Impulse Inputs and Jump Conditions

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The idea

The Dirac delta is an idealized impulse with specified area, not an ordinary function with a finite value at its center. It models a short input whose duration is negligible compared with the system’s response time. Its effect follows by integrating the equation across the impulse.

Visual guide

VISUAL GUIDEAn impulse creates a jump followed by free decay

Before the impulse
After the impulse

For y′ + y = 2δ(t − 1), the state jumps from 0 to 2 at t = 1, then decays as 2e⁻⁽ᵗ⁻¹⁾. The vertical arrow represents the impulse’s effect, not a finite-height graph of the delta distribution.

Method and assumptions

For an impulse at $a>0$ , $\mathcal L[\delta(t-a)]=e^{-as}$ . In $y'+ky=J\delta(t-a)$ , integration over a shrinking interval around $a$ gives $y(a^+)-y(a^-)=J$ . For a mass equation, a force impulse changes momentum rather than position.

Worked example: a first-order jump

For $y'+y=2\delta(t-1)$ , $y(0)=0$ , the response is $2H(t-1)e^{-(t-1)}$ . It jumps from $0$ to $2$ at $t=1$ , then decays according to the unforced equation.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

In y′+ay=Jδ(t−t₀), with finite one-sided values of y, what jump does the impulse produce?

y(t₀⁺)−y(t₀⁻)=aJ

y(t₀⁺)−y(t₀⁻)=J

No jump

Hint 1 · Find a starting point

Integrate over an interval shrinking around t₀.

Hint 2 · Take the next step

The integral of ay tends to zero, while the impulse has area J.

Show the reasoning

Answer: y(t₀⁺)−y(t₀⁻)=J

The derivative integral is the jump in y, so that jump equals J.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a struck oscillator

For $x''+x=J\delta(t-a)$ with zero initial state, displacement remains continuous and velocity jumps by $J$ . The response is $JH(t-a)\sin(t-a)$ . Its value at the impulse is zero, while its right velocity is $J$ .

Interpreting the result

If the leading coefficient is a mass $m$ , the velocity jump is $J/m$ . A finite pulse approximation must preserve impulse area when its width shrinks. Its peak height alone does not determine the limiting effect.

Practice

What is the velocity jump for $m=2,J=6$ ?
Transform $3\delta(t-4)$ .
Why is a unit-height narrowing pulse not a unit impulse?

Show worked solutions

$\Delta v=J/m=3$ .
$3e^{-4s}$ .
Its area tends to zero. A unit-area rectangle of width $\epsilon$ needs height $1/\epsilon$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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