From F(s) to f(t): Laplace, animated

Every pole of F(s) becomes one term of f(t). Watch partial fractions split F(s) one pole at a time, each pole's colour turn into its time-domain mode, and the modes add up to f(t). Drag the poles, apply shift, delay, scaling or integration, and see when the final value theorem lies.

Start from a GATE-style F(s), or drag its poles and zeros below

Apply a property

$F(s)=\dfrac{s+3}{(s+1)(s+2)}$

Two simple real poles: two decaying exponentials, the slower one (−1) lasting longest.

s-plane (drag a pole or zero; drop a pole on another to make it repeated)

Time domain: each pole's mode (thin) and their sum f(t) (thick)

Step 1. $F(s)$ has poles at $-1$ , $-2$ . **Each pole becomes one term of $f(t)$ .** A pole's real part sets how fast its term grows or dies (left of the $j\omega$ axis it decays, the further left the faster); its imaginary part sets how fast it oscillates.

Partial fractions → standard pairs

\dfrac{2}{s+1}

⟷

2e^{-t}

dies out, τ = 1

-\dfrac{1}{s+2}

⟷

-e^{-2t}

dies out, τ = 0.5

f(t)=2e^{-t}-e^{-2t}

Initial and final value theorems

lim s→∞ sF(s)1

f(0⁺) from f(t)1 ✓

lim s→0 sF(s)0

f(∞) from f(t)0 ✓

The final value theorem needs every pole of sF(s) strictly inside the left half plane.

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