Convolution and Impulsive Sources

MATH 341 — Notes 8

University of Scranton

2026-06-16

Goals

What We Will Cover

  1. Convolution \((x*y)(t) = \int_0^t x(\tau)\,y(t-\tau)\,d\tau\) — definition and direct computation

  2. Convolution property \(\mathscr{L}[x*y] = X(s)\,Y(s)\) — inverting products of transforms

  3. Commutativity \(x*y = y*x\) — choosing the easier arrangement

  4. Arbitrary forcing — solution formulas via convolution integrals

  5. Impulsive sources — the physical motivation for \(\delta_a(t)\)

  6. Sifting property and \(\mathscr{L}[\delta_a(t)] = e^{-as}\)

  7. Transfer functions and input–output systems

Note

Sections 3.3 and 3.4 of (Logan 2015).

Tip

New vocabulary: convolution, sifting property, delta function, generalized function, transfer function, impulse response.

Section 3.3

The Convolution Property

Products of Transforms — The Key Question

We know Laplace transforms respect sums: \[\mathscr{L}[x+y] = X(s) + Y(s).\]

Natural question: what about products of functions?

\[\mathscr{L}[x(t)\,y(t)] = \;?\]

Warning

\[\mathscr{L}[x(t)\,y(t)] \neq X(s)\,Y(s)\] The transform of a product is not the product of the transforms. (Exercise 2, §3.3 asks for a concrete counterexample.)

The right question is the inverse one: what function has transform \(X(s)\,Y(s)\)?

Definition: Convolution

Definition

Let \(x(t)\) and \(y(t)\) be defined on \([0,\infty)\). The convolution of \(x\) and \(y\), denoted \(x * y\), is \[ (x * y)(t) = \int_0^t x(\tau)\,y(t - \tau)\,d\tau. \]

How to read the integral: for fixed \(t\), substitute \(\tau\) for the argument of \(x\) and \(t-\tau\) for the argument of \(y\) (flipping and shifting \(y\)), then integrate from \(0\) to \(t\).

Why This Structure? — Memory and History

Tip

Think of the integral as a weighted accumulation of history: it combines the past values of \(x\) against a time-reversed, shifted copy of \(y\).

This memory structure arises naturally whenever a system’s current output depends on all of its past inputs — exactly the behavior encoded in a forced differential equation.

The key question the convolution answers:

Given two transforms \(X(s)\) and \(Y(s)\), what time-domain function has transform equal to their product \(X(s)\,Y(s)\)?

The answer — the Convolution Theorem — is on the next slide.

The Convolution Property

Theorem: Convolution Property

If \(\mathscr{L}[x] = X(s)\) and \(\mathscr{L}[y] = Y(s)\), then \[\mathscr{L}[x * y] = X(s)\,Y(s).\]

Equivalently, \[\mathscr{L}^{-1}[X(s)\,Y(s)] = (x * y)(t) = \int_0^t x(\tau)\,y(t-\tau)\,d\tau.\]

Why this matters: solving a DE via Laplace transforms almost always produces a product of transforms \(X(s) = K(s)\,F(s)\). The convolution property tells us exactly how to invert it.

Proof Sketch

Start from the definition, pull \(e^{-st}\) inside, then interchange the order of integration. The region \(0 \le \tau \le t < \infty\) is the same as \(0 \le \tau < \infty\), \(\tau \le t < \infty\):

\[ \mathscr{L}\!\left[\int_0^t x(\tau)\,y(t-\tau)\,d\tau\right] = \int_0^\infty\!\left(\int_\tau^\infty y(t-\tau)\,e^{-st}\,dt\right)x(\tau)\,d\tau. \]

Substitute \(r = t-\tau\) in the inner integral:

\[ = \int_0^\infty\!\left(\int_0^\infty y(r)\,e^{-sr}\,dr\right)e^{-s\tau}\,x(\tau)\,d\tau = \left(\int_0^\infty e^{-s\tau}x(\tau)\,d\tau\right)\!\left(\int_0^\infty y(r)e^{-sr}\,dr\right) = X(s)\,Y(s). \;\square \]

Commutativity of Convolution

Remark 3.23

\[(x * y)(t) = (y * x)(t).\] Convolution is commutative. (Proved by a change of variables — Exercise 5, §3.3.)

