1. Apparently this question’s answer says this some static systems have memory especially those that hysteresis: Confusion about 'memoryless' meaning

  2. So the word static to me comes from static equilibrium of a system. There are courses in mechanical / civil engineering that study the balance of forces and moments when there is no motion or constant velocity motion, but most of all that the inertial forces are zero i.e. the acceleration is zero (second time rate of change of position). They also go further and study stress and strain in the body in follow on courses that assume static equilibrium.

  3. In many real scenarios, the system is not actually static but in quasi-static equilibrium where for example the loading onto the system is ever so slowly applied that the system’s responds like it is static, where the inertial forces are zero.

  4. One feature of static equilibrium in most mechanical systems is that when the dynamics are turned off then the motion returns to static equilibrium. Think of a spring mass damper system where the weight is balanced by a static spring force, that static position is where the dynamics oscillate about. So a related topic, and to me the exact same, is the Zero State Response (zero initial conditions i.e. zero initial displacement and zero initial velocity), where the spring mass damper system is solely driven by an external input, but once that input dies off the system returns to the IC which is just the static equilibrium state.

So from 3 and 4, while having zero inertial forces (really the second time rate of change of displacement), I have come to believe that systems in a state of quasi-static equilibrium are just systems that are not influenced by previous states.

For example using continuous state space representation, in that quasi-static equilibrium scenario, the state variables like

  • state-displacement (or strain state) is equal to the output-displacement
  • state-velocity (or time rate of change of strain state) is zero
  • output-velocity is not zero nor is it equal to state-velocity as that is zero. Therefore output-velocity is driven solely by the time rate of change of the loading
  • there is no dynamics (transient nor steady state) from systems state variables
  • once the load is removed the displacement stays at that quasi-static equilibrium position like the Zero State Response like the mass spring damper whose dynamics are turned off and rest at static equilibrium position. But here the system has continuous quasi-static equilibrium positions.
  1. State Variables for the System in Quasi-Static Equilibrium written in state space notation:

[x;dx/dt] := [strain state; time rate of strain state]

[dx/dt; d2x/dt2] := [time rate of change of strain state, second time rate of change of strain state]

Assign for quasi-static equilibrium:

dx/dt = 0 (this state has to be zero otherwise it influences the time rate of change of strain-output)

d2x/dt2 = 0 (has to be zero for basic static equilibrium)


[y; dy/dt] := [strain-output; time rate of strain-output]

which is solely driven by the input and so is a Relativity Zero State Response at each time increment.


Quasi-static equilibrium are just systems whose output-motion is driven solely by input and (like zero state response by definition sets IC to zero), where the states previous has no effect on current output-response.

So in that zero state response reference, when the dynamics are turned off then the motion will stop and the output will return to the IC which is essentially just like a static equilibrium position; that same position about which the dynamics had been oscillating about prior to being turned off.

Going back to the quasi-statics, in that case there is no system dynamics, maybe input dynamics since the input is time varying u and du/dt,but as far as the system is concerned it’s states and outputs are in quasi-static equilibrium. We would see that the previous state variates are zero for the dx/dt and x is not zero but acts like a relative zero point as this the x position acts like a quasi-static equilibrium point so that the output will return to. Prior to returning to that quasi-static equilibrium point, these states remain zero for the current time instant when the output and inputs (u, du/dt) are processed/applied to the system. If after that instant had passed that the input process were to be turned off, then the output would fall back to the zero states dx/dt = 0 and back to x which acts like a relative zero, and so therefore is like a quasi-static equilibrium position.

For example Second Order Mechanical System whose state variables are x, dx/dt are set to zero, then looking at state space control:

States are x, dx/dt, d2x/dt2 Inputs are u, du/dt Outputs are y, dy/dt

[dx/dt; d2x/dt2] = A*[x;dx/dt] + B*[u;du/dt]

[0;0] =A*[x;0] + B*[u;du/dt]

Where A=[a11,a12;a21,a22] is a 2x2 state transition matrix Where B=[b11,b12;b21,b22] is a 2x2 matrix

What that attempts to show is that the system is 1) it is at least a type of static system because d2x/dt2 is zero, 2) it is quasi-static because dx/dt is zero, but not x, 3) u and du/dt are not zero so the load is slowly applied which drives the output of the system, even those the states are zero.

