Can I use Sympy to deduce the telegraphers equation?












1














I was trying to deduce the Telegrapher's equation using SymPy. But I couldn't handle nor extract the equation's different parts, so I finally did it manually. Can this be done using SymPy ? I've tried to use subs but I didn't achieve good results.



Thanx in advance! The following code was my attempt to solve the problem:



from sympy import *

# init_printing(use_unicode=True, wrap_line=False) 

V, I = symbols('V I', cls=Function)

z = Symbol('z') 

t = Symbol('t') 

R, L, C, G = symbols('R L C G', positive=True) 

eq1 = Derivative(V(z, t),z, 1) + R * I(z, t)+L* Derivative(I(z,t),t,1) 

# Derivative(V(z, t), z) + L*Derivative(I(z, t), t) + R*I(z, t) 

eq2 = Derivative(I(z,t),z,1)+G*V(z,t)+C*Derivative(V(z,t),t,1) 

# Derivative(I(z, t), z) + C*Derivative(V(z, t), t) + G*V(z, t) 

eq3 = eq1.diff(z,1) 

#eq3 = L*Derivative(I(z, t), t, z) + R*Derivative(I(z, t), z) +  Derivative(V(z, t), z, z) 

eq4 = eq2.diff(z,1) 

#eq4 = C*Derivative(V(z, t), t, z) + G*Derivative(V(z, t), z) +  Derivative(I(z, t), z, z) 



# Then part of eq1 is replaced in eq4 

# (-1*(R*I(z,t)+L*I(z,t).diff(t,1))) 

eq5 = C*Derivative(V(z, t), t, z) + G*(-1* (R*I(z,t)+L*I(z,t).diff(t,1))) + Derivative(I(z, t), z, z) 



# And similarly for eq2 in eq3 

# (-1*(G*V(z,t)+C*V(z,t).diff(t,1))) 

eq6 = L*Derivative(I(z, t), t, z) + R*(-1* (G*V(z,t)+C*V(z,t).diff(t,1))) + Derivative(V(z, t), z, z) 



# the derivative in time of eq1 

eq7 = eq1.diff(t,1)



# eq7 = L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t) + Derivative(V(z, t), t, z) 



# Derivative(V(z, t), t, z) = -1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) 



# the derivative in time of eq2

eq8 = eq2.diff(t,1)



# eq8 = C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t) + Derivative(I(z, t), t, z)



# Derivative(I(z, t), t, z) = -1* (C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) 



# Replace eq 7 in eq5 

Telegraphers_eq1 = C*-1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) + G*(-L*Derivative(I(z, t), t) - R*I(z, t)) + Derivative(I(z, t), z, z) 



# Replace eq 8 in eq6 

Telegraphers_eq2 = L*-1*(C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) + R*(-C*Derivative(V(z, t), t) - G*V(z, t)) + Derivative(V(z, t), z, z)


cheers,
Pedro






python sympy cas equation differential-equations







share|improve this question





























    1














    I was trying to deduce the Telegrapher's equation using SymPy. But I couldn't handle nor extract the equation's different parts, so I finally did it manually. Can this be done using SymPy ? I've tried to use subs but I didn't achieve good results.



    Thanx in advance! The following code was my attempt to solve the problem:



    from sympy import *
    
# init_printing(use_unicode=True, wrap_line=False) 
    
V, I = symbols('V I', cls=Function)
    
z = Symbol('z') 
    
t = Symbol('t') 
    
R, L, C, G = symbols('R L C G', positive=True) 
    
eq1 = Derivative(V(z, t),z, 1) + R * I(z, t)+L* Derivative(I(z,t),t,1) 
    
# Derivative(V(z, t), z) + L*Derivative(I(z, t), t) + R*I(z, t) 
    
eq2 = Derivative(I(z,t),z,1)+G*V(z,t)+C*Derivative(V(z,t),t,1) 
    
# Derivative(I(z, t), z) + C*Derivative(V(z, t), t) + G*V(z, t) 
    
eq3 = eq1.diff(z,1) 
    
#eq3 = L*Derivative(I(z, t), t, z) + R*Derivative(I(z, t), z) +  Derivative(V(z, t), z, z) 
    
eq4 = eq2.diff(z,1) 
    
