Intuition The big picture (WHY these properties exist)The Laplace transform 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 algebra . But computing that integral every time is painful. These four properties are shortcuts : if you know the transform of a building-block function, you can get transforms of shifted , scaled , and combined functions without integrating again . That is the whole 80/20 of Laplace tables — almost every entry is one of these rules applied to 1 , e a t , sin , cos , t n 1, e^{at}, \sin, \cos, t^n 1 , e a t , sin , cos , t n
For constants a , b a,b a , b f , g f,g f , g F ( s ) , G ( s ) F(s),G(s) F ( s ) , G ( s ) L { a f ( t ) + b g ( t ) } = a F ( s ) + b G ( s ) \mathcal{L}\{a f(t)+b g(t)\}=a F(s)+b G(s) L { a f ( t ) + b g ( t )} = a F ( s ) + b G ( s )
WHY it's true (derive from scratch): The Laplace transform is an integral, and integration is linear.
L { a f + b g } = ∫ 0 ∞ e − s t ( a f ( t ) + b g ( t ) ) d t = a ∫ 0 ∞ e − s t f d t ⏟ F ( s ) + b ∫ 0 ∞ e − s t g d t ⏟ G ( s ) . \mathcal{L}\{af+bg\}=\int_0^\infty e^{-st}\big(af(t)+bg(t)\big)\,dt
=a\underbrace{\int_0^\infty e^{-st}f\,dt}_{F(s)}+b\underbrace{\int_0^\infty e^{-st}g\,dt}_{G(s)}. L { a f + b g } = ∫ 0 ∞ e − s t ( a f ( t ) + b g ( t ) ) d t = a F ( s ) ∫ 0 ∞ e − s t f d t + b G ( s ) ∫ 0 ∞ e − s t g d t . Why this step? Pull constants out and split the sum — both are legal because the integral operator is linear. Done.
Worked example Transform of3 sin 2 t − 4 3\sin 2t - 4 3 sin 2 t − 4 L { sin 2 t } = 2 s 2 + 4 \mathcal{L}\{\sin 2t\}=\dfrac{2}{s^2+4} L { sin 2 t } = s 2 + 4 2 L { 1 } = 1 s \mathcal{L}\{1\}=\dfrac1s L { 1 } = s 1 L { 3 sin 2 t − 4 } = 3 ⋅ 2 s 2 + 4 − 4 ⋅ 1 s = 6 s 2 + 4 − 4 s . \mathcal{L}\{3\sin 2t-4\}=3\cdot\frac{2}{s^2+4}-4\cdot\frac1s=\frac{6}{s^2+4}-\frac4s. L { 3 sin 2 t − 4 } = 3 ⋅ s 2 + 4 2 − 4 ⋅ s 1 = s 2 + 4 6 − s 4 . Why this step? Linearity lets us handle each term separately and just scale — no new integral.
Definition First Shift TheoremIf L { f ( t ) } = F ( s ) \mathcal{L}\{f(t)\}=F(s) L { f ( t )} = F ( s ) e a t e^{at} e a t shifts s s s :
L { e a t f ( t ) } = F ( s − a ) . \mathcal{L}\{e^{at}f(t)\}=F(s-a). L { e a t f ( t )} = F ( s − a ) .
WHY / HOW (derive):
L { e a t f ( t ) } = ∫ 0 ∞ e − s t e a t f ( t ) d t = ∫ 0 ∞ e − ( s − a ) t f ( t ) d t . \mathcal{L}\{e^{at}f(t)\}=\int_0^\infty e^{-st}e^{at}f(t)\,dt
=\int_0^\infty e^{-(s-a)t}f(t)\,dt. L { e a t f ( t )} = ∫ 0 ∞ e − s t e a t f ( t ) d t = ∫ 0 ∞ e − ( s − a ) t f ( t ) d t . Why this step? Combine the exponentials: e − s t e a t = e − ( s − a ) t e^{-st}e^{at}=e^{-(s-a)t} e − s t e a t = e − ( s − a ) t exactly the definition of F F F s s s s − a s-a s − a F ( s − a ) F(s-a) F ( s − a )
means
Multiplying f f f time world = sliding the whole transform sideways in the s world. "Time-domain multiply by e a t e^{at} e a t a a a
L { e 3 t cos 2 t } \mathcal{L}\{e^{3t}\cos 2t\} L { e 3 t cos 2 t } L { cos 2 t } = s s 2 + 4 = F ( s ) \mathcal{L}\{\cos 2t\}=\dfrac{s}{s^2+4}=F(s) L { cos 2 t } = s 2 + 4 s = F ( s ) s s s s − 3 s-3 s − 3 L { e 3 t cos 2 t } = F ( s − 3 ) = s − 3 ( s − 3 ) 2 + 4 . \mathcal{L}\{e^{3t}\cos 2t\}=F(s-3)=\frac{s-3}{(s-3)^2+4}. L { e 3 t cos 2 t } = F ( s − 3 ) = ( s − 3 ) 2 + 4 s − 3 . Why this step? The e 3 t e^{3t} e 3 t a = 3 a=3 a = 3 s → s − 3 s\to s-3 s → s − 3 everywhere , including the denominator.
