2E2 Tutorial Sheet 5 First Term, Solutions1
8 November 2002
1. (2) Find the convolution (f ∗ g)(t) when f (t) = t, g(t) = e2t (t ≥ 0).
Solution: From the definition of convolutions
Z t
Z t
(f ∗ g)(t) =
f (τ )g(t − τ ) dτ =
τ e2(t−τ ) dτ
Z0 t
Z t0
τ e2t e−2τ dτ = e2t
=
τ e−2τ dτ
0
Use integration by parts with
0
U = τ,
dV = e−2τ dτ
1
dU = dτ, V = − e−2τ
2
Z
t
2t
2t
]t0
Z
t
[U V −
V dU
0
h
Z t
τ −2τ it
1 −2τ
2t
e
−
− e
− e
dτ
2
2
0
0
Z
1 t −2τ
t −2t
2t
e
dτ
e
− e +0+
2
2 0
t
t e2t
1 −2τ
− +
− e
2
2
2
0
2t
1 −2t 1
t e
− e +
− +
2
2
2
2
t 1 1
− − + e2t
2 4 4
= e
U dV = e
0
=
=
=
=
=
2. (2) Use the convolution theorem to find the function f (t) with
L(f ) =
1
.
− 4)
s2 (s
(1)
1
Solution: We know L(t)) = s12 and L(e4t ) = s−4
. From the convolution theorem, we
see
1
L(f ) = 2
= L(t)L(e4t ) = L(t ∗ e4t )
s (s − 4)
so that f (t) is the convolution t ∗ e4t .
Z
f (t) =
1
t
τ e4(t−τ ) dτ
0
Conor Houghton, houghton@maths.tcd.ie, see also http://www.maths.tcd.ie/~houghton/2E2.html
1
Z
=
Use integration by parts with
t
4t −4τ
τe e
dτ = e
0
4t
Z
t
τ e−4τ dτ
0
U = τ,
dV = e−4τ dτ
1
dU = dτ, V = − e−4τ
4
Z
t
4t
]t0
4t
Z
t
V dU
[U V −
0
h
Z t
τ −4τ it
1 −4τ
4t
e
− e
dτ
−
− e
4
4
0
0
Z
1 t −4τ
t −4t
4t
e
dτ
e
− e +0+
4
4 0
t
t e4t
1 −4τ
− +
− e
4
4
2
0
4t
t e
1 −4t 1
− +
− e +
4
4
4
4
1
1 4t
t
+ e
− −
4 16 16
= e
U dV = e
0
=
=
=
=
=
3. (4) Use the formula for the Laplace transform of a periodic function with period c:
Z c
1
f (t)e−st dt
(2)
L(f ) =
−cs
1−e
0
to find the Laplace transform of a half-rectified wave
sin t sin t > 0
f (t) =
0
sin t ≤ 0
(3)
1
0
2
4
6
8
10
12
14
16
18
t
This is the form a AC current has after going through a diode.
So we substitute this into the formula
Z 2π
Z π
1
1
−st
L(f ) =
f (t)e dt =
sin te−st dt
1 − e−2πs 0
1 − e−2πs 0
(4)
We need to do the integral. There are two obvious ways, the first is to split the sine
into exponentials
Z π
Z π
Z π
1
(i−s)t
−(i+s)t
−st
e
dt −
e
dt
sin te dt =
2i
0
0
0
1
1
1
(i−s)π
−(i+s)π
=
e
−1 +
e
−1
(5)
2i i − s
i+s
2
Now, we use
eiπ = e−iπ = −1
(6)
and
1
1 −i − s
s+i
=
=− 2
i−s
i − s −i − s
s +1
1
1 −i + s
s−i
=
= 2
i+s
i + s −i + s
s +1
to get
Z
or
L(f ) =
π
sin te−st dt =
0
1 + e−sπ
1 + s2
(7)
(8)
1 1 + e−sπ
1
1
=
s2 + 1 1 − e−2sπ
s2 + 1 1 − e−sπ
(9)
where the final equality uses
1 − e−2sπ = 1 − e−sπ
1 + e−sπ
The other way to do the integral is to integrate by parts. Briefly, write
Z
Z π
1 π
−st
cos te−st dt
I=
sin te dt = −
s
0
0
1
1 −πs
1
+1 + I
= − − e
s
s
s
and solve for I to get the answer given at (8) above.
3
(10)
(11)