

In [3]:

    
import numpy as np
import sympy as sy
#import control.matlab as cm
sy.init_printing()

First order hold sampling of a lead compensator

\begin{equation} F(s) = K\frac{s+b}{s+a} \end{equation}



In [4]:

    
h, b, a,K = sy.symbols('h, b, a, K', real=True, positive=True)
s, z = sy.symbols('s, z', real=False)



In [5]:

    
F = K*(s+b)/(s+a)
U = F/s/s
Up = sy.apart(U, s)
Up









    Out[5]:





$$\frac{K b}{a s^{2}} - \frac{K \left(a - b\right)}{a^{2} \left(a + s\right)} + \frac{K \left(a - b\right)}{a^{2} s}$$



In [6]:

    
from sympy.integrals.transforms import inverse_laplace_transform
from sympy.abc import t



In [7]:

    
u = sy.simplify(inverse_laplace_transform(Up, s, t))
u









    Out[7]:





$$\frac{K \theta\left(t\right)}{a^{2}} \left(a b t e^{a t} + a e^{a t} - a - b e^{a t} + b\right) e^{- a t}$$

Sampling and taking the z-transform of the step-response

\begin{equation} Y(z) = \frac{1}{\lambda} \left( \frac{z}{z-\mathrm{e}^{\lambda h}} - \frac{z}{z-1} \right). \end{equation}

Dividing by the z-transform of the input signal

\begin{equation} H(z) = \frac{z-1}{z}Y(z) = \frac{1}{\lambda} \left( \frac{ \mathrm{e}^{\lambda h} - 1 }{ z - \mathrm{e}^{\lambda h} } \right) \end{equation}

Verifying for specific value of lambda



In [9]:

    
import control.matlab as cm
lam = -0.5
h = 0.1
G = cm.tf([1], [1, -lam])
Gd = cm.c2d(G, h)
Hd = 1/lam * cm.tf([np.exp(lam*h)-1],[1, np.exp(lam*h)])



In [10]:

    
print(Gd)
print(Hd)









    



  0.09754
----------
z - 0.9512

dt = 0.1


  0.09754
----------
s + 0.9512



In [ ]: