Section 13.2 Unit Step Function Variants

The basic unit step $u(t)$ switches a function on at $t=0\text{.}$ But what if you need the switch to flip later, flip off, or even turn on for just a short while? This paragraph introduces three key variants:

$u_c(t)\text{:}$ switches ON at a chosen time $c\text{.}$
$1 - u_c(t)\text{:}$ switches OFF at time $c\text{.}$
$u_c(t) - u_d(t)\text{:}$ switches ON for a window from $c$ to $d\text{.}$

Together, these act like a complete toolkit for modeling stepwise ON/OFF behavior in real systems.

Subsection Shifted Unit Step Function

Checkpoint 263. Prep-Questions.

(a) 📖❓ True or False: Shifting a Function.

The graph of $\cos(t - 3)$ is the same as the graph of $\cos(t)\ $ shifted right $3$ units.

True.
False.

(b) 📖❓ When Does It Switch?

Sometimes you need the step to turn ON at a different time. The shifted unit step function, written $u_c(t)\text{,}$ is simply the basic step shifted right (if $c \gt 0$) or left (if $c \lt 0$):

📜 Shifted Unit Step Function.

\begin{equation*} u_c(t) = u(t - c) = \left\{ \begin{array}{ll} 1, \amp t \ge c \\ 0, \amp t \lt c \end{array} \right. \end{equation*}

Before $t=c\text{,}$ $u_c(t)$ is OFF; after $t=c\text{,}$ it turns ON.

¹³

$u_c(t)$ is not a new function; it is just the shorthand for a shifted version of $u(t)\text{.}$

$Graph of the shifted unit step u-sub-c of t: 0 for t less than c, 1 for t at least c, with the jump marked at t = c.$

Exploring the Effect of $c$.

Consider the product of the parabolic function and the shifted unit step function

\begin{equation*} \left(\frac{1}{5}t^2 - 1\right) u_c(t)\text{.} \end{equation*}

Use the tool to see how changing $c$ affects the ON/OFF behavior of the parabola.

An interactive plot of the parabola one-fifth t squared minus 1 multiplied by the shifted unit step u-sub-c of t; the reader drags the switch point c along the t-axis and hovers over the labels to highlight each graph and see how c turns the parabola on and off.

Hover over the labels $g(t)\text{,}$ $\left(\frac15 t^2 - 1\right)$ and $u_c(t)$ to highlight its graph.
Drag the value of $c\text{,}$ initially at $(0,0)\text{,}$ along the $t$ axis.

Checkpoint 264. 📖❓ Effect of $u_c(t)$ on a Function.

What does multiplying a function $f(t)$ by $u_{5}(t)$ do?

It forces $f(t)$ OFF before $t = 5\text{,}$ then switches it ON at $t = 5\text{.}$
Exactly! Multiplying by $u_5(t)$ means the function is 0 before $t = 5$ and normal after.
It multiplies $f(t)$ by $5\text{.}$
Not quite. $u_5(t)$ is not a scaling factor; it’s a switching function.
It adds $5$ to $f(t)\text{.}$
Nope, $u_5(t)$ does not alter the function’s output directly. It controls when it turns on.
It delays $f(t)$ by 5 seconds.
Close, but that’s what happens when you write $f(t - 5)\text{.}$ Multiplying by $u_5(t)$ just turns it ON at $t = 5\text{.}$

So, to activate a function at $t = c\text{,}$ simply multiply it by $u_c(t)\text{.}$ This switch-like behavior will be useful later when we write piecewise functions in terms of $u_c\text{.}$

Subsection Reversed Unit Step Function

Checkpoint 265. Prep-Questions.

(a) 📖❓ When Does It Switch?

(b) 📖❓ Understanding ON/OFF Language.

