Difficulty: Intermediate to Advanced | Prerequisites: Core Laplace transform pairs (see companion document), integration by parts, basic ODE setup.
This is the second of two study-notes documents on the Laplace transform table. It covers the operational properties (time shifting, derivatives, integration, convolution) and special functions (Heaviside step, Dirac delta, Gamma function) that let you solve piecewise and higher-order ODE problems.
The operational properties of the Laplace transform are what make it powerful beyond simple table lookups. Time shifting (Heaviside functions) lets you handle piecewise inputs that switch on or off at a given time. The derivative and integral properties turn differential equations into polynomial algebra in s. The convolution theorem connects multiplication in the s-domain to a specific integral in the t-domain, which is essential for systems analysis.
Heaviside step function (unit step function)
Written u_c(t) or u(t - c). It equals 0 for t < c and 1 for t >= c. Think of it as a switch that turns on at time c. It is the standard way to write piecewise-defined forcing functions in Laplace problems.
Dirac delta function
Written delta(t - c). It is zero everywhere except at t = c, where it is "infinitely tall" with a total area of 1. In simple terms, it models an instantaneous impulse, like a hammer strike at time c. Its Laplace transform is simply e^(-cs).
Convolution
The integral (f * g)(t) = integral from 0 to t of f(t - tau) g(tau) d(tau). It describes how two functions "blend" over time. In the s-domain, convolution in time becomes simple multiplication: L{f * g} = F(s) G(s).
Second Shifting Theorem (time-shifting property)
L{u_c(t) f(t - c)} = e^(-cs) F(s). Multiplying by a delayed step function and shifting the argument of f corresponds to multiplying by e^(-cs) in the s-domain. This is how you handle inputs that start at a time other than zero.
Derivative property
L{f'(t)} = s F(s) - f(0). Each derivative in the time domain multiplies by s and subtracts an initial condition. This is the property that converts differential equations into algebraic equations.
Multiplication by t^n
L{t^n f(t)} = (-1)^n F^(n)(s), where F^(n) denotes the nth derivative of F with respect to s. Multiplying by t in time corresponds to differentiating (with a sign flip) in the s-domain.
The Heaviside function u_c(t) = u(t - c) is your tool for "turning on" a function at time c. Its transform is straightforward: L{u_c(t)} = e^(-cs) / s.
The real power comes from the Second Shifting Theorem:
L{u_c(t) f(t - c)} = e^(-cs) F(s). You delay f by c units and multiply by the step. In the s-domain, this just tacks on an exponential factor e^(-cs).
L{u_c(t) g(t)} = e^(-cs) L{g(t + c)}. An alternative form when g is not already written as a shifted function. You substitute t + c into g, take the transform, then multiply by e^(-cs).
The Dirac delta models an instantaneous impulse of unit strength at time c.
L{delta(t - c)} = e^(-cs). No denominator, no polynomial, just the exponential. When c = 0, L{delta(t)} = 1.
This is the simplest transform to remember and one of the most useful in applications. Any time a problem describes a sudden force, voltage spike, or impulse load, the Dirac delta is the model.
L{e^(ct) f(t)} = F(s - c). Multiplying by an exponential in the time domain shifts the argument of F in the s-domain. This was covered in the companion document for specific pairs, but it applies to any f(t) whose transform F(s) you already know.
L{t^n f(t)} = (-1)^n F^(n)(s), for n = 1, 2, 3, ...
Each factor of t corresponds to one differentiation with respect to s, with alternating signs. So L{t f(t)} = -F'(s), and L{t^2 f(t)} = F''(s).
L{f(t)/t} = integral from s to infinity of F(u) du.
Dividing by t in time corresponds to integrating the transform from s to infinity. This is less commonly tested but appears in problems involving integral representations.
L{integral from 0 to t of f(v) dv} = F(s) / s.
Integrating in time just divides by s. This is the counterpart to the derivative property: derivatives multiply by s, integrals divide by s.
L{integral from 0 to t of f(t - tau) g(tau) d(tau)} = F(s) G(s).
The convolution integral in the time domain becomes multiplication in the s-domain. This is central to systems theory: the output of a linear system is the convolution of its impulse response with the input, which in the s-domain is just the product of transfer function and input transform.
If f(t + T) = f(t), then L{f(t)} = [integral from 0 to T of e^(-st) f(t) dt] / (1 - e^(-sT)).
For a function that repeats with period T, you only need to compute the Laplace integral over one period and divide by (1 - e^(-sT)). This handles square waves, sawtooth waves, and other periodic inputs.
These are the properties that make Laplace transforms useful for solving ODEs.
L{f'(t)} = s F(s) - f(0). First derivative: multiply by s, subtract the initial value.
L{f''(t)} = s^2 F(s) - s f(0) - f'(0). Second derivative: multiply by s^2, subtract initial terms.
L{f^(n)(t)} = s^n F(s) - s^(n-1) f(0) - s^(n-2) f'(0) - ... - f^(n-1)(0). The general nth-derivative formula. Each successive term drops one power of s and picks up the next initial condition.
The Gamma function extends the factorial to non-integer (and even complex) arguments. It appears in the transform of t^p when p is not a positive integer.
Definition
Gamma(t) = integral from 0 to infinity of e^(-x) x^(t-1) dx.
Key identities
Gamma(n + 1) = n! for any positive integer n.
Gamma(p + 1) = p * Gamma(p) for any p. This is the recursive property, analogous to n! = n * (n-1)!.
