Page 62 - MATLAB an introduction with applications

MATLAB Basics ——— 47

x′ = Ax + Bu
y = Cx + Du
The function has four output matrices:
[A, B, C, D] = zp2ss(z, p, k) Determines the matrices A, B, C, and D of the control-canonical form state-
space equations. p is a column vector of the pole locations of the zero-pole-
gain transfer function, z is a matrix of the corresponding zero locations, having
one column for each output of a multi-output system, k is the gain of the zero-
pole-gain transfer function. In the case of a single-output system, z is a
column vector of the zero locations corresponding to the pole locations of
vector p.
Example Problems and Solutions
Example E1.7: Consider the function
ns
()
() =
Hs
()
ds
2
3
4
where n(s) = s + 6s + 5s + 4s + 3
4
5
d(s) = s + 7s + 6s + 5s + 4s + 7
3
2
(a) Find n(–10), n(–5), n(–3) and n(–1)
(b) Find d(–10), d(–5), d(–3) and d(–1)
(c) Find H(–10), H(–5), H(–3) and H(–1)
Solution:
(a) >> n=[1 6 5 4 3]; % n=s^4+6s^3+5s^2+4s+3
>> d=[1 7 6 5 4 7]; % d=s^5+7s^4+6s^3+5s^2+4s+7
>> n2=polyval(n,[–10])
n2=4463
>> nn10=polyval(n,[–10])
nn10=4463
>> nn5=polyval(n,[–5])
nn5=–17
>> nn3=polyval(n,[–3])
nn3=–45
>> nn1=polyval(n,[–1])
nn1=–1
(b) >> dn10=polyval(d,[–10])
dn10=–35533
>> dn5=polyval(d,[–5])
dn5=612
>> dn3=polyval(d,[–3])

F:\Final Book\Sanjay\IIIrd Printout\Dt. 10-03-09