What does it mean when we say a function $g(t)$ is "OFF" for $t \gt 4\text{?}$

That $g(t)$ becomes negative for $t \gt 4\text{.}$
OFF doesn’t mean negative. It means the function is multiplied by $0\text{,}$ it’s gone.
That $g(t)$ stops increasing after $t = 4\text{.}$
Not quite. Saying a function is OFF means it’s set to 0, not just leveled off.
That $g(t)$ is multiplied by 0 for all $t \gt 4\text{.}$
Exactly. OFF means the function is inactive because it’s being multiplied by 0.
That $g(t)$ is undefined for $t \gt 4\text{.}$
Nope, OFF doesn’t mean undefined. The function is still defined, but multiplication by $0$ turns it off.

What if you need the opposite behavior — a switch that turns a function OFF at $t=c\text{?}$

This reverse behavior can be built using the unit step function itself. Since $u_c(t)$ jumps from 0 to 1 at $t = c\text{,}$ we can reverse the switch with the subtraction: $1-u_c(t)\text{.}$

📜 Reversed Unit Step Function.

\begin{equation*} 1 - u_c(t) = \left\{ \begin{array}{ll} 1, \amp t \lt c \\ 0, \amp t \ge c \end{array} \right. \end{equation*}

Before $t=c\text{,}$ $u_c(t)$ is ON; after $t=c\text{,}$ it turns OFF.

$Graph of the reversed unit step 1 minus u-sub-c of t: 1 for t less than c, 0 for t at least c, with the drop marked at t = c.$

🌌 Example 266. Switching OFF an Exponential.

Describe the product below and express it in piecewise form.

\begin{equation*} h(t) = 2e^{-0.5t^2}\left(1 - u_1(t)\right) \end{equation*}

Solution.

Here, the exponential function $2e^{-0.5t^2}$ is ON before $t = 1$ and switches OFF afterward. In piecewise form, this looks like:

\begin{equation*} h(t) = \left\{ \begin{array}{ll} 2e^{-0.5t^2}, \amp t \lt 1 \\ 0, \amp t \ge 1 \end{array} \right. \end{equation*}

$Graph of h of t equals 2 e to the minus 0.5 t squared times 1 minus u-sub-1 of t: bell curve for t less than 1, then 0.$

Checkpoint 267. 📖❓ Reversed Switch Logic.

Which of the following is equivalent to the function

\begin{equation*} f(t) = \begin{cases} t,\amp t \lt -2 \\ 0,\amp t \ge -2 \end{cases} \end{equation*}

$t \cdot u_{-2}(t)$
This would set $t$ to 1 at $t = -2\text{,}$ but we want it to turn OFF there.
$t \cdot (1 - u_{-2}(t))$
Correct! $1 - u_{-2}(t)$ is ON before $t = -2$ and OFF afterward, matching the piecewise definition.
$(t - 2) \cdot u_{-2}(t)$
This starts a different function at $t = -2\text{.}$ It’s neither $t$ nor OFF before then.
$t + 2$
This is just a shifted version of $t\text{,}$ not a piecewise function that shuts off.

With both $u_c(t)$ and $1 - u_c(t)$ in hand, you can now switch functions ON or OFF exactly when you need. Next, we will cover the last switch, which turns functions ON for a limited window of time over a finite interval.

Subsection Windowed Unit Step Function

Checkpoint 268. Prep-Questions.

(a) 📖❓ True or False: Interval Notation.

The interval $[2, 5)$ includes $t = 2$ but not $t = 5\text{.}$

True.
The square bracket means included and the parenthesis means excluded, so $t = 2$ is in the interval, but $t = 5$ is not.
False.
The square bracket means included and the parenthesis means excluded, so $t = 2$ is in the interval, but $t = 5$ is not.

(b) 📖❓ When Is the Window ON?

The final variant combines the first two ideas to create a “time window” where a function is ON for a limited interval, then turns OFF again.

We can build this behavior by subtracting two step functions:

$u_c(t)$ turns ON at $t = c\text{.}$
$u_d(t)$ turns ON at $t = d\text{,}$ flipping us back to OFF since we are subtracting.

So the difference $u_c(t) - u_d(t)$ is the ON-window we need.