Gamma(1/2) = sqrt(pi). This is the identity that makes L{sqrt(t)} work out to sqrt(pi) / (2 s^(3/2)).
p(p+1)(p+2)...(p+n-1) = Gamma(p+n) / Gamma(p). This is the rising factorial, sometimes called the Pochhammer symbol.
You will rarely need to compute Gamma integrals by hand. The exam use case is recognising when a transform involves Gamma(p+1) in place of n! and substituting the correct value, especially Gamma(1/2) = sqrt(pi) and Gamma(3/2) = sqrt(pi)/2.
The Heaviside function models any input that switches on at a set time: a valve opening in a fluid system, a voltage source connecting to a circuit at t = 5 seconds, a load being applied to a beam. The Dirac delta models impacts and impulses: a hammer blow on a structure, a lightning strike on a power line, a short burst of current. The convolution theorem is the backbone of signal processing and control engineering, where the output of any linear time-invariant system is the convolution of the input with the system's impulse response. The derivative properties are why Laplace transforms exist in engineering curricula at all: they turn a second-order ODE with initial conditions (describing a spring-mass-damper system, an RLC circuit, or a feedback controller) into a polynomial equation you can solve with partial fractions.
Forgetting to shift the function argument with the Heaviside. L{u_c(t) f(t - c)} = e^(-cs) F(s) requires that f is shifted by exactly c. If the problem gives you u_c(t) g(t) (where g is not already shifted), you must rewrite g(t) as a function of (t - c) or use the alternative form with g(t + c).
Dropping initial-condition terms from the derivative formula. L{f''(t)} = s^2 F(s) - s f(0) - f'(0). Every initial-condition term matters. Leaving out the -s f(0) or the -f'(0) makes the entire solution wrong.
Thinking convolution is commutative in the way ordinary multiplication is. Convolution is commutative (f * g = g * f), so the order does not matter, but students sometimes confuse the convolution integral's limits or bounds. The integral runs from 0 to t, not from 0 to infinity.
Confusing e^(-cs)/s (Heaviside) with e^(-cs) (Dirac delta). The step function transform has a 1/s in it; the delta does not. Forgetting the 1/s means you are modelling an impulse when you meant a step, or vice versa.
The derivative properties (first and second) are tested on nearly every Laplace-method ODE problem. You will be expected to set up s^2 F(s) - s f(0) - f'(0) = ... from a second-order ODE, solve for F(s), then invert using the table.
The Heaviside step function appears whenever a problem describes a piecewise or "switched" input: "a force of 10 N is applied at t = 3 seconds." Expect at least one such problem on an exam.
The convolution theorem is sometimes tested directly ("express the inverse transform of F(s)G(s) as a convolution integral") and sometimes indirectly (in systems or transfer-function problems where you multiply transforms).
The Dirac delta is a quick-mark problem: if you know L{delta(t - c)} = e^(-cs), you can write down the answer in one step.
True or false: L{delta(t - 4)} = e^(-4s) / s. False. There is no 1/s. The correct answer is simply e^(-4s). The 1/s belongs to the Heaviside function.
Fill in the blank: L{f'(t)} = s F(s) - ___. Answer: f(0).
True or false: L{integral from 0 to t of f(v) dv} = s F(s). False. Integration in time divides by s, not multiplies. The correct formula is F(s) / s.
Fill in the blank: L{u_c(t) f(t - c)} = ___ F(s). Answer: e^(-cs).
True or false: The convolution integral runs from 0 to infinity. False. It runs from 0 to t.
Q: Solve y'' + 4y = delta(t - pi), with y(0) = 0 and y'(0) = 0, by taking Laplace transforms.
A: Taking the transform: s^2 Y(s) + 4 Y(s) = e^(-pi s). So Y(s) = e^(-pi s) / (s^2 + 4). The inverse of 1/(s^2 + 4) is (1/2) sin(2t). Applying the Second Shifting Theorem: y(t) = (1/2) u_pi(t) sin(2(t - pi)).
Q: Find L{u_3(t) (t - 3)^2}.
A: This is u_c(t) f(t - c) with c = 3 and f(t) = t^2. So L = e^(-3s) L{t^2} = e^(-3s) * 2/s^3.
Q: Express L^(-1){1 / [s(s^2 + 1)]} using the convolution theorem.
A: Write 1/[s(s^2 + 1)] = (1/s) * (1/(s^2 + 1)) = L{1} * L{sin(t)}. So the inverse is the convolution: integral from 0 to t of sin(tau) d(tau) = 1 - cos(t).
Q: Given L{f(t)} = F(s), what is L{t f(t)}?
A: L{t f(t)} = -F'(s). You differentiate F with respect to s and negate.
Q: A piecewise function is defined as g(t) = 0 for t < 2, and g(t) = t for t >= 2. Write g(t) using Heaviside notation and find its Laplace transform.
A: g(t) = u_2(t) * t = u_2(t) [(t - 2) + 2]. So g(t) = u_2(t)(t - 2) + 2 u_2(t). Taking the transform: L{g} = e^(-2s) / s^2 + 2 e^(-2s) / s.
The derivative and initial-condition properties connect directly to your ODE course material: every initial-value problem you solve by Laplace transforms uses these. The convolution theorem links forward to signals and systems courses, where it underpins filtering, frequency response, and transfer-function analysis. The Heaviside and Dirac delta functions reappear in any course that models switched or impulsive inputs, from circuit theory (step response, impulse response) to structural dynamics (sudden loading). The Gamma function connects to probability and statistics, where it defines the Gamma distribution and appears in the normalisation of the chi-squared distribution.
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