线性控制系统工程 M09
Linear Control Systems Engineering
Homework Solutions—Module9
9-1 Solution:
(a) CE: s 4+3s3+s2+3s+1=0
Routh’s array:
The sign of first column changed twice, indicating 2 CL poles in RHP. The system is unstable.
(b) CE: s 3+10s2+16s+160=0
Routh’s array:
s s s s
The s 1 row is an all-zero row. To cope with this, we need to construct an auxiliary equation using elements of the preceding row (the s 2 row):
10s 2+160=0
Its derivative is:
20s=0
Fill the all-zero row with coefficients of this derivative equation, and the Routh’s array can now be completed as:
s s s s There is no sign changes in the first column, indicating no CL poles in RHP.
(关于原点对称), and these ‘symmetrical poles’ are the roots of the auxiliary equation 10s 2+160=0. Solving this equation gives:
s=± 4j
Now that the system has a pair of CL poles on the imaginary axis, it is only critically stable .
9-2 Solution: Closed-loop transfer function:
10(1+s ) C =3 2R s +(1+10) s +10s +10
Characteristic equation:
s 3+(1+10τ) s 2+10s +10=0
Routh’s array:
Apparently τ must be positive to insure stability.