How To Obtain The Z-Transform Y(z) Response? #1218

oneEclipseUser1 · 2026-06-03T14:09:05Z

oneEclipseUser1
Jun 3, 2026

Hello,

I am attempting to obtain the Y(z) z-transform discrete response of a transfer function to a unit response as shown in the following screenshot. Does the control library provide a method for calculating it without resorting to long hand paper calculations as shown in the following screenshot?

import sympy

z, q = sympy.symbols('z, q')

Gz = (0.3678*z + 0.2644) / (z**2 - 1.3678*z + 0.3678)
print('The transfer function: \n', Gz)

def qsubs(fz):
    return fz.subs({z: q**-1})

new_G = qsubs(Gz)
print(qsubs(Gz))

Answered by murrayrm

Jun 3, 2026

I just tried using vectors and unfortunately it didn't work (maybe my approach). Is there a method so that I can simplify it such that I can create a function where I can pass in the numerator and denominator vectors as arguments without using the 'z' variables?

sys = ct.TransferFunction(
    [0.3678, 0.2644],             # numerator polynomial
    [ 1.    , -1.3678,  0.3678],  # denominator polynomial
    dt=True)                      # discrete time

View full answer

oneEclipseUser1 · 2026-06-03T14:57:06Z

oneEclipseUser1
Jun 3, 2026
Author

Ok, I think that I figured it out.

import sympy

z, q = sympy.symbols('z, q')
Gz = (0.3678*z + 0.2644) / (z**2 - 1.3678*z + 0.3678)

def qsubs(fz):
    return fz.subs({z: q**-1})

new_G = qsubs(Gz)

N = 4

qpoly = sympy.Poly(new_G.series(q, 0, N).removeO(), q)
responses = list(reversed(qpoly.all_coeffs()))

print('\nY(z) = ')
for index, value in enumerate(responses):
    print(f'z^-{index}: {round(value, 4)}')

The printout is:

Y(z) = 
z^-0: 0.0
z^-1: 0.3678
z^-2: 0.7675
z^-3: 0.9145

... which aligns with the coefficients obtained via the manually worked out results.

0 replies

murrayrm · 2026-06-03T16:04:21Z

murrayrm
Jun 3, 2026
Maintainer

Your solution looks fine, but doesn't actually use python-control. Here's how to do that using the controls toolbox:

import control as ct
z = ct.tf('z')
sys = (0.3678/z**-1 + 0.2644) / (1/z**-2 - 1.3678/z**-1 + 0.3678)
response = sys.impulse_response()
print(response.outputs)

which returns

[0.         0.3678     0.76747684 0.91447798 0.968545   0.98843085
 0.99574487 0.99843496 0.99942438 0.99978829 0.99992213 0.99997136
 0.99998947 0.99999613 0.99999858 0.99999948 0.99999981 0.99999993
 0.99999997 0.99999999 1.         1.         1.         1.
 1.         1.        ]

1 reply

oneEclipseUser1 Jun 3, 2026
Author

Oh, ok, even better as it is more concise! 😊

By the way, the solution that I posted above is a slight tweek from the tutorial as stated in the control library docs here:
https://dynamics-and-control.readthedocs.io/en/latest/1_Dynamics/9_Sampled_systems/The%20z%20transform.html

Is there an expanded control library tutorial available when working with z-transforms as they apply to control theory? I am just getting familiar with this method and I do see huge potential vs. the analog circuit implementation method.

I just tried using vectors and unfortunately it didn't work (maybe my approach). Is there a method so that I can simplify it such that I can create a function where I can pass in the numerator and denominator vectors as arguments without using the 'z' variables?

murrayrm · 2026-06-03T16:32:16Z

murrayrm
Jun 3, 2026
Maintainer

I just tried using vectors and unfortunately it didn't work (maybe my approach). Is there a method so that I can simplify it such that I can create a function where I can pass in the numerator and denominator vectors as arguments without using the 'z' variables?

sys = ct.TransferFunction(
    [0.3678, 0.2644],             # numerator polynomial
    [ 1.    , -1.3678,  0.3678],  # denominator polynomial
    dt=True)                      # discrete time

3 replies

sawyerbfuller Jun 3, 2026
Collaborator

To get more in depth than what you can find in the control systems docs, you might want to consult a discrete-time controls text. I teach a class from Fadali and Visioli https://www.sciencedirect.com/book/monograph/9780128144336/digital-control-engineering

It covers topics like converting between discrete and continuous-time representations of dynamics, the z-transform, and control system design in the z-plane rather than the s-plane, and discrete-time systems in state-space form.

oneEclipseUser1 Jun 3, 2026
Author

To get more in depth than what you can find in the control systems docs, you might want to consult a discrete-time controls text. I teach a class from Fadali and Visioli https://www.sciencedirect.com/book/monograph/9780128144336/digital-control-engineering

It covers topics like converting between discrete and continuous-time representations of dynamics, the z-transform, and control system design in the z-plane rather than the s-plane, and discrete-time systems in state-space form.

Hello,

I see. Thank you for passing along this book lead. I am currently on the last chapter of Modern Control Systems by Bishop/Dorf 13th Ed. which is an introduction to 'Digital Control Systems'. I see that there are many benefits over the hardware implementation version (in my humble opinion, i.e., parameter values don't change over time as you would generally see with resistors and capacitors (plus tolerances), easy to modify on the fly without having to redesign/re-spin the board, etc.).

The book that you're recommending, is this like an industry standard for digital design? Does it cover/teach practical examples on how to go from circuit / system analysis (i.e., get the plant/system model) -> compensator design -> z-transform -> embedded application SW? That is, from initial to final application?

oneEclipseUser1 Jun 3, 2026
Author

I just tried using vectors and unfortunately it didn't work (maybe my approach). Is there a method so that I can simplify it such that I can create a function where I can pass in the numerator and denominator vectors as arguments without using the 'z' variables?
sys = ct.TransferFunction(
    [0.3678, 0.2644],             # numerator polynomial
    [ 1.    , -1.3678,  0.3678],  # denominator polynomial
    dt=True)                      # discrete time

Doh!😉 I forgot about the 'dt' argument and setting it to "True". I actually went over that yesterday.

Ok, awesome! Thank you a bunch. 😊👌

How To Obtain The Z-Transform Y(z) Response? #1218

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