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12.3.4. Heaviside’s Expansion Formula (for distinct poles)

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Session 1: Introduction to Heaviside’s Expansion Formula

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Sarah
SarahInstructor

Today, we will explore Heaviside’s Expansion Formula, a powerful technique used in calculating inverse Laplace transforms of rational functions with distinct poles. Can anyone tell me what we understand by 'poles' in this context?

Noah
Noah

Are they the values of 's' that make the denominator zero?

Sarah
SarahInstructor

Exactly! Each distinct pole corresponds to a solution of the denominator set to zero. This is critical for our formula because it dictates how we construct our inverse function.

Isabella
Isabella

How does the formula incorporate these poles?

Sarah
SarahInstructor

Great question! The formula sums up contributions from each pole, weighted by how they interact with each other, giving us a complete view of the function's behavior. Remember: P.E. (Poles contribute to the function and we Weight their contributions).

Session 2: Understanding the Formula

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Robert
RobertInstructor

Let's look at the formula: L−1{F(s)}=∑i=1nP(ai)eait∏j≠i(ai−aj)L^{-1} \{ F(s) \} = \sum_{i=1}^{n} \frac{P(a_i)e^{a_it}}{\prod_{j\neq i}(a_i - a_j)}. What does every part stand for?

Akash
Akash

P(a_i) seems to represent the polynomial evaluated at the pole?

Robert
RobertInstructor

Correct! And what about ∏j≠i(ai−aj)\prod_{j\neq i}(a_i - a_j)?

Ananya
Ananya

It’s the product of the differences of the pole with all other poles? Right?

Robert
RobertInstructor

Exactly! This represents the scaling or normalization needed for each pole's contribution to the overall function, reinforcing our understanding of its impact.

Noah
Noah

Why do we need to sum these contributions?

Robert
RobertInstructor

Because the overall function could behave differently depending on the interactions of the poles, leading to a richer solution.

Session 3: Example Application of Heaviside's Formula

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Sarah
SarahInstructor

Let's apply Heaviside’s formula with an example: Find the inverse Laplace transform of F(s)=3s+4(s−1)(s−2)F(s) = \frac{3s + 4}{(s-1)(s-2)}. Who can begin this process?

Isabella
Isabella

First, we identify the poles: s = 1 and s = 2.

Sarah
SarahInstructor

Correct! Now, evaluate P(s)P(s) at these poles. What do we find?

Akash
Akash

At s = 1, P(1) = 3(1) + 4 = 7, and at s = 2, P(2) = 3(2) + 4 = 10.

Sarah
SarahInstructor

Good! Now plug these values into the formula. What will the contributions look like?

Ananya
Ananya

We'll have L−1{F(s)}=7et(1−2)+10e2t(2−1)=−7et+10e2tL^{-1} \{ F(s) \} = \frac{7e^{t}}{(1-2)} + \frac{10e^{2t}}{(2-1)} = -7e^{t} + 10e^{2t}.

Sarah
SarahInstructor

Excellent! This process illustrates how the formula allows us to derive the time-domain function systematically.