Looking at the output equation

[y; dy/dt] = C*[x;dx/dt] + D*[u; du/dt] [y; dy/dt] = C*[x; 0] + D*[u; du/dt]

where C=[c11,c12;c21,c22] a 2x2 matrix where D=[d11,d12;d21,d22] a 2x2 matrix

Now what this attempts to show is that the output is not zero, even thought the all the states except x are zero


  1. Does my state space representation look correct for quasi-static response due to time dependent loading?

  2. Does this explain memoryless systems?

It seems to makes sense and to explain Quasistatics. Thanks

———— EDIT / PARTIAL ANSWER after comments with @aconcernedcitizen:

  1. how do I obtain quasi-statics from dynamics formulation in state space?

I assert that the State Space Formulation uses dynamical information, so my system state variables: x, dx/dt, and d2x/dt2 become x_dynamic, dx/dt_dynamic, d2x/dt2_dynamic.

Written out in the first vector state space equations it would be:


Inputs u, du/dt, and matrices A and B

B=[b11, b12; b21, b22] which is still satisfies the LTI condition, BUT is still able to switched OFF so in that sense is a function of which-process time, i.e. if dynamics are turned off then B=[0,0;0,0].

But that does NOT mean that u nor du/dt are turned off (zero) necessarily.

I assert that x=x_dynamic + x_static.

-If fully still-static (dead) then x=x_static, and x_dynamic ARE zero and that includes x_dynamic’s time derivatives. (Also probably for this case du/dt and dy/dt should be zero. The input/output would not be zero/zero 0/0 as because input=u,output=y=x_static)

-if QAUSI-STATIC then x=x_quasistatic which can vary with time, but still x_dynamic and it’s time derivatives are zero. But dy/dt is not necessarily zero, nor is du/dt necessarily zero.

I assign the OUTPUT process variables y and dy/dt to be equal to the x_quasistatic, and dx/dt_quasistatic respectively.

So the second “state space formulation” for the OUTPUTS is:




  • 1
    $\begingroup$ If a 2nd order system (for example) will reach and stay in steady-state it will be because of the states. $\endgroup$ Commented Jun 2, 2022 at 22:13
  • 1
    $\begingroup$ It is unclear what you are asking. Could you please edit your question so that somewhere in it (probably at the very beginning or the very end) is an actual question, i.e. a sensible sentence ending with a question mark that forms the actual question you want answered. $\endgroup$
    – TimWescott
    Commented Jun 3, 2022 at 2:29
  • $\begingroup$ @TimWescott Ok I added more info and questions $\endgroup$
    – L92MD14
    Commented Jun 5, 2022 at 19:43
  • $\begingroup$ @aconcernedcitizen ok but the output can be driven by the input $\endgroup$
    – L92MD14
    Commented Jun 5, 2022 at 20:31
  • $\begingroup$ @L92MD14 In that case it will be a part of a whole system ith feedback, so you can't just consider the forward path, you have to account for all the paths. $\endgroup$ Commented Jun 6, 2022 at 6:57

2 Answers 2


You're confusing the meanings of "static" and "has memory".

Static, in the sense of "static equilibrium" just means the system is sitting still and will remain sitting still. A pen balanced on its tip is not in static equilibrium -- once it has fallen over and stopped moving, it is. In practical terms it may also mean it's moving so slowly that you can ignore any dynamic behavior.

"Memory" means that the system has the capacity to "remember" it's states. So, any mechanical system with mass, any electrical system that has storage elements (either discrete capacitors and inductors, or just wires that form transmission lines), etc., has "memory".

Just the action of expressing a system in state space, i.e. $\dot{\mathbf x} = f(\mathbf x, \mathbf u)$, is saying the system has memory -- the fact that you need to express the system as a differential equation implies that there are integrators, and integrators are a memory element.

Perhaps the confusion is arising from true static analysis, where you start by ignoring all of the storage elements. I.e., in mechanical static analysis you don't worry about the mass of any member -- they're just lines on a page, and various anchors. Ditto, doing DC analysis of an electronic circuit starts by ignoring the storage (memory) elements -- capacitors are considered to be open circuits, inductors are considered to be shorts.

When you make those assumptions then your system either becomes memoryless (or is never given memory). So only in that sense a static system is memoryless. But -- that sort of "static" is different from a system in static equalibrium.