Practical Consequence

You can choose which function to shift when setting up the integral. Always pick the arrangement that makes the integration easier.

In Example 3.22 below, shifting the constant \(1\) is far simpler than shifting \(t^2\).

Example 3.22 — Computing \(1 * t^2\) Directly

Find the convolution \(1 * t^2\).

Apply the definition with \(x(\tau) = 1\), \(y(t-\tau) = (t-\tau)^2\):

\[ 1 * t^2 = \int_0^t 1\cdot(t-\tau)^2\,d\tau = \int_0^t(t^2-2t\tau+\tau^2)\,d\tau = t^2\cdot t - 2t\!\left(\frac{t^2}{2}\right)+\frac{t^3}{3} = \frac{t^3}{3}. \]

Verify via commutativity — shift the simpler function instead:

\[t^2 * 1 = \int_0^t \tau^2\cdot 1\,d\tau = \frac{t^3}{3}. \checkmark\]

The second calculation is shorter because shifting the constant \(1\) changes nothing. Commutativity always gives you this option. \(\square\)

Example 3.24 — Inverse Transform via Convolution

Find \(\mathscr{L}^{-1}\!\left[\dfrac{3}{s(s^2+9)}\right]\).

We can use partial fractions, but convolution is more direct. Factor into recognizable table entries:

\[\frac{3}{s(s^2+9)} = \underbrace{\frac{1}{s}}_{\mathscr{L}^{-1}\;\to\; 1}\cdot\underbrace{\frac{3}{s^2+9}}_{\mathscr{L}^{-1}\;\to\;\sin 3t}\]

Apply the convolution property: \[\mathscr{L}^{-1}\!\left[\frac{3}{s(s^2+9)}\right] = 1 * \sin 3t = \int_0^t \sin 3\tau\,d\tau\]

Evaluate: \[= \left[-\frac{1}{3}\cos 3\tau\right]_0^t = \boxed{\frac{1}{3}(1-\cos 3t).} \;\square\]

Show the code 
s_v, t_v = sym.symbols('s t', positive=True)
X_24 = sym.Integer(3)/(s_v*(s_v**2+9))
x_24 = sym.inverse_laplace_transform(X_24, s_v, t_v)
display(Math(r"\mathscr{L}^{-1}\!\left[\frac{3}{s(s^2+9)}\right] = " + sym.latex(sym.simplify(x_24))))

\(\displaystyle \mathscr{L}^{-1}\!\left[\frac{3}{s(s^2+9)}\right] = \frac{1}{3} - \frac{\cos{\left(3 t \right)}}{3}\)

Example 3.25 — DE with Arbitrary Forcing

Solve \(x'' + k^2 x = f(t)\), \(x(0)=x_0\), \(x'(0)=x_1\), for any input \(f(t)\).

Transform and solve for \(X(s)\): \[X(s) = x_0\,\frac{s}{s^2+k^2} + x_1\,\frac{1}{s^2+k^2} + \frac{F(s)}{s^2+k^2}.\]

Invert — table for the first two terms, convolution on the last:

\[\boxed{x(t) = x_0\cos kt + \frac{x_1}{k}\sin kt + \frac{1}{k}\int_0^t f(\tau)\sin k(t-\tau)\,d\tau.}\]

Note

This formula is valid for any forcing \(f(t)\) — we never need to know the explicit form of \(f\) to write it down. The convolution integral absorbs all the complexity of the input into a single expression. \(\square\)

Integral Equations (§3.3, Exercises 14–17)

The convolution property also solves integral equations — equations where the unknown \(x(t)\) appears under an integral. An important class:

\[x(t) = f(t) + \int_0^t k(t-\tau)\,x(\tau)\,d\tau \qquad (k = \text{kernel}).\]

Take \(\mathscr{L}\) of both sides and recognise the convolution:

\[X(s) = F(s) + K(s)\,X(s) \implies X(s) = \frac{F(s)}{1-K(s)}.\]

Then invert \(X(s)\) to find \(x(t)\).

