Inverse Laplace Transform

Returning from s-Domain to Time Domain

Learning Objectives:
  • Master partial fraction decomposition for 3 cases: distinct real roots, repeated real roots, and complex conjugate pairs
  • Use Heaviside's method for quickly finding coefficients
  • Apply the inverse transform table to recover time-domain solutions
  • Connect pole locations to time-domain behavior

After solving in the s-domain, we need to return to the time domain. The key technique is partial fraction decomposition — breaking a complex rational function into simple pieces we can look up in our transform table. This converts the difficult inverse problem into a sum of elementary pieces.

"The important thing in science is not so much to obtain new facts as to discover new ways of thinking about them."

— William Lawrence Bragg (1890–1971)

وَقُل رَّبِّ زِدْنِي عِلْمًا
"And say, 'My Lord, increase me in knowledge.'"
— Quran, Surah Ta-Ha 20:114

Theory

The Inverse Transform

If \(\mathcal{L}\{f(t)\} = F(s)\), then \(f(t) = \mathcal{L}^{-1}\{F(s)\}\). The inverse is LINEAR:
$$\mathcal{L}^{-1}\{c_1 F(s) + c_2 G(s)\} = c_1 f(t) + c_2 g(t)$$

Partial Fractions — The Three Cases

Case 1: Distinct Real Roots (Non-repeated Linear Factors)

$$\frac{P(s)}{(s-r_1)(s-r_2)\cdots(s-r_n)} = \frac{A_1}{s-r_1} + \frac{A_2}{s-r_2} + \cdots + \frac{A_n}{s-r_n}$$
Engineering Insight: Each distinct real pole at \(s = r_k\) contributes a term \(A_k e^{r_k t}\) to the time response. Negative poles → decaying exponentials (stable), positive poles → growing exponentials (unstable).

Case 2: Repeated Real Roots

$$\frac{P(s)}{(s-r)^n} = \frac{A_1}{s-r} + \frac{A_2}{(s-r)^2} + \cdots + \frac{A_n}{(s-r)^n}$$
Caution: Repeated roots generate terms like \(t^n e^{rt}\) — the polynomial factor \(t\) causes the response to grow initially before exponential decay takes over (if \(r < 0\)). This is why repeated poles are harder to handle in control systems.

Case 3: Complex/Irreducible Quadratic Factors

$$\frac{P(s)}{(s^2 + bs + c)} = \frac{As + B}{s^2 + bs + c}$$
Complete the square: \(s^2 + bs + c = (s + b/2)^2 + (c - b^2/4)\) and use the inverse formulas for \(e^{at}\cos(\omega t)\) and \(e^{at}\sin(\omega t)\).
Engineering Insight: Complex poles at \(s = -\alpha \pm j\omega\) produce oscillatory responses \(e^{-\alpha t}(A \cos(\omega t) + B \sin(\omega t))\). The real part \(\alpha\) controls decay rate (damping), the imaginary part \(\omega\) controls oscillation frequency.

Heaviside's Cover-up Method (for Case 1)

For distinct linear factors, the coefficient \(A_k\) is found by:
$$A_k = \left. \frac{P(s)}{Q(s)/(s-r_k)} \right|_{s=r_k}$$
"Cover up" the factor \((s-r_k)\) in the denominator and evaluate at \(s = r_k\).

Step-by-Step Method

  1. Factor the denominator \(Q(s)\) completely into linear and/or irreducible quadratic factors
  2. Identify the case: distinct real roots, repeated roots, or complex conjugate pairs
  3. Set up the partial fraction form according to your case
  4. Find coefficients (Heaviside's method for Case 1, algebra for Cases 2–3)
  5. Look up each term in the inverse Laplace transform table
  6. Combine results: \(f(t) =\) sum of all inverse terms

Worked Examples

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

Solution:

This is a direct match in the transform table. We have \(\mathcal{L}\{e^{at}\} = \frac{1}{s-a}\), so:

$$\mathcal{L}^{-1}\left\{\frac{5}{s-3}\right\} = 5e^{3t}$$

Find: \(\mathcal{L}^{-1}\left\{\frac{2s+1}{s^2+4s+13}\right\}\)

Solution:

Complete the square: \(s^2 + 4s + 13 = (s+2)^2 + 9\)

Rewrite the numerator to separate it into forms we recognize:

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

Using the tables with \(a = -2\) and \(\omega = 3\):

$$f(t) = 2e^{-2t}\cos(3t) - e^{-2t}\sin(3t)$$

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

Solution:

Set up partial fractions (Case 1):

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

Using Heaviside's method:

$$A = \left.\frac{3s+5}{s+2}\right|_{s=-1} = \frac{-3+5}{-1+2} = \frac{2}{1} = 2$$ $$B = \left.\frac{3s+5}{s+1}\right|_{s=-2} = \frac{-6+5}{-2+1} = \frac{-1}{-1} = 1$$

Therefore:

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

Find: \(\mathcal{L}^{-1}\left\{\frac{s^2+9s+2}{(s-1)(s+2)(s-3)}\right\}\)

Solution:

Set up partial fractions:

$$\frac{s^2+9s+2}{(s-1)(s+2)(s-3)} = \frac{A}{s-1} + \frac{B}{s+2} + \frac{C}{s-3}$$

Using Heaviside at each root:

$$A = \left.\frac{s^2+9s+2}{(s+2)(s-3)}\right|_{s=1} = \frac{1+9+2}{(3)(-2)} = \frac{12}{-6} = -2$$ $$B = \left.\frac{s^2+9s+2}{(s-1)(s-3)}\right|_{s=-2} = \frac{4-18+2}{(-3)(-5)} = \frac{-12}{15} = -\frac{4}{5}$$ $$C = \left.\frac{s^2+9s+2}{(s-1)(s+2)}\right|_{s=3} = \frac{9+27+2}{(2)(5)} = \frac{38}{10} = \frac{19}{5}$$

Therefore:

$$f(t) = -2e^{t} - \frac{4}{5}e^{-2t} + \frac{19}{5}e^{3t}$$

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

Solution:

Set up partial fractions (Case 2):

$$\frac{2s-1}{(s+1)^3} = \frac{A_1}{s+1} + \frac{A_2}{(s+1)^2} + \frac{A_3}{(s+1)^3}$$

Rewrite the numerator as: \(2s - 1 = 2(s+1) - 3\)

This gives:

$$\frac{2(s+1) - 3}{(s+1)^3} = \frac{2}{(s+1)^2} - \frac{3}{(s+1)^3}$$

Using inverse table entries \(\mathcal{L}^{-1}\{\frac{1}{(s-a)^n}\} = \frac{t^{n-1}}{(n-1)!}e^{at}\):

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

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

Solution:

Set up partial fractions (mixed: repeated \(s^2\) and simple):

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

Multiply through by \(s^2(s+2)\):

$$1 = A \cdot s(s+2) + B(s+2) + C \cdot s^2$$

Setting \(s = 0\): \(1 = 2B \Rightarrow B = \frac{1}{2}\)

Setting \(s = -2\): \(1 = 4C \Rightarrow C = \frac{1}{4}\)

Setting \(s = 1\): \(1 = 3A + 3B + C \Rightarrow 1 = 3A + \frac{3}{2} + \frac{1}{4} \Rightarrow A = -\frac{1}{4}\)

Therefore:

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

Find: \(\mathcal{L}^{-1}\left\{\frac{3s+7}{s^2+2s+5}\right\}\)

Solution:

Complete the square: \(s^2 + 2s + 5 = (s+1)^2 + 4\)

$$\frac{3s+7}{(s+1)^2+4} = \frac{3(s+1) + 4}{(s+1)^2+4} = \frac{3(s+1)}{(s+1)^2+4} + \frac{4}{(s+1)^2+4}$$

With \(a = -1\) and \(\omega = 2\):

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

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

Solution:

Set up partial fractions with both real and complex terms:

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

Multiply through by \((s+1)(s^2+4)\):

$$s = A(s^2+4) + (Bs+C)(s+1)$$

Setting \(s = -1\): \(-1 = A(5) \Rightarrow A = -\frac{1}{5}\)

Expanding: \(s = As^2 + 4A + Bs^2 + Bs + Cs + C\)

Comparing coefficients:

  • \(s^2\): \(0 = A + B \Rightarrow B = \frac{1}{5}\)
  • \(s^0\): \(0 = 4A + C \Rightarrow C = \frac{4}{5}\)

Therefore:

$$f(t) = -\frac{1}{5}e^{-t} + \frac{1}{5}\cos(2t) + \frac{2}{5}\sin(2t)$$

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.

