The Laplace Transform of step functions (Sect. 6.3).
I
Overview and notation.
I
The definition of a step function.
I
Piecewise discontinuous functions.
I
The Laplace Transform of discontinuous functions.
I
Properties of the Laplace Transform.
Overview and notation. Overview: The Laplace Transform method can be used to solve constant coefficients differential equations with discontinuous source functions.
Notation: If L[f (t)] = F (s), then we denote L−1 [F (s)] = f (t).
Remark: One can show that for a particular type of functions f , that includes all functions we work with in this Section, the notation above is well-defined.
Example 1 From the Laplace Transform table we know that L e at = . s − a h 1 i −1 Then also holds that L C = e at . s −a
The Laplace Transform of step functions (Sect. 6.3).
I
Overview and notation.
I
The definition of a step function.
I
Piecewise discontinuous functions.
I
The Laplace Transform of discontinuous functions.
I
Properties of the Laplace Transform.
The definition of a step function. Definition A function u is called a step function at t = 0 iff holds ( 0 for t < 0, u(t) = 1 for t > 0.
Example Graph the step function values u(t) above, and the translations u(t − c) and u(t + c) with c > 0. Solution: u(t)
u(t − c)
1
1
0
t
u(t + c)
0
1
c
t
−c
0
t
C
The definition of a step function. Remark: Given any function values f (t) and c > 0, then f (t − c) is a right translation of f and f (t + c) is a left translation of f .
Example at
f(t)
a(t−c)
f(t)= e
f(t)
1
1
0
f(t)
0
t
at
f(t)
f(t)= u(t)e
1
0
f(t)= e
c
t
a(t−c)
f(t)= u(t−c) e
1
t
0
c
t
The Laplace Transform of step functions (Sect. 6.3).
I
Overview and notation.
I
The definition of a step function.
I
Piecewise discontinuous functions.
I
The Laplace Transform of discontinuous functions.
I
Properties of the Laplace Transform.
Piecewise discontinuous functions. Example Graph of the function b(t) = u(t − a) − u(t − b), with 0 < a < b. Solution: The bump function b can be graphed as follows:
u(t −a)
u(t −b)
1
1
0
a
b
0
t
a
b
t
b(t)
1
0
a
b
t
C
Piecewise discontinuous functions. Example Graph of the function f (t) = e at u(t − 1) − u(t − 2) . Solution: y
f(t)=e
at
[ u ( t −1 ) − u ( t −2 ) ] e
1
at
[ u ( t −1 ) − u ( t −2 ) ]
1
2
t
Notation: The function values u(t − c) are denoted in the textbook as uc (t).
The Laplace Transform of step functions (Sect. 6.3).
I
Overview and notation.
I
The definition of a step function.
I
Piecewise discontinuous functions.
I
The Laplace Transform of discontinuous functions.
I
Properties of the Laplace Transform.
The Laplace Transform of discontinuous functions. Theorem Given any real number c, the following equation holds, L[u(t − c)] =
e −cs , s
s > 0.
Proof: Z
∞
L[u(t − c)] =
e
−st
Z u(t − c) dt =
N→∞
e −st dt,
c
0
L[u(t − c)] = lim −
∞
e −cs 1 −Ns e − e −cs = , s s
e −cs We conclude that L[u(t − c)] = . s
s > 0.
The Laplace Transform of discontinuous functions. Example Compute L[3u(t − 2)]. e −2s L[3u(t − 2)] = 3 L[u(t − 2)] = 3 . s
Solution:
3e −2s We conclude: L[3u(t − 2)] = . s
C
Example −1
Compute L
Solution:
h e 3s i s
−1
L
.
h e 3s i s −1
We conclude: L
−1
=L
h e 3s i s
h e −(−3)s i s
= u(t − (−3)).
= u(t + 3).
The Laplace Transform of step functions (Sect. 6.3).
I
Overview and notation.
I
The definition of a step function.
I
Piecewise discontinuous functions.
I
The Laplace Transform of discontinuous functions.
I
Properties of the Laplace Transform.
C
Properties of the Laplace Transform. Theorem (Translations) If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds L[u(t − c)f (t − c)] = e −cs F (s),
s > a.
Furthermore, L[e ct f (t)] = F (s − c),
s > a + c.
Remark: I I
L translation (uf ) = (exp) L[f ] . L (exp) (f ) = translation L[f ] .
Equivalent notation: I
L[u(t − c)f (t − c)] = e −cs L[f (t)],
I
L[e ct f (t)] = L[f ](s − c).
