The transform of a derivative is the transform of the function multiplied by s, less the function's value at the start: ℒ[y′] = s·Y − y(0). Applying it twice gives ℒ[y″] = s²·Y − s·y(0) − y′(0). The rule is proved from the product rule and the fundamental theorem, and the lab checks it on a signal by transforming the derivative directly and comparing. The initial values come with the rule, which is what makes the method of the next lesson work.
Core ideas
Three ideas: the rule, where the initial value comes from, and why it is easy to forget.
If the transform of y is Y(s), then the transform of y′ is s·Y(s) less the value of y at t = 0. In t, the operation is a derivative. In s, it is a multiplication and a subtraction. That exchange is the whole reason the transform is useful for differential equations.
The transform integrates from t = 0. When the product rule is integrated over that interval, the endpoint at t = 0 leaves behind y(0). The starting value is not added by hand afterward: it is part of the rule, and an equation transformed with the rule already contains it.
A reader who forgets the correction and writes ℒ[y′] = s·Y gets the right answer for every signal that starts at zero, sin(t) and t² among them. That is why the slip survives: it fails only when the starting value is not zero, and then it is wrong by exactly that value.
The material
The rule for a first derivative, a proof that uses nothing beyond the product rule and the fundamental theorem, the rule for a second derivative, and the lab's check.
Throughout, a capital letter stands for the transform of the lower-case one: Y(s) is ℒ[y] and F(s) is ℒ[f]. The rules are stated for a signal y and its derivatives, at values of s large enough that the transforms exist.
ℒ[y′] = s·Y − y(0), for s large enough that the transforms exist and e^(−sT)·y(T) shrinks toward zero as T grows.
The product rule gives (e^(−st)·y)′ = −s·e^(−st)·y + e^(−st)·y′. Integrate both sides from 0 to T. The fundamental theorem says the left side is e^(−sT)·y(T) − y(0).
On the right, the first integral is −s times the integral that defines Y, and the second is the integral that defines ℒ[y′]. Let T grow: the term e^(−sT)·y(T) goes to zero, which leaves −y(0) = −s·Y + ℒ[y′]. Solving for ℒ[y′] gives the rule.
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The condition in the theorem is not decoration. It holds for every signal in the table, at every s above the signal's floor, and the lab only transforms signals from the table. The proof shows where the initial value enters: it is the value of the left side at the lower limit.
ℒ[y″] = s²·Y − s·y(0) − y′(0).
Apply the first rule to the function y′: ℒ[y″] = s·ℒ[y′] − y′(0).
Then replace ℒ[y′] by s·Y − y(0): ℒ[y″] = s·(s·Y − y(0)) − y′(0) = s²·Y − s·y(0) − y′(0).
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Take y = e^(2t), so y′ = 2e^(2t) and y(0) = 1. The transform of y is 1/(s − 2), and the transform of y′ is twice that, 2/(s − 2), by linearity.
The rule says s·Y − y(0) = s/(s − 2) − 1. Over a common denominator that is (s − (s − 2))/(s − 2) = 2/(s − 2). The two sides agree, and the lab's first preset prints equal.
The same check on y = sin(t) and on y = t² also prints equal, and for both of them the starting value is zero. Then the rule reads ℒ[y′] = s·Y, and a reader who had forgotten the correction would not see anything wrong. Only a signal with a non-zero start, such as e^(2t), makes the forgotten term visible: dropping it leaves s/(s − 2), which is 1 too large.
For y = t² the transform is 2/s³, and y(0) = 0. The derivative is 2t, whose transform is 2/s². The rule gives s·(2/s³) − 0 = 2/s².
Both sides are the same fraction, and the lab prints equal.
What the lab does and does not do. It computes ℒ[y′] by differentiating the signal and transforming the result, and computes s·Y − y(0) separately, as exact fractions. Showing them equal on three signals is a demonstration; the proof above is the reason it holds for every signal.
Interactive
Transforms, partial fractions and solutions, exactly
Method
Transform the signal, differentiate it and transform that, then compare against the rule.
Find Y(s) = ℒ[y] from the table, and record y(0).
Compute y′ in t, then find ℒ[y′] from the table. This is the left-hand side of the rule.
Multiply Y by s, subtract y(0), and combine into one fraction. This is the right-hand side.
If the two fractions are the same, the rule holds on this signal. If they differ by a constant, the starting value was left out or mis-signed.
The first preset. The left side is computed straight from the derivative; the right side from the rule.
The subtraction of 1 is what makes the match work. Without it the right side is s/(s − 2), which is 1 more than 2/(s − 2) at every s.
For a second derivative the same check needs two corrections, and the next lesson uses both: s·y(0) and y′(0).
You should be able to state both rules, say where the initial value comes from, transform a signal and its derivative separately, and show by exact fractions that the two sides of the rule agree; and say why the slip of leaving the initial value out is invisible when it is zero.
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