Questions: Unit Step Function and Piecewise-Defined Forcing

Question 1

Using the second shifting theorem, what is the Laplace transform of f(t) = sin(t − 3) · u(t − 3)?

• A. e^{−3s} / (s² + 1)
• B. sin(3) · e^{−3s} / (s² + 1)
• C. e^{3s} / (s² + 1)
• D. 1 / (s² + 1)

Answer: (A)

The second shifting theorem states L{f(t − a) · u(t − a)} = e^{−as} · F(s), where F(s) = L{f(t)}. Here f(t) = sin(t), so F(s) = 1/(s² + 1), and a = 3. The result is e^{−3s} · (1/(s² + 1)). The factor e^{−3s} encodes the time delay of 3 in the s-domain. Options B and C reflect common errors: confusing the time-shift with evaluating sin at the activation point, or getting the sign of the exponent wrong.

Question 2

A student wants the Laplace transform of g(t) = sin(t) · u(t − 3) — a sine wave that switches on at t = 3 but continues as sin(t), not as sin(t − 3). They write e^{−3s} · L{sin(t)} = e^{−3s}/(s² + 1). Is this correct?

• A. Yes — the step function u(t − 3) always contributes e^{−3s} regardless of the argument of sin
• B. No — the second shifting theorem requires f(t − 3) · u(t − 3), not f(t) · u(t − 3). The student must rewrite sin(t) as sin((t − 3) + 3) and expand before applying the theorem
• C. No — you cannot take the Laplace transform of a product involving a step function
• D. Yes, but only for t > 3 where sin(t) is defined

Answer: (B)

This is the most common error with the second shifting theorem. The theorem applies to f(t − a) · u(t − a), where the argument of f and the activation point of u match. Writing f(t) · u(t − a) is a different function — it switches on at t = a but continues with the unshifted argument. To apply the theorem, the student must rewrite: sin(t) = sin((t − 3) + 3) = sin(t − 3)cos(3) + cos(t − 3)sin(3), then apply the theorem to each term.

Question 3

The Laplace transform of u(t − a) is e^{−as}/s, which means a time delay of a in the time domain corresponds to multiplication by e^{−as} in the s-domain.

• T. True
• F. False

Answer: True

This is correct and is the foundational fact behind the entire unit step / piecewise-forcing framework. The exponential e^{−as} is the s-domain signature of a time delay: any function that 'starts at time a' rather than time 0 will have its transform multiplied by e^{−as}. When you see e^{−as} · F(s) in the s-domain, you immediately know the inverse is f(t − a) · u(t − a) — a copy of f shifted right by a, active only for t ≥ a.

Question 4

If e^{−2s} · F(s) appears in the s-domain, the inverse Laplace transform is f(t) · u(t − 2), where f(t) is the function whose transform is F(s).

• T. True
• F. False

Answer: False

This is false — the inverse is f(t − 2) · u(t − 2), not f(t) · u(t − 2). The time-shift must be applied to the argument of f as well. f(t − 2) · u(t − 2) is a copy of f shifted right by 2 units, turned on at t = 2. Writing f(t) · u(t − 2) would be f evaluated at the original t but switched on at t = 2 — a different function that does not correspond to e^{−2s} · F(s) under the second shifting theorem.

Question 5

Why must a forcing term be written as f(t − a) · u(t − a) rather than f(t) · u(t − a) in order to directly apply the second shifting theorem? What goes wrong if you use f(t) · u(t − a) instead?

Model answer: The second shifting theorem is L{f(t − a) · u(t − a)} = e^{−as} · F(s). The argument of f must be (t − a) to match the activation time of the step function. If you write f(t) · u(t − a), you are describing a genuinely different function — f with its original, unshifted argument, just switched on at t = a. This does not satisfy the hypothesis of the theorem, so applying it directly yields an incorrect result. To proceed, you must first rewrite f(t) in terms of (t − a) using algebra (e.g., expanding f((t − a) + a)), and only then can the theorem be applied to each resulting term.

The theorem is a precise statement about a specific form. Many errors in this topic come from treating u(t − a) as a generic 'delay operator' that can be applied to any expression, rather than as a function that shifts the entire expression including the argument. The mismatch between f(t) and u(t − a) is invisible to a student who isn't thinking carefully about what the two functions are actually computing.