#eq4 = C*Derivative(V(z, t), t, z) + G*Derivative(V(z, t), z) +  Derivative(I(z, t), z, z) 
    

    
# Then part of eq1 is replaced in eq4 
    
# (-1*(R*I(z,t)+L*I(z,t).diff(t,1))) 
    
eq5 = C*Derivative(V(z, t), t, z) + G*(-1* (R*I(z,t)+L*I(z,t).diff(t,1))) + Derivative(I(z, t), z, z) 
    

    
# And similarly for eq2 in eq3 
    
# (-1*(G*V(z,t)+C*V(z,t).diff(t,1))) 
    
eq6 = L*Derivative(I(z, t), t, z) + R*(-1* (G*V(z,t)+C*V(z,t).diff(t,1))) + Derivative(V(z, t), z, z) 
    

    
# the derivative in time of eq1 
    
eq7 = eq1.diff(t,1)
    

    
# eq7 = L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t) + Derivative(V(z, t), t, z) 
    

    
# Derivative(V(z, t), t, z) = -1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) 
    

    
# the derivative in time of eq2
    
eq8 = eq2.diff(t,1)
    

    
# eq8 = C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t) + Derivative(I(z, t), t, z)
    

    
# Derivative(I(z, t), t, z) = -1* (C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) 
    

    
# Replace eq 7 in eq5 
    
Telegraphers_eq1 = C*-1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) + G*(-L*Derivative(I(z, t), t) - R*I(z, t)) + Derivative(I(z, t), z, z) 
    

    
# Replace eq 8 in eq6 
    
Telegraphers_eq2 = L*-1*(C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) + R*(-C*Derivative(V(z, t), t) - G*V(z, t)) + Derivative(V(z, t), z, z)


    cheers,
    Pedro






    python sympy cas equation differential-equations







    share|improve this question



























      1












      1








      1







      I was trying to deduce the Telegrapher's equation using SymPy. But I couldn't handle nor extract the equation's different parts, so I finally did it manually. Can this be done using SymPy ? I've tried to use subs but I didn't achieve good results.



      Thanx in advance! The following code was my attempt to solve the problem:



      from sympy import *
      
# init_printing(use_unicode=True, wrap_line=False) 
      
V, I = symbols('V I', cls=Function)
      
z = Symbol('z') 
      
t = Symbol('t') 
      
R, L, C, G = symbols('R L C G', positive=True) 
      
eq1 = Derivative(V(z, t),z, 1) + R * I(z, t)+L* Derivative(I(z,t),t,1) 
      
# Derivative(V(z, t), z) + L*Derivative(I(z, t), t) + R*I(z, t) 
      
eq2 = Derivative(I(z,t),z,1)+G*V(z,t)+C*Derivative(V(z,t),t,1) 
      
# Derivative(I(z, t), z) + C*Derivative(V(z, t), t) + G*V(z, t) 
      
eq3 = eq1.diff(z,1) 
      
#eq3 = L*Derivative(I(z, t), t, z) + R*Derivative(I(z, t), z) +  Derivative(V(z, t), z, z) 
      
eq4 = eq2.diff(z,1) 
      
#eq4 = C*Derivative(V(z, t), t, z) + G*Derivative(V(z, t), z) +  Derivative(I(z, t), z, z) 
      

      
# Then part of eq1 is replaced in eq4 
      
# (-1*(R*I(z,t)+L*I(z,t).diff(t,1))) 
      
eq5 = C*Derivative(V(z, t), t, z) + G*(-1* (R*I(z,t)+L*I(z,t).diff(t,1))) + Derivative(I(z, t), z, z) 
      

      
# And similarly for eq2 in eq3 
      
# (-1*(G*V(z,t)+C*V(z,t).diff(t,1))) 
      
eq6 = L*Derivative(I(z, t), t, z) + R*(-1* (G*V(z,t)+C*V(z,t).diff(t,1))) + Derivative(V(z, t), z, z) 
      

      
# the derivative in time of eq1 
      
eq7 = eq1.diff(t,1)
      

