Solutions 4.4-Page 299
Problem 3
Find the convolution f (t ) ∗ g (t ) .
f (t ) = g (t ) = sin t
t
Eq.3 states that ( f ∗ g )(t ) = ∫ f (τ ) g (t − τ )dτ . Direct substitution into Eq.3 yields:
0
τ
( f ∗ g )(t ) = ∫ sin τ (sin(t − τ ))dτ . To evaluate the integral, the trigonometric identity
0
1
[cos( A − B) − cos( A + B)] can be used. Substitution of this identity into
2
the integral yields:
sin A sin B =
τ
( f ∗ g )(t ) = ∫ sin τ (sin(t − τ ))dτ =
0
1 t
1 t
[
cos(τ − t + τ ) − cos(τ + t − τ )]dτ = ∫ [cos(2τ − t ) − cos(t )]dτ
∫
2 0
2 0
Evaluating the integral yields:
1 t
[cos(2τ − t ) − cos(t )]dτ = 1 12 sin(2τ −t ) −τ
∫
0
2
2
[
Simplifying yields:
( f ∗ g )(t ) =
sin t − t cos t
2
cos t ]
t
0
=
1 1
[( sin t − t cos t ) − ( 12 sin(−t ) )]
2 2
Problem 10
Apply the convolution theorem to find the inverse Laplace transforms of the functions.
F (s) =
1
s (s + k 2 )
2
2
Using Figure 4.1.2 on pg.268 to find the inverse Laplace functions,


1

1
1
sin kt 1 τ
= L -1  2 ⋅ 2
L -1  F ( s ) = 2 2
=t∗
= ∫ τ (sin k (t − τ ))dτ . This
2 
2 
k 0
k
s (s + k ) 

 s (s + k ) 
integral can be evaluated by parts, by using a computer program like Mathematica, or a
calculator. Using the TI-89 calculator, the integral evaluates to
L -1 {F ( s )} =
kt − sin kt
k3
Problem 11
Apply the convolution theorem to find the inverse Laplace transforms of the functions.
s2
F (s) = 2
( s + 4) 2
Using Figure 4.1.2 on pg.268 to find the inverse Laplace functions,


 s
s2
s 
L -1  F ( s ) = 2
= L -1  2
⋅ 2
 = cos 2t ∗ cos 2t =
2 
( s + 4) 
 ( s + 4) ( s + 4) 

τ
∫
0
cos 2τ (cos 2(t − τ ))dτ .
To evaluate the integral, the trigonometric identity
1
coA cos B = [cos( A + B) + cos( A − B)] can be used. Substitution of this identity into the
2
integral yields:
τ
( f ∗ g )(t ) = ∫ cos 2τ (cos 2(t − τ ))dτ =
0
1 t
1 t
[
cos(2τ + 2t − 2τ ) + cos(2τ − 2t + 2τ )] dτ = ∫ [cos(2t ) + cos(4τ − 2t )] dτ
∫
2 0
2 0
Evaluating the integral yields:
1

cos 2t + sin(4τ − 2t ) _ 
4


1 t
1 
[
cos(2t ) + cos(4τ − 2t )]dτ =
τ
∫
2 0
2
Simplifying yields
L -1 {F ( s )} =
sin 2t + 2t cos 2t
4
t
0