Difficulty: Intermediate | Prerequisites: Calculus I (integration), basic complex numbers, familiarity with differential equations.
This is the first of two study-notes documents covering the standard Laplace transform table used in Calculus II for Engineers. This document covers the core transform pairs you need to recognise and apply. The companion document covers advanced operations (Heaviside, Dirac delta, convolution, derivatives in the s-domain) and the Gamma function.
Laplace transforms convert time-domain functions into s-domain (frequency-domain) expressions, which turns differential equations into algebraic ones you can solve with basic algebra. The standard table of transform pairs is essentially a dictionary: if you know f(t), you look up F(s), and vice versa. Memorising the core pairs and a handful of properties lets you solve most ODE problems on an exam without re-deriving anything from the integral definition.
Laplace transform
The integral operation that maps a time-domain function f(t) to a complex-frequency-domain function F(s), defined by F(s) = L{f(t)} = integral from 0 to infinity of e^(-st) f(t) dt. In simple terms, it converts calculus problems (derivatives, integrals) into algebra problems (polynomials in s).
Inverse Laplace transform
The reverse operation, written f(t) = L^(-1){F(s)}, that recovers the original time-domain function from its s-domain expression. Think of it as looking up F(s) in the table and reading off the corresponding f(t).
s-domain (frequency domain)
The domain in which Laplace-transformed functions live. The variable s is complex (s = sigma + j*omega). In simple terms, when you "take the Laplace transform," you move from t-world to s-world.
t-domain (time domain)
The original domain of the function, where the independent variable is time t and t >= 0. This is the world your physical problem lives in.
Transfer function
The ratio of output to input in the s-domain, commonly written H(s) = Y(s)/X(s). Think of it as the system's "recipe" for turning any input into its output, expressed entirely in s.
Gamma function
An extension of the factorial to non-integer values, defined by the integral from 0 to infinity of e^(-x) x^(t-1) dx. For positive integers, it reduces to the familiar factorial: Gamma(n+1) = n!. It appears in the transform of t^p when p is not a positive integer.
These are the pairs you will use most often. Almost every exam problem starts or ends with one of these.
1 maps to 1/s. The simplest pair. The constant function 1 (for t >= 0) transforms to 1/s.
e^(at) maps to 1/(s - a). The exponential shift: the pole in the s-domain sits at s = a. If a is negative you get exponential decay; if positive, exponential growth.
t^n (n = 1, 2, 3, ...) maps to n! / s^(n+1). Powers of t pull down factorials. So t maps to 1/s^2, t^2 maps to 2/s^3, t^3 maps to 6/s^4, and so on.
t^p (p > -1, not necessarily integer) maps to Gamma(p+1) / s^(p+1). The generalisation of the previous pair using the Gamma function. When p is a positive integer this reduces to the factorial form.
sqrt(t) maps to sqrt(pi) / (2 s^(3/2)). A special case of the t^p pair with p = 1/2, since Gamma(3/2) = sqrt(pi)/2.
t^(n - 1/2) (n = 1, 2, 3, ...) maps to [1 * 3 * 5 * ... * (2n - 1)] sqrt(pi) / (2^n s^(n + 1/2)). The half-integer powers; these appear less often but show up in heat-equation problems.
The sign in the denominator is the giveaway: s^2 + a^2 for "normal" trig, s^2 - a^2 for hyperbolic trig. Mixing these up is the single most common table-lookup error.
sin(at) maps to a / (s^2 + a^2). Numerator is the frequency a.
cos(at) maps to s / (s^2 + a^2). Numerator is s.
sinh(at) maps to a / (s^2 - a^2). Same form as sin, but with a minus in the denominator.
cosh(at) maps to s / (s^2 - a^2). Same form as cos, but with a minus in the denominator.
Recall the definitions: cosh(t) = (e^t + e^(-t))/2 and sinh(t) = (e^t - e^(-t))/2.
Multiplying by t in the time domain corresponds (roughly) to differentiating with respect to s in the s-domain. These pairs are commonly tested because they require you to recognise the squared denominator.
t sin(at) maps to 2as / (s^2 + a^2)^2.
t cos(at) maps to (s^2 - a^2) / (s^2 + a^2)^2.
These come up when partial-fraction decomposition produces repeated quadratic factors. The numerator pattern distinguishes them.
sin(at) - at cos(at) maps to 2a^3 / (s^2 + a^2)^2.
sin(at) + at cos(at) maps to 2as^2 / (s^2 + a^2)^2.
cos(at) - at sin(at) maps to s(s^2 - a^2) / (s^2 + a^2)^2.
cos(at) + at sin(at) maps to s(s^2 + 3a^2) / (s^2 + a^2)^2.