Section 3.4

Impulsive Sources

Motivation — Forces Acting in an Instant

Some forcing terms act at a single instant of time:

  • Mass struck by a swift hammer blow
  • Medication injection idealized as instantaneous
  • Electrical switch closed for only an instant → impulsive voltage
  • Shock absorber hitting a bump in the road

We need a mathematical model for a force that:

  • Acts at a single point \(t = a\)
  • Still has nonzero total effect (unit area)

This leads to the delta function \(\delta_a(t)\).

A Naive Definition Fails

Consider the damped mass \(mv' + v = f(t)\), \(v(0)=0\). By Laplace / convolution:

\[v(t) = \frac{1}{m}\int_0^t e^{-(t-\tau)/m}\,f(\tau)\,d\tau. \tag{3.9}\]

Naive definition: set \(\delta_a(t) = 1\) if \(t=a\), zero otherwise.

Problem: substituting into (3.9) gives \(v(t) = 0\) for all \(t\). The integrand is nonzero at only one point, which does not contribute to the integral.

This is physically wrong

A particle struck by a hammer that never moves is nonsensical. The pointwise definition fails. We need a fundamentally different approach.

The Physical Intuition: Impulse = Change in Momentum

In physics, impulse = change in momentum: \(\Delta p = f(t)\,\Delta t\).

For a unit impulse centered at \(t=a\) over interval \((a-\varepsilon/2,\,a+\varepsilon/2)\): \[\Delta p = \int_{a-\varepsilon/2}^{a+\varepsilon/2} f(t)\,dt = 1 \qquad \text{for every } \varepsilon > 0.\]

Model the impulse by the rectangular function

\[f_\varepsilon(t) = \begin{cases} 1/\varepsilon, & a-\varepsilon/2 < t < a+\varepsilon/2 \\ 0, & \text{otherwise.}\end{cases}\]

Property Value
Height \(1/\varepsilon\)
Width \(\varepsilon\)
Area \(1\) (always)

As \(\varepsilon\to 0\): taller, narrower — area always 1.

The Rectangular Approximation — Visual

Figure 1

Tip

Each rectangle has area exactly 1, regardless of \(\varepsilon\). As \(\varepsilon\to 0\) the rectangle becomes the idealized delta function \(\delta_a(t)\).

The Delta Function — Sifting Property

The unit impulse \(\delta_a(t)\) is a generalized function — it cannot be defined pointwise, only through how it acts under integration:

Definition: Sifting Property

\[\int_0^\infty \delta_a(t)\,\phi(t)\,dt = \phi(a)\] for any continuous function \(\phi(t)\).

Integrating \(\delta_a\) against \(\phi\) picks out the value of \(\phi\) at \(t=a\).

Tip

This is the definition of the delta function — not a derived property. It replaces the failed pointwise definition by specifying how \(\delta_a\) acts, not what it equals.

Sifting Property — Variable Limit and Verification

Over a variable upper limit: \[\int_0^t \delta_a(\tau)\,\phi(t-\tau)\,d\tau = H(t-a)\,\phi(a).\]

The integral is zero for \(t < a\) (the impulse hasn’t fired yet) and equals \(\phi(a)\) for \(t \geq a\).

Returning to the velocity problem: for the damped mass \(mv'+v = \delta_a(t)\), \(v(0)=0\), we had: \[v(t) = \frac{1}{m}\int_0^t e^{-(t-\tau)/m}\,\delta_a(\tau)\,d\tau = \frac{H(t-a)}{m}\,e^{-(t-a)/m}. \checkmark\]

Note

Zero before the impulse, jumps to \(1/m\) at \(t=a\), then decays — exactly what a sharp blow should produce. The naive pointwise definition gave zero; the sifting property gives the correct physics.

