0
\$\begingroup\$

I have a discrete linear system that reads as follows: $$ x_{i+1} = A x_{i-1} + Bu_{i-1} \\ y_{i} = C x_i + Du_{i} $$ The first equation is supposed to denote the discrete time evolution of some system (a continuous system sampled periodically and the index then corresponds to the sample number) and the latter equation is supposed to denote the output an experimenter reads. The second summand denotes a control function applied to the system.

I am given to understand this is something standard in EE and control theory.

My question is the following: How can I derive the transfer function of the theory using what apparently is called a discrete z-transform (and is it true this is an analogy to Laplace transform)?

I apologise if the question is trivial but my background is in differential topology and not engineering.

\$\endgroup\$

1 Answer 1

3
\$\begingroup\$

I'm just a student in EE but I'll try to give it my best shot. Z-Transform in discrete time domain can be defined as what Laplace Transform is in continuous time domain. It is a detailed topic, but basically a function's Z-Transform is found by using the following formula:

$$\mathcal{Z}\{f_i\} = F(z) = \sum_{n=-\infty}^\infty f_nz^{-n}$$

As far as I can see, the system in this example is time shifted since a default state-space representation would look like this:

$$x_{i+1} = Ax_i+Bu_i$$ $$y_i=Cx_i+Du_i$$

The time shifting property of Z-Transform states that \$\mathcal{Z}\{f_{i-i_0}\}=F(z)z^{-i_0}\$

Let \$\mathcal{Z}\{x_i\}=X(z), \mathcal{Z}\{u_i\}=U(z), \mathcal{Z}\{y_i\}=Y(z)\$

Take the Z-transform of both equations.

$$zX(z)=z^{-1}AX(z)+z^{-1}BU(z)$$ $$Y(z) = CX(z)+DU(z)$$

Multiply both sides by z and leave X(z) alone in the first equation to obtain the equation below.

$$X(z)=(z^2I-A)^{-1}BU(z)$$

Substitute the first equation into the second one:

$$Y(z)=C(z^2I-A)^{-1}BU(z)+DU(z)$$

Therefore, the transfer function is obtained as

$$\frac{Y(z)}{U(z)}=C(z^2I-A)^{-1}B+D$$

If the time shift was not present, the transfer function would've been \$\frac{Y(z)}{U(z)}=C(zI-A)^{-1}B+D\$

I hope this helps.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Thanks, this is a nice and well detailed derivation. The final result is a bit weird though since there is a dependence on B. What if B=D=0 in the, i.e., there is no control? Then the transfer then seems to be 0. \$\endgroup\$
    – Marion
    Sep 28, 2021 at 19:43
  • \$\begingroup\$ If B and D are equal to zero, then it means that the system does not depend on an input at all, and our TF becomes equal to 0. That means the system's output is independent of the chosen input. \$\endgroup\$ Sep 29, 2021 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.