Two facts make the transform usable. It is linear: the transform of c·f + d·g is c·ℒ[f] + d·ℒ[g]. And a table of five families — powers, exponentials, cosines, sines, and a power times an exponential — covers every signal the rest of the course uses. Together they transform a sum of terms one term at a time and combine the result into a single fraction. They do not turn a product of signals into a product of transforms.
Core ideas
Three ideas: what linearity says, what the table says, and the product that the transform does not respect.
To transform 3t² − 2e^(−t), transform each term separately, keep the constants 3 and −2 where they are, and add: 3·ℒ[t²] − 2·ℒ[e^(−t)]. This works because the transform is an integral, and an integral of a sum is the sum of the integrals, with a constant factor taken outside.
Powers tⁿ, exponentials e^(at), cosines cos(bt), sines sin(bt) and a power times an exponential t·e^(at). The cosine and sine entries share the denominator s² + b², and the entry for t·e^(at) is the entry for t with s replaced by s − a. A signal outside these families is outside this course, apart from the switch that the last module adds.
Linearity is about sums and constants only. Multiplying two signals does something else to their transforms, and the shortest test is the signal 1·1: its transform is 1/s, while ℒ[1]·ℒ[1] is 1/s².
The material
A theorem, a proof from the integral, the table, and a worked case that shows linearity doing its job and the product refuting itself.
For signals f and g and constants c and d, at every s where both transforms exist, ℒ[c·f + d·g] = c·ℒ[f] + d·ℒ[g].
The weighted signal is e^(−st)·(c·f + d·g) = c·e^(−st)·f + d·e^(−st)·g.
The integral from 0 to T of a sum is the sum of the integrals, and a constant factor comes outside, because the integral is a limit of sums of the form rate times step. Let T grow in each term and the right side becomes c·ℒ[f] + d·ℒ[g].
□
The first lesson derived 1/s and 1/(s − a). The rest of the table is stated here as claims. The lab computes every transform by exactly these rules, so it cannot confirm them; what it offers is evidence, in its status line, where the truncated integrals of the weighted signal close in on the value the fraction gives at one s. The entry for tⁿ has n! on top, the product n·(n − 1)·⋯·1, so ℒ[t] = 1/s², ℒ[t²] = 2/s³ and ℒ[t³] = 6/s⁴.
The last entry is worth a second look. Put a = 0 in ℒ[t·e^(at)] = 1/(s − a)² and it becomes 1/s², which is ℒ[t]. Multiplying the signal by e^(at) has replaced s by s − a. The same replacement turns 1/s into 1/(s − a), the second entry, and that pattern is the reason the table needs only these families.
Transform f = 3t² − 2e^(−t). By linearity it is 3·ℒ[t²] − 2·ℒ[e^(−t)]. From the table, ℒ[t²] = 2/s³ and ℒ[e^(−t)] = 1/(s + 1), because a = −1.
So the transform is 3·(2/s³) − 2·(1/(s + 1)) = 6/s³ − 2/(s + 1). Over a common denominator s³·(s + 1) the numerator is 6·(s + 1) − 2·s³ = 6s + 6 − 2s³, and the lab prints the single fraction in lowest terms.
Sines and cosines work the same way, with one new feature: their denominator s² + b² has no real zero. For cos(2t) the lab prints s/(s² + 4). The poles of that fraction are the numbers where the denominator vanishes, ±2i, and the lab lists them in its second tile. That frequency 2 in the signal reappears, as the square root of 4, in the transform.
Linearity does not say anything about a product of two signals. Take f = g = 1. The product f·g is 1, and ℒ[1] = 1/s. The product of the transforms is (1/s)·(1/s) = 1/s², a different function. The transform of a product is a different and harder operation, which this course does not need.
The signal t·e^(3t) is a product of t and e^(3t). Its transform is the table entry 1/(s − 3)².
The product of the two separate transforms is ℒ[t]·ℒ[e^(3t)] = (1/s²)·(1/(s − 3)) = 1/(s²·(s − 3)), which has a pole at 0 that the real transform does not. The lab's third preset prints the correct one, and a poles tile that agrees: the only pole is at 3.
A table does the work of a computation, so check one entry before trusting it. The lab does that at every keystroke: it transforms what you type by the table and linearity and prints a reduced fraction, so a slip in your own arithmetic shows up as a mismatch.
Interactive
Transforms, partial fractions and solutions, exactly
Method
Split, look up, scale, add, combine. Combine last: the table's entries are simpler uncombined.
Each term must be a constant times one table family: 3t², −2e^(−t), 5·cos(2t). If a term is not, it is outside the table.
Match the term to its entry. Name n, a or b for each, and watch the signs: e^(−t) has a = −1.
Multiply each transform by the constant in front of its term. The constant is outside the transform, not inside the fraction.
Write the sum with each fraction left as it is. The result is already correct.
Multiply each numerator by the factors its denominator lacks. Compare with the lab, which prints the reduced form.
The first preset. Two terms, two table entries.
The lab prints (−2s³ + 6s + 6)/(s³(s + 1)), which is the last line with the numerator written from the highest power down; its fourth tile keeps the term-by-term form 6/s³ − 2/(s + 1).
A check that costs nothing: at a large s the 6/s³ term is tiny and the −2/(s + 1) term dominates, so the fraction should be close to −2/s. Both forms agree.
You should be able to scale each term by its constant, look up the entry with the right sign of a and the right factorial, add the fractions over a common denominator, and say why the transform of a product of signals is not the product of their transforms.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.