The Laplace transform turns a function of time, f(t), into a function of a new variable, s. Fix s, multiply the signal by e^(−st), add up the area under the product from t = 0 onward, and that is one number; doing it for every s gives the transform F(s). For constants, powers and exponentials the result is a rational function of s, which the lab prints exactly. Two entries are worked out here: ℒ[1] = 1/s and ℒ[e^(at)] = 1/(s − a).
Core ideas
Three ideas, and the first is the one the rest of the course leans on: the transform is a whole function, not an answer.
Pick a value of s. Multiply the signal by e^(−st), which shrinks it faster the larger s is, and add up the area under the product from t = 0 onward. That gives one number. Choose a different s and the number changes. The numbers taken together, one for each s, are the function F(s), and that function is the transform. The signal lives in t; its transform lives in s.
For f = e^(2t) the weighted signal is e^(−(s − 2)t). If s > 2 that shrinks toward zero and the area is finite. At s = 2 it is the constant 1 forever, so the area has no end, and below 2 it grows. So 1/(s − 2) is the transform only for s > 2. Every entry in the table comes with a floor of this kind, and the lab shows its integrals at a value of s above it.
Constants, powers of t, exponentials, sines and cosines all transform into rational functions of s: a polynomial over a polynomial. That is why the lab can print every transform exactly, and why the equations of the next lessons become algebra on fractions.
The material
The definition, the two computations that give the first table entries, and the one habit of mind the rest of the course needs: the answer is a function.
For a function f defined for t ≥ 0, its transform is ℒ[f](s) = ∫₀^∞ e^(−st)·f(t) dt, for every s at which the integral has a finite value.
The integral to infinity means the integrals from 0 to T, as T grows. The result is a function of s, and it is written F(s) when the signal is called f.
The definition is an instruction, and the quickest way to see what it does is to follow it on the simplest signal there is. Take f = 1. The weighted signal is e^(−st), and a function whose rate is −s times itself has the antiderivative −e^(−st)/s, as you can check by taking its derivative. So the integral from 0 to T is (1 − e^(−sT))/s.
At s = 1 the lab prints those three integrals in its status line. They are rounded, which is why each carries ≈, and they are closing in on 1. The reason is the term e^(−sT): for positive s it shrinks toward zero as T grows, so the integral approaches 1/s. That is a claim the three numbers support and the next theorem proves for every exponential.
For s > 0, ℒ[1] = 1/s. For every constant a and every s > a, ℒ[e^(at)] = 1/(s − a).
The second statement contains the first, with a = 0. The weighted signal is e^(−st)·e^(at) = e^(−(s − a)t), and its antiderivative is −e^(−(s − a)t)/(s − a), because the derivative of e^(−kt) is −k·e^(−kt) with k = s − a.
So the integral from 0 to T is (1 − e^(−(s − a)T))/(s − a). When s > a the exponent is negative and the term e^(−(s − a)T) shrinks toward zero as T grows, which leaves 1/(s − a).
□
The signal grows, and the transform exists anyway, because the weight shrinks faster: at s = 3 the product is e^(−t), and its integrals are the same three numbers as before.
The lab prints 1/(s − 2), and at s = 3 that is 1/(3 − 2) = 1, which is where those integrals are heading. Take s = 1 instead and the formula gives 1/(1 − 2) = −1, an impossible area for a positive function, because the integral has no finite value there.
The signal f = t is a straight line, and the lab prints ℒ[t] = 1/s². That one is a claim here and not a computation: finding the antiderivative of t·e^(−st) needs a rule this course does not develop. What the lab can do is show the truncated integrals closing in on the value of 1/s², which at s = 1 is 1: for T = 1, 2, 5 they are ≈ 0.264241, ≈ 0.593994 and ≈ 0.959572, rounded.
Read the three entries together: 1/s, 1/s² and 1/(s − a). Each is a fraction, each has a denominator that vanishes at one value of s, and that value, 0 or a, is exactly the floor under s. The lab's second tile lists these values under the name they carry for the rest of the course: the poles of the fraction, the values of s at which its denominator is zero. The denominators are going to matter more than anything else in this course.
One habit before the lab. When someone writes ℒ[1] = 1/s, the right side is a function: at s = 1 it is 1, at s = 2 it is 1/2 and at s = 4 it is 1/4. The transform of one signal is never a number. The equations of the next lessons are equations between functions of s, and they are solved by algebra only because a function of s can be added, multiplied and divided like a fraction.
The lab keeps its two tiers apart. The transform it prints in the first tile is an exact rational function in lowest terms. The integrals in the status line are floating-point sums, rounded to six figures and marked with ≈; they are evidence for the claim, and they close in on it without ever proving it.
Interactive
Transforms, partial fractions and solutions, exactly
Method
Weight, combine, integrate to T, let T grow. The last step is the one that gives the floor on s.
Multiply the signal by e^(−st). For e^(at) the product is e^(−st)·e^(at).
Add the exponents: e^(−st)·e^(at) = e^(−(s − a)t). Now the integrand is a single exponential with the rate −(s − a).
The antiderivative of e^(−kt) is −e^(−kt)/k, so the integral is (1 − e^(−kT))/k with k = s − a.
If k > 0 the term e^(−kT) shrinks toward zero and the integral approaches 1/k. If k ≤ 0 it does not, and the transform does not exist at that s.
Type the signal and read the first tile. The status line shows the integrals at three limits, rounded, closing in on the value of your fraction.
Follow the steps for f = e^(2t), so a = 2. The lab's second preset is this signal.
The lab prints the same fraction. At s = 3 it also prints the integrals from 0 to T for T = 1, 2, 5, rounded, and they are the same three numbers as for the constant at s = 1, because the weighted signal is e^(−t) in both. Each one sits a little below 1, as 1 − e^(−T) does.
Notice what the floor did. The formula 1/(s − 2) is a fraction, and it has a value at s = 1. The integral does not. The fraction is the transform only where the integral is finite, which is s > 2.
You should be able to say that the result is a function of s and not a number, state for which s each entry holds, integrate the weighted exponential from 0 to T and let T grow, and read a transform off the lab as an exact fraction with the truncated integrals marked as rounded.
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