Useful when your problem involves a sine or cosine with a constant phase offset b.
sin(at + b) maps to [s sin(b) + a cos(b)] / (s^2 + a^2).
cos(at + b) maps to [s cos(b) - a sin(b)] / (s^2 + a^2).
When an exponential multiplies a trig or hyperbolic function, you replace s with (s - a) everywhere in the trig transform. This is the First Shifting Theorem in action.
e^(at) sin(bt) maps to b / [(s - a)^2 + b^2].
e^(at) cos(bt) maps to (s - a) / [(s - a)^2 + b^2].
e^(at) sinh(bt) maps to b / [(s - a)^2 - b^2].
e^(at) cosh(bt) maps to (s - a) / [(s - a)^2 - b^2].
t^n e^(at) (n = 1, 2, 3, ...) maps to n! / (s - a)^(n+1). Again, the First Shifting Theorem: take the t^n pair and replace every s with (s - a).
f(ct) maps to (1/c) F(s/c). If you compress or stretch the time axis by a factor c, the s-domain scales inversely.
Confusing + a^2 with - a^2 in the denominator. The denominator s^2 + a^2 belongs to sin and cos. The denominator s^2 - a^2 belongs to sinh and cosh. Swapping the sign gives you the wrong function entirely. If you remember only one thing from this section, remember this.
Forgetting the numerator difference between sin and cos. sin(at) puts the constant a in the numerator. cos(at) puts s in the numerator. Students often write s for both or a for both.
Misapplying the s-shift. The First Shifting Theorem says L{e^(at) f(t)} = F(s - a). The shift replaces s with (s - a), not (s + a). A positive exponent a shifts the pole to the right.
Using n! when the Gamma function is needed. The pair t^n with n! / s^(n+1) only works for non-negative integers. For fractional powers like t^(1/2), you must use Gamma(p+1) / s^(p+1). The factorial is a special case of the Gamma function, not the other way round.
Laplace transforms are the standard tool for solving linear constant-coefficient ODEs in engineering courses. Every circuit-analysis, control-systems, and vibrations problem you will meet later relies on fluency with this table. On a Calculus II exam, you can expect at least one problem that requires you to:
Recognise a standard pair from the table and write down F(s) or f(t) directly.
Apply the First Shifting Theorem (exponential times a function) without hesitation.
Distinguish between trig and hyperbolic forms by the sign in the denominator.
The pairs for 1, e^(at), t^n, sin(at), and cos(at) appear on virtually every exam. The exponential-trig combinations e^(at) sin(bt) and e^(at) cos(bt) are close behind. If you have these committed to memory, you can handle partial-fraction results quickly and focus your time on the harder setup work.
True or false: L{sin(at)} = a / (s^2 - a^2). False. The correct denominator is s^2 + a^2. The minus sign belongs to sinh.
Fill in the blank: L{e^(3t)} = 1 / (s - ___). Answer: 3.
True or false: L{t^2} = 2 / s^2. False. L{t^2} = 2! / s^3 = 2/s^3. The exponent of s is n+1, not n.
Fill in the blank: L{e^(at) cos(bt)} = (s - a) / [(s - ___)^2 + b^2]. Answer: a.
True or false: The Gamma function satisfies Gamma(1/2) = sqrt(pi). True.
Q: Find L{5e^(-2t) sin(3t)}.
A: Apply the exponential-trig pair with a = -2 and b = 3. L{e^(at) sin(bt)} = b / [(s - a)^2 + b^2]. So the answer is 5 * 3 / [(s + 2)^2 + 9] = 15 / (s^2 + 4s + 13).
Q: What is the inverse Laplace transform of 6 / s^4?
A: Use L{t^n} = n! / s^(n+1). Here n+1 = 4, so n = 3 and n! = 6. The 6 in the numerator and the 6 from 3! cancel, giving f(t) = t^3.
Q: Find L{t cos(2t)}.
A: The pair is L{t cos(at)} = (s^2 - a^2) / (s^2 + a^2)^2. With a = 2: F(s) = (s^2 - 4) / (s^2 + 4)^2.
Q: Determine L^(-1){(s - 1) / [(s - 1)^2 + 16]}.
A: This matches the form (s - a) / [(s - a)^2 + b^2] with a = 1 and b = 4. The inverse is e^t cos(4t).
Q: A student claims L{cosh(5t)} = s / (s^2 + 25). What is the error?
A: The denominator should be s^2 - 25, not s^2 + 25. The plus sign belongs to cos, not cosh. The correct transform is s / (s^2 - 25).
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