2E2 Tutorial Sheet 5 First Term, Solutions1
Z
=
Use integration by parts with
8 November 2002
dV = e
=
=
=
=
=
=
dτ
1 −2τ
V =− e
2
1
.
s2 (s − 4)
=
=
=
3. (4) Use the formula for the Laplace transform of a periodic function with period c:
Z c
1
L(f ) =
f (t)e−st dt
(2
1 − e−cs 0
to find the Laplace transform of a half-rectified wave
sin t sin t > 0
f (t) =
0
sin t ≤ 0
0
1
(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
2
4
6
8
10
12
14
16
18
t
so that f (t) is the convolution t ∗ e4t .
Z
(3
1
1
. From the convolution theorem, we
Solution: We know L(t)) = s12 and L(e4t ) = s−4
see
1
L(f ) = 2
= L(t)L(e4t ) = L(t ∗ e4t )
s (s − 4)
f (t) =
Z t
U dV = e4t [U V ]t0 −
V dU
0
0h
it Z t 1
τ
e4t − e−4τ −
− e−4τ dτ
4
4
0
0
Z
t −4t
1 t −4τ
4t
− e +0+
e
e
dτ
4
4 0
t
1 −4τ
t e4t
− e
− +
4
4
2
0
t e4t
1
1
− +
− e−4t +
4
4
4
4
1
1 4t
t
+ e
− −
4 16 16
−2τ
2. (2) Use the convolution theorem to find the function f (t) with
L(f ) =
τ e−4τ dτ
0
dV = e−4τ dτ
1
dU = dτ, V = − e−4τ
4
Z
=
dU = dτ,
Z t
Z t
e2t
U dV = e2t [U V ]t0 −
V dU
0
0h
it Z t 1
τ
e2t − e−2τ −
− e−2τ dτ
2
2
0
0
Z
1 t −2τ
t −2t
2t
e
dτ
− e +0+
e
2
2 0
t
t e2t
1 −2τ
− +
− e
2
2
2
0
t e2t
1
1
− +
− e−2t +
2
2
2
2
t 1 1 2t
− − + e
2 4 4
=
t
t
0
U = τ,
0
Z
= e4t
Solution: From the definition of convolutions
Z t
Z t
f (τ )g(t − τ ) dτ =
τ e2(t−τ ) dτ
(f ∗ g)(t) =
0
0
Z t
Z t
=
τ e2t e−2τ dτ = e2t
τ e−2τ dτ
Use integration by parts with
τ e4t e−4τ dτ = e4t
U = τ,
1. (2) Find the convolution (f ∗ g)(t) when f (t) = t, g(t) = e2t (t ≥ 0).
0
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 sin
into exponentials
Z π
Z π
Z π
1
sin te−st dt =
e(i−s)t dt −
e−(i+s)t dt
2i
0
0
0
1
1
1
(i−s)π
e
e−(i+s)π − 1
−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 −i + s
s−i
1
=
= 2
i+s
i + s −i + s
s +1
to get
Z
π
sin te−st dt =
0
or
L(f ) =
s2
1 + e−sπ
1 + s2
(7)
(8)
1
1 1 + e−sπ
1
= 2
+ 1 1 − e−2sπ
s + 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 π
Z
1 π
I=
sin te−st dt = −
cos te−st dt
s 0
0
1
1
1
= − − e−πs + 1 + I
s
s
s
and solve for I to get the answer given at (8) above.
3
(10)
(11)