Chapter 3 · Laplace Transforms

Inverse Laplace Transform

From s-Domain Back to Time Domain

Dr. Mohamed Mabrok · Qatar University

The Big Picture: Laplace Transform Journey

Definition
(Direct)Done ✓

Laplace Transform

Inverse Transform
(This Lecture!)

Solving ODEs
Step Functions

Return Journey

      We've learned to go from $f(t) \to F(s)$. Now we learn the inverse route: $F(s) \to f(t)$. This is the key to solving differential equations!

Definition: Inverse Laplace Transform

Notation & Definition

      $$\mathcal{L}^{-1}\{F(s)\} = f(t)$$
    
      The inverse Laplace transform "undoes" the forward transform. If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}^{-1}\{F(s)\} = f(t)$.

Linearity Property

      $$\mathcal{L}^{-1}\{aF(s) + bG(s)\} = af(t) + bg(t)$$
    
      The inverse transform is linear: we can split sums and pull out constants.

Key Strategy

      Use the transform table "in reverse"! Match the s-domain function $F(s)$ to known forms, then read off the corresponding time-domain function $f(t)$.

Essential Inverse Transform Pairs

$F(s)$ (s-domain)	$f(t) = \mathcal{L}^{-1}\{F(s)\}$ (time-domain)
$\frac{1}{s}$	$1$
$\frac{1}{s^2}$	$t$
$\frac{n!}{s^{n+1}}$	$t^n$
$\frac{1}{s-a}$	$e^{at}$
$\frac{b}{s^2+b^2}$	$\sin(bt)$
$\frac{s}{s^2+b^2}$	$\cos(bt)$
$\frac{b}{(s-a)^2+b^2}$	$e^{at}\sin(bt)$
$\frac{s-a}{(s-a)^2+b^2}$	$e^{at}\cos(bt)$

Memorize These!

      These are the fundamental building blocks. You'll use them constantly to solve more complex problems through decomposition.

Three Key Techniques for Finding Inverse Transforms

Direct Lookup

Simple cases

Match $F(s)$ directly to the standard table and read off $f(t)$. Fast but only works for recognizable forms.

Partial Fractions

Rational functions

Break $F(s) = \frac{P(s)}{Q(s)}$ into simpler pieces, each matching the table. Three cases: distinct roots, repeated roots, complex roots.

Completing the Square

Irreducible quadratics

For denominators like $(s-a)^2+b^2$, complete the square to match exponential-oscillation forms.

Decision Tree

      1) Look for direct table match. 2) If rational function, use partial fractions. 3) If irreducible quadratic, complete the square.

Partial Fractions Case 1

Distinct Real Roots

Decomposition Form

      $$\frac{P(s)}{(s-a)(s-b)(s-c)\cdots} = \frac{A}{s-a} + \frac{B}{s-b} + \frac{C}{s-c} + \cdots$$
    
      Each distinct linear factor becomes its own fraction with a constant numerator.

Worked Example

      Find $\mathcal{L}^{-1}\left\{\frac{5s+3}{(s+1)(s-2)}\right\}$

STEP 1 Decompose

\frac{5s+3}{(s+1)(s-2)} = \frac{A}{s+1} + \frac{B}{s-2}

STEP 2 Find A and B (cover-up method)

A = \frac{5(-1)+3}{(-1)-2} = \frac{-2}{-3} = \frac{2}{3}$$ $$B = \frac{5(2)+3}{(2)+1} = \frac{13}{3}

STEP 3 Invert each term using the table

\mathcal{L}^{-1}\left\{\frac{2/3}{s+1} + \frac{13/3}{s-2}\right\} = \frac{2}{3}e^{-t} + \frac{13}{3}e^{2t}

Partial Fractions Case 2

Repeated Roots

Decomposition Form

      $$\frac{P(s)}{(s-a)^n} = \frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \cdots + \frac{A_n}{(s-a)^n}$$
    
      Repeated roots generate multiple terms with increasing powers in the denominator.

Worked Example

      Find $\mathcal{L}^{-1}\left\{\frac{2s+1}{(s-1)^3}\right\}$

STEP 1 Decompose with powers

\frac{2s+1}{(s-1)^3} = \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{C}{(s-1)^3}

STEP 2 Find coefficients (multiply by $(s-1)^3$ and equate)

2s+1 = A(s-1)^2 + B(s-1) + C$$ $$\text{At } s=1: \quad 3 = C \quad \text{; At } s=0: \quad 1 = A - B + C

STEP 3 Invert using $\mathcal{L}^{-1}\{n!/(s-a)^{n+1}\} = t^n e^{at}$

f(t) = 2e^t + 3te^t + \frac{3}{2}t^2e^t

Partial Fractions Case 3

Complex/Irreducible Quadratic Roots

Complex Denominator

      $$\frac{P(s)}{(s^2+bs+c)} = \frac{As+B}{s^2+bs+c}$$
    
      Complex roots (irreducible quadratic) require a linear numerator $As+B$.