Definition Second Shift TheoremLet u ( t − a ) u(t-a) u ( t − a ) unit step (0 for t < a t<a t < a t ≥ a t\ge a t ≥ a a ≥ 0 a\ge0 a ≥ 0 L { f ( t − a ) u ( t − a ) } = e − a s F ( s ) . \mathcal{L}\{f(t-a)\,u(t-a)\}=e^{-as}F(s). L { f ( t − a ) u ( t − a )} = e − a s F ( s ) .
WHY / HOW (derive):
L { f ( t − a ) u ( t − a ) } = ∫ 0 ∞ e − s t f ( t − a ) u ( t − a ) d t = ∫ a ∞ e − s t f ( t − a ) d t . \mathcal{L}\{f(t-a)u(t-a)\}=\int_0^\infty e^{-st}f(t-a)u(t-a)\,dt
=\int_a^\infty e^{-st}f(t-a)\,dt. L { f ( t − a ) u ( t − a )} = ∫ 0 ∞ e − s t f ( t − a ) u ( t − a ) d t = ∫ a ∞ e − s t f ( t − a ) d t . Why this step? The step function kills everything below t = a t=a t = a a a a τ = t − a \tau=t-a τ = t − a t = τ + a t=\tau+a t = τ + a d t = d τ dt=d\tau d t = d τ 0 → ∞ 0\to\infty 0 → ∞
= ∫ 0 ∞ e − s ( τ + a ) f ( τ ) d τ = e − a s ∫ 0 ∞ e − s τ f ( τ ) d τ = e − a s F ( s ) . =\int_0^\infty e^{-s(\tau+a)}f(\tau)\,d\tau=e^{-as}\int_0^\infty e^{-s\tau}f(\tau)\,d\tau=e^{-as}F(s). = ∫ 0 ∞ e − s ( τ + a ) f ( τ ) d τ = e − a s ∫ 0 ∞ e − s τ f ( τ ) d τ = e − a s F ( s ) . Why this step? The substitution rewrites the delayed f f f f f f e − a s e^{-as} e − a s
Intuition Time-delay = exponential tagDelaying a signal by a a a doesn't change its shape (F ( s ) F(s) F ( s ) e − a s e^{-as} e − a s e − a s e^{-as} e − a s a a a
L { ( t − 2 ) 2 u ( t − 2 ) } \mathcal{L}\{(t-2)^2\,u(t-2)\} L {( t − 2 ) 2 u ( t − 2 )} f ( t ) = t 2 f(t)=t^2 f ( t ) = t 2 f ( t − 2 ) = ( t − 2 ) 2 f(t-2)=(t-2)^2 f ( t − 2 ) = ( t − 2 ) 2 F ( s ) = L { t 2 } = 2 s 3 F(s)=\mathcal{L}\{t^2\}=\dfrac{2}{s^3} F ( s ) = L { t 2 } = s 3 2 L { ( t − 2 ) 2 u ( t − 2 ) } = e − 2 s ⋅ 2 s 3 . \mathcal{L}\{(t-2)^2u(t-2)\}=e^{-2s}\cdot\frac{2}{s^3}. L {( t − 2 ) 2 u ( t − 2 )} = e − 2 s ⋅ s 3 2 . Why this step? The argument is already in the form ( t − a ) (t-a) ( t − a ) a = 2 a=2 a = 2 e − a s F ( s ) e^{-as}F(s) e − a s F ( s )
Definition Scaling propertyFor c > 0 c>0 c > 0 L { f ( c t ) } = 1 c F ( s c ) . \mathcal{L}\{f(ct)\}=\frac1c\,F\!\left(\frac sc\right). L { f ( c t )} = c 1 F ( c s ) .
