Laplace Frequency-Shift Transform: Shift the Transform Argument

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 Laplace transform of $e^{at}f(t)$ by the already-known transform $F(s)$ with its argument shifted to $s-a$.

The invariant: multiplying the time-domain function by $e^{at}$ does not subtract $a$ from the transform value; it changes the transform’s input. The equivalent transform-domain expression is $F(s-a)$, provided $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists.

Pattern:

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

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

The illegal route treats the exponential as a subtraction outside the transform. The legal route shifts the argument of the whole known transform.

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Conditions of Applicability

Condition: $F(s)=\mathcal{L}\{f(t)\}$; shifted transform exists

This guide uses the ordinary one-sided Laplace transform introduced in Laplace Transform Definition. The symbol $F(s-a)$ means “take the known formula for $F(s)$ and substitute $s-a$ everywhere the transform argument appears.”

Before applying, check: identify the base function $f(t)$ and its transform $F(s)$ first, then verify that multiplying by $e^{at}$ gives the time-domain expression in front of you.

If the condition is violated: the shifted expression may not be the Laplace transform of the given time-domain function.

• The exponential must multiply the whole base function $f(t)$.
• The shift is $s-a$, so $e^{3t}$ gives $F(s-3)$ and $e^{-2t}$ gives $F(s+2)$.
• The shifted transform must exist in the working region of $s$.
• This rule is different from the time-shift transform, which handles delayed functions and unit steps.

In practice, this means the shifted expression must have a valid region of convergence after the substitution; the algebraic formula alone is not the whole transform claim.

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

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Common Failure Modes

Failure mode: treat $e^{at}$ as a factor that stays outside the transform or as a subtraction from the final value -> the transform argument is not shifted, so the table entry no longer matches the time-domain product.

Debug: first name the base transform $F(s)$, then rewrite every occurrence of its argument as $s-a$.

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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 $F(s-a)$ mean when $F(s)$ is a whole formula rather than a single symbol?
• Why does $e^{-2t}f(t)$ shift the argument to $s+2$ instead of $s-2$?

For the Principle

• What fast check tells you that a term has the form $e^{at}f(t)$ rather than a delayed unit-step form?
• Why must the shifted transform exist after replacing $s$ by $s-a$?

Between Principles

• How does this move connect the table-based Laplace Transform Definition with the return step in Inverse Laplace Transform Relation?

Generate an Example

• Write one valid frequency-shift transform and one near miss where the exponential factor is not multiplying the whole base function.

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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: _____Multiplying a time-domain function by e to the a t shifts its Laplace transform argument from s to s minus a.

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

State the canonical condition: _____$F(s)=\mathcal{L}\{f(t)\};\, \text{shifted transform exists}$

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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 $f(t)=\sin 2t$, transform $e^{-3t}f(t)$ using the known base transform.

Step	Expression	Operation
0	$f(t)=\sin 2t$	Start with the base time function.
1	$F(s)=\mathcal{L}\{\sin 2t\}=\frac{2}{s^2+4}$	Identify the base transform $F(s)$.
2	$\mathcal{L}\{e^{-3t}\sin 2t\}=F(s-(-3))$	Apply the frequency-shift rule with $a=-3$.
3	$\mathcal{L}\{e^{-3t}\sin 2t\}=F(s+3)=\frac{2}{(s+3)^2+4}$	Substitute $s+3$ into the whole base transform.

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Drills

Forward Step

Apply the frequency-shift transform once. Assume $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists.

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

Reveal

Use $a=2$, so the shifted argument is $s-2$:

$$\mathcal{L}\{e^{2t}f(t)\}=F(s-2)=\frac{1}{s-6}.$$

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Apply the frequency-shift transform once. Assume $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists.

$$F(s)=\frac{s}{s^2+9},\qquad \mathcal{L}\{e^{-t}f(t)\}$$

Reveal

Here $a=-1$, so replace $s$ by $s+1$ in the whole formula:

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

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Reject or complete the step. Assume $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists.

$$F(s)=\frac{1}{s^2+4},\qquad \mathcal{L}\{e^{5t}f(t)\}\Longrightarrow \frac{1}{s^2+4}-5$$

Reveal

Reject the step. The rule shifts the argument inside $F$, not the value outside it:

$$\mathcal{L}\{e^{5t}f(t)\}=F(s-5)=\frac{1}{(s-5)^2+4}.$$

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Apply the frequency-shift transform to a known table entry. Assume the shifted transform exists.

$$\mathcal{L}\{e^{4t}\cos 3t\}$$

Reveal

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

$$\mathcal{L}\{e^{4t}\cos 3t\}=F(s-4)=\frac{s-4}{(s-4)^2+9}.$$

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

A. $e^{2t}\sin t$ with $F(s)=\mathcal{L}\{\sin t\}$
B. $e^{2t}+\sin t$ with $F(s)=\mathcal{L}\{\sin t\}$
C. $u(t-2)f(t-2)$ with $F(s)=\mathcal{L}\{f(t)\}$

Reveal

A is eligible.

