Unit Step Function and the Time Shifting Theorem

Real inputs switch on and off. A force acts for a minute and is removed; a voltage is applied and then cut. The unit step function $u (t - c)$ is the algebraic switch that describes such behavior, turning a term off up to $t = c$ and on afterward. Combined with the time shifting theorem, it makes discontinuous forcing terms as easy to transform as smooth ones.

Quick Reference

Item	Statement
Unit step	u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases}
Shifted step	u(t-c) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases}, also written $u_{c} (t)$
Transform	$ℒ {u (t - c)} = \frac{e^{- c s}}{s}$
Special case	$ℒ {u (t)} = ℒ {1} = \frac{1}{s}$
Two-piece function	$f = g [u (t) - u (t - c)] + h u (t - c)$
"Window" on $[a, b)$	$u (t - a) - u (t - b)$
Time shifting theorem	$ℒ {f (t - c) u (t - c)} = e^{- c s} F (s)$ , $c > 0$
Inverse form	$ℒ^{- 1} {e^{- c s} F (s)} = f (t - c) u (t - c)$
Contrast	$s$ -shifting: $ℒ {e^{c t} f (t)} = F (s - c)$

Why a Step Function

In many applications, inputs or driving forces act for some time only. An external force is applied to a body for a minute and is then completely removed; a constant voltage is applied to a circuit and then reduced to zero for all subsequent times. Such situations arise constantly.

The unit step function can suppress a given term up to a certain value of $t$ and then insert it for all larger $t$ . It provides a powerful tool for solving differential equations with complicated right-hand sides.

The unit step function, also called the Heaviside step function, is defined by

u(t) = \left\{ \begin{aligned} &0, && \text{if } t < 0 \\ &1, && \text{if } t \geq 0 \end{aligned} \right.

In words, $u (t)$ is zero when its input is negative and one when its input is positive or zero.

Graph of the unit step function: zero along the negative t-axis, jumping to the value 1 at t = 0 and staying there. — The unit step function $u (t)$ . It is the simplest possible switch: off before the origin, on from the origin onward.

It follows from the definition that

u(t-c) = \left\{ \begin{aligned} &0, && \text{if } t < c \\ &1, && \text{if } t \geq c \end{aligned} \right.

Note that $u (t - c)$ is a unit step shifted $c$ units to the right. The argument $t - c$ becomes non-negative exactly when $t \geq c$ , which is where the step turns on. It is sometimes denoted $u_{c} (t)$ .

Graph of the shifted unit step function u(t - c): zero until t = c, then jumping to 1. — The shifted step $u (t - c)$ , which switches on at $t = c$ .

Sketch the graph of $f (t) = 1 - u (t - c)$ .

Solution

f(t) = 1 - u(t-c) = \left\{ \begin{aligned} &1, && \text{if } t < c \\ &0, && \text{if } t \geq c \end{aligned} \right.

Subtracting the step from $1$ reverses it: instead of switching on at $c$ , the function switches off at $c$ . This is how a term is removed after a given time.

Graph of one minus the shifted unit step: constant at 1 until t = c, then dropping to zero. — The function $1 - u (t - c)$ , a switch that turns off at $t = c$ .

Expressing Piecewise-Defined Functions

The unit step function writes piecewise-defined functions in a compact form, which makes their Laplace transforms much quicker to evaluate.

f(t) = \left\{ \begin{aligned} &g(t), && \text{if } 0 \leq t < c \\ &h(t), && \text{if } c \leq t < \infty \end{aligned} \right.

then

f (t) = g (t) [u (t) - u (t - c)] + h (t) u (t - c) .

The bracket $u (t) - u (t - c)$ is a window: it equals $1$ on $[0, c)$ and $0$ elsewhere, so it selects the first piece. The final term selects the second piece. More generally, $u (t - a) - u (t - b)$ is the window that is $1$ exactly on $[a, b)$ .

Write f(t) = \begin{cases} t^{2} & 0 \leq t < 3 \\ 5 & t \geq 3 \end{cases} using unit step functions.

Solution

Applying the formula with $g (t) = t^{2}$ , $h (t) = 5$ , and $c = 3$ ,

f (t) = t^{2} [u (t) - u (t - 3)] + 5 u (t - 3) = t^{2} + (5 - t^{2}) u (t - 3)

for $t \geq 0$ , where $u (t) = 1$ throughout the region of interest.

The second form is often the more useful one. It reads: start with $t^{2}$ , then at $t = 3$ add the correction $5 - t^{2}$ that converts it into $5$ .

