Digital Signal Processing Reference
In-Depth Information
Thus,
Y(z)
is given by
0.1z
2
(z - 0.9)(z - 1)
=
z
z - 1
-
0.9z
z - 0.9
,
Y(z) =
and from Table 11.2, the inverse transform yields
y[n] = 1 - 0.9(0.9)
n
= 1 - 0.9
n+ 1
,
for The partial-fraction expansion can be verified by modifying the MATLAB pro-
gram in Section 11.2.
The
z
-transform solution gives
n G 0.
y[0] = 0.1;
this value is also found from the initial-value
property:
0.1z
2
(z - 0.9)(z - 1)
= 0.1.
y[0] =
lim
z:
q
Y(z) =
lim
z:
q
As a final point, we verify the first three values of
y
[
n
] by the iterative solution of the
filter difference equation:
y[n] = 0.9y[n - 1] + 0.1x[n];
n = 0:
y[0] = 0.9(0) + 0.1(1) = 0.1;
n = 1:
y[1] = 0.9(0.1) + 0.1(1) = 0.19;
n = 2:
y[2] = 0.9(0.19) + 0.1(1) = 0.271.
The solution yields
y[n] = 1 - 0.9
n+ 1
,
n = 0:
y[0] = 1 - 0.9 = 0.1,
n = 1:
y[1] = 1 - 0.81 = 0.19,
and
n = 2:
y[2] = 1 - 0.729 = 0.271,
and these values are verified. The values can also be verified with the following MATLAB
program:
ynminus1 = 0;
xn = 1;
for n = 0:2;
yn = 0.9*ynminus1 + 0.1*xn
ynminus1 = yn;
end
result:yn=0.1 0.19 0.271
In this program,
yn = y[n], ynminus1 = y[n - 1],
and
xn = x[n].
A S
IMULINK
simulation
that also verifies the response is illustrated in Figure 11.6.
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