Digital Signal Processing Reference
In-Depth Information
h
(
t
)
t
0
Figure 12-3
Impulse function.
Examination of equation (12-7) reveals the convolution operation to be an integral
equation that may be difficult to solve. A simpler alternative is to perform the
operation in the frequency domain, where it becomes a simple multiplication:
Y(f)
=
H(f)X(f)
(12-9)
where
Y(f),H(f)
, and
X(f )
are the frequency-domain representations of
y(t), h(t)
, and
x(t)
.
We obtain the frequency-domain forms by applying the Fourier transform:
∞
h(t)e
2
πjf t
dt
H(f)
=
(12-10)
t
=−∞
where
f
is the frequency in hertz. Notice that the units of
H(f)
will be the units
of
h(t)
multiplied by seconds. For example, if
h(t)
is dimensionless,
H(f)
has
units of seconds. Restoring results back to the time domain requires application
of the inverse Fourier transform:
∞
H(f)e
−
2
πjf t
df
h(t)
=
(12-11)
f
=−∞
The time-frequency equivalence and the relationships between input and output
are summarized in Figure 12-4. Note that the frequency-domain representa-
tion of the impulse response
H(f)
is known as the
transfer function
. In gen-
eral, the frequency-domain representations will be complex quantities, whereas
time-domain quantities are real.
Example 12-1
Transfer Function for a PCB Differential Transmission-Line Pair
Figure 12-5 shows the transfer function for a 0.381-m-long differential pair on a
printed circuit that has the following characteristics at a reference frequency,
f
0
,
of 1 GHz:
3
.
299
3
.
299
nH
/
cm
1
.
098
1
.
098
pF
/
cm
0
.
407
0
.
085
L
=
C
=
0
.
407
0
.
085
509
.
160
.
63
60
.
64
509
.
1
m
/
cm
0
.
131
0
.
131
mS
/
cm
0
.
012
R
=
G
=
0
.
012
Search WWH ::
Custom Search