Digital Signal Processing Reference
In-Depth Information
(4) Shift the time-inverted impulse function
h
(
−τ
) by a selected value of
“
t
.” The resulting function represents
h
(
t
− τ
).
(5) Multiply function
x
(
τ
)by
h
(
t
− τ
) and plot the product function
x
(
τ
)
h
(
t
− τ
).
(6) Calculate the total area under the product function
x
(
τ
)
h
(
t
− τ
) by inte-
grating it over
τ =
[
−∞, ∞
].
(7) Repeat steps 4
−
6 for different values of
t
to obtain
y
(
t
) for all time,
−∞ ≤
t
≤∞
.
Example 3.7
Repeat Example 3.6 and determine the zero-state response of the system using
the graphical convolution method.
Solution
Functions
x
(
τ
)
=
exp(
−τ
)
u
(
τ
)
,
h
(
τ
)
=
exp(
−
2
τ
)
u
(
τ
), and
h
(
−τ
)
=
exp(
−
2
τ
)
u
(
−τ
) are plotted, respectively, in Figs. 3.7(a)-(c). The function
h
(
t
− τ
)
=
h
(
−
(
τ −
t
)) is obtained by shifting
h
(
−τ
) by time
t
. We consider
the following two cases of
t
.
Case 1
For
t
<
0, the waveform
h
(
t
− τ
) is on the left-hand side of the vertical
axis. As is apparent in Fig. 3.7(e), waveforms for
h
(
t
− τ
) and
x
(
τ
)donot
overlap. In other words,
x
(
τ
)
h
(
t
− τ
)
=
0 for all
τ
, hence
y
(
t
)
=
0.
Case 2
For
t
≥
0, we see from Fig. 3.7(f ) that the non-zero parts of
h
(
t
− τ
)
and
x
(
τ
) overlap over the duration
t
=
[0
,
t
]. Therefore,
t
t
−
2
t
+τ
d
τ
−
2
t
e
τ
d
τ
−
2
t
[e
t
−
t
−
2
t
.
y
(
t
)
=
e
=
e
=
e
−
1]
=
e
−
e
0
0
Combining the two cases, we obtain
0
t
<
0
y
(
t
)
=
−
t
−
2
t
e
−
e
t
≥
0
,
which is equivalent to
−
t
−
2
t
)
u
(
t
)
.
y
(
t
)
=
(e
−
e
The output
y
(
t
) of the LTIC system is plotted in Fig. 3.7(g).
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