Lecture 17: Unit Step Function and its Laplace
Transform (§6.3)
April 17, 2012 (Tue)
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Unit Step Function (Heaviside Function)
I
The unit step function (or Heaviside function) is defined by
0 if t < c
uc = u(t − c) =
1 if t ≥ c
where c > 0.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Unit Step Function (Heaviside Function)
I
I
The unit step function (or Heaviside function) is defined by
0 if t < c
uc = u(t − c) =
1 if t ≥ c
where c > 0.
Examples.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Unit Step Function (Heaviside Function)
I
I
The unit step function (or Heaviside function) is defined by
0 if t < c
uc = u(t − c) =
1 if t ≥ c
where c > 0.
Examples.
1. Sketch the graph of y = h(t), where
h(t) = uπ (t) − u2π (t),
t ≥ 0.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Unit Step Function (Heaviside Function)
I
I
The unit step function (or Heaviside function) is defined by
0 if t < c
uc = u(t − c) =
1 if t ≥ c
where c > 0.
Examples.
1. Sketch the graph of y = h(t), where
h(t) = uπ (t) − u2π (t),
t ≥ 0.
2. Express the function
f (t) =
2 if 1 ≤ t < 4
0 otherwise
using the unit step function.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Unit Step Function (Heaviside Function)
I
I
The unit step function (or Heaviside function) is defined by
0 if t < c
uc = u(t − c) =
1 if t ≥ c
where c > 0.
Examples.
1. Sketch the graph of y = h(t), where
h(t) = uπ (t) − u2π (t),
t ≥ 0.
2. Express the function
f (t) =
2 if 1 ≤ t < 4
0 otherwise
using the unit step function.
3. Describe the function given in the figure using the unit step
function.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Practice
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Practice: Express f (t) in terms

1

cos(2t)
f (t) =

2
of the unit function uc (t)
if 0 ≤ t < π
if π ≤ t < 3π
otherwise
Lecture 17: Unit Step Function and its Laplace Transform (§6.
The Laplace Transform of the Unit Step Function
I
The Laplace transform of the unit step dunction is, for c > 0,
L{uc (t)} =
e −cs
for s > 0
s
Lecture 17: Unit Step Function and its Laplace Transform (§6.
The Laplace Transform of the Unit Step Function
I
The Laplace transform of the unit step dunction is, for c > 0,
L{uc (t)} =
I
e −cs
for s > 0
s
Example: Find L{f (t)}, where
2 if 1 ≤ t < 4
f (t) =
0 otherwise
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Translation of a Function
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Translation of a function.
u(t − c)f (t − c) =
0
if t < c
f (t − c) if t ≥ c
Thus, if f (t) represents, say a signal, then u(t − c)f (t − c),
with c > 0, represents the same signal, but delayed by c units
of time.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Translation of a Function
I
Translation of a function.
u(t − c)f (t − c) =
0
if t < c
f (t − c) if t ≥ c
Thus, if f (t) represents, say a signal, then u(t − c)f (t − c),
with c > 0, represents the same signal, but delayed by c units
of time.
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Example: Let f (t) = 5 sin t and g (t) = u2 (t)f (t − 2).
Sketch the graphs of two functions on the same window.
Lecture 17: Unit Step Function and its Laplace Transform (§6.
The Laplace Transform of a translate of a function
I
If c > 0, then
L{u(t − c)f (t − c)} = e −cs F (s),
where F (s) = L{f (t)}. Equivalently,
L−1 {e −cs F (s)} = u(t − c)f (t − c).
Lecture 17: Unit Step Function and its Laplace Transform (§6.
The Laplace Transform of a translate of a function
I
If c > 0, then
L{u(t − c)f (t − c)} = e −cs F (s),
where F (s) = L{f (t)}. Equivalently,
L−1 {e −cs F (s)} = u(t − c)f (t − c).
I
Example: Find L{f (t)}, where
sin t
if 0 ≤ t < π/4
f (t) =
sin t + cos(t − π/4) t ≥ π/4
Lecture 17: Unit Step Function and its Laplace Transform (§6.
The Laplace Transform of a translate of a function
I
If c > 0, then
L{u(t − c)f (t − c)} = e −cs F (s),
where F (s) = L{f (t)}. Equivalently,
L−1 {e −cs F (s)} = u(t − c)f (t − c).
I
Example: Find L{f (t)}, where
sin t
if 0 ≤ t < π/4
f (t) =
sin t + cos(t − π/4) t ≥ π/4
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Example: Find the Laplace transform
L{tu(t − π)}
Lecture 17: Unit Step Function and its Laplace Transform (§6.
The Laplace Transform of a translate of a function
I
If c > 0, then
L{u(t − c)f (t − c)} = e −cs F (s),
where F (s) = L{f (t)}. Equivalently,
L−1 {e −cs F (s)} = u(t − c)f (t − c).
I
Example: Find L{f (t)}, where
sin t
if 0 ≤ t < π/4
f (t) =
sin t + cos(t − π/4) t ≥ π/4
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Example: Find the Laplace transform
L{tu(t − π)}
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Example: Find the inverse transform of
F (s) =
1 − e −2s
.
s2
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Translation in the Laplace Transform
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If F (s) = L{f (t)} exists for s > a ≥ 0, and if c is a constant,
then
L{e ct f (t)} = F (s − c), s > a + c.
Conversely, if f (t) = L−1 {F (s)}, then
L−1 {F (s − c)} = e ct f (t).
Lecture 17: Unit Step Function and its Laplace Transform (§6.
Translation in the Laplace Transform
I
If F (s) = L{f (t)} exists for s > a ≥ 0, and if c is a constant,
then
L{e ct f (t)} = F (s − c), s > a + c.
Conversely, if f (t) = L−1 {F (s)}, then
L−1 {F (s − c)} = e ct f (t).
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Example: Find the inverse transform of
G (s) =
1
.
s 2 − 4s + 5
Lecture 17: Unit Step Function and its Laplace Transform (§6.