      
# eq7 = L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t) + Derivative(V(z, t), t, z) 
      

      
# Derivative(V(z, t), t, z) = -1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) 
      

      
# the derivative in time of eq2
      
eq8 = eq2.diff(t,1)
      

      
# eq8 = C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t) + Derivative(I(z, t), t, z)
      

      
# Derivative(I(z, t), t, z) = -1* (C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) 
      

      
# Replace eq 7 in eq5 
      
Telegraphers_eq1 = C*-1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) + G*(-L*Derivative(I(z, t), t) - R*I(z, t)) + Derivative(I(z, t), z, z) 
      

      
# Replace eq 8 in eq6 
      
Telegraphers_eq2 = L*-1*(C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) + R*(-C*Derivative(V(z, t), t) - G*V(z, t)) + Derivative(V(z, t), z, z)


      cheers,
      Pedro






      python sympy cas equation differential-equations







      share|improve this question















      I was trying to deduce the Telegrapher's equation using SymPy. But I couldn't handle nor extract the equation's different parts, so I finally did it manually. Can this be done using SymPy ? I've tried to use subs but I didn't achieve good results.



      Thanx in advance! The following code was my attempt to solve the problem:



      from sympy import *
      
# init_printing(use_unicode=True, wrap_line=False) 
      
V, I = symbols('V I', cls=Function)
      
z = Symbol('z') 
      
t = Symbol('t') 
      
R, L, C, G = symbols('R L C G', positive=True) 
      
eq1 = Derivative(V(z, t),z, 1) + R * I(z, t)+L* Derivative(I(z,t),t,1) 
      
# Derivative(V(z, t), z) + L*Derivative(I(z, t), t) + R*I(z, t) 
      
eq2 = Derivative(I(z,t),z,1)+G*V(z,t)+C*Derivative(V(z,t),t,1) 
      
# Derivative(I(z, t), z) + C*Derivative(V(z, t), t) + G*V(z, t) 
      
eq3 = eq1.diff(z,1) 
      
#eq3 = L*Derivative(I(z, t), t, z) + R*Derivative(I(z, t), z) +  Derivative(V(z, t), z, z) 
      
eq4 = eq2.diff(z,1) 
      
#eq4 = C*Derivative(V(z, t), t, z) + G*Derivative(V(z, t), z) +  Derivative(I(z, t), z, z) 
      

      
# Then part of eq1 is replaced in eq4 
      
# (-1*(R*I(z,t)+L*I(z,t).diff(t,1))) 
      
eq5 = C*Derivative(V(z, t), t, z) + G*(-1* (R*I(z,t)+L*I(z,t).diff(t,1))) + Derivative(I(z, t), z, z) 
      

      
# And similarly for eq2 in eq3 
      
# (-1*(G*V(z,t)+C*V(z,t).diff(t,1))) 
      
eq6 = L*Derivative(I(z, t), t, z) + R*(-1* (G*V(z,t)+C*V(z,t).diff(t,1))) + Derivative(V(z, t), z, z) 
      

      
# the derivative in time of eq1 
      
eq7 = eq1.diff(t,1)
      

      
# eq7 = L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t) + Derivative(V(z, t), t, z) 
      

      
# Derivative(V(z, t), t, z) = -1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) 
      

      
# the derivative in time of eq2
      
eq8 = eq2.diff(t,1)
      

      
# eq8 = C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t) + Derivative(I(z, t), t, z)
      

      
# Derivative(I(z, t), t, z) = -1* (C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) 
      

      
# Replace eq 7 in eq5 
      
Telegraphers_eq1 = C*-1*(L*Derivative(I(z, t), t, t) + R*Derivative(I(z, t), t)) + G*(-L*Derivative(I(z, t), t) - R*I(z, t)) + Derivative(I(z, t), z, z) 
      

      
# Replace eq 8 in eq6 
      
Telegraphers_eq2 = L*-1*(C*Derivative(V(z, t), t, t) + G*Derivative(V(z, t), t)) + R*(-C*Derivative(V(z, t), t) - G*V(z, t)) + Derivative(V(z, t), z, z)


      cheers,
      Pedro






      python sympy cas equation differential-equations




      python sympy cas equation differential-equations






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      edited Nov 22 at 19:11

























      asked Nov 21 at 20:43









      Pedro S.

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