Delayed Inputs and Convolution

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The idea

The step $H(t-a)$ switches on at time $a$ . A delayed waveform $H(t-a)f(t-a)$ transforms to $e^{-as}F(s)$ . The argument inside $f$ must be shifted as well; $H(t-a)f(t)$ is generally a different signal.

Visual guide

VISUAL GUIDEA delayed input produces a delayed continuous response

Step input
Response

The input switches from 0 to 1 at t = 2. For y′ + y = H(t − 2), the state stays zero beforehand and then approaches 1 as 1 − e⁻⁽ᵗ⁻²⁾. The response has no jump, but its slope changes at the switch.

Method and assumptions

For zero initial data, a linear time-invariant system with impulse response $h$ responds to input $u$ through convolution: $y(t)=\int_0^t h(t-\tau)u(\tau)d\tau$ . The Laplace transform of convolution is the product $H(s)U(s)$ . Each past input contributes a delayed copy of the impulse response.

Worked example: a delayed switch

For $y'+y=H(t-2)$ and $y(0)=0$ , the transform is $Y=e^{-2s}/[s(s+1)]$ . Therefore $y=H(t-2)[1-e^{-(t-2)}]$ . It stays zero until the input begins, then approaches one.

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.

What is ℒ{u(t−a)f(t−a)} for a>0 and F(s)=ℒ{f}?

eᵃˢF(s)

F(s−a)

e⁻ᵃˢF(s)

Hint 1 · Find a starting point

A delay shifts the start of the signal to t=a.

Hint 2 · Take the next step

Substitute τ=t−a in the transform integral.

Show the reasoning

Answer: e⁻ᵃˢF(s)

The exponential factor splits as e⁻ᵃˢe⁻ˢτ, giving e⁻ᵃˢF(s).

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

Worked example: a ramp through a filter

For $y'+y=t$ , $y(0)=0$ , the impulse response is $h=e^{-t}$ . Convolution gives $y=\int_0^t e^{-(t-\tau)}\tau d\tau=t-1+e^{-t}$ . Differentiating confirms $y'+y=t$ .

Interpreting the result

An ordinary step in forcing usually leaves the state continuous while changing its derivative. An impulse can cause a jump instead. Separate the forcing’s discontinuities from the response’s continuity requirements.

Practice

Transform $H(t-3)(t-3)$ .
Write a unit pulse active from $1$ to $2$ .
Find the zero-state response of $y'+y=1$ .

Show worked solutions

$e^{-3s}/s^2$ .
$H(t-1)-H(t-2)$ .
Convolution with $e^{-t}$ gives $1-e^{-t}$ .

Further study

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

MAKE IT YOURS

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