Properties of the Laplace Transform. Example Compute L u(t − 2) sin(a(t − 2)) . a , L[u(t − c)f (t − c)] = e −cs L[f (t)]. 2 2 s +a a . L u(t − 2) sin(a(t − 2)) = e −2s L[sin(at)] = e −2s 2 s + a2 a We conclude: L u(t − 2) sin(a(t − 2)) = e −2s 2 . C s + a2
Solution: L[sin(at)] =
Example Compute L e 3t sin(at) . Solution: Recall: L[e ct f (t)] = L[f ](s − c). a We conclude: L e 3t sin(at) = , with s > 3. (s − 3)2 + a2
C
Properties of the Laplace Transform. Example ( Find the Laplace transform of f (t) =
0,
t < 1,
(t 2 − 2t + 2), t > 1.
Solution: Using step function notation, f (t) = u(t − 1)(t 2 − 2t + 2). Completing the square we obtain, t 2 − 2t + 2 = (t 2 − 2t + 1) − 1 + 2 = (t − 1)2 + 1. f(t)
This is a parabola t 2 translated to the right by 1 and up by one. This is a discontinuous function.
1
0
1
t
Properties of the Laplace Transform. Example ( Find the Laplace transform of f (t) =
0,
t < 1,
(t 2 − 2t + 2), t > 1. Solution: Recall: f (t) = u(t − 1) (t − 1)2 + 1 . This is equivalent to f (t) = u(t − 1) (t − 1)2 + u(t − 1). Since L[t 2 ] = 2/s 3 , and L[u(t − c)g (t − c)] = e −cs L[g (t)], then L[f (t)] = L[u(t − 1) (t − 1)2 ] + L[u(t − 1)] = e −s e −s We conclude: L[f (t)] = 3 2 + s 2 . s
2 −s 1 + e . s3 s C
Properties of the Laplace Transform. Remark: The inverse of the formulas in the Theorem above are: L−1 e −cs F (s) = u(t − c) f (t − c), L−1 F (s − c) = e ct f (t).
Example h e −4s i Find L . s2 + 9 h −4s i 1 h 3 i −1 e −1 −4s Solution: L = L e . s2 + 9 3 s2 + 9 h a i −1 = sin(at). Then, we conclude that Recall: L s 2 + a2 h −4s i 1 −1 e L = u(t − 4) sin 3(t − 4) . s2 + 9 3 −1
C
Properties of the Laplace Transform. Example (s − 2) i Find L . (s − 2)2 + 9 h s i −1 −1 = cos(at), Solution: L L F (s − c) = e ct f (t). 2 2 s +a h (s − 2) i −1 = e 2t cos(3t). We conclude: L 2 (s − 2) + 9 −1
h
Example −1
Find L
h 2e −3s i s2 − 4
.
a i Solution: Recall: L = sinh(at) s 2 − a2 and L−1 e −cs F (s) = u(t − c) f (t − c). −1
h
C
Properties of the Laplace Transform. Example −1
Find L
h 2e −3s i s2 − 4
.
Solution: Recall: h a i −1 = sinh(at), L s 2 − a2 −1
L
−1
We conclude: L
L−1 e −cs F (s) = u(t − c) f (t − c).
h 2e −3s i s2 − 4
h 2e −3s i s2 − 4
−1
=L
h
e
−3s
2 i . s2 − 4
= u(t − 3) sinh 2(t − 3) .
Properties of the Laplace Transform. Example −1
Find L
h
e −2s i . s2 + s − 2
Solution: Find the roots of the denominator: ( √ s+ = 1, 1 s± = −1 ± 1 + 8 ⇒ 2 s− = −2. Therefore, s 2 + s − 2 = (s − 1) (s + 2). Use partial fractions to simplify the rational function: b 1 a 1 = + , = (s − 1) (s + 2) s2 + s − 2 (s − 1) (s + 2) 1 (a + b) s + (2a − b) = a(s + 2) + b(s − 1) = . s2 + s − 2 (s − 1) (s + 2)
C
Properties of the Laplace Transform. Example e −2s i . Find L s2 + s − 2 (a + b) s + (2a − b) 1 = Solution: Recall: 2 s +s −2 (s − 1) (s + 2) 1 1 a + b = 0, 2a − b = 1, ⇒ a = , b = − . 3 3 h e −2s i 1 h 1 i 1 −1 h −2s 1 i −1 −1 −2s = L − L . L e e s2 + s − 2 3 s −1 3 s +2 h 1 i −1 Recall: L = e at , L−1 e −cs F (s) = u(t − c) f (t − c), s −a h e −2s i 1 1 −1 (t−2) L = u(t − 2) e − u(t − 2) e −2(t−2) . 2 s +s −2 3 3 h e −2s i 1 h i −1 (t−2) −2(t−2) = u(t − 2) e −e Hence: L . C s2 + s − 2 3 −1
h