Laplace Transform of the Unit Step Function

ℒ {u (t - c)} = \frac{e^{- c s}}{s} (c \geq 0)

Calculation

\begin{aligned} \mathscr{L}\left\{u(t-c)\right\} &= \int_0^{\infty}e^{-st}u(t-c)\,dt \\ &= \int_0^{c}e^{-st}(0)\,dt + \int_c^{\infty}e^{-st}(1)\,dt \\ &= \lim_{R\to\infty}\int_c^{R}e^{-st}\,dt \\ &= \lim_{R\to\infty}\left[-\frac{e^{-st}}{s}\right]_{t=c}^{t=R} \\ &= \frac{e^{-cs}}{s} \end{aligned}

The step function did nothing except move the lower limit of integration from $0$ to $c$ , which is exactly what produces the factor $e^{- c s}$ .

Since $u (t) = 1$ throughout $[0, \infty)$ , which is the entire interval of integration,

ℒ {u (t)} = ℒ {1} = \frac{1}{s},

the case $c = 0$ of the general formula.

The Time Shifting Theorem

The meaning of

f (t - c) u (t - c)

Let $f (t)$ be the input to a system and let $c > 0$ . Suppose the input is delayed by $c$ seconds, after which it begins in exactly the same fashion as before, as when one waits a few seconds before turning on a generator. Then the input is described by

f_c(t) = \left\{ \begin{aligned} &0, && \text{if } 0 \leq t < c \\ &f(t-c), && \text{if } c \leq t < \infty \end{aligned} \right.

The graph of $f_{c} (t)$ is obtained by translating the graph of $f (t)$ to the right by $c$ units and making it identically zero between $t = 0$ and $t = c$ . We call $f_{c}$ the shifted function.

The unit step function describes this translation compactly:

f_{c} (t) = f (t - c) u (t - c)

provided $f (t)$ is defined for $- c \leq t < 0$ . The values of $f$ on that range have no bearing on the result, since they are multiplied by zero; they merely need to exist so that $f (t - c)$ is defined for all $t \geq 0$ .

Two stacked panels. The upper shows f(t) and its right-translate f(t-c), with an arrow marking the shift of c units. The lower shows f(t-c)u(t-c), identically zero until t = c and then following the translated curve. — Panel (a): the original $f (t)$ and its translate $f (t - c)$ , shifted right by $c$ . Panel (b): the product $f (t - c) u (t - c)$ , which discards everything before $t = c$ and starts the translated signal there.

The next figure compares three closely related functions. Imagine a signal generator feeding a system.

If the generator is on and connected from the beginning, the input is $f (t)$ .
If the generator has been running from the beginning but is not connected until $t = c$ , the input is $f (t) u (t - c)$ : the signal is already partway through its cycle when it arrives.
If the generator is off but connected, and is switched on at $t = c$ , the input is $f (t - c) u (t - c)$ : the signal starts from its beginning at time $c$ .

Three stacked panels comparing f(t), f(t)u(t-c) which is f truncated before c, and f(t-c)u(t-c) which is f translated to begin at c. — (a) $f (t)$ , running from the start. (b) $f (t) u (t - c)$ , the same signal with everything before $t = c$ deleted, so it arrives mid-cycle. (c) $f (t - c) u (t - c)$ , the signal delayed so that it *begins* at $t = c$ . Only the third has a clean transform, and the distinction between (b) and (c) is the most common source of error in this section.

Previously we saw that multiplying $f (t)$ by $e^{c t}$ translates $F (s)$ along the $s$ -axis, giving $ℒ {f (t) e^{c t}} = F (s - c)$ , the Exponential Scaling or $s$ -shifting theorem. Dually, if $f (t)$ is shifted along the $t$ -axis by $c > 0$ , the transform $F (s)$ is multiplied by $e^{- c s}$ .

Time Shifting ( $t$ -Shifting) Theorem, also called the second shifting theorem. If $ℒ {f (t)} = F (s)$ and $c > 0$ , then

ℒ {f (t - c) u (t - c)} = e^{- c s} F (s) .