B is not eligible as a single frequency-shift step from $F(s)=\mathcal{L}\{\sin t\}$, because $e^{2t}$ is added, not multiplying the whole base function. You would handle it with linearity and a separate transform.

C is a time-delay/unit-step form, so it belongs to the time-shift rule instead of the frequency-shift rule.

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Action Labels

What was done between these two steps? Assume $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists.

$$F(s)=\frac{1}{s^2+1} \quad \Longrightarrow \quad \mathcal{L}\{e^{3t}f(t)\}=\frac{1}{(s-3)^2+1}$$

Reveal

The Laplace Frequency-Shift Transform was applied with $a=3$: replace the argument $s$ in $F(s)$ by $s-3$.

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What condition licenses this transition?

$$\mathcal{L}\{e^{-2t}f(t)\}=F(s+2)$$

Reveal

The condition is that $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists. Since $a=-2$, the argument becomes $s-a=s+2$.

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

$$\mathcal{L}\{e^{2t}+f(t)\}\Longrightarrow F(s-2)$$

Reveal

The exponential is not multiplying $f(t)$. The frequency-shift rule applies to $e^{at}f(t)$, not to a sum $e^{at}+f(t)$.

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Name the move in this chain. Assume the shifted transform exists.

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

Reveal

The frequency-shift rule was applied with $f(t)=t^2$, $F(s)=\frac{2}{s^3}$, and $a=-3$.

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Transition Identification

Where does the frequency-shift transform enter this chain?

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

Reveal

It enters in the second transition, where $e^{2t}\sin t$ is transformed as $F(s-2)$. The first transition only identifies the base transform, and the final transition substitutes into the formula.

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What is missing from this worked chain?

$$F(s)=\frac{s}{s^2+16} \quad \Longrightarrow \quad \mathcal{L}\{e^{-5t}f(t)\}=\frac{s}{s^2+16}$$

Reveal

The chain is missing the argument shift. Since $a=-5$,

$$\mathcal{L}\{e^{-5t}f(t)\}=F(s+5)=\frac{s+5}{(s+5)^2+16}.$$

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $F(s)=\mathcal{L}\{t\sin 2t\}$ with $F(s)=\frac{4s}{(s^2+4)^2}$, reach $\mathcal{L}\{e^{3t}t\sin 2t\}$ using Laplace Frequency-Shift Transform.

Full solution

Step	Expression	Move
0	$F(s)=\frac{4s}{(s^2+4)^2}$	Start from the known base transform for $f(t)=t\sin 2t$.
1	$\mathcal{L}\{e^{3t}t\sin 2t\}=F(s-3)$	Apply the frequency-shift rule with $a=3$.
2	$\mathcal{L}\{e^{3t}t\sin 2t\}=\frac{4(s-3)}{((s-3)^2+4)^2}$	Substitute $s-3$ into every occurrence of the transform argument.

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Related Guides

• Differential Equations Subdomain - Return to the transform-and-boundary-methods lane.
• Laplace Transform Definition - Review what $F(s)$ means before shifting its argument.
• Inverse Laplace Transform Relation - See how shifted transform-domain expressions return to time-domain functions.
• Laplace Derivative Transform - Connect transform-domain algebra with initial-value differential equations.
• Principle Structures - Separate the rule name, condition, formula, and legal move while studying.

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FAQ

What is Laplace Frequency-Shift Transform?

Laplace Frequency-Shift Transform is the rule $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$ when $F(s)=\mathcal{L}\{f(t)\}$ and the shifted transform exists. It says that multiplying by an exponential in time shifts the transform argument.

When is Laplace Frequency-Shift Transform valid?

It is valid when $F(s)$ is the Laplace transform of the base function $f(t)$ and the shifted transform exists. In a table-based problem, first identify the transform of $f(t)$, then substitute $s-a$ into the whole formula.

Why is the shift s minus a?

In the Laplace integral, $e^{-st}$ and $e^{at}$ combine as $e^{-(s-a)t}$. That is why multiplying by $e^{at}$ in the time domain changes the transform argument to $s-a$.

Is this the same as time shifting?

No. Frequency shifting handles multiplication by $e^{at}$. Time shifting handles delayed functions such as $u(t-a)f(t-a)$ and creates an exponential factor in the transform domain.

What is the most common mistake?

The most common mistake is shifting only one visible term or subtracting $a$ outside the formula. The rule is $F(s-a)$, so every occurrence of the transform argument inside $F$ changes consistently.

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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 base transform, check that the exponential multiplies the whole time function, retrieve the rule, and explain why every occurrence of the transform argument shifts. That makes transform-table work less like pattern matching and more like a checkable sequence of decisions.

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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