Practice 1: Find $\mathcal{L}^{-1}\left\{\dfrac{2s+7}{s^2+4s+13}\right\}$ ▼

Solution

Step 1: Complete the square in the denominator:

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

Step 2: Rewrite the fraction:

$$\frac{2s+7}{(s+2)^2+9}$$

Step 3: Separate the numerator to match shift forms. Rewrite $2s+7 = 2(s+2) + 3$:

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

Step 4: Apply inverse transform with first shifting theorem:

$$\mathcal{L}^{-1}\left\{\frac{2(s+2)}{(s+2)^2+9}\right\} = 2e^{-2t}\cos(3t)$$
$$\mathcal{L}^{-1}\left\{\frac{3}{(s+2)^2+9}\right\} = \frac{3}{3}e^{-2t}\sin(3t) = e^{-2t}\sin(3t)$$

Answer: $\boxed{2e^{-2t}\cos(3t) + e^{-2t}\sin(3t)}$

Practice 2: Find $\mathcal{L}^{-1}\left\{\dfrac{5s - 1}{s^2 - 6s + 10}\right\}$ ▼

Solution

Step 1: Complete the square:

$$s^2 - 6s + 10 = (s-3)^2 + 1$$

Step 2: Rewrite the numerator: $5s - 1 = 5(s-3) + 14$:

$$\frac{5(s-3)+14}{(s-3)^2+1} = \frac{5(s-3)}{(s-3)^2+1} + \frac{14}{(s-3)^2+1}$$

Step 3: Apply inverse transform:

$$\mathcal{L}^{-1}\left\{\frac{5(s-3)}{(s-3)^2+1}\right\} = 5e^{3t}\cos(t)$$
$$\mathcal{L}^{-1}\left\{\frac{14}{(s-3)^2+1}\right\} = 14e^{3t}\sin(t)$$

Answer: $\boxed{5e^{3t}\cos(t) + 14e^{3t}\sin(t)}$

Practice 3: Find $\mathcal{L}^{-1}\left\{\dfrac{s+3}{(s+1)(s^2+4)}\right\}$ using partial fractions ▼

Solution

Step 1: Set up partial fractions:

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

Step 2: Multiply both sides by $(s+1)(s^2+4)$:

$$s+3 = A(s^2+4) + (Bs+C)(s+1)$$

Step 3: Find coefficients:

  • At $s = -1$: $2 = A(1+4) = 5A$, so $A = \dfrac{2}{5}$
  • At $s = 0$: $3 = 4A + C = \dfrac{8}{5} + C$, so $C = \dfrac{7}{5}$
  • At $s = 1$: $4 = 5A + 2(B+C) = 2 + 2B + \dfrac{14}{5} = \dfrac{24}{5} + 2B$, so $B = -\dfrac{2}{5}$

Step 4: Decompose:

$$\frac{2/5}{s+1} + \frac{-2s/5 + 7/5}{s^2+4} = \frac{2}{5} \cdot \frac{1}{s+1} - \frac{2}{5} \cdot \frac{s}{s^2+4} + \frac{7}{5} \cdot \frac{1}{s^2+4}$$

Step 5: Apply inverse transforms:

$$\mathcal{L}^{-1}\left\{\frac{2}{5(s+1)}\right\} = \frac{2}{5}e^{-t}$$
$$\mathcal{L}^{-1}\left\{-\frac{2}{5} \cdot \frac{s}{s^2+4}\right\} = -\frac{2}{5}\cos(2t)$$
$$\mathcal{L}^{-1}\left\{\frac{7}{5} \cdot \frac{1}{s^2+4}\right\} = \frac{7}{5} \cdot \frac{1}{2}\sin(2t) = \frac{7}{10}\sin(2t)$$

Answer: $\boxed{\dfrac{2}{5}e^{-t} - \dfrac{2}{5}\cos(2t) + \dfrac{7}{10}\sin(2t)}$

Interactive Visualizer: Poles and Time Response

Tip: The pole locations tell you EVERYTHING about the system's time response. Left half-plane = stable (decaying), right half-plane = unstable (growing), imaginary axis = oscillation frequency.

Practice Quiz

Question 1: What is \(\mathcal{L}^{-1}\left\{\frac{3}{s+4}\right\}\)?

Question 2: For \(F(s) = \frac{1}{(s+1)(s+3)}\), what is the partial fraction form?

Question 3: For \(F(s) = \frac{2s+3}{s^2+6s+13}\), what is the first step?

Question 4: Using Heaviside's method for \(F(s) = \frac{s}{(s-1)(s+2)}\), what is \(A\) (coefficient of \(\frac{1}{s-1}\))?

Question 5: A system with poles at \(s = -2 \pm 3j\) will exhibit...

Question 6: What is \(\mathcal{L}^{-1}\left\{\frac{1}{(s+2)^3}\right\}\)?