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Section 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 subsection introduces three key variants:
\(u_c(t)\) โ switches ON at a chosen time \(c\text{.}\)
\(1 - u_c(t)\) โ switches OFF at time \(c\text{.}\)
\(u_c(t) - u_d(t)\) โ 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 246 . Prep-Questions.
(a) ๐โ 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.
Replacing \(t\) with \(t - 3\) in shifts the graph of ANY function 3 units to the right, not just \(\cos(t)\text{.}\)
False.
Replacing \(t\) with \(t - 3\) in shifts the graph of ANY function 3 units to the right, not just \(\cos(t)\text{.}\)
(b) ๐โ 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
\(c \lt 0\) ):
๐ 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.
๐ฎโ Interactive Question 247 . 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.
Instructions.
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.
(a) ๐โ 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, itโs a switching function.
It adds \(5\) to \(f(t)\text{.}\)
Nope, \(u_5(t)\) does not alter the functionโs output directly. It controls when it turns on.
It delays \(f(t)\) by 5 seconds.
Close, but thatโs 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 248 . Prep-Questions.
(a) ๐โ When Does It Switch?
(b) ๐โ 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 doesnโt mean negative. It means the function is multiplied by \(0\text{,}\) itโs gone.
That
\(g(t)\) stops increasing after
\(t = 4\text{.}\)
Not quite. Saying a function is OFF means itโs zeroed out, 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 itโs being multiplied by 0.
That
\(g(t)\) is undefined for
\(t \gt 4\text{.}\)
Nope, OFF doesnโt mean undefined. The function is still defined, but turned off by multiplication with \(0\text{.}\)
What if you need the opposite behavior โ 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:
\(1-u_c(t)\text{.}\)
๐ 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.
๐ Example 249 . 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*}
Checkpoint 250 . ๐โ 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*}
This would turn \(t\) ON at \(t = -2\text{,}\) but here 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 \(t = -2\text{.}\) Itโs neither \(t\) nor OFF before then.
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 that turns ON functions for a limited window of time, over a finite interval.
Subsection Windowed Unit Step Function
Checkpoint 251 . Prep-Questions.
(a) ๐โ 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) ๐โ When Is the Window ON?
The final variant combines the first two ideas to create a โtime windowโ 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.
๐ 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.
This โwindowedโ switch models limited bursts of activity โ like a force that acts for a few seconds, then disappears.
๐ฎโ Interactive Question 252 . Exploring the Effect of \(c\) and \(d\) .
\(\longleftarrow\ \text{Drag}\ c\ \text{here.}\)
\(\longleftarrow\ \text{Drag}\ d\ \text{here.}\)
\(\longleftarrow\ \text{Watch how}\ u_c-u_d\ \text{changes.}\)
(a) ๐โ 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!
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.
๐ Example 253 . 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
๐ฎ Interactive 254 . 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.
Together, these three step function variants give you complete control over when a function is active. In the next section, weโll combine them to rewrite entire piecewise functions in a single, neat formula.
Subsection ๐ค Wrap-Up
๐๏ธ \(\textbf{Key Takeaways...}\)
Check Your Understanding.
Checkpoint 255 . ๐ค๐ญ Unit Step Function Variants Reading Questions.
(a) ๐ค๐ญ 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 โ \(u_c(t)\) equals 0 before \(t=c\) and 1 afterward.
It switches
\(f(t)\) OFF at
\(t=c\text{.}\)
Not quite โ that would be \(1-u_c(t)\text{.}\)
It reverses the direction of
\(f(t)\) at
\(t=c\text{.}\)
No โ thereโs no โreverseโ here, just ON/OFF switching.
It multiplies
\(f(t)\) by
\(t\) after
\(t=c\text{.}\)
No โ \(u_c(t)\) is either 0 or 1; it doesnโt add factors of \(t\text{.}\)
You have attempted
of
activities on this page.
