Lecture notes - School Of Electrical & Electronic Engineering - USM

Thus, the system is not stable.
1.3 Example 2
By using RH criterion, determine the stability of a
digital system whose CE is given as P (z) = z 3 +
3.3z 2 + 3z + 0.8 = 0.
1.3.1 Solution to Example 2
Transforming the CE P (z) into w-domain by using
bilinear transformation z = w+1
w−1 , gives:
8.1w 3 + 0.9w 2 − 0.9w − 0.1 = 0 (3)
The RH tabulation for the transformed CE 3 is as
follows:
w 3 8.1 -0.9
w 2 0.9 -0.1
w 1 0 0
w 0
The routh test could not be continued in the usual
manner since row w 1 contains all zeros. The tabulation
is proceeded by the auxillary equation from the
coeffecients in the w 2 row:
A(w) = 0.9w 2 − 0.1 = 0 (4)
Taking the derivative of A(w) in equation 4 with
respect to w , gives:
A(w)
dw
= 1.8w (5)
The coeffecient in the w 1 row are then filled with
A(w)
the coeffecient of
dw
, and the Routh tabulation
then becomes:
w 3 8.1 -0.9
w 2 0.9 -0.1
w 1 1.8 0
w 0 -0.1
From the Routh table, it can be see that there is
one sign change, which implies that the CE has one
root in the right half of the w-plane, or P (z) has one
root outside the unit circle in the z-plane.
1.4 Example 3
By using Routh-Hurwitz stability criterion, determine
the range of the gain K and the sampling period
T, so that the digital system whose CE is given as
P (z) = z 3 + a 2 z 2 + a 1 z + a 0 is asymptotically stable.
The coeffecients of the CE are as follows:
a 2 = 111.6T 2 + 16.74T − 3 (6)
a 1 = 1.395 ∗ 10 −4 KT 3 − 33.48T + 3 (7)
a 0 = 1.395 ∗ 10 −4 KT 3
− 16.74T − 111.6T 3 − 1 (8)
1.4.1 Solution to example 3
Transforming the CE P(z) into w-domain by using
the bilinear transformation z = w+1
w−1 , gives
where,
A 3 w 3 + A 2 w 2 + A 1 w + A 0 = 0, (9)
A 3 = a 2 + a 1 + a 0 + 1 = 2.79 ∗ 10 −4 KT 3 (10)
A 2 = a 2 − a 1 + 3a 0 + 3 = 446.4T 2
− 5.58 ∗ 10 −4 KT 3 (11)
A 1 = −446.4T 2 + 66.96T
+ 2.79 ∗ 10 −4 KT 3 (12)
A 0 = −a 2 + a 1 − a 0 + 1 = 8 − 66.96T (13)
2