Heaviside Function
Summarize
Introduction
The Heaviside step function is $0$ to the left of the origin and $1$ to the right. With $H(0)=\frac{1}{2},$ its piecewise definition is:
$$ H(x) = \begin{cases} 0 & \text{if } x < 0 \\[6pt] \dfrac{1}{2} & \text{if } x = 0 \\[8pt] 1 & \text{if } x > 0 \end{cases} \quad \forall \ x \in \mathbb{R} $$
For example, the definition gives:
$$ H(-3) = 0 \qquad H(0) = \frac{1}{2} \qquad H(5) = 1 $$
When $x$ denotes time, $H(x)$ models a quantity that is $0$ for $x<0$ and has the constant value $1$ for $x>0,$ such as the normalised voltage of a circuit switched on at time zero.
The graph has two horizontal half-lines, one at height $0$ for $x<0$ and one at height $1$ for $x>0,$ together with the isolated point $\left(0, \frac{1}{2}\right)$ midway between them. Neither horizontal half-line meets the $y$-axis.
The value of $H(0)$ depends on the convention. The choice $H(0)=0$ makes $H$ continuous from the left at the jump, while the choice $H(0)=1$ makes it continuous from the right. The value $\frac{1}{2}$ adopted here keeps the identity $H(x)+H(-x)=1$ valid on the whole real line and makes $H$ the pointwise limit of the rescaled sigmoid $\sigma(kx)$ as $k \to +\infty.$
Properties
- Domain: $\mathbb{R}$
- Range: $\left\{ 0,\ \dfrac{1}{2},\ 1 \right\}$
- The function is bounded, since $0 \leq H(x) \leq 1$ for every $x \in \mathbb{R}.$
- The function is non-decreasing on $\mathbb{R}$ and is constant on each of the intervals $(-\infty, 0)$ and $(0, +\infty).$
- The function is neither even nor odd; it satisfies $H(x)+H(-x)=1$ for every $x \in \mathbb{R}.$
- The function has a jump discontinuity of amplitude $1$ at $x=0$ and is continuous at every other point.
- The derivative exists and equals zero at each point of the two open half-lines. At the origin the derivative does not exist.
- The function is idempotent away from the origin, since $H(x)^2 = H(x)$ whenever $x \neq 0.$
- The function is Riemann integrable on every bounded interval, since it is bounded and its only discontinuity is at $x=0.$
The two one-sided limits at the origin are:
$$ \begin{align} \lim_{x \to 0^-} H(x) &= 0 \\[6pt] \lim_{x \to 0^+} H(x) &= 1 \end{align} $$
Their values differ, so the two-sided limit $\lim_{x \to 0} H(x)$ does not exist and $H$ has a jump discontinuity at the origin. The jump has amplitude $1-0=1.$
Limits, derivatives, and integrals
Since $H$ is constant on each half-line, its limits at infinity are:
$$ \begin{align} \lim_{x \to -\infty} H(x) &= 0 \\[6pt] \lim_{x \to +\infty} H(x) &= 1 \end{align} $$
The lines $y=0$ and $y=1$ are horizontal asymptotes. The graph coincides with $y=0$ on $(-\infty,0)$ and with $y=1$ on $(0,+\infty).$
Since $H$ is constant on each open half-line, its derivative is:
$$ \frac{d}{dx} H(x) = 0 \quad \text{for } x \neq 0 $$
At $x=0$ the classical derivative does not exist, since a function that is not continuous at a point cannot be differentiable there. In the sense of distributions, the derivative of the Heaviside function is the Dirac delta:
$$ \frac{d}{dx} H(x) = \delta(x) $$
The Dirac delta is supported at the origin and has total mass $1,$ equal to the amplitude of the jump. Equivalently, $H$ is a distributional primitive of $\delta.$ This identity does not determine the value at the origin, because changing a function at one point does not change its associated distribution.
The sign function jumps from $-1$ to $1$ at the origin. Its jump has amplitude $2,$ and its distributional derivative is $2\delta(x).$
On each open half-line, $xH(x)+c$ is a classical antiderivative of $H.$ No classical antiderivative exists on any open interval containing the origin, since Darboux's theorem gives every derivative the intermediate value property while $H$ has a jump. The accumulated integral from the origin is:
$$ \int_0^x H(t) \ dt = xH(x) $$
The definite integral over an interval $[a,b]$ measures the part of that interval lying to the right of the origin. For $a
$$ \int_a^b H(x) \ dx = \begin{cases} 0 & \text{if } b \leq 0 \\[6pt] b & \text{if } a < 0 < b \\[6pt] b-a & \text{if } 0 \leq a \end{cases} $$
In the middle case, $H$ is $0$ on $[a,0)$ and $1$ on $(0,b].$ Its value at $0$ does not affect the integral, which is the length of $[0,b],$ namely $b.$
The ramp function
The accumulated integral above is the ramp function:
$$R(x) = xH(x)$$
The graph is the negative half of the $x$-axis joined to the bisector $y=x$ in the first quadrant. In circuit models, $R$ describes a voltage that grows at a constant rate after the switch closes, whereas $H$ has an instantaneous jump.
