The Unit Step Function and Inverse Laplace Transforms with e^(−as)

When you feed a piecewise input into a differential equation, 90% of the work is writing it with unit step function laplace transforms. The remaining 10% is inverting the e−as term that appears in the transform. If you forget to multiply the inverse by u(t−a), you get a function that exists for negative time, physically wrong for any causal system. This covers the Heaviside function, the second shift theorem, and the exact inverse of delayed transforms, using Kreyszig ch. 6.3-6.4 as the reference.

The unit step function u(t−a) is zero for t < a and one for t ≥ a. Oliver Heaviside used it to model switching on a circuit at a given time. The transform pair is ℒ{u(t−a)} = e−as/s for a ≥ 0, sourced from Boyce & DiPrima §6.2 and Kreyszig ch. 6.3.

A common mistake is writing u(t−a) without the −a offset in the argument. The function u(t) alone switches at t = 0. To delay that switch to t = 3, write u(t−3). The argument inside the step is the time at which the function becomes active.

Failure case: a student writes u(t+2) thinking it shifts right. It shifts left, active for t ≥ −2, which is meaningless for a time-domain signal starting at zero.

The second shift theorem states: ℒ{u(t−c) f(t−c)} = e−cs F(s). In words: a delay of c seconds in the time function multiplies its Laplace transform by e−cs. The inverse direction is where most errors happen: ℒ−1{e−asF(s)} = f(t−a) u(t−a).

Source: Boyce & DiPrima §6.2, Kreyszig ch. 6.3. The theorem is also called the time-shift theorem. Do not confuse it with the frequency-shift theorem, which multiplies f(t) by eat.

Why The Unit Step Must Be There

If you invert e−asF(s) and write only f(t−a), you have defined f(t−a) for t < a, which is before the input was applied. In engineering, no response exists before the cause. Multiplying by u(t−a) forces the inverse to be zero for t < a, matching causality. Lerch's theorem guarantees uniqueness for continuous functions, but at the jump discontinuity at t = a, the step function is what makes the inverse physically correct.

The procedure has three steps: identify F(s) by removing the exponential; find f(t) = ℒ−1{F(s)} using a standard pair table or partial fractions; then replace t with t−a and multiply by u(t−a). Partial-fraction decomposition is needed when F(s) does not match a table entry directly. The Heaviside cover-up method from Boyce & DiPrima §6.3 handles distinct real poles quickly.

Worked Example 1: Simple Exponential Term

Find ℒ−1{e−2s / (s−4)}. Remove e−2s: F(s) = 1/(s−4). From the table, ℒ−1{1/(s−4)} = e4t. Replace t with t−2 and multiply by u(t−2): result is e4(t−2) u(t−2).

Common mistake: writing e4t u(t−2) instead of shifting the exponent. The exponent must be 4(t−2), not 4t. Check by evaluating at t = 3: the correct value is e4, the wrong value is e12, a factor of e8 off.

Worked Example 2: Partial Fractions With A Delay

Find ℒ−1{e−3s / (s2 + 5s + 6)}. Factor denominator: (s+2)(s+3). Remove exponential: F(s) = 1/[(s+2)(s+3)]. Decompose: A/(s+2) + B/(s+3). Heaviside cover-up: A = 1/(−2+3) = 1, B = 1/(−3+2) = −1. So F(s) = 1/(s+2) − 1/(s+3). Inverse: f(t) = e−2t − e−3t. Apply delay: ℒ−1 = [e−2(t−3) − e−3(t−3)] u(t−3).

Failure case: forgetting to shift the exponentials inside the brackets. The −2t becomes −2(t−3).

Worked Example 3: Piecewise Input With Two Steps

A voltage is zero for t < 1, 5 V for 1 ≤ t < 4, and zero again for t ≥ 4. Write as 5 u(t−1) − 5 u(t−4). Transform: 5 e−s/s − 5 e−4s/s. Invert: 5 u(t−1) − 5 u(t−4), the original function is recovered exactly, confirming the pair. This works because ℒ−1{e−as/s} = u(t−a).

Any piecewise function defined by intervals can be written as a sum of step functions. For each interval [a, b) where the function equals g(t), add g(t) [u(t−a) − u(t−b)]. If the final interval extends to infinity, use u(t−a) alone.

Example: a function that is 2 for 0 ≤ t < 3, t for 3 ≤ t < 7, and 0 for t ≥ 7. Write: 2 u(t) − 2 u(t−3) + t u(t−3) − t u(t−7). The t in the middle interval must be expressed relative to the step: t u(t−3) = (t−3+3) u(t−3) = (t−3) u(t−3) + 3 u(t−3). This uses the second shift theorem in reverse: to transform, you need f(t−a) form.

Source: Boyce & DiPrima ch. 6 exercises. The method is essential for the solve differential equations laplace transform approach because the transform of a step-wise input is a sum of exponentials times rational functions.

The Dirac delta δ(t−a) is an impulse at t = a. Its Laplace transform is e−as, since ℒ{δ(t−a)} = e−as for a ≥ 0. The unit impulse response of a system is the inverse of its transfer function when the input transform is 1.

The Dirac delta is used in impulse response analysis and in the study of Green's functions. It is not a function in the ordinary sense but a distribution; for Laplace transform work, the pair δ(t) ↔ 1 is sufficient. Confusion arises when students try to invert e−as and expect a smooth function instead of an impulse.

Five errors account for nearly all failures in problems involving the unit step function and delayed inverse:

1. Writing f(t−a) without the unit step factor. The result is defined for negative time, violating causality.
2. Forgetting to shift the argument inside the inverse function. ℒ−1{e−asF(s)} yields f(t−a) u(t−a), not f(t) u(t−a).
3. Using the wrong sign in the exponential of the second shift theorem. ℒ{u(t−a) f(t−a)} = e−as F(s), not e+as.
4. Confusing the second shift theorem with the first. The first shift theorem (frequency shift) gives ℒ{eat f(t)} = F(s−a).
5. Mishandling the sign in u(t−a) when a is negative. For a < 0, u(t−a) is active for t ≥ a, which includes negative time; this is used only in theoretical contexts.

Source: Kreyszig ch. 6.3-6.4 problem sets. Cross-check with the inverse laplace transform formula explained section of your textbook.

Graphing the shifted inverse clarifies the effect of the delay. Consider f(t) = sin(t) for t ≥ 0. The delayed version f(t−2) u(t−2) = sin(t−2) u(t−2) is zero until t = 2, then follows the sine wave shifted right by 2 seconds. The amplitude, frequency, and phase relative to the shifted origin are unchanged; only the start time moves.

Failure case: plotting sin(t) u(t−2) instead. That graph shows the sine wave starting at t = 2 but with the phase of the original sin(t), not sin(t−2). The value at t = 2 would be sin(2) ≈ 0.909 instead of sin(0) = 0, a clear discontinuity.