Digital Signal Processing Reference
In-Depth Information
2
s
8
Transfer Fcn
Step Input
Scope
yout
To Workspace
(a)
3
2.5
2
1.5
1
0.5
0
0
0.1
0.2
0.3
Time
(b)
0.4
0.5
0.6
Figure 7.9
Simulink simulation for
Example 7.12.
Note from the circuit that the initial current is
i(0) = 0.
The Laplace-transform solu-
i(0
+
) = 0;
tion gives
this value is also found from the initial-value property:
24
s + 8
= 0.
i(0
+
) =
lim
s:
q
sI(s) =
lim
s:
q
Hence, the current does not change instantaneously. The solution can be verified by the sub-
stitution of
i(t)
into the loop differential equation.
This system was simulated with S
IMULINK
. The block diagram from the simulation is
given in Figure 7.9(a), and the response is given in Figure 7.9(b). We see that the system time
constant is 0.125 s. Hence, the transient part of the response becomes negligible after approx-
imately four times this time constant, or 0.5 s. Figure 7.9(b) shows this. In addition, the final
value of
i(t)
is 3, which is also evident in the figure.
■
If the numerator and denominator polynomials in (7.49) are presented in
product-of-sums form, the transfer function is shown as
K(s
-
z
1
)(s
-
z
2
)
Á
(s
-
z
m
)
(s - p
1
)(s - p
2
)
Á
(s - p
n
)
H(s) =
.
(7.50)
>
In (7.50), where is the coefficient of the highest-order power of
s
in
the numerator polynomial, as shown in (7.49). In the transfer functions of many phys-
ical systems,
K = b
m
a
n
,
b
m
m 6 n
[i.e., in application of (7.49),
b
n
and often some of the other b
i