The ramp function is continuous at the origin, since both one-sided limits are zero and $R(0)=0.$ Differentiating on each half-line gives $R'(x)=0$ for $x<0$ and $R'(x)=1$ for $x>0.$ The two one-sided derivatives at the origin differ, so $R$ has a corner point there. Thus $R'(x)=H(x)$ for every $x\neq 0.$
The ramp function has an equivalent expression in terms of the absolute value:
$$R(x) = \frac{x+|x|}{2}$$
For $x>0$ the numerator equals $2x$ and the quotient equals $x.$ For $x<0$ we have $|x|=-x,$ so the numerator equals $x-x=0.$ At the origin both expressions vanish, and the two formulas agree on all of $\mathbb{R}.$
Relationship with the sign function and the absolute value
The Heaviside function and the sign function differ by an affine transformation of their values. Adding $1$ to $\mathrm{sgn}(x)$ and halving the result gives:
$$H(x) = \frac{1+\mathrm{sgn}(x)}{2}$$
Solving the relation for $\mathrm{sgn}(x)$ gives:
$$\mathrm{sgn}(x) = 2H(x)-1$$
At the origin, where $H(0)=\frac{1}{2},$ the two identities reduce to $\frac{1}{2}=\frac{1+0}{2}$ and $0=2\cdot\frac{1}{2}-1.$
Substituting the second identity into $|x| = x\mathrm{sgn}(x)$ gives an expression for the absolute value in terms of $H:$
$$|x| = x\left(2H(x)-1\right)$$
For $x>0$ the bracket equals $1$ and the product equals $x,$ while for $x<0$ the bracket equals $-1$ and the product equals $-x.$
Translating the step and writing piecewise functions compactly
Replacing $x$ by $x-a$ moves the jump from the origin to the point $a:$
$$ H(x-a) = \begin{cases} 0 & \text{if } x < a \\[6pt] \dfrac{1}{2} & \text{if } x = a \\[8pt] 1 & \text{if } x > a \end{cases} $$
The translated step is $0$ for $x
$$f(x) = f_1(x) + \left(f_2(x)-f_1(x)\right)H(x-a)$$
It equals $f_1(x)$ for $x
$$ f(x) = \begin{cases} 2 & \text{if } x < 3 \\[6pt] x^2 & \text{if } x > 3 \end{cases} $$
For $x\neq 3,$ it can be written as $f(x) = 2 + (x^2-2)H(x-3).$ At $x=3$ this expression has the value $\frac{11}{2},$ while the piecewise definition above does not specify a value.
Subtracting one translated step from another isolates a bounded interval. For $a
$$p(x) = H(x-a)-H(x-b)$$
The function $p$ equals $1$ on $(a,b)$ and $0$ for $x
Because $H(0)=\frac{1}{2},$ we have $p(a)=p(b)=\frac{1}{2}.$ The function $p$ is the rectangular pulse of unit height supported on $[a,b].$ Multiplying a formula by $p$ preserves it on $(a,b),$ makes it zero outside $[a,b],$ and halves its values at the endpoints. Away from the junctions, a function with formulas $f_1,$ $f_2,$ and $f_3$ on $(-\infty,a),$ $(a,b),$ and $(b,+\infty)$ can be written as:
$$f(x) = f_1(x) + \left(f_2(x)-f_1(x)\right)H(x-a) + \left(f_3(x)-f_2(x)\right)H(x-b)$$
At $a$ and $b,$ this formula assigns the averages of the adjacent values. Any other values at the junctions must be specified separately.
If $t$ denotes time, a switch that changes a voltage from $0$ to $120$ volts at $t=5$ is modelled by:
$$V(t) = 120H(t-5)$$
The voltage is $0$ for $t<5,$ is $120$ volts for $t>5,$ and equals $60$ volts at $t=5$ under the adopted convention.
A voltage that rises at a constant rate from $0$ to $120$ volts over the first $60$ seconds and then stays there is modelled by a linear combination of two ramps:
$$V(t) = 2tH(t)-2(t-60)H(t-60)$$
For $0
Laplace transform of the Heaviside function
For real $s,$ the Laplace transform of a function $f$ defined for $t \geq 0$ is the improper integral below, whenever it converges:
$$\mathcal{L}\{f(t)\}(s) = \int_0^{+\infty} f(t)e^{-st} \ dt$$
For the translated step with $a \geq 0,$ the factor $H(t-a)$ is $0$ for $t
$$\mathcal{L}\{H(t-a)\}(s) = \int_a^{+\infty} e^{-st} \ dt$$
Integrating up to a finite bound $M$ and letting $M \to +\infty$ gives, for $s>0:$
$$ \begin{align} \int_a^{M} e^{-st} \ dt &= \frac{e^{-as}-e^{-Ms}}{s} \\[6pt] \lim_{M \to +\infty} \frac{e^{-as}-e^{-Ms}}{s} &= \frac{e^{-as}}{s} \end{align} $$
The transform of the translated step is therefore:
$$\mathcal{L}\{H(t-a)\}(s) = \frac{e^{-as}}{s} \quad \text{for } s>0$$
Setting $a=0$ recovers the transform of the step at the origin, $\mathcal{L}\{H(t)\}(s)=\frac{1}{s}.$ The restriction $s>0$ is needed for the improper integral to converge, since $e^{-Ms}$ tends to zero only when $s$ is positive.
For $a \geq 0,$ the same computation applied to a product yields the shifting rule. If $F(s)=\mathcal{L}\{f(t)\}(s),$ the substitution $u=t-a$ in the defining integral gives:
$$\mathcal{L}\{H(t-a)f(t-a)\}(s) = e^{-as}F(s)$$
Thus $H(t-a)f(t-a)$ is the input $f$ delayed by $a,$ and its Laplace transform is $e^{-as}F(s).$