Solution of ECE 316 Test #7 S04
A system has a transfer function, H( s) = 10
s2 − 16
. Three realizations are illustrated below, canonical,
s( s2 + 4 s + 3)
cascade and parallel. Find the values of all the gains, K. (Some may be zero.)
s2 − 16
Y( s)
⇒ 10( s2 − 16) X( s) = ( s3 + 4 s2 + 3s) Y( s)
= 10 3
Canonical:
s + 4 s2 + 3s
X( s)
s3 Y1 ( s) = X( s) − 4 s2 Y1 ( s) − 3s Y1 ( s) and Y( s) = 10( s2 − 16) Y1 ( s)
160
70
−
−
75
−53.33 −11.67
75
3
Parallel:
+
H( s) =
+ 6 +
=
+
s
s + 3 s +1
s
s+3
s +1
1
1
s− 4 s+ 4
s+ 4 s− 4
Cascade:
×
=L
×
= 10 ×
×
H( s) = 10 ×
×
s
s + 3 s +1
s
s +1 s + 3
K c21
1
s
K c22
K c23
X(s)
1
s
1
s
1
s
Y(s)
Y1 (s)
K p11
K p12
1
s
X(s)
K p21
K p22
K c13
.
1
s
K c12
K c11
X(s)
1
s
K d11
K c11 = 0
K c 21 = 10
K c12 = 3
K c 22 = 0
Y(s)
K p31
K p32
K d12
1
s
K d21
K d22
1
s
K d32
Y(s)
K d31
K c13 = 4
K c 23 = − 160
In the parallel case, the order of the subsystems does not matter but the coefficients must occur in these pairs.
K p11 = − 53.33 K p12 = 0
K p 21 = − 11.67 K p 22 = 3
K p 31 = 75
K p 32 = 1
In the cascade case, the coefficients in the left-hand column can be in any order and the -4 and 4 in the righthand column can also be reversed. However, because of the way the system is drawn, K d 32 must be 10.
K d12 = − 4
K d11 = 0
K d 21 = 3
K d 22 = 4
K d 31 = 1
K d 32 = 10