Laplace Time-Shift Transform: Delay with Unit Steps

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Guides | FAQ

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The Principle

The move: replace the one-sided Laplace transform of a delayed unit-step term $u(t-a)f(t-a)$ by $e^{-as}F(s)$.

The invariant: the transform-domain expression represents the same forcing, but the start time $a$ is now carried by the exponential factor $e^{-as}$.

Pattern:

$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$$

Legal route	Illegal route
$F(s)=\frac{1}{s^2},\ a=3\Longrightarrow \mathcal{L}\{u(t-3)(t-3)\}=e^{-3s}\frac{1}{s^2}$	$F(s)=\frac{1}{s^2},\ a=3\not\Longrightarrow \mathcal{L}\{u(t-3)t\}=e^{-3s}\frac{1}{s^2}$

The illegal route has a unit step, but the function after the switch is $t$, not $t-3$. The theorem is licensed by the delayed argument $f(t-a)$, not by the unit step alone.

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Conditions of Applicability

Condition: $a>0$; $F(s)=\mathcal{L}\{f(t)\}$; $delayed forcing defined for t\geq a$

This guide uses the one-sided Laplace transform from Laplace Transform Definition. The unit step $u(t-a)$ turns the forcing on at time $a$, while $f(t-a)$ restarts the base function’s clock at that same time.

Before applying, check: identify the base function $f(t)$, confirm its transform $F(s)$, then verify that the delayed expression uses $f(t-a)$ after the unit step turns on.

If the condition is violated: the exponential factor may describe the wrong time-domain forcing.

• The delay must be positive in this one-sided transform setting.
• The expression must be a product of the unit step and the shifted function $f(t-a)$.
• A term like $u(t-a)f(t)$ is not automatically eligible; first rewrite the post-switch formula as a function of $t-a$ if possible.
• This rule is different from Laplace Frequency-Shift Transform, which handles multiplication by $e^{at}$ instead of delayed unit steps.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: treat every $u(t-a)$ term as a time-shift transform → the exponential factor gets attached to the wrong base transform when the post-switch formula is not $f(t-a)$.

Debug: after the switch, set $\tau=t-a$ and ask whether the remaining formula is exactly $f(\tau)$.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What does the unit step $u(t-a)$ do before and after time $a$?
• Why does the formula need $f(t-a)$ rather than $f(t)$ for this clean transform rule?

For the Principle

• What fast check tells you whether a delayed forcing term is ready for the time-shift transform?
• Why does a delay in time become multiplication by $e^{-as}$ in the transform domain?

Between Principles

• How does this move differ from Laplace Frequency-Shift Transform?

Generate an Example

• Write one valid delayed unit-step term and one near miss where the unit step is present but the shifted argument is not.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____A delayed unit-step function u of t minus a times f of t minus a transforms into e to the minus a s times the original transform F of s.

Write the canonical pattern: _____$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$

State the canonical condition: _____$a>0; F(s)=\mathcal{L}\{f(t)\}; delayed forcing defined for t\geq a$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to learn from worked examples.)

Procedure Walkthrough

Starting from the delayed forcing $u(t-2)(t-2)^2$, transform it using the known base transform.

Step	Expression	Operation
0	$f(t)=t^2$	Identify the base function whose clock starts at zero.
1	$F(s)=\mathcal{L}\{t^2\}=\frac{2}{s^3}$	Find the base transform.
2	$u(t-2)(t-2)^2=u(t-a)f(t-a)$ with $a=2$	Check the delayed argument form.
3	$\mathcal{L}\{u(t-2)(t-2)^2\}=e^{-2s}F(s)=\frac{2e^{-2s}}{s^3}$	Apply the time-shift transform.

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Drills

Forward Step

Apply the time-shift transform once. Assume $a>0$, $F(s)=\mathcal{L}\{f(t)\}$, and the delayed forcing is defined for $t\geq a$.

$$F(s)=\frac{1}{s-4},\qquad \mathcal{L}\{u(t-3)f(t-3)\}$$

Reveal

Use $a=3$:

$$\mathcal{L}\{u(t-3)f(t-3)\}=e^{-3s}F(s)=\frac{e^{-3s}}{s-4}.$$

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Apply the time-shift transform once. Assume the condition holds.

$$\mathcal{L}\{u(t-5)\sin(t-5)\}$$

Reveal

Here $f(t)=\sin t$, $F(s)=\frac{1}{s^2+1}$, and $a=5$:

$$\mathcal{L}\{u(t-5)\sin(t-5)\}=\frac{e^{-5s}}{s^2+1}.$$

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Reject or complete the step. Assume one-sided Laplace transforms are being used.

$$\mathcal{L}\{u(t-4)t^2\}\Longrightarrow e^{-4s}\frac{2}{s^3}$$

Reveal

Reject the step. The term after the switch is $t^2$, not $(t-4)^2$. The direct rule would apply to

$$\mathcal{L}\{u(t-4)(t-4)^2\}=e^{-4s}\frac{2}{s^3}.$$

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Apply the time-shift transform once. Assume the condition holds.

$$\mathcal{L}\{u(t-1)\cos(3(t-1))\}$$

Reveal

Use $f(t)=\cos 3t$, $F(s)=\frac{s}{s^2+9}$, and $a=1$:

$$\mathcal{L}\{u(t-1)\cos(3(t-1))\}=e^{-s}\frac{s}{s^2+9}.$$

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Which expressions are eligible for this time-shift rule? Assume the base transforms exist.

A. $u(t-2)(t-2)^3$
B. $u(t-2)t^3$
C. $e^{2t}f(t)$
D. $u(t-6)e^{-(t-6)}$

Reveal

A and D are eligible for the direct time-shift rule.