WHY / HOW (derive):
L { f ( c t ) } = ∫ 0 ∞ e − s t f ( c t ) d t . \mathcal{L}\{f(ct)\}=\int_0^\infty e^{-st}f(ct)\,dt. L { f ( c t )} = ∫ 0 ∞ e − s t f ( c t ) d t . Substitute τ = c t \tau=ct τ = c t t = τ / c t=\tau/c t = τ / c d t = d τ / c dt=d\tau/c d t = d τ / c c > 0 c>0 c > 0
= ∫ 0 ∞ e − s τ / c f ( τ ) d τ c = 1 c ∫ 0 ∞ e − ( s / c ) τ f ( τ ) d τ = 1 c F ( s c ) . =\int_0^\infty e^{-s\tau/c}f(\tau)\,\frac{d\tau}{c}=\frac1c\int_0^\infty e^{-(s/c)\tau}f(\tau)\,d\tau=\frac1c F\!\left(\frac sc\right). = ∫ 0 ∞ e − s τ / c f ( τ ) c d τ = c 1 ∫ 0 ∞ e − ( s / c ) τ f ( τ ) d τ = c 1 F ( c s ) . Why this step? Squeezing time by c c c s s s 1 / c 1/c 1/ c 1 / c 1/c 1/ c
L { sin 3 t } \mathcal{L}\{\sin 3t\} L { sin 3 t } f ( t ) = sin t f(t)=\sin t f ( t ) = sin t F ( s ) = 1 s 2 + 1 F(s)=\dfrac1{s^2+1} F ( s ) = s 2 + 1 1 c = 3 c=3 c = 3 L { sin 3 t } = 1 3 F ( s 3 ) = 1 3 ⋅ 1 ( s / 3 ) 2 + 1 = 1 3 ⋅ 9 s 2 + 9 = 3 s 2 + 9 . \mathcal{L}\{\sin 3t\}=\frac13 F\!\left(\frac s3\right)=\frac13\cdot\frac{1}{(s/3)^2+1}=\frac13\cdot\frac{9}{s^2+9}=\frac{3}{s^2+9}. L { sin 3 t } = 3 1 F ( 3 s ) = 3 1 ⋅ ( s /3 ) 2 + 1 1 = 3 1 ⋅ s 2 + 9 9 = s 2 + 9 3 . Why this step? Matches the table value 3 s 2 + 9 \dfrac{3}{s^2+9} s 2 + 9 3
Recall Predict before peeking:L { e − t t 3 } = ? \mathcal{L}\{e^{-t}t^3\}=? L { e − t t 3 } = ? Forecast: L { t 3 } = 3 ! s 4 = 6 s 4 \mathcal{L}\{t^3\}=\dfrac{3!}{s^4}=\dfrac{6}{s^4} L { t 3 } = s 4 3 ! = s 4 6 a = − 1 a=-1 a = − 1 s → s + 1 s\to s+1 s → s + 1 Verify: 6 ( s + 1 ) 4 \dfrac{6}{(s+1)^4} ( s + 1 ) 4 6
Common mistake Mixing up the two shiftsWrong feeling: "Both have an exponential, so both give e − a s e^{-as} e − a s feels right because both rules involve e e e Fix: First shift multiplies time by e a t e^{at} e a t s : F ( s − a ) F(s-a) F ( s − a ) delays time with u ( t − a ) u(t-a) u ( t − a ) e − a s e^{-as} e − a s s . One moves the function, the other tags it.
Common mistake Forgetting to write everything in terms of( t − a ) (t-a) ( t − a ) Wrong: L { t 2 u ( t − 2 ) } = e − 2 s ⋅ 2 s 3 \mathcal{L}\{t^2 u(t-2)\}=e^{-2s}\cdot\frac{2}{s^3} L { t 2 u ( t − 2 )} = e − 2 s ⋅ s 3 2 Why it tempts: You see u ( t − 2 ) u(t-2) u ( t − 2 ) Fix: The rule needs f ( t − a ) f(t-a) f ( t − a ) t 2 = ( ( t − 2 ) + 2 ) 2 = ( t − 2 ) 2 + 4 ( t − 2 ) + 4 t^2=((t-2)+2)^2=(t-2)^2+4(t-2)+4 t 2 = (( t − 2 ) + 2 ) 2 = ( t − 2 ) 2 + 4 ( t − 2 ) + 4 delayed-argument form gets e − a s F ( s ) e^{-as}F(s) e − a s F ( s )
Common mistake Dropping the1 / c 1/c 1/ c Wrong: L { f ( c t ) } = F ( s / c ) \mathcal{L}\{f(ct)\}=F(s/c) L { f ( c t )} = F ( s / c ) Fix: The Jacobian d t = d τ / c dt=d\tau/c d t = d τ / c 1 c \frac1c c 1 1 c F ( s / c ) \frac1c F(s/c) c 1 F ( s / c )
Statement of linearity of Laplace transform L { a f + b g } = a F ( s ) + b G ( s ) \mathcal{L}\{af+bg\}=aF(s)+bG(s) L { a f + b g } = a F ( s ) + b G ( s ) ; true because integration is linear.