Laplace Transform of \(\delta_a(t)\)

Apply the sifting property with \(\phi(t) = e^{-st}\):

\[\mathscr{L}[\delta_a(t)] = \int_0^\infty \delta_a(t)\,e^{-st}\,dt = e^{-as}.\]

Key Formulas

\[\mathscr{L}[\delta_a(t)] = e^{-as}, \quad a\ge 0.\] \[\mathscr{L}[\delta_0(t)] = 1. \qquad \mathscr{L}^{-1}[1] = \delta_0(t).\]

Tip

Notice: \(\mathscr{L}[\delta_a(t)] = e^{-as}\) is exactly the same factor that the switching property produces for time-delayed Heaviside terms. The inversion procedure is identical — a factor \(e^{-as}\) always signals a delay of \(a\), whether it came from a Heaviside function or a delta function.

Alternate Derivation — Limiting Argument

We can also compute \(\mathscr{L}[\delta_a(t)]\) by transforming the rectangular approximation \(f_\varepsilon\) and taking \(\varepsilon\to 0\):

\[\mathscr{L}[f_\varepsilon(t)] = e^{-as}\,\frac{\sinh(\varepsilon s/2)}{\varepsilon s/2} \xrightarrow[\varepsilon\to 0]{\text{l'Hôpital}} e^{-as}\cdot 1 = e^{-as}. \checkmark\]

This confirms that the distributional definition (sifting property) and the limiting definition (rectangles shrinking to a spike) give the same Laplace transform.

Note

The limiting argument is the most physically transparent justification: as the impulse gets sharper and shorter while keeping unit area, its transform converges to \(e^{-as}\). The sifting property then provides the rigorous mathematical foundation.

Solving IVPs with Delta Forcing — The Method

Procedure: identical to previous Laplace methods.

  1. Transform both sides. Replace \(\delta_a(t)\) with \(e^{-as}\) directly.
  2. Solve for \(X(s)\) algebraically.
  3. Invert — the \(e^{-as}\) factor is handled by the switching property, exactly as with \(\mathscr{L}[H(t-a)f(t-a)] = e^{-as}F(s)\).

Tip

The only new ingredient is the formula \(\mathscr{L}[\delta_a(t)] = e^{-as}\). Everything else — partial fractions, the switching property, the table — works exactly as before.

Example 3.26 — IVP with a Unit Impulse

Solve \(x''+x' = \delta_2(t)\), \(x(0)=x'(0)=0\).

Step 1 — Transform (\(\mathscr{L}[\delta_2(t)] = e^{-2s}\)): \[s^2X + sX = e^{-2s} \implies X(s) = \frac{e^{-2s}}{s(s+1)}.\]

Step 2 — Base inverse transform (partial fractions or table): \[\mathscr{L}^{-1}\!\left[\frac{1}{s(s+1)}\right] = 1 - e^{-t}.\]

Step 3 — Switching property (\(e^{-2s}\) signals a delay of 2): \[x(t) = \bigl(1-e^{-(t-2)}\bigr)\,H(t-2).\]

Region \(x(t)\) Meaning
\(0\le t < 2\) \(0\) System at rest
\(t=2\) impulse fires Activated
\(t\to\infty\) \(\to 1\) Approaches steady state

Example 3.26 — Solution Plot

Figure 2

Input–Output Systems

Transfer Functions

Many engineering problems fit the input–output framework:

\[\underbrace{f(t)}_{\text{input}}\;\longrightarrow\;\underbrace{ax''+bx'+cx=f(t),\;x(0)=x'(0)=0}_{\text{system}}\;\longrightarrow\;\underbrace{x(t)}_{\text{output (forced response)}}\]

In the \(s\)-domain: \(X(s) = K(s)\,F(s)\), where

Transfer Function

\[K(s) = \frac{1}{as^2+bs+c}.\] Knowing \(K(s)\) gives complete knowledge of the system’s response to any input.

