

In [2]:

    
import sympy as sy
import numpy as np
sy.init_printing(use_latex=True)

Problem 1

(a)

Zero-order-hold sampling



In [6]:

    
s,z,h = sy.symbols('s,z,h')
sy.apart(1/(s*(s+1)*(s-1)))









    Out[6]:





$$\frac{1}{2 s + 2} + \frac{1}{2 s - 2} - \frac{1}{s}$$



In [14]:

    
num = (0.5 * (sy.exp(h) + sy.exp(-h)) - 1)*(z+1)
den = (z**2 - (sy.exp(h) + sy.exp(-h))*z + 1)
H = num/den
print num









    



(z + 1)*(0.5*exp(h) - 1 + 0.5*exp(-h))



In [15]:

    
H.subs(h, 0.1)









    Out[15]:





$$\frac{0.00500416805580362 z + 0.00500416805580362}{z^{2} - 2.01000833611161 z + 1}$$

(b)

Backward difference approximation of given continuous-time controller



In [36]:

    
s,z = sy.symbols('s, z')
h = sy.symbols('h', positive=True)
F = (16*s+1)/(100*s+1)
H = sy.simplify(F.subs(s, (z-1)/(z*h)))
print H









    



(h*z + 16*z - 16)/(h*z + 100*z - 100)



In [ ]:



In [6]:

    
p1,p2,p3,p4 = sy.symbols('p1, p2, p3, p4') 

sy.expand((z-0.7+sy.I*0.1)*(z-0.7-sy.I*0.1))

Problem 2

Calculate the pulse transfer function from the disturbance d to the output y



In [10]:

    
B = 0.6*z + 0.5
A = z*(z**2 - 1.9*z + 0.9)
S, R = sy.symbols('S, R')

H_dy = (B/A) / (1 + (B/A)*(S/R))
sy.simplify(H_dy)









    Out[10]:





$$\frac{R \left(0.6 z + 0.5\right)}{R z \left(z^{2} - 1.9 z + 0.9\right) + S \left(0.6 z + 0.5\right)}$$



In [13]:

    
H_dys = sy.simplify(H_dy)
s=sy.latex(H_dys)
print s









    



\frac{R \left(0.6 z + 0.5\right)}{R z \left(z^{2} - 1.9 z + 0.9\right) + S \left(0.6 z + 0.5\right)}

Problem 3

Find the diagonal form of the system in problem 2, but without the time delay



In [18]:

    
H = sy.simplify(z*B/A)
sy.apart(H)









    Out[18]:





$$- \frac{10.4}{1.0 z - 0.9} + \frac{11.0}{z - 1}$$

Set up the state-space model. Make sure it is correct.



In [20]:

    
Phi = sy.Matrix([[0.9, 0], [0, 1]])
Gamma = sy.Matrix([[1],[1]])
Cm = sy.Matrix([[-10.4, 11.0]])

Htest = Cm*(z*sy.eye(2)-Phi).inv()*Gamma
Htest









    Out[20]:





$$\left[\begin{matrix}- \frac{10.4}{z - 0.9} + \frac{11.0}{z - 1}\end{matrix}\right]$$



In [21]:

    
H









    Out[21]:





$$\frac{0.6 z + 0.5}{z^{2} - 1.9 z + 0.9}$$

(c) Finding the feedback gain



In [26]:

    
l1, l2 = sy.symbols('l1, l2')
L = sy.Matrix([[l1, l2]])
sy.factor((z*sy.eye(2) - (Phi - Gamma*L)).det(), z)









    Out[26]:





$$1.0 \left(- 1.0 l_{1} - 0.9 l_{2} + 1.0 z^{2} + z \left(1.0 l_{1} + 1.0 l_{2} - 1.9\right) + 0.9\right)$$



In [38]:

    
sy.simplify(sy.expand((z-0.6+sy.I*0.3)*(z-0.6-sy.I*0.3)))









    Out[38]:





$$z^{2} - 1.2 z + 0.45$$

Check with matlab



In [1]:

    
%load_ext pymatbridge









    



Starting MATLAB on ZMQ socket ipc:///tmp/pymatbridge-4d29d37c-5632-4ec9-913b-670fa8e16c20
Send 'exit' command to kill the server
.....MATLAB started and connected!






    



/home/kjartan/anaconda2/lib/python2.7/site-packages/IPython/nbformat.py:13: ShimWarning: The `IPython.nbformat` package has been deprecated. You should import from nbformat instead.
  "You should import from nbformat instead.", ShimWarning)



In [23]:

    
%%matlab  --size 800,400

%Mean arteriar pressure, automatic anasthesia model
% Plant
G = tf([1], [120 1 0]);

a = 1/100;
b = 1/160;
F0 = zpk([-b], [-a], a/b);

K = 1e-2;
F = K*F0;

Gc = feedback(G*F, 1);
step(Gc, 1000)



In [30]:

    
%%matlab --size 800,400 -o L,y,t
h = 40.0;
Phi = [0.9 0; 0 1]; 
Gamma = [1;1]; 
C = [-10.4  11.0]; 
D = 0;
sys = ss(Phi, Gamma, C, D, h);
L = place(Phi, Gamma, [0.6+i*0.3 0.6-i*0.3]);
sys_cl = ss(Phi-Gamma*L, Gamma, C, D, h);
[y, t] = step(sys_cl);



In [34]:

    
import matplotlib.pyplot as plt
%matplotlib inline
plt.figure()
plt.plot(t, y)









    Out[34]:





[<matplotlib.lines.Line2D at 0x7fcfdf6fe250>]



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