Step Functions, Impulse Functions, and the Delta Function
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Step Functions, Impulse Functions, and the Delta Function Department of Mathematics and Statistics September 7, 2012 Methods of Applied Calculus (JMU) Math 337 September 7, 2012 1/9
Transcript of Step Functions, Impulse Functions, and the Delta Function
Step Functions, Impulse Functions,and the Delta Function
Department of Mathematics and Statistics
September 7, 2012
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 1 / 9
The Unit Step Function
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 2 / 9
The Unit Step Function
Definition
The unit step function or Heaviside function is a function of the form
ua(t) =
{
0, t < a,1, t ≥ a,
where a > 0.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 2 / 9
The Unit Step Function
Definition
The unit step function or Heaviside function is a function of the form
ua(t) =
{
0, t < a,1, t ≥ a,
where a > 0.
y
ta
1
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The Laplace Transform of the Unit Step Function
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The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
∫
∞
a
e−st dt
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
∫
∞
a
e−st dt
= limb→∞
−1
se−st
∣
∣
∣
∣
b
a
=
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 3 / 9
The Laplace Transform of the Unit Step Function
The Laplace transform of ua is
L(ua(t)) =
∫
a
0e−st · 0 dt +
∫
∞
a
e−st dt =
∫
∞
a
e−st dt
= limb→∞
−1
se−st
∣
∣
∣
∣
b
a
=1
se−as .
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The Difference Between Two Unit Step Functions
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The Difference Between Two Unit Step Functions
u1(t) − u2(t)
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 4 / 9
The Difference Between Two Unit Step Functions
u1(t) − u2(t)
y
t
1
1 2 3
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 4 / 9
The Difference Between Two Unit Step Functions
u1(t) − u2(t)
y
t
1
1 2 3
L(u1(t) − u2(t)) = 1se−s − 1
se−2s .
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The Unit Step Function Combined with Other Functions
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The Unit Step Function Combined with Other Functions
Consider
g(t) =
{
0, t < a,f (t − a), t ≥ a,
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
The Unit Step Function Combined with Other Functions
Consider
g(t) =
{
0, t < a,f (t − a), t ≥ a,
We may write g(t) = ua(t)f (t − a).
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
The Unit Step Function Combined with Other Functions
Consider
g(t) =
{
0, t < a,f (t − a), t ≥ a,
We may write g(t) = ua(t)f (t − a).
t
y
1 2 3 4 5 6 7
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 5 / 9
L(g(t))
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L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
∫
∞
0e−s(τ+a)f (τ) dτ
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
∫
∞
0e−s(τ+a)f (τ) dτ = e−as
∫
∞
0e−sτ f (τ) dτ
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
L(g(t))
L(g(t)) = L(ua(t)f (t − a)) =
∫
∞
0e−stua(t)f (t − a) dt
=
∫
∞
a
e−st f (t − a) dt.
Substituting τ = t − a:
∫
∞
0e−s(τ+a)f (τ) dτ = e−as
∫
∞
0e−sτ f (τ) dτ = e−asL(f (t)).
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 6 / 9
The kth Unit Impulse Function
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The kth Unit Impulse Function
Definition
The kth unit impulse function is
δa,k(t) =
{
k, a < t < a + 1k,
0, 0 ≤ t ≤ a or t ≥ a + 1k,
where a > 0.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 7 / 9
The kth Unit Impulse Function
Definition
The kth unit impulse function is
δa,k(t) =
{
k, a < t < a + 1k,
0, 0 ≤ t ≤ a or t ≥ a + 1k,
where a > 0.
y
t
10
1 2 3
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The Laplace Transform of the Unit Impulse Function
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The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt =
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
= e−sa ·1 − e−s/k
s/k.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
= e−sa ·1 − e−s/k
s/k.
By L’Hopital’s rule: limk→∞
L(δa,k(t)) =
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Laplace Transform of the Unit Impulse Function
Note that
∫
∞
0δa,k(t) dt = 1, and that
L(δa,k(t)) =
∫
a+1k
a
k e−st dt = −k ·e−s
“
a+1k
”
− e−sa
s
= e−sa ·1 − e−s/k
s/k.
By L’Hopital’s rule: limk→∞
L(δa,k(t)) = e−sa.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 8 / 9
The Dirac Delta Function
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The Dirac Delta Function
Note that limk→∞
L(δ0,k(t)) = 1.
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The Dirac Delta Function
Note that limk→∞
L(δ0,k(t)) = 1.
We let limk→∞
L(δ0,k(t)) = L(δ(t)), where δ(t) is called the Dirac delta
function or delta function.
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 9 / 9
The Dirac Delta Function
Note that limk→∞
L(δ0,k(t)) = 1.
We let limk→∞
L(δ0,k(t)) = L(δ(t)), where δ(t) is called the Dirac delta
function or delta function.
So L(δ(t)) = 1 and L−1(1) = δ(t).
Methods of Applied Calculus (JMU) Math 337 September 7, 2012 9 / 9
