Inverse Laplace Formula Sheet

INVERSE LAPLACE TRANSFORM
STANDARD FORMS

SJMATHTUBE | Exam-ready inverse Laplace formula sheet for engineering mathematics

1 Meaning of Inverse Laplace Transform Start here

Laplace transform changes \(f(t)\) into \(F(s)\). Inverse Laplace transform brings \(F(s)\) back to \(f(t)\).

Laplace Transform \(\mathcal{L}\{f(t)\}=F(s)\)

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Inverse Laplace Transform \(\mathcal{L}^{-1}\{F(s)\}=f(t)\)

In inverse Laplace, we usually start with an expression in \(s\) and convert it back to a function of \(t\).

2 Basic Inverse Standard Forms Core

\[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\}=1 \]

\[ \mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at} \]

\[ \mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\}=e^{-at} \]

3 Trigonometric Inverse Forms Plus sign

\[ \mathcal{L}^{-1}\left\{\frac{a}{s^{2}+a^{2}}\right\}=\sin at \]

\[ \mathcal{L}^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\}=\cos at \]

\[ \mathcal{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\}=\frac{1}{a}\sin at \]

If the numerator is \(1\) instead of \(a\), the answer becomes \(\frac{1}{a}\sin at\).

4 Hyperbolic Inverse Forms Minus sign

\[ \mathcal{L}^{-1}\left\{\frac{a}{s^{2}-a^{2}}\right\}=\sinh at \]

\[ \mathcal{L}^{-1}\left\{\frac{s}{s^{2}-a^{2}}\right\}=\cosh at \]

\[ \mathcal{L}^{-1}\left\{\frac{1}{s^{2}-a^{2}}\right\}=\frac{1}{a}\sinh at \]

5 Power of \(s\) Form Useful

\[ \mathcal{L}^{-1}\left\{\frac{1}{s^{n}}\right\}=\frac{t^{n-1}}{(n-1)!} \]

This form is useful when the denominator is a pure power of \(s\).

6 First Shifting Theorem Change in s

Forward form

\[ \text{If }\mathcal{L}\{f(t)\}=F(s),\text{ then }\mathcal{L}\{e^{at}f(t)\}=F(s-a) \]

Inverse form

\[ \mathcal{L}^{-1}\{F(s-a)\}=e^{at}\mathcal{L}^{-1}\{F(s)\} \]

\[ \mathcal{L}^{-1}\{F(s+a)\}=e^{-at}\mathcal{L}^{-1}\{F(s)\} \]

First shifting changes \(s\) inside \(F(s)\).

7 Second Shifting Theorem Unit step

Forward form

\[ \mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s) \]

Inverse form

\[ \mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a) \]

Second shifting usually appears when \(e^{-as}\) is multiplied with \(F(s)\).

8 Convolution Theorem Product form

\[ \mathcal{L}^{-1}\{F(s)G(s)\}=\mathcal{L}^{-1}\{F(s)\} * \mathcal{L}^{-1}\{G(s)\} \]

Use this only when the expression is naturally a product of two Laplace forms.

▶ Learn Inverse Laplace with Videos Playlist

Use this video playlist with the formula sheet for Laplace Transform, Inverse Laplace Transform, shifting theorems, and exam-style problems.

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▶ PART 1 Introduction

PART 2 Basics

INVERSE Standard problems

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9 Memory Aid Remember fast

\(\frac{1}{s}\) This gives \(1\).

\(\frac{1}{s-a}\) This gives \(e^{at}\).

\(\frac{1}{s+a}\) This gives \(e^{-at}\).

Plus and minus \(s^{2}+a^{2}\) points to trig. \(s^{2}-a^{2}\) points to hyperbolic.

First shifting Look for \(F(s-a)\) or \(F(s+a)\).

Second shifting Look for \(e^{-as}F(s)\).

10 Common Mistake Careful

First shifting is not second shifting \(F(s-a)\) is first shifting. \(e^{-as}F(s)\) is second shifting.

Check the numerator \(\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\}\) is \(\frac{1}{a}\sin at\), not \(\sin at\).

INVERSE LAPLACE TRANSFORM STANDARD FORMS

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INVERSE LAPLACE TRANSFORM
STANDARD FORMS