The unit step u(t − c) is 0 before t = c and 1 from then on. Multiplying a signal by it switches the signal on at c, and the transform of the switched-on, shifted signal is e^(−cs)·F(s). To solve an equation with a switched forcing, invert the ordinary fraction as before, then replace t by t − c and multiply by the step. The solution is written in pieces, one for each stretch of time.
Core ideas
Three ideas: what a step is, what the shift rule does to a transform, and why an effect cannot come before its cause.
The function u(t − c) is 0 for every t before c and 1 for every t from c on. Multiplying a signal by it blanks the signal before c. Two steps with a minus sign between them, u(t − 1) − u(t − 3), give a pulse that is on from 1 up to 3.
If f has the transform F(s), the copy of f that starts at c has the transform e^(−cs)·F(s). The fraction F is exactly what it was, and the delay sits entirely in the factor e^(−cs). Inverting reverses it: remove the factor, invert the fraction, shift.
The solution on t < c depends only on what the equation and the forcing were doing on t < c. Before the switch the forcing is the old one, so the solution is the old one. Whatever happens after c is invisible there.
The material
The shift rule, a proof of it, and three equations: a switch-on, a pulse, and a second-order case.
For a constant c ≥ 0, u(t − c) = 0 when t < c and u(t − c) = 1 when t ≥ c.
The product u(t − c)·f(t − c) is the signal f shifted to the right by c: it is 0 up to t = c, and from there it repeats f as it was from t = 0.
If ℒ[f] = F(s), then for every c > 0, ℒ[u(t − c)·f(t − c)] = e^(−cs)·F(s).
The signal is zero before c, so the integral starts at t = c and contains f(t − c). Write t = c + τ. Then e^(−st) = e^(−cs)·e^(−sτ), and τ runs from 0 onward. Renaming the variable this way is a shift of the time axis, and all it uses is that sliding a graph along the axis does not change the area under it.
The constant e^(−cs) comes outside the integral, and what is left is the integral of e^(−sτ)·f(τ) from 0 onward, which is F(s). So the result is e^(−cs)·F(s).
□
Putting f = 1 gives the case that matters most for forcing: ℒ[u(t − c)] = e^(−cs)/s. So the equation y′ + y = u(t − 2) transforms to (s + 1)·Y = e^(−2s)/s, and Y = e^(−2s)/(s·(s + 1)). The factor e^(−2s) is a passenger. Partial fractions are done on 1/(s·(s + 1)) = 1/s − 1/(s + 1), giving f = 1 − e^(−t), and then the shift is applied once, at the end.
Take y′ + y = u(t − 1) − u(t − 3) from y(0) = 0. The forcing is on from t = 1 up to t = 3. The transform is Y = (e^(−s) − e^(−3s))/(s·(s + 1)), which is the same fraction 1/s − 1/(s + 1) times two different delays.
The solution has three stretches. Before t = 1 nothing has happened and y = 0. From 1 to 3 it is the charging curve 1 − e^(−(t − 1)). From 3 on, the second delay subtracts a copy that starts at 3, which leaves e^(−(t − 3)) − e^(−(t − 1)): the output decays, since the forcing is off again. The two pieces meet at t = 3, where both are 1 − e^(−2).
For y″ + 4y = u(t − 1) from rest, the transform is (s² + 4)·Y = e^(−s)/s. The fraction 1/(s·(s² + 4)) decomposes as (1/4)/s − (1/4)·s/(s² + 4): A = 1/4 at s = 0, and the s² terms give B = −1/4. Its inverse is (1/4)·(1 − cos(2t)), and the shift gives the solution (1/4)·(1 − cos(2(t − 1))) from t = 1 on. It sits at 0 until the switch and then oscillates between 0 and 1/2, with its first minimum exactly at the switch.
A limit of the lab, stated where it applies. The shift c must be a positive rational number. The equation y″ + y = u(t − π) is a common textbook example, but π is not rational and the lab does not take it. The method is the same; only the typed input is restricted.
Interactive
Transforms, partial fractions and solutions, exactly
Method
Transform with the delay in the factor, invert the fraction, shift, and write the pieces.
Replace each u(t − c) by e^(−cs)/s. Keep the factors e^(−cs) as they are.
Divide by the characteristic polynomial. Y is a sum of terms e^(−cs)·F(s) with ordinary fractions F.
Decompose each F and read the table backwards. Do not touch the factor e^(−cs) yet.
For each term, replace t by t − c and multiply by u(t − c).
List the solution on each stretch of time. It must be 0 where nothing has yet happened, and each piece must meet the next.
The first preset: the forcing switches on at t = 2.
Nothing happens until t = 2, and then the charging curve starts from 0 as if time began at the switch. The pieces meet: at t = 2 the second piece is 1 − e⁰ = 0. The lab prints the same two pieces.
Compare with the pulse preset on the stretch from 1 to 3: there the solution is 1 − e^(−(t − 1)), the same curve with the switch at 1. The switch-off at 3 has no effect before t = 3.
You should be able to transform a unit step, factor the delay out of Y, invert the remaining fraction, apply the shift, and check that the pieces start from zero and meet at the switch.
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