Proof

\begin{aligned} \mathscr{L}\left\{f(t-c)u(t-c)\right\} &= \int_0^{\infty}e^{-st}f(t-c)u(t-c)\,dt \\ &= \int_0^{c}e^{-st}f(t-c)\underbrace{u(t-c)}_{0}\,dt + \int_c^{\infty}e^{-st}f(t-c)\underbrace{u(t-c)}_{1}\,dt \\ &= \int_c^{\infty}e^{-st}f(t-c)\,dt \end{aligned}

Substitute $t = τ + c$ , so $τ = t - c$ and $d τ = d t$ . When $t = c$ we have $τ = 0$ , so the lower limit becomes zero:

\begin{aligned} \mathscr{L}\left\{f(t-c)u(t-c)\right\} &= \int_0^{\infty}e^{-s(\tau+c)}f(\tau)\,d\tau \\ &= \int_0^{\infty}e^{-cs}e^{-s\tau}f(\tau)\,d\tau \\ &= e^{-cs}\underbrace{\int_0^{\infty}e^{-s\tau}f(\tau)\,d\tau}_{F(s)} \\ &= e^{-cs}F(s) \end{aligned}

The constant $e^{- c s}$ came out of the integral because it does not involve $τ$ .

The two shifting theorems side by side.

$s$ -shifting: multiplying by $e^{c t}$ in time shifts the transform, $F (s) \to F (s - c)$ .

$t$ -shifting: delaying in time multiplies the transform, $F (s) \to e^{- c s} F (s)$ .

Both are used constantly, and confusing them is easy. The presence of a factor $e^{- c s}$ multiplying the transform always signals a time delay; an $s$ appearing only as $s - c$ always signals an exponential factor in $t$ .

Worked Examples

Find $ℒ {f (t)}$ where f(t) = \begin{cases} 0 & 0 \leq t < 2 \\ (t-2)^{3} & t \geq 2 \end{cases}.

Solution

This is already in the exact form $f (t - c) u (t - c)$ with $c = 2$ and the underlying function $t^{3}$ :

f (t) = (t - 2)^{3} u (t - 2) .

Since $ℒ {t^{3}} = \frac{6}{s^{4}}$ , the time shifting theorem gives

ℒ {f (t)} = \frac{6 e^{- 2 s}}{s^{4}} .

Find $ℒ {t^{2} u (t - 1)}$ .

Solution

This is not in the form $f (t - c) u (t - c)$ , because the argument of the power is $t$ rather than $t - 1$ . Rewrite $t^{2}$ in powers of $t - 1$ :

t^{2} = {[(t - 1) + 1]}^{2} = (t - 1)^{2} + 2 (t - 1) + 1 .

Therefore

t^{2} u (t - 1) = [(t - 1)^{2} + 2 (t - 1) + 1] u (t - 1),

and each term is now correctly shifted. Applying the theorem term by term with $ℒ {t^{2}} = \frac{2}{s^{3}}$ , $ℒ {t} = \frac{1}{s^{2}}$ , $ℒ {1} = \frac{1}{s}$ ,

ℒ {t^{2} u (t - 1)} = e^{- s} (\frac{2}{s^{3}} + \frac{2}{s^{2}} + \frac{1}{s}) .

Rewriting in powers of $t - c$ is the standard preparation step whenever the function multiplying the step is not already shifted.

Find $ℒ^{- 1} {\frac{e^{- 3 s}}{s^{2} + 4}}$ .

Solution

Ignore the exponential first. Since $\frac{1}{s^{2} + 4} = \frac{1}{2} \cdot \frac{2}{s^{2} + 4}$ ,

f (t) = ℒ^{- 1} {\frac{1}{s^{2} + 4}} = \frac{1}{2} \sin 2 t .

The factor $e^{- 3 s}$ then delays this by $3$ :

ℒ^{- 1} {\frac{e^{- 3 s}}{s^{2} + 4}} = \frac{1}{2} \sin (2 (t - 3)) u (t - 3) .

Note that every $t$ in $f$ becomes $t - 3$ , including the one inside the sine.

Solve y' + y = f(t) with $y (0) = 0$ , where f(t) = \begin{cases} 1 & 0 \leq t < 2 \\ 0 & t \geq 2 \end{cases}.

Solution

Write the forcing function with steps: $f (t) = 1 - u (t - 2)$ , so

F (s) = \frac{1}{s} - \frac{e^{- 2 s}}{s} .

Transforming the equation, $s Y + Y = F (s)$ , so

Y (s) = \frac{1}{s (s + 1)} (1 - e^{- 2 s}) .

Since $\frac{1}{s (s + 1)} = \frac{1}{s} - \frac{1}{s + 1}$ , the inverse of the first piece is $1 - e^{- t}$ . Writing $g (t) = 1 - e^{- t}$ ,

y (t) = g (t) - g (t - 2) u (t - 2) = (1 - e^{- t}) - (1 - e^{- (t - 2)}) u (t - 2) .