Worked Example

      Find $\mathcal{L}^{-1}\left\{\frac{3s+7}{s^2+4s+13}\right\}$

STEP 1 Complete the square in denominator

$$s^2+4s+13 = (s+2)^2+9$$

STEP 2 Rewrite numerator in terms of $(s+2)$

$$3s+7 = 3(s+2)+1 = 3(s+2) + 1$$

STEP 3 Split and match to table forms

= 3 \cdot \frac{s+2}{(s+2)^2+9} + \frac{1}{3} \cdot \frac{3}{(s+2)^2+9}

STEP 4 Invert using exponential-oscillation forms

f(t) = 3e^{-2t}\cos(3t) + \frac{1}{3}e^{-2t}\sin(3t)

Quick Shortcut: The Heaviside Cover-Up Method

The Technique

      For distinct linear factors in $\frac{P(s)}{(s-a_1)(s-a_2)\cdots}$:
    
      $$\text{To find } A_k: \text{ Cover up } (s-a_k) \text{ and evaluate at } s=a_k$$

Worked Example

      Find $\mathcal{L}^{-1}\left\{\frac{s+5}{(s+1)(s+3)}\right\}$

STEP 1 For A: Cover $(s+1)$, substitute $s=-1$

A = \left.\frac{s+5}{s+3}\right|_{s=-1} = \frac{-1+5}{-1+3} = \frac{4}{2} = 2

STEP 2 For B: Cover $(s+3)$, substitute $s=-3$

B = \left.\frac{s+5}{s+1}\right|_{s=-3} = \frac{-3+5}{-3+1} = \frac{2}{-2} = -1

STEP 3 Invert

f(t) = 2e^{-t} - e^{-3t}

Completing the Square: Systematic Approach

Method

      When denominator has an irreducible quadratic $(s^2+bs+c)$ or $(s-a)^2+b^2$:
    
      1) Complete the square in the denominator

      2) Rewrite the numerator in matching terms

      3) Split into cosine and sine parts

      4) Apply exponential-oscillation inverse formulas

Worked Example

      Find $\mathcal{L}^{-1}\left\{\frac{2s-3}{s^2-6s+25}\right\}$

STEP 1 Complete the square

$$s^2-6s+25 = (s-3)^2 - 9 + 25 = (s-3)^2+16$$

STEP 2 Rewrite numerator: $2s-3 = 2(s-3) + 3$

\frac{2(s-3)+3}{(s-3)^2+16} = \frac{2(s-3)}{(s-3)^2+16} + \frac{3}{(s-3)^2+16}

STEP 3 Scale sine term and invert

= 2\cdot\frac{s-3}{(s-3)^2+16} + \frac{3}{4}\cdot\frac{4}{(s-3)^2+16}

f(t) = 2e^{3t}\cos(4t) + \frac{3}{4}e^{3t}\sin(4t)

Full Worked Example

Three Distinct Real Roots

Find $\mathcal{L}^{-1}\left\{\frac{s^2+6s+2}{(s-1)(s+2)(s+3)}\right\}$

STEP 1 Decompose

\frac{s^2+6s+2}{(s-1)(s+2)(s+3)} = \frac{A}{s-1} + \frac{B}{s+2} + \frac{C}{s+3}

STEP 2 Find A, B, C using cover-up

A = \frac{1+6+2}{(1+2)(1+3)} = \frac{9}{12} = \frac{3}{4}$$ $$B = \frac{4-12+2}{(-2-1)(-2+3)} = \frac{-6}{-3} = 2$$ $$C = \frac{9-18+2}{(-3-1)(-3+2)} = \frac{-7}{4}

STEP 3 Invert each term

f(t) = \frac{3}{4}e^t + 2e^{-2t} - \frac{7}{4}e^{-3t}

Full Worked Example

Mixed Real & Repeated Roots

Find $\mathcal{L}^{-1}\left\{\frac{4s+5}{(s+1)^2(s-2)}\right\}$

STEP 1 Decompose with repeat factor

\frac{4s+5}{(s+1)^2(s-2)} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{s-2}

STEP 2 Find coefficients

\text{At } s=-1: \quad B = \frac{-4+5}{-1-2} = -\frac{1}{3}$$ $$\text{At } s=2: \quad C = \frac{8+5}{(2+1)^2} = \frac{13}{9}$$ $$\text{Equating coeff of } s: \quad A + C = 0 \Rightarrow A = -\frac{13}{9}

