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I was parsing the forum when I saw this post surging out of the depths of this forum like an old Kraken. The problem is quite simple.

You have a continuous time state space model :

$$ \begin{split} \dot{X} &= AX+BU\\ Y &= CX \end{split} \tag{1}\label{eq1} $$ The state equation giving in Laplace :

$$(sI-A)X(s) = BU(s) \tag{2}\label{eq2}$$

Then compute the equivalent discrete time model by using the Bilinear transformation :

$$s = \alpha \frac{z-1}{z+1}\tag{3}\label{eq3}$$

Let us focus on the state transition matrix. After some tedious calculations they get :

$$A_d = (\alpha I-A)^{-1}(\alpha I+A)\tag{4}\label{eq4}$$

The thing is that I know another way to solve this. You just solve the linear differential equation given by \ref{eq4}. An exact solution is, between instants $kT$ and $(k+1)T$ :

$$ X((k+1)T) = e^{AT}X(kT) + e^{AT}\int_{kT}^{(k+1)T} e^{-A\tau}BU(\tau)d\tau \tag{5}\label{eq5} $$

clearly we see that, no matter what assumption we make to compute $\int_{kT}^{(k+1)T} e^{-A\tau}BU(\tau)d\tau$, even if we use the the trapezoidal rule, we will have $$A_d = e^{AT}\tag{6}\label{eq6}$$

here comes my question. I always heard that Tustin's rule was equivalent to using the trapezoidal rule to integrate the differential equation. Since clearly we have :

$$ (\alpha I-A)^{-1}(\alpha I+A) \neq e^{AT}$$

it must be more complicated than that. Is it because I didn't discretize the differential equation \ref{eq5}? Is there a bilinear transform that is equivalent to solving \ref{eq5}? What do we exactly assume when we use Tustin's bilinear transformation?

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What do we exactly assume when we use Tustin's bilinear transformation?

I'm not sure what you mean by "trapezoidal rule", but Tustin's approximation is the $z$-domain way to use a discrete time trapezoidal approximation for integration -- i.e., that $$\int x(t) dt \simeq \frac T 2 \sum x(nT) + x\left((n-1)T\right)$$

If you derive the $z$ transform by starting with the declaration that $z = e^{sT}$, then for $sT \ll 1$,

$$\begin{align} z &= e^{sT} \\ \\ &= \frac{e^{sT/2}}{e^{-sT/2}} \\ \\ &\approx \frac{1+\frac{sT}{2}}{1-\frac{sT}{2}} \\ \end{align} $$

Solve for $s$.

$$\frac T 2 \frac{z + 1}{z - 1} \simeq s^{-1}$$ (try this out with the first term of the Taylor's expansion of $z = e^{sT}$)

That is what we assume when we use the Tustin's approximation.

Is there a bilinear transform that is equivalent to solving (4)?

Nope. It's called Tustin's approximation for a reason.

For the general case where $U(t)$ varies arbitrarily between sampling instants, there really isn't an exact conversion from the continuous-time to the discrete-time (although they'll all have $A_d = e^{AT}$).

When $U(t)$ is some known then you can find it -- in fact I used to use this when doing state space control. But if you don't have control of $U(t)$ between sampling intervals, then you just need to accept that you're going to approximate things.

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  • $\begingroup$ Thank you for your answer! I had a mismatch between my equation references and my labels. my 4s were supposed to be 5s but you didn't let that distract you. I think I have a better grasp of what was troubling me. I'll take some time to put it down on paper and I'll come back then. EDIT : I also found this post about bilinear transform $\endgroup$
    – NokiYola
    Commented Apr 17, 2023 at 9:29
  • $\begingroup$ If you decide that this is the answer to your question, check it as such -- it'll help others when they're asking the same thing and find your question. $\endgroup$
    – TimWescott
    Commented Apr 17, 2023 at 15:34
  • $\begingroup$ In the end my problem was that Tustin's formula is equivalent to the trapezoidal approximation in the context of solving $\dot{X}=f(X,t)$ we integrate $X(t)=X(0)+\int f(X,t)dt$ and approximate the integrale with the trapezoidal rule. In the case of my equation \ref{eq5} what was done was, with $f(X,t) = AX+BU$ exact integration for the $AX$ term and an approximation for $BU$. But if you want Tustin's formula to be equivalent to trapezoidal integration you need to integrate the whole f with trapezoidal approximation. $\endgroup$
    – NokiYola
    Commented Apr 19, 2023 at 7:20

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