# What exactly are the assumptions behind Tustin's formula? Application on state space models

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?

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.

• 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 Commented Apr 17, 2023 at 9:29
• 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. Commented Apr 17, 2023 at 15:34
• 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. Commented Apr 19, 2023 at 7:20