109
S-FUNCTIONS
WITH
COMPLICATED ARGUMENTS
t-
<
0
<
t+
otherwise
.
lt
dt
s(t)f(t)
=
{
i")
(5.23)
Remember, by definition, anything that behaves like a 6-function inside an integral
is a &function,
so
S(-t)
=
i3(t).
This implies the &function is an even function.
5.3.2
S(at)
Now consider another simple variation
6(at),
where
a
is any positive constant. Again,
start with a sifting integral in the form
(5.24)
With a variable change
oft'
=
at,
this
becomes
so
that
t-
<
0
<
t+
otherwise
l-'+
dt
f(t>&(at)
=
{
L(0)"
(5.26)
Notice again, this is just the definition of the function
6(t)
multiplied by the constant
l/a,
so
Wx)
=
S(x)/a
a
>
0.
(5.27)
This derivation was made assuming a positive
a.
The same manipulations can be
performed with a negative
a,
and combined with the previous result, to obtain the
more general expression
6(ax)
=
6(n)/lal.
(5.28)
Notice that
our
first example,
6(t)
=
8(-t),
can
be derived from Equation
5.28
by
setting
a
=
-
1.
5.3.3
S(tZ
-
2)
As
a bit more complicated argument for the Dirac &function, consider
6(t2
-
aL).
The argument
of
this function goes to zero when
t
=
fa
and
t
=
-a,
which seems
to imply two &functions.
To
test this theory, place the function in the sifting integral
lr+
dt
f(t)6(t2
-
a2).
(5.29)