By the convolution property, the time-domain forced response is: \[x(t) = k(t)*f(t) = \int_0^t k(t-\tau)\,f(\tau)\,d\tau, \qquad k(t)=\mathscr{L}^{-1}[K(s)].\]

Time Domain vs. Transform Domain

Time domain

\[x(t) = k(t) * f(t)\]

Convolution — weighted accumulation of past inputs against the system kernel \(k\)

Transform domain

\[X(s) = K(s)\,F(s)\]

Multiplication — simple algebra

\[\text{Convolution in time} \;\longleftrightarrow\; \text{Multiplication in } s.\]

This is one of the deepest reasons Laplace transforms are so powerful for engineering: a complicated time-domain operation becomes trivial algebra in the \(s\)-domain.

Example 3.27 — General Impulse Response

If \(f(t) = \delta_{t_0}(t)\), then \(F(s) = e^{-t_0 s}\) and:

Transform domain: \(X(s) = K(s)\,e^{-t_0 s}\).

Time domain (sifting property applied to the convolution): \[x(t) = k(t)*\delta_{t_0}(t) = \int_0^t k(t-\tau)\,\delta_{t_0}(\tau)\,d\tau = H(t-t_0)\,k(t-t_0). \;\square\]

Note

The impulse response is a time-shifted copy of \(k(t)\), switched on at \(t=t_0\). This is why \(k(t)\) is called the impulse response function of the system.

Example 3.28 — Damped Oscillator Impulse Response

Solve \(x''+2x'+5x = \delta_{t_0}(t)\), \(x(0)=x'(0)=0\).

Transfer function: \[K(s) = \frac{1}{s^2+2s+5} = \frac{1}{2}\cdot\frac{2}{(s+1)^2+4}\]

so \(k(t) = \mathscr{L}^{-1}[K(s)] = \tfrac{1}{2}\,e^{-t}\sin 2t\).

Impulse response (by Example 3.27): \[x(t) = H(t-t_0)\,\frac{1}{2}\,e^{-(t-t_0)}\sin 2(t-t_0). \;\square\]

Zero until \(t=t_0\); then damped oscillation for \(t>t_0\). See Figure 3.10 in the text for \(t_0=2\).

Figure 3

Summary

Key Formulas — Convolution

Concept Formula
Definition \((x*y)(t) = \displaystyle\int_0^t x(\tau)\,y(t-\tau)\,d\tau\)
Convolution property \(\mathscr{L}[x*y] = X(s)\,Y(s)\)
Inverse form \(\mathscr{L}^{-1}[X(s)\,Y(s)] = (x*y)(t)\)
Commutativity \((x*y)(t) = (y*x)(t)\)
Arbitrary forcing \(x(t) = \frac{1}{k}\int_0^t f(\tau)\sin k(t-\tau)\,d\tau + \cdots\)

Key Formulas — Delta Function and Transfer Function

Concept Formula
Sifting property \(\displaystyle\int_0^\infty \delta_a(t)\,\phi(t)\,dt = \phi(a)\)
\(\mathscr{L}\) of delta \(\mathscr{L}[\delta_a(t)] = e^{-as}\)
Inverse \(\mathscr{L}^{-1}[e^{-as}] = \delta_a(t)\)
Transfer function \(K(s) = 1/(as^2+bs+c)\)
Forced response (s-domain) \(X(s) = K(s)\,F(s)\)
Forced response (time) \(x(t) = k(t)*f(t)\)

Common Pitfalls

  • \(\mathscr{L}[x(t)y(t)] \neq X(s)Y(s)\) — the transform of a product is not the product of the transforms. The convolution \(\mathscr{L}^{-1}[X(s)Y(s)] = (x*y)(t)\) is something entirely different.
  • \(\delta_a(t)\) is not an ordinary function. Never try to evaluate it pointwise. It is defined only through the sifting property.
  • When using commutativity, check both arrangements of the convolution integrand and pick the one that is easier to integrate.
  • A factor \(e^{-as}\) in \(X(s)\) always signals a time delay of \(a\) — handle it with the switching property, whether it came from a Heaviside function or a delta function.

References

Logan, J David. 2015. A First Course in Differential Equations, Third Edition.