Laplace Transforms The definition The Laplace transform turns a function of time f ( t ) f(t) f ( t ) s s s t t t s s s
L { f ( t ) } = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\, dt L { f ( t )} = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t Standard transforms and properties Common Laplace transforms and properties f(t) F(s) 1 1 1 1 s \dfrac{1}{s} s 1 t n t^n t n n ! s n + 1 \dfrac{n!}{s^{n+1}} s n + 1 n ! t 3 → 6 s 4 t^3 \to \dfrac{6}{s^4} t 3 → s 4 6 e a t e^{at} e a t 1 s − a \dfrac{1}{s-a} s − a 1 sin a t \sin at sin a t a s 2 + a 2 \dfrac{a}{s^2+a^2} s 2 + a 2 a cos a t \cos at cos a t s s 2 + a 2 \dfrac{s}{s^2+a^2} s 2 + a 2 s e a t f ( t ) e^{at}f(t) e a t f ( t ) F ( s − a ) F(s-a) F ( s − a ) f ( t − a ) u ( t − a ) f(t-a)u(t-a) f ( t − a ) u ( t − a ) a > 0 a > 0 a > 0 e − a s F ( s ) e^{-as}F(s) e − a s F ( s ) t f ( t ) t\,f(t) t f ( t ) − F ′ ( s ) -F'(s) − F ′ ( s ) f ′ ( t ) f'(t) f ′ ( t ) s F ( s ) − f ( 0 ) sF(s) - f(0) s F ( s ) − f ( 0 ) f ′ ′ ( t ) f''(t) f ′′ ( t ) s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 ) s^2F(s) - sf(0) - f'(0) s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 )
The transform is linear, so it can be applied term by term.
Worked example: first shifting From L { sin 4 t } = 4 s 2 + 16 \mathcal{L}\{\sin 4t\} = \dfrac{4}{s^2+16} L { sin 4 t } = s 2 + 16 4 s s s s − 3 s-3 s − 3 L { e 3 t sin 4 t } = 4 ( s − 3 ) 2 + 16 \mathcal{L}\{e^{3t}\sin 4t\} = \dfrac{4}{(s-3)^2+16} L { e 3 t sin 4 t } = ( s − 3 ) 2 + 16 4
Similarly, L { t sin t } = − d d s ( 1 s 2 + 1 ) = 2 s ( s 2 + 1 ) 2 \mathcal{L}\{t\sin t\} = -\dfrac{d}{ds}\left(\dfrac{1}{s^2+1}\right) = \dfrac{2s}{(s^2+1)^2} L { t sin t } = − d s d ( s 2 + 1 1 ) = ( s 2 + 1 ) 2 2 s
Worked example: a first-order ODE Solve d y d t + 2 y = e − t \dfrac{dy}{dt} + 2y = e^{-t} d t d y + 2 y = e − t y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0
s Y − 0 + 2 Y = 1 s + 1 ⇒ Y = 1 ( s + 1 ) ( s + 2 ) = 1 s + 1 − 1 s + 2 ⇒ y ( t ) = e − t − e − 2 t sY - 0 + 2Y = \dfrac{1}{s+1} \;\Rightarrow\; Y = \dfrac{1}{(s+1)(s+2)} = \dfrac{1}{s+1} - \dfrac{1}{s+2} \;\Rightarrow\; y(t) = e^{-t} - e^{-2t} s Y − 0 + 2 Y = s + 1 1 ⇒ Y = ( s + 1 ) ( s + 2 ) 1 = s + 1 1 − s + 2 1 ⇒ y ( t ) = e − t − e − 2 t Check: y ( 0 ) = 0 y(0) = 0 y ( 0 ) = 0 y ′ + 2 y = ( − e − t + 2 e − 2 t ) + ( 2 e − t − 2 e − 2 t ) = e − t y' + 2y = (-e^{-t}+2e^{-2t}) + (2e^{-t}-2e^{-2t}) = e^{-t} y ′ + 2 y = ( − e − t + 2 e − 2 t ) + ( 2 e − t − 2 e − 2 t ) = e − t
Worked example: a second-order ODE Solve y ′ ′ + 3 y ′ + 2 y = 0 y'' + 3y' + 2y = 0 y ′′ + 3 y ′ + 2 y = 0 y ( 0 ) = 1 y(0) = 1 y ( 0 ) = 1 y ′ ( 0 ) = 0 y'(0) = 0 y ′ ( 0 ) = 0
( s 2 Y − s ) + 3 ( s Y − 1 ) + 2 Y = 0 ⇒ Y = s + 3 ( s + 1 ) ( s + 2 ) = 2 s + 1 − 1 s + 2 (s^2Y - s) + 3(sY - 1) + 2Y = 0 \;\Rightarrow\; Y = \dfrac{s+3}{(s+1)(s+2)} = \dfrac{2}{s+1} - \dfrac{1}{s+2} ( s 2 Y − s ) + 3 ( s Y − 1 ) + 2 Y = 0 ⇒ Y = ( s + 1 ) ( s + 2 ) s + 3 = s + 1 2 − s + 2 1 So y ( t ) = 2 e − t − e − 2 t y(t) = 2e^{-t} - e^{-2t} y ( t ) = 2 e − t − e − 2 t Check: y ( 0 ) = 2 − 1 = 1 y(0) = 2 - 1 = 1 y ( 0 ) = 2 − 1 = 1 y ′ ( 0 ) = − 2 + 2 = 0 y'(0) = -2 + 2 = 0 y ′ ( 0 ) = − 2 + 2 = 0
Initial and final value theorems f ( 0 + ) = lim s → ∞ s F ( s ) f(0^+) = \lim_{s\to\infty} sF(s) f ( 0 + ) = lim s → ∞ s F ( s ) f ( ∞ ) = lim s → 0 s F ( s ) f(\infty) = \lim_{s\to0} sF(s) f ( ∞ ) = lim s → 0 s F ( s ) f f f s F ( s ) sF(s) s F ( s ) F ( s ) = 1 s ( s + 2 ) F(s) = \dfrac{1}{s(s+2)} F ( s ) = s ( s + 2 ) 1 f ( ∞ ) = lim s → 0 1 s + 2 = 1 2 f(\infty) = \lim_{s\to0}\dfrac{1}{s+2} = \dfrac{1}{2} f ( ∞ ) = lim s → 0 s + 2 1 = 2 1 f ( t ) = 1 2 − 1 2 e − 2 t f(t) = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t} f ( t ) = 2 1 − 2 1 e − 2 t
Common mistakes Dropping initial conditions in L { y ′ } = s Y − y ( 0 ) \mathcal{L}\{y'\} = sY - y(0) L { y ′ } = s Y − y ( 0 ) Writing L { t n } = n s n + 1 \mathcal{L}\{t^n\} = \dfrac{n}{s^{n+1}} L { t n } = s n + 1 n It's n ! n! n ! Using the final value theorem when f f f For sin t \sin t sin t Mixing up the two shifting theorems: e a t e^{at} e a t s s s e − a s e^{-as} e − a s Where this comes up in exams Laplace transforms are listed in the GATE 2027 syllabi for EC (solving network equations), EE and IN (Signals and Systems), and ME (Differential Equations).
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Fit it into your week with the free timetable generator . Mark it off as you revise in the syllabus tracker . See how long you have left: GATE 2027 countdown Continue learning Laplace transforms and Fourier series are the two core tools for moving a problem out of the time domain.