Laplace Transforms of (t − 2)u(t − 2) and tu(t − 2) − 2u(t − 2)

The two expressions differ only in how they are written: factoring out the unit step shows that they represent the same function. The shift in the step function delays the ramp by 2, so the second shifting theorem applies directly.

The expressions are identical

Distribute the factor in the first expression, or factor the second one:

$$(t-2)u(t-2)=t u(t-2)-2u(t-2).$$

Here $u(t-2)$ is the unit step shifted to $t=2$. Under the common convention, it is zero for $t<2$ and one for $t>2$. Its value at exactly $t=2$ does not affect the Laplace integral.

The function is therefore zero before $t=2$ and equals $t-2$ after that time. In terms of the shifted variable $t-2$, it is a ramp that begins at zero at $t=2$.

Apply the second shifting theorem

For the unilateral Laplace transform, the second shifting theorem states that

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

where $F(s)=\mathcal{L}\{f(t)\}$. In this problem, take $a=2$ and $f(t)=t$. Since $\mathcal{L}\{t\}=1/s^2$, the transform is

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

The shift theorem is sometimes misapplied because the expression inside the step is written using the original variable $t$. The function multiplying $u(t-2)$ must be expressed in terms of the shifted variable $t-2$. Here that function is exactly $f(t-2)=t-2$, so $f(x)=x$ and its transform is $1/s^2$.

Verify by computing the integral

The unilateral Laplace transform is defined by

$$\mathcal{L}\{g(t)\}=\int_0^\infty e^{-st}g(t)\,dt.$$

Because the step is zero before 2, the integral starts at 2:

$$\begin{aligned} \mathcal{L}\{(t-2)u(t-2)\} &=\int_2^\infty e^{-st}(t-2)\,dt\\ &=e^{-2s}\int_0^\infty e^{-s\tau}\tau\,d\tau\\ &=e^{-2s}\frac{1}{s^2}. \end{aligned}$$

The substitution is $\tau=t-2$. This direct calculation confirms the shift-rule result. The integral converges for $\operatorname{Re}(s)>0$.

A useful check using the second form

You can also transform the two terms separately. The theorem gives

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

For $t u(t-2)$, rewrite $t=(t-2)+2$. Then

$$\begin{aligned} \mathcal{L}\{t u(t-2)\} &=\mathcal{L}\{(t-2)u(t-2)\}+2\mathcal{L}\{u(t-2)\}\\ &=\frac{e^{-2s}}{s^2}+\frac{2e^{-2s}}{s}. \end{aligned}$$

Subtracting the transform of $2u(t-2)$ cancels the extra term:

$$\mathcal{L}\{t u(t-2)-2u(t-2)\} =\left(\frac{e^{-2s}}{s^2}+\frac{2e^{-2s}}{s}\right)-\frac{2e^{-2s}}{s} =\frac{e^{-2s}}{s^2}.$$

This calculation is especially helpful if you are asked to transform the two terms separately rather than recognize the shifted ramp.

Common mistakes

• Using $1/s^2$ without the delay factor. The ramp begins at $t=2$, so its transform includes $e^{-2s}$.
• Treating $t u(t-2)$ as the shifted ramp. It is not: after $t=2$, it equals $t$, not $t-2$. Its transform includes the additional term $2e^{-2s}/s$.
• Forgetting to subtract the step term. In $t u(t-2)-2u(t-2)$, that subtraction cancels the additional term and leaves the shifted ramp transform.
• Changing the integration lower limit incorrectly. In a direct integral, the step makes the integrand zero until $t=2$, so integrate from 2 rather than from 0.

For a general delayed ramp $(t-a)u(t-a)$, the same reasoning gives $e^{-as}/s^2$.