B has a unit step, but the post-switch formula is not written as $f(t-2)$. C belongs to the frequency-shift rule, not the time-shift rule.

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Action Labels

What was done between these two steps? Assume the condition holds.

$$F(s)=\frac{2}{s^3} \quad \Longrightarrow \quad \mathcal{L}\{u(t-3)(t-3)^2\}=e^{-3s}\frac{2}{s^3}$$

Reveal

The Laplace Time-Shift Transform was applied with $a=3$: the delayed term $u(t-3)f(t-3)$ became $e^{-3s}F(s)$.

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What condition licenses this transition?

$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$$

Reveal

The condition is $a>0$, $F(s)=\mathcal{L}\{f(t)\}$, and the delayed forcing is defined for $t\geq a$.

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A student claims this transition uses the time-shift rule. What is wrong?

$$\mathcal{L}\{u(t-2)\sin t\}\Longrightarrow e^{-2s}\frac{1}{s^2+1}$$

Reveal

The shifted argument is missing. The displayed rule would be licensed for $u(t-2)\sin(t-2)$, not for $u(t-2)\sin t$.

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Name the move in this chain. Assume the condition holds.

$$f(t)=e^{-t} \quad \Longrightarrow \quad F(s)=\frac{1}{s+1} \quad \Longrightarrow \quad \mathcal{L}\{u(t-7)e^{-(t-7)}\}=\frac{e^{-7s}}{s+1}$$

Reveal

The final transition uses the Laplace Time-Shift Transform with $a=7$.

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Transition Identification

Where does the time-shift transform enter this chain?

$$f(t)=t \quad \Longrightarrow \quad F(s)=\frac{1}{s^2} \quad \Longrightarrow \quad \mathcal{L}\{u(t-4)(t-4)\}=e^{-4s}F(s) \quad \Longrightarrow \quad \frac{e^{-4s}}{s^2}$$

Reveal

It enters in the second transition, where the delayed time-domain expression becomes $e^{-4s}F(s)$. The first transition only identifies the base transform, and the final transition substitutes the formula for $F(s)$.

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What is missing from this worked chain?

$$\mathcal{L}\{u(t-3)(t-3)^2\} \quad \Longrightarrow \quad \frac{2e^{-3s}}{s^3}$$

Reveal

The chain skips the base transform and condition check. A clearer legal chain is

$$f(t)=t^2,\quad F(s)=\frac{2}{s^3},\quad a=3,$$

then

$$\mathcal{L}\{u(t-3)(t-3)^2\}=e^{-3s}F(s)=\frac{2e^{-3s}}{s^3}.$$

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $f(t)=t\cos 2t$ and $F(s)=\frac{s^2-4}{(s^2+4)^2}$, reach $\mathcal{L}\{u(t-3)(t-3)\cos(2(t-3))\}$ using Laplace Time-Shift Transform.

Full solution

Step	Expression	Move
0	$f(t)=t\cos 2t$	Start from the base function.
1	$F(s)=\frac{s^2-4}{(s^2+4)^2}$	Use the given base transform.
2	$u(t-3)(t-3)\cos(2(t-3))=u(t-a)f(t-a)$ with $a=3$	Check the delayed argument form.
3	$\mathcal{L}\{u(t-3)(t-3)\cos(2(t-3))\}=e^{-3s}F(s)$	Apply the time-shift transform.
4	$\mathcal{L}\{u(t-3)(t-3)\cos(2(t-3))\}=e^{-3s}\frac{s^2-4}{(s^2+4)^2}$	Substitute the known $F(s)$.

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Related Guides

• Differential Equations Subdomain - Return to the transforms and boundary-methods lane.
• Laplace Transform Definition - Review what $F(s)$ means before applying shift rules.
• Laplace Frequency-Shift Transform - Compare time delay with exponential multiplication in time.
• Inverse Laplace Transform Relation - See how transform-domain expressions return to time-domain solutions.
• Principle Structures - Separate the rule name, formula, condition, and legal move while studying.

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FAQ

What is Laplace Time-Shift Transform?

Laplace Time-Shift Transform is the rule $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$. It says that delaying a function with a unit step turns into multiplication by an exponential factor in the transform domain.

When is Laplace Time-Shift Transform valid?

It is valid when $a>0$, $F(s)=\mathcal{L}\{f(t)\}$, and the delayed forcing is defined for $t\geq a$. In practice, the expression must contain both the unit step $u(t-a)$ and the shifted function $f(t-a)$.

Is this the same as multiplying by e to the a t?

No. Multiplying by $e^{at}$ uses the frequency-shift rule and changes the transform argument to $s-a$. Delaying a function with $u(t-a)f(t-a)$ uses the time-shift rule and multiplies the transform by $e^{-as}$.

Why is f of t minus a required?

The argument $t-a$ restarts the base function’s clock at the switch time. Without that restart, the post-switch formula is a different function, so the transform may need a rewrite before the time-shift theorem applies.

What is the most common mistake?

The most common mistake is seeing $u(t-a)$ and immediately multiplying the old transform by $e^{-as}$. First check that the rest of the term is written as $f(t-a)$ for the same $f$ whose transform is $F(s)$.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium trains this as a condition-first transform move: identify the switch time, name the base function, check the shifted argument, and retrieve the exponential factor. The Unisium Study System pairs that habit with retrieval practice, self-explanation, and compact problem-solving chains so delayed forcing terms do not collapse into pattern matching.

Ready to practice differential equations with structure? Check access and join the Unisium waitlist or explore the complete framework in Masterful Learning.

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