In [1]:

    

import sympy
from sympy.utilities.lambdify import lambdify
import numpy.linalg as la
import numpy as np
from sympy import init_printing
init_printing()


    

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z, r,tau=sympy.symbols('z r tau')


    

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t=1.-r**2


    

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e=sympy.exp


    

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dim=2


    

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T_denom = (1.-r**2* e(z)**2)


    

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T_denom


    

    
Out[23]:





$$- r^{2} e^{2 z} + 1.0$$

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## transfer function in the program
T_program = -r*sympy.eye(dim) + ((1.-r**2)/T_denom) * sympy.Matrix([[r*e(z)**2,e(z)],[e(z),r*e(z)**2]])


    

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## coefficients of transfer between the two internal nodes
M1 = sympy.Matrix([[0,r],[r,0]])


    

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## from inputs to internal nodes
M2 = sympy.Matrix([[t,0],[0,t]])


    

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## from internal nodes to outputs
M3 = sympy.Matrix([[0,t],[t,0]])


    

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## from inputs to outputs. Note the phase shift 
M4 = sympy.Matrix([[-r,0],[0,-r]])


    

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M = np.vstack([np.hstack([M1,M2]),np.hstack([M3,M4])])


    

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print M


    

    




[[0 r -r**2 + 1.0 0]
 [r 0 0 -r**2 + 1.0]
 [0 -r**2 + 1.0 -r 0]
 [-r**2 + 1.0 0 0 -r]]

In [56]:

    

E = sympy.Matrix([[e(z),0],[0,e(z)]])
E


    

    
Out[56]:





$$\left[\begin{matrix}e^{z} & 0\\0 & e^{z}\end{matrix}\right]$$

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sympy.eye(2)-M1*E


    

    
Out[57]:





$$\left[\begin{matrix}1 & - r e^{z}\\- r e^{z} & 1\end{matrix}\right]$$

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##inverse of sympy.eye(2)-M1*E
inv = sympy.Matrix([[1,r*e(z)],[r*e(z),1]])/(1-r**2*e(2*z))


    

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##derived transfer function
T = M3*E*inv*M2+M4
T


    

    
Out[72]:





$$\left[\begin{matrix}\frac{r \left(- r^{2} + 1.0\right)^{2} e^{2 z}}{- r^{2} e^{2 z} + 1} - r & \frac{\left(- r^{2} + 1.0\right)^{2} e^{z}}{- r^{2} e^{2 z} + 1}\\\frac{\left(- r^{2} + 1.0\right)^{2} e^{z}}{- r^{2} e^{2 z} + 1} & \frac{r \left(- r^{2} + 1.0\right)^{2} e^{2 z}}{- r^{2} e^{2 z} + 1} - r\end{matrix}\right]$$

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## transfer function in code
T_program


    

    
Out[75]:





$$\left[\begin{matrix}\frac{r \left(- r^{2} + 1.0\right) e^{2 z}}{- r^{2} e^{2 z} + 1.0} - r & \frac{\left(- r^{2} + 1.0\right) e^{z}}{- r^{2} e^{2 z} + 1.0}\\\frac{\left(- r^{2} + 1.0\right) e^{z}}{- r^{2} e^{2 z} + 1.0} & \frac{r \left(- r^{2} + 1.0\right) e^{2 z}}{- r^{2} e^{2 z} + 1.0} - r\end{matrix}\right]$$

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