Step Functions
Differential Equations · Section 6.3 · generated practice set
The unit step as a switch: writing piecewise forcing in step form, and the second shifting theorem both ways.
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Try one now
Find $\mathcal{L}\left\{u_{2}(t)\,\cos(3(t-2))\right\}$, where $u_{2}(t)$ is the unit step turning on at $t = 2$.
Solution
- The second shifting theorem: $\mathcal{L}\left\{u_c(t)f(t-c)\right\} = e^{-cs}F(s)$, where $F = \mathcal{L}\{f\}$.
- Identify $f$ by stripping the shift. Here $c = 2$.
- Transform that $f$ from the table first, then multiply by $e^{-2s}$.
- The exponent is $-cs$ — negative, and multiplying $s$, not a bare constant.
- Answer: $\dfrac{s\,e^{-2s}}{s^2 + 9}$.
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Method
The unit step $u_c(t)$ is 0 before $t = c$ and 1 from $t = c$ onward. It is the switch that lets a single formula describe a piecewise input.
Writing a piecewise function with steps
Start with the value on the first interval, then add the jump at each breakpoint:
$$f(t) = \begin{cases} f_1, & 0 \le t < c \\ f_2, & t \ge c\end{cases} \quad\Longrightarrow\quad f(t) = f_1 + (f_2 - f_1)u_c(t).$$
The coefficient on each step is the change, never the new value. For a pulse that is on between $a$ and $b$ and off outside, the answer is $u_a(t) - u_b(t)$.
The second shifting theorem
$$\mathcal{L}\left\{u_c(t)f(t-c)\right\} = e^{-cs}F(s), \qquad \mathcal{L}^{-1}\left\{e^{-cs}F(s)\right\} = u_c(t)f(t-c).$$
Two directions of the same fact. Going forward, the function must be written as $f(t-c)$ — shifted — before you may use it. If it is not, force it: for $u_2(t),t^2$, write $t^2 = (t-2)^2 + 4(t-2) + 4$ first.
Going backward, strip the $e^{-cs}$, invert what remains to get $f(t)$, then replace every $t$ by $t - c$ and attach $u_c(t)$.
Useful special case: $\mathcal{L}\{u_c(t)\} = \dfrac{e^{-cs}}{s}$.
Definitions and theorems
Worked example
Write $f(t) = \begin{cases} 0, & 0 \le t < 3 \\ t, & t \ge 3\end{cases}$ in step form and transform it.
- The jump at $t = 3$ is from 0 to $t$, so $f(t) = u_3(t)\cdot t$.
- That is not yet in the form $u_3(t)g(t-3)$ — the theorem needs the argument shifted.
- Force the shift: $t = (t-3) + 3$, so $f(t) = u_3(t)\left[(t-3) + 3\right]$.
- Now $g(t) = t + 3$, and $G(s) = \dfrac{1}{s^2} + \dfrac{3}{s}$.
- $\mathcal{L}\{f\} = e^{-3s}\left(\dfrac{1}{s^2} + \dfrac{3}{s}\right)$.
Common mistakes
- Using the new value as the step coefficient. Going from 2 to 7 at $t = c$ gives $2 + 5u_c(t)$, not $2 + 7u_c(t)$.
- Applying the theorem to an unshifted function. $\mathcal{L}\{u_2(t)t^2\} \neq e^{-2s}\cdot\frac{2}{s^3}$. Rewrite $t^2$ in powers of $(t-2)$ first.
- Sign of the exponent. It is $e^{-cs}$ with $c > 0$; a positive exponent is not a Laplace transform of anything well behaved.
- Forgetting to re-attach $u_c(t)$ on the inverse. Without it the answer is wrong for all $t < c$, where the true value is 0.
- Shifting only some of the $t$'s. Every $t$ in $f$ becomes $t - c$, including inside exponentials and trig arguments.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.