule: $h(0) = 0$ and $h'(0) = \tfrac12\cdot 2 = 1$, so $h$ is the motion after a unit kick of the velocity.

Step 4 — read it: the oscillation has angular frequency $2$, and its envelope $\pm\tfrac12e^{-t}$ halves every $\ln 2 \approx 0.69$ time units.

Change variables to $t = \tau + \sigma$ (total time) and $\tau$: for each fixed $t$, $\tau$ ranges over $[0, t]$, giving $\int_0^\infty e^{-st}\left[\int_0^t f(\tau)g(t-\tau)\,d\tau\right]dt = \mathcal L\{f * g\}$. A Fubini swap turns a product into a convolution. $\blacksquare$

Step 1 — write the forcing with steps: $f = 1 - u(t - \pi)$, so $\mathcal L\{f\} = \dfrac1s - \dfrac{e^{-\pi s}}{s}$.

Step 2 — transform: $(s^2 + 1)Y = \dfrac{1 - e^{-\pi s}}{s}$, so $Y = \dfrac{1 - e^{-\pi s}}{s(s^2+1)}$.

Step 3 — partial fractions on $\dfrac{1}{s(s^2+1)} = \dfrac1s - \dfrac{s}{s^2+1}$, which inverts to $g(t) = 1 - \cos t$.

Step 4 — the $e^{-\pi s}$ copy is the same response delayed: $y = g(t) - g(t-\pi)u(t-\pi)$.

Step 5 — read it: for $t < \pi$, $y = 1 - \cos t$ (pushed from rest); for $t \ge \pi$, $y = (1 - \cos t) - (1 - \cos(t - \pi)) = -2\cos t$ — the force released exactly when the oscillator was at its far point, leaving it swinging with amplitude $2$ forever. Piecewise forcing, zero case-splitting.

Step 1 — transform: $(s^2 + 3s + 2)H = 1$, so $H(s) = \dfrac{1}{(s+1)(s+2)} = \dfrac{1}{s+1} - \dfrac{1}{s+2}$: impulse response $h(t) = e^{-t} - e^{-2t}$.

Step 2 — for general forcing, $y = h * f$: $y(t) = \displaystyle\int_0^t \left(e^{-\tau} - e^{-2\tau}\right)e^{-(t-\tau)}\,d\tau$.

Step 3 — simplify: $e^{-t}\displaystyle\int_0^t\left(1 - e^{-\tau}\right)d\tau = e^{-t}\left(t - 1 + e^{-t}\right)$.

Step 4 — so $y = te^{-t} - e^{-t} + e^{-2t}$; check: $y(0) = 0$ ✓, and the $te^{-t}$ term is exactly what resonance-style forcing at a homogeneous rate produces. One impulse response, every forcing solved by an integral.