First shift theorem L { e a t f ( t ) } = F ( s − a ) \mathcal{L}\{e^{at}f(t)\}=F(s-a) L { e a t f ( t )} = F ( s − a ) — multiply by
e a t e^{at} e a t in time ⇒ replace
s s s by
s − a s-a s − a .
Second shift theorem L { f ( t − a ) u ( t − a ) } = e − a s F ( s ) \mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s) L { f ( t − a ) u ( t − a )} = e − a s F ( s ) — delay by
a a a ⇒ multiply transform by
e − a s e^{-as} e − a s .
Scaling property L { f ( c t ) } = 1 c F ( s / c ) \mathcal{L}\{f(ct)\}=\frac1c F(s/c) L { f ( c t )} = c 1 F ( s / c ) for
c > 0 c>0 c > 0 .
Why does the second shift need a unit step? u ( t − a ) u(t-a) u ( t − a ) makes the integral start at
t = a t=a t = a , enabling substitution
τ = t − a \tau=t-a τ = t − a to factor out
e − a s e^{-as} e − a s .
L { e 3 t cos 2 t } \mathcal{L}\{e^{3t}\cos 2t\} L { e 3 t cos 2 t } s − 3 ( s − 3 ) 2 + 4 \dfrac{s-3}{(s-3)^2+4} ( s − 3 ) 2 + 4 s − 3 .
Key difference between first and second shift First shifts in
s s s -domain (
F ( s − a ) F(s-a) F ( s − a ) ); second tags a delay in
t t t -domain (
e − a s F ( s ) e^{-as}F(s) e − a s F ( s ) ).
L { ( t − 2 ) 2 u ( t − 2 ) } \mathcal{L}\{(t-2)^2u(t-2)\} L {( t − 2 ) 2 u ( t − 2 )} e − 2 s ⋅ 2 s 3 e^{-2s}\cdot\frac{2}{s^3} e − 2 s ⋅ s 3 2 .
Recall Feynman: explain to a 12-year-oldImagine a machine that turns wiggly time-pictures into simple s-recipes. Linearity: if you add two pictures, the machine just adds their recipes. First shift: if you make a picture grow with e a t e^{at} e a t Second shift: if you start a picture late by a a a e − a s e^{-as} e − a s Scaling: if you fast-forward the picture, the recipe stretches the other way and shrinks a bit. No re-integrating needed — just remember the four tricks.
"L-Eat-Late-Squeeze"
L inearity (split & scale) · E at e a t e^{at} e a t s−a (first shift) · Late u ( t − a ) u(t-a) u ( t − a ) e − a s e^{-as} e − a s Squeeze time f ( c t ) f(ct) f ( c t ) 1 c F ( s / c ) \frac1c F(s/c) c 1 F ( s / c )
step limits + substitute tau
multiply by e^at gives F of s-a
delay by unit step u of t-a
Laplace transform integral
Intuition Hinglish mein samjhoDekho, Laplace transform ek machine hai jo time-domain ke functions ko s s s linearity, first shift, second shift, scaling — shortcuts hain. Inse 80% table-entries banti hain bina naya integral kiye.
Linearity simple hai: integration linear hota hai, isliye L { a f + b g } = a F + b G \mathcal{L}\{af+bg\}=aF+bG L { a f + b g } = a F + b G First shift mein agar time mein e a t e^{at} e a t s s s s → s − a s\to s-a s → s − a e 3 t cos 2 t e^{3t}\cos 2t e 3 t cos 2 t cos 2 t \cos 2t cos 2 t s s s s − 3 s-3 s − 3
Second shift thoda alag hai — yahan function ko late start karaate hain unit step u ( t − a ) u(t-a) u ( t − a ) e − a s e^{-as} e − a s ( t − a ) (t-a) ( t − a ) Scaling mein f ( c t ) f(ct) f ( c t ) 1 c F ( s / c ) \frac1c F(s/c) c 1 F ( s / c ) 1 c \frac1c c 1
Sabse common galti: pehla aur doosra shift mix karna. Dono mein exponential dikhta hai, isliye confusion hoti hai. Bas yaad rakho — first shift = s s s F ( s − a ) F(s-a) F ( s − a ) , second shift = e − a s e^{-as} e − a s . Mnemonic "L-Eat-Late-Squeeze" se sab order mein aa jaata hai.
