How to invert a matrix with transfer functions in python?-CodePudding

I know that in matlab I can do the following:

s = tf('s')
G11 = (s   1)/(s   2)
G12 = 1/(2*s   1)
G21 = 1/(3*s   1)
G22 = 1/(4*s   1)

A = [G11 G12; G21, G22]
Ai = inv(A)
bode(A)

and it will work just fine. In python, I tried to do something similar:

import control as co
import numpy as np

s = co.tf('s')
G11 = (s   1)/(s   2)
G12 = 1/(2*s   1)
G21 = 1/(3*s   1)
G22 = 1/(4*s   1)

A = np.array([[G11, G12], [G21, G22]])
Ai = np.linalg.inv(A)
co.bode(A)

But this doesnt work - numpy doesnt know how to invert this matrix.

Is there a good way to do this in python? I know that I can use scipy with s being a symbol, but I think that doesnt help me when using the others tools in the control toolbox.

Edit:

numpy returns the following error:

---------------------------------------------------------------------------
UFuncTypeError                            Traceback (most recent call last)
<ipython-input-1-ec46afd90eb6> in <module>
     10 
     11 A = np.array([[G11, G12], [G21, G22]])
---> 12 Ai = np.linalg.inv(A)
     13 co.bode(A)

<__array_function__ internals> in inv(*args, **kwargs)

/usr/local/lib/python3.7/dist-packages/numpy/linalg/linalg.py in inv(a)
    543     signature = 'D->D' if isComplexType(t) else 'd->d'
    544     extobj = get_linalg_error_extobj(_raise_linalgerror_singular)
--> 545     ainv = _umath_linalg.inv(a, signature=signature, extobj=extobj)
    546     return wrap(ainv.astype(result_t, copy=False))
    547 

UFuncTypeError: Cannot cast ufunc 'inv' input from dtype('O') to dtype('float64') with casting rule 'same_kind'

CodePudding user response：

Numpy (hint: it's right in the name) is a numerics library only; it does not do symbolic math. Sympy (also in the name) does symbolic math, so use it:

import sympy

s = sympy.Symbol('s', imaginary=True)
g11 = (s   1)/(s   2)
g12 = 1/(2*s   1)
g21 = 1/(3*s   1)
g22 = 1/(4*s   1)
A = sympy.Matrix((
    (g11, g12),
    (g21, g22),
))
sympy.pprint(A.inv())

with output

⎡    3       2                       3       2              ⎤
⎢ 6⋅s    17⋅s    11⋅s   2      - 12⋅s  - 31⋅s  - 15⋅s - 2   ⎥
⎢ ───────────────────────      ──────────────────────────   ⎥
⎢     3      2                      3      2                ⎥
⎢  6⋅s    7⋅s  - 3⋅s - 1         6⋅s    7⋅s  - 3⋅s - 1      ⎥
⎢                                                           ⎥
⎢     3       2                 4       3       2           ⎥
⎢- 8⋅s  - 22⋅s  - 13⋅s - 2  24⋅s    50⋅s    35⋅s    10⋅s   1⎥
⎢─────────────────────────  ────────────────────────────────⎥
⎢     3      2                      3      2                ⎥
⎣  6⋅s    7⋅s  - 3⋅s - 1         6⋅s    7⋅s  - 3⋅s - 1      ⎦

CodePudding user response：

It looks like control.tf returns an object of class control.TransferFunction. This is different from the MATLAB version that returns a symbolic function object.

By looking through the documentation, I don’t see a built-in way to convert a control.TransferFunction object to a symbolic function object, but I did see there are the num and den methods, you could construct a symbolic function using those values. And then you can apply the answer by Reinderien.