📜 Windowed Unit Step Function.

\begin{equation*} u_c(t) - u_d(t) = \left\{ \begin{array}{ll} 1, \amp c \le t \lt d \\ 0, \amp \text{otherwise} \end{array} \right. \end{equation*}

Between $t=c$ and $t=d\text{,}$ $u_c(t)$ is ON and OFF otherwise.

$Graph of the windowed step u-sub-c of t minus u-sub-d of t: 0, then 1 on c to d, then 0 again, shown with c = 2 and d = 5.$

This “windowed” switch models limited bursts of activity (like a force that acts for a few seconds, then disappears).

Effect of $c$ and $d$.

Drag $c$ and d along the t-axis in the first two plots to see how u-sub-c of t and u-sub-d of t combine so that their difference forms an ON-window between c and d.

Checkpoint 269. 📖❓ Selecting the ON-Interval Expression.

Which expression represents a function, $f(t)\text{,}$ that is active (ON) on the interval $[2, 5)$ and $0$ elsewhere?

$\ f(t)\left(u_2(t) - u_5(t)\right)$
Correct!
$\ 2f(t) + 5f(t)$
Incorrect
$\ f(t)\left(u_5(t) - u_2(t)\right)$
Incorrect
$\ f(t)\ u_5(t) - u_2(t) $
Incorrect

As with previous step functions, this step function switches ON any function over the interval $[c,d)$ and OFF everywhere else.

🌌 Example 270. Switching ON a Sine Wave for a While.

Write the function that equals $\sin t + 2$ only on $[\frac{\pi}{2}, 2\pi)$ and is $0$ elsewhere, using step functions.

Solution.

Since this function is $\sin t + 2$ only on a finite window, we multiply it by the following difference of shifted unit step functions:

\begin{equation*} g(t) = \left(\sin t + 2\right) \cdot \left(u_{\sfrac{\pi}{2}}(t) - u_{2\pi}(t)\right)\text{.} \end{equation*}

In piecewise form, this function is

\begin{equation*} g(t) = \left\{ \begin{array}{ll} \sin t + 2, \amp \sfrac{\pi}{2} \le t \lt 2\pi \\ 0, \amp \text{otherwise} \end{array} \right. \end{equation*}

and its graph is given by

$Graph of g of t equals sine t plus 2, times u-sub-pi-over-2 of t minus u-sub-2pi of t: raised sine on that window, else 0.$

Explore the Product $f(t) \cdot (u_c(t)-u_d(t))$.

Instructions:

Hover over $g(t)$ to hightlight the graph of $(\sin t + 2)(u_c(t) - u_d(t))\text{.}$
Hovering over $(\sin t + 2)$ or $(u_c(t) - u_d(t))$ highlights their individual graph.
Drag the value of $c$ and $d$ along the $t$ axis.

An interactive plot of the quantity sine t plus 2 multiplied by the window u-sub-c of t minus u-sub-d of t; the reader drags c and d along the t-axis and hovers over the labels to watch the windowed sine burst change width.

Together, these three step function variants give you complete control over when a function is active. In the next section, we’ll combine them to rewrite entire piecewise functions in a single, neat formula.

Checkpoint 271. 🤔💭 Unit Step Function Variants Reading Questions.

(a) 📖❓ What Does $u_c(t)$ Do?

The unit step function $u_c(t)$ is multiplied by another function $f(t)\text{.}$ What does $u_c(t)$ do in this context?

It switches $f(t)$ ON at $t=c$ and keeps it ON forever.
Correct — $u_c(t)$ equals 0 before $t=c$ and 1 afterward.
It switches $f(t)$ OFF at $t=c\text{.}$
Not quite — that would be $1-u_c(t)\text{.}$
It reverses the direction of $f(t)$ at $t=c\text{.}$
No — there’s no “reverse” here, just ON/OFF switching.
It multiplies $f(t)$ by $t$ after $t=c\text{.}$
No — $u_c(t)$ is either 0 or 1; it doesn’t add factors of $t\text{.}$

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