STEP 3 Invert using $\mathcal{L}^{-1}\{1/(s-a)^2\} = te^{at}$

f(t) = -\frac{13}{9}e^{-t} - \frac{1}{3}te^{-t} + \frac{13}{9}e^{2t}

Full Worked Example

Mixed Real & Complex Roots

Find $\mathcal{L}^{-1}\left\{\frac{s+3}{(s+1)(s^2+4)}\right\}$

STEP 1 Decompose with linear and quadratic

\frac{s+3}{(s+1)(s^2+4)} = \frac{A}{s+1} + \frac{Bs+C}{s^2+4}

STEP 2 Find A, B, C

A = \frac{-1+3}{(-1)^2+4} = \frac{2}{5}$$ $$s+3 = A(s^2+4) + (Bs+C)(s+1)$$ $$\text{Coeff of } s^2: 0 = A + B \Rightarrow B = -\frac{2}{5}$$ $$\text{Coeff of } s^0: 3 = 4A + C \Rightarrow C = 3 - \frac{8}{5} = \frac{7}{5}

STEP 3 Invert each part

f(t) = \frac{2}{5}e^{-t} - \frac{2}{5}\cos(2t) + \frac{7}{10}\sin(2t)

Full Worked Example

Pure Complex Conjugate Pairs

Find $\mathcal{L}^{-1}\left\{\frac{5s}{(s^2+1)(s^2+9)}\right\}$

STEP 1 Decompose with two quadratic factors

\frac{5s}{(s^2+1)(s^2+9)} = \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+9}

STEP 2 Find A, B, C, D by substitution and comparison

5s = (As+B)(s^2+9) + (Cs+D)(s^2+1)$$ $$\text{Coeff of } s^3: 0 = A + C$$ $$\text{Coeff of } s: 5 = 9A + C \Rightarrow A = \frac{5}{8}, C = -\frac{5}{8}$$ $$B = D = 0

STEP 3 Invert

f(t) = \frac{5}{8}\cos(t) - \frac{5}{8}\cos(3t)

Full Worked Example

Higher-Order with Multiple Poles

Find $\mathcal{L}^{-1}\left\{\frac{3}{s^4(s+2)}\right\}$

STEP 1 Decompose with powers of s and simple pole

\frac{3}{s^4(s+2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s^4} + \frac{E}{s+2}

STEP 2 Find coefficients

\text{At } s=-2: \quad E = \frac{3}{(-2)^4} = \frac{3}{16}$$ $$3 = A(s+2)s^3 + B(s+2)s^2 + C(s+2)s + D(s+2) + Es^4$$ $$\text{Equating by powers: } A = -\frac{3}{32}, B = \frac{3}{16}, C = -\frac{3}{8}, D = \frac{3}{4}

STEP 3 Invert using table formulas

f(t) = -\frac{3}{32} + \frac{3}{16}t - \frac{3}{16}t^2 + \frac{1}{4}t^3 + \frac{3}{16}e^{-2t}

Common Pitfalls to Avoid

⚠️ Critical Errors

Numerator degree ≥ denominator degree: Always do polynomial long division first! Otherwise partial fractions fails.
Wrong partial fraction form for repeated roots: Repeated root $r$ of multiplicity $n$ needs ALL powers: $\frac{A_1}{s-r} + \frac{A_2}{(s-r)^2} + \cdots + \frac{A_n}{(s-r)^n}$
Forgetting to complete the square: For irreducible quadratics, completing the square is essential to match the table forms!
Sign errors in numerator rewriting: When rewriting $As + B$ after completing the square, double-check: $s + 3 = (s-(-3))$, not $(s+3)$.
Confusing the constant in cover-up: For $\frac{P(s)}{(s-a)Q(s)}$, the residue at $s=a$ is $\frac{P(a)}{Q(a)}$, not $\frac{P(a)}{(a-a)Q(a)}$!

Sanity Check

      Always verify your answer: take the Laplace transform of $f(t)$ and check that you get back $F(s)$. This catches errors immediately!

Key Takeaways & Summary

Direct Lookup

Memorize table of 8 basic pairs. Use for simple forms that match directly.

Partial Fractions

Three cases: distinct real roots, repeated roots, and complex/irreducible quadratics.

Completing Square

Essential for exponential-oscillation forms. Match denominator to $(s-a)^2+b^2$.

Workflow

      1) Check degree (numerator < denominator?)

      2) Try direct table lookup

      3) If rational function: use partial fractions

      4) For quadratic factors: complete the square

      5) Invert each piece using the table

      6) Combine and verify

Next Up

      Now that we can invert Laplace transforms, we can solve differential equations! The method: Take $\mathcal{L}$ of the ODE, solve in s-domain (algebra!), then use $\mathcal{L}^{-1}$ to get the solution in t-domain.