Explicitly,

y(t) = \begin{cases} 1 - e^{-t} & 0 \leq t < 2 \\ e^{-(t-2)} - e^{-t} & t \geq 2 \end{cases}

The solution rises toward $1$ while the input is on, then decays once it is switched off. Note that $y$ is continuous at $t = 2$ even though the forcing is not: both branches give $1 - e^{- 2}$ there. Solving this by the methods of Chapter 4 would have required treating the two intervals separately and matching by hand.

Exercises

Express f(t) = \begin{cases} 0 & 0 \leq t < 1 \\ t & 1 \leq t < 4 \\ 0 & t \geq 4 \end{cases} using step functions and find its transform.

Solution

Using a window, $f (t) = t [u (t - 1) - u (t - 4)]$ .

Rewrite each piece in shifted form: $t u (t - 1) = [(t - 1) + 1] u (t - 1)$ and $t u (t - 4) = [(t - 4) + 4] u (t - 4)$ . Therefore

ℒ {f} = e^{- s} (\frac{1}{s^{2}} + \frac{1}{s}) - e^{- 4 s} (\frac{1}{s^{2}} + \frac{4}{s}) .

Find $ℒ^{- 1} {\frac{e^{- 2 s}}{s^{3}}}$ .

Solution

Since $ℒ^{- 1} {\frac{1}{s^{3}}} = \frac{t^{2}}{2}$ , delaying by $2$ gives

\frac{(t - 2)^{2}}{2} u (t - 2) .

Find $ℒ {\cos t \cdot u (t - π)}$ .

Solution

The cosine is not shifted, so rewrite it. Using $\cos t = \cos ((t - π) + π) = - \cos (t - π)$ ,

\cos t \cdot u (t - π) = - \cos (t - π) u (t - π),

which is now in standard form. Therefore

ℒ {\cos t \cdot u (t - π)} = - \frac{s e^{- π s}}{s^{2} + 1} .

Frequently Asked Questions

What is the unit step function used for?

To switch terms on and off at prescribed times. Multiplying a term by $u (t - c)$ suppresses it before $t = c$ and admits it afterward, which is how inputs that start, stop, or change form are written as a single formula rather than a case list.

How do I write a piecewise function using steps?

Multiply each piece by the window that selects its interval. The window for $[a, b)$ is $u (t - a) - u (t - b)$ , and the window for $[c, \infty)$ is $u (t - c)$ . Adding the products gives a single expression valid for all $t \geq 0$ .

What is the difference between

f (t) u (t - c)

and

f (t - c) u (t - c)

The first truncates $f$ , deleting it before $t = c$ but leaving the rest where it was; the second translates $f$ so that it begins at $t = c$ . Only the second has the clean transform $e^{- c s} F (s)$ . The comparison figure above shows all three cases side by side.

What do I do when the function is not already shifted?

Rewrite it in powers of $t - c$ , or use a trigonometric addition formula. For $t^{2} u (t - 1)$ , expand $t^{2} = (t - 1)^{2} + 2 (t - 1) + 1$ . For $\cos t \cdot u (t - π)$ , use $\cos t = - \cos (t - π)$ . Then apply the theorem to each term.

How do I recognize a time shift in an inverse transform?

By the factor $e^{- c s}$ multiplying the transform. Set it aside, invert what remains to get $f (t)$ , then replace every $t$ by $t - c$ and multiply by $u (t - c)$ . Forgetting the step function is the most common mistake, and it produces an answer that is wrong for $t < c$ .

How do the two shifting theorems differ?

$s$ -shifting concerns $e^{c t} f (t)$ , giving $F (s - c)$ : an exponential in time translates the transform. $t$ -shifting concerns $f (t - c) u (t - c)$ , giving $e^{- c s} F (s)$ : a delay in time multiplies the transform by an exponential. The roles of "multiply" and "translate" are exactly swapped between the two.

Does the value of

u

at the jump matter?

No. Defining $u (0)$ as $0$ , $1$ , or $\frac{1}{2}$ changes nothing, because the Laplace transform is an integral and cannot detect the value at a single point, as Lerch's theorem in Section 7.3 makes precise.

The Unit Step Function (Heaviside Function). Time Shifting Theorem

Quick Reference

Why a Step Function

Expressing Piecewise-Defined Functions

Laplace Transform of the Unit Step Function

The Time Shifting Theorem

Worked Examples

Exercises

Frequently Asked Questions

Quick Links

Social Media