From Simple English Wikipedia, the free encyclopedia
The Heaviside step function, using the half-maximum convention
The Heaviside function , H is a no-continuous function whose value is zero for negative argument and one for positive argument.
The function is used in the mathematics of control theory to represent a signal that switches on at a specified time and stays switched on indefinitely. It was named after the Englishman Oliver Heaviside .
The Heaviside function is the integral of the Dirac delta function : H ′ = δ . This is sometimes written as
H
(
x
)
=
∫
−
∞
x
δ
(
t
)
d
t
{\displaystyle H(x)=\int _{-\infty }^{x}{\delta (t)}\mathrm {d} t}
We can also define an alternative form of the Heaviside step function as a function of a discrete variable n :
H
[
n
]
=
{
0
,
n
<
0
1
,
n
≥
0
{\displaystyle H[n]={\begin{cases}0,&n<0\\1,&n\geq 0\end{cases}}}
where n is an integer .
Or
H
(
x
)
=
lim
z
→
x
−
(
(
|
z
|
/
z
+
1
)
/
2
)
{\displaystyle H(x)=\lim _{z\rightarrow x^{-}}((|z|/z+1)/2)}
The discrete-time unit impulse is the first difference of the discrete-time step
δ
[
n
]
=
H
[
n
]
−
H
[
n
−
1
]
.
{\displaystyle \delta \left[n\right]=H[n]-H[n-1].}
This function is the cumulative summation of the Kronecker delta :
H
[
n
]
=
∑
k
=
−
∞
n
δ
[
k
]
{\displaystyle H[n]=\sum _{k=-\infty }^{n}\delta [k]\,}
where
δ
[
k
]
=
δ
k
,
0
{\displaystyle \delta [k]=\delta _{k,0}\,}
is the discrete unit impulse function .
Representations
Often an integral representation of the Heaviside step function is useful:
H
(
x
)
=
lim
ϵ
→
0
+
−
1
2
π
i
∫
−
∞
∞
1
τ
+
i
ϵ
e
−
i
x
τ
d
τ
=
lim
ϵ
→
0
+
1
2
π
i
∫
−
∞
∞
1
τ
−
i
ϵ
e
i
x
τ
d
τ
.
{\displaystyle H(x)=\lim _{\epsilon \to 0^{+}}-{1 \over 2\pi \mathrm {i} }\int _{-\infty }^{\infty }{1 \over \tau +\mathrm {i} \epsilon }\mathrm {e} ^{-\mathrm {i} x\tau }\mathrm {d} \tau =\lim _{\epsilon \to 0^{+}}{1 \over 2\pi \mathrm {i} }\int _{-\infty }^{\infty }{1 \over \tau -\mathrm {i} \epsilon }\mathrm {e} ^{\mathrm {i} x\tau }\mathrm {d} \tau .}
H (0)
The value of the function at 0 can be defined as H (0) = 0, H (0) = ½ or H (0) = 1.
H
(
x
)
=
1
+
sgn
(
x
)
2
=
{
0
,
x
<
0
1
2
,
x
=
0
1
,
x
>
0.
{\displaystyle H(x)={\frac {1+\operatorname {sgn}(x)}{2}}={\begin{cases}0,&x<0\\{\frac {1}{2}},&x=0\\1,&x>0.\end{cases}}}
See also