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Laplace Transform Table: The Most Commonly Used Pairs

Q: When can the final value theorem be used?

f(infty)=lim_{sto0}sF(s) is valid only when all poles of sF(s) lie in the left half of the s plane. If there are poles on the imaginary axis or in the right half plane (for example a sine that oscillates forever) the formula gives a wrong answer.

Q: How do I remember the whole table?

Remember three main pairs first: 1/s, 1/(s+a) and omega/(s^2+omega^2). The rest follow from properties: multiplying by e^{-at} replaces s with s+a, and multiplying by t means differentiating with respect to s and changing the sign.

A summary of the transform pairs and properties used in almost every circuit and control problem. Every pair is the unilateral Laplace transform ( $t \ge 0$ ), with tips so you do not have to memorise the whole table.

By Ajarn ModUpdated 16 Oct 2026

Table of basic transform pairs

In the table, $a$ and $\omega$ are real constants, $n$ is a positive integer, and every time function is understood to be multiplied by $u(t)$ .

#	$f(t)$ , $t \ge 0$	$F(s)$
1	$\delta(t)$ (unit impulse)	$1$
2	$1$ or $u(t)$	$\dfrac{1}{s}$
3	$t$	$\dfrac{1}{s^2}$
4	$t^n$	$\dfrac{n!}{s^{n+1}}$
5	$e^{-at}$	$\dfrac{1}{s+a}$
6	$t\,e^{-at}$	$\dfrac{1}{(s+a)^2}$
7	$t^n e^{-at}$	$\dfrac{n!}{(s+a)^{n+1}}$
8	$\sin\omega t$	$\dfrac{\omega}{s^2+\omega^2}$
9	$\cos\omega t$	$\dfrac{s}{s^2+\omega^2}$
10	$e^{-at}\sin\omega t$	$\dfrac{\omega}{(s+a)^2+\omega^2}$
11	$e^{-at}\cos\omega t$	$\dfrac{s+a}{(s+a)^2+\omega^2}$
12	$\sinh at$	$\dfrac{a}{s^2-a^2}$
13	$\cosh at$	$\dfrac{s}{s^2-a^2}$
14	$t\sin\omega t$	$\dfrac{2\omega s}{(s^2+\omega^2)^2}$
15	$t\cos\omega t$	$\dfrac{s^2-\omega^2}{(s^2+\omega^2)^2}$

Table of commonly used properties

Let $\mathcal{L}\{f(t)\}=F(s)$ and $\mathcal{L}\{g(t)\}=G(s)$ .

Property	Time domain	$s$ domain
Linearity	$af(t)+bg(t)$	$aF(s)+bG(s)$
Frequency shift (multiplying by an exponential)	$e^{-at}f(t)$	$F(s+a)$
Time shift (delay)	$f(t-T)\,u(t-T)$ , $T>0$	$e^{-sT}F(s)$
First derivative	$f'(t)$	$sF(s)-f(0)$
Second derivative	$f''(t)$	$s^2F(s)-sf(0)-f'(0)$
Integral	$\displaystyle\int_0^t f(\tau)\,d\tau$	$\dfrac{F(s)}{s}$
Multiplication by $t$	$t\,f(t)$	$-\dfrac{dF}{ds}$
Convolution	$(f*g)(t)$	$F(s)\,G(s)$
Initial value	$f(0^+)$	$\displaystyle\lim_{s\to\infty}sF(s)$
Final value	$f(\infty)$	$\displaystyle\lim_{s\to0}sF(s)$ (when all poles of $sF$ are in the left half plane)

Tips for remembering the table without rote learning

Three pairs are enough: $1/s$ , $1/(s+a)$ , $\omega/(s^2+\omega^2)$ . The rest can be built from them.
Seeing $(s+a)$ in place of $s$ means multiplying by $e^{-at}$ in the time domain. For example, replacing $s \to s+a$ in $\omega/(s^2+\omega^2)$ gives pair 10.
Sine versus cosine: sine has $\omega$ in the numerator (since $\sin 0 = 0$ there is no $s$ term), while cosine has $s$ in the numerator (since $\cos 0 = 1$ ).
Multiplying by $t$ means differentiating with respect to $s$ and changing the sign, for example $-\dfrac{d}{ds}\dfrac{1}{s+a}=\dfrac{1}{(s+a)^2}$ gives pair 6.

Using the tip

Find $\mathcal{L}\{t\cos\omega t\}$ from pair 9

Multiplying by $t$ means $-\dfrac{d}{ds}$
$-\frac{d}{ds}\frac{s}{s^2+\omega^2} = -\frac{(s^2+\omega^2) - s(2s)}{(s^2+\omega^2)^2} = \frac{s^2-\omega^2}{(s^2+\omega^2)^2}$

This matches pair 15 in the table ✓

Keep learning in the library

The full lessons are in Thai, with English technical terms and the same formulas.

Frequently asked questions

When can the final value theorem be used?

$f(\infty)=\lim_{s\to0}sF(s)$ is valid only when all poles of $sF(s)$ lie in the left half of the $s$ plane. If there are poles on the imaginary axis or in the right half plane (for example a sine that oscillates forever) the formula gives a wrong answer.

Does this table apply for t less than 0?

No. This table is for the unilateral Laplace transform, which treats the function as zero for negative time. Every function in the table therefore has an implied factor of $u(t)$ .

How do I remember the whole table?

Remember three main pairs first: $1/s$ , $1/(s+a)$ and $\omega/(s^2+\omega^2)$ . The rest follow from properties: multiplying by $e^{-at}$ replaces $s$ with $s+a$ , and multiplying by $t$ means differentiating with respect to $s$ and changing the sign.