  • $\begingroup$ Thanks for the response, but I am sorry to disagree with you that I am confusing static and memoryless. I am not confusing them (though I admit that maybe I don’t understand their context - I will screenshot the section soon), as according to PHILLIPS4e Digital Control System Analysis & Design Chapter 10, it defines a system that is Static to be one whose states are memoryless. $\endgroup$
    – L92MD14
    Commented Jun 5, 2022 at 22:53
  • $\begingroup$ Ok I actually see what you are saying - sorry I hadn’t realized that when I wrote my previous comment. So it is a true static analysis which where the confusion comes up, between a true static analysis which is in a sense memoryless. Though do you think I should post a follow on question about Quasi-statics represented in State Space as I showed above? If you look up there we see that the system is a sort of static (quasi-static) because dx/dt is zero but not x and also the outputs are not zero, and so are driven by the time varying inputs u & du/dt. This is Quasi-static due to slow loading $\endgroup$
    – L92MD14
    Commented Jun 5, 2022 at 23:09
  • $\begingroup$ @L92MD14 If a zeroth order system is defined as static that's because it has absolutely zero dynamics involved. Its output will always be K times input, with K being some constant. It's a glorified wire. $\endgroup$ Commented Jun 6, 2022 at 6:59
  • $\begingroup$ @aconcernedcitizen yes so to get a second order system that is mass•d2x/dt2 + damping•dx/dt + stiffness•x + mass•gravity = external forcing function can be brought into state space after noting that mass•gravity is balanced out by a stiffness•common_static_deflection term. Then if the input, which is the external forcing function (u and du/dt), is ever so slowly applied, then we can assume d2x/dt2 and dx/dt are zero. Yet the outputs and the state position are NOT zero: dy/dt nor y nor x. So I can’t see that as still a zeroth order system. There are rate terms dy/dt & also y & x. $\endgroup$
    – L92MD14
    Commented Jun 6, 2022 at 16:00

@Tim Wescott's answer is already addressing this issue but, if you still have problems with it then why not concoct a test case and see, practically, what happpens when you're removing the states?

Here is a 2nd order system, which is this, for simplicity:

$$H(s)=\dfrac{1}{s^2+s+1} \tag{1}\label{1}$$

Its step response is:

$$h(t)=1-\text{e}^{-\frac{t}{2}}\left[\dfrac{1}{\sqrt3}\sin\left(\dfrac{\sqrt3}{2}t\right)+\cos\left(\dfrac{\sqrt3}{2}t\right)\right] \tag{2}\label{2}$$

At ~8.75 s the response gets below 1% variation and stays there. After ~12.7 s it gets below 1‰. it's safe to say that at the 15 s mark the system is well in its steady-state. Still, nota bene(!), due to the $\text{e}^{-t/2}$ term the system is never trully flat. But, for the sake of the explanation...

If what you're saying is true then, if we were to remove the states at the 15 s mark, the system should maintain its previous output indefinitely, perfectly flat. Let's test that:

2nd order dynamic states

G1, C1, G2, C2 form $\eqref{1}$. The switches are there to connect/disconnect the capacitors. They work in antiphase: when the type a switches are ON, the type b are OFF, and vice-versa. The capacitors are connected to ground either through a 1 mΩ (the Ron for the type a switch), which is virtually grounded, or through a GΩ, which is virtually floating (Roff for type a). Roff for the type b switches are also 1 GΩ, but their Ron is 1 Ω, which means that the current sources wil be virtual repeaters (no states, unity gain). It's not exactly pretty, but it does the job.

In short, the current sources have either a 1 F capacitor, or a 1 Ω resistor as loads, depending on the controlling voltage V2 -- which acts as a switch at the 15 s mark.

What you see plotted at the bottom is the output (blue trace), dropping to 0.333 V after the 15 s mark. At the top, the voltage across G1 (red) showing a step decrease in amplitude, while the black trace is the voltage across the capacitor, showing that it retains its state, unperturbed. That is its memory effect is alive and well.

The input is the Heaviside function (with a 1 ms rise time, good enough for this case) and, sure enough,for the first 15 s of the timespan the output is exactly as predicted in $\eqref{2}$. But as soon as the capacitors are switched for their resistive counterparts, everything goes wrong.

Why? Let's see:

The transfer function is $\eqref{1}$ when the capacitors are active, but when the resistors are active it becomes this:

"2nd order" no states

It's not exactly a transfer function since there are no states so it is, at best, a diagram representation of an equation but, for the sake of it, let's consider it a transfer function. If you'll determine it's I/O function it will be ($p$ being the node at the output of the first gain):

$$\begin{align} p&=x-y \\ y&=p-y=x-y-y \\ y+y+y&=x \\ 3y&=x \\ \Rightarrow \\ \dfrac{y}{x}&=\dfrac13 \tag{3}\label{3} \end{align}$$

...and this is exactly the value that you see after the 15 s mark. You might say at this point: "why not 'coerce' the value of the gains so that the output ends up being unity?". Let's solve for the gain then: instead of $1$ let's use $K$:

$$\begin{align} p&=K(x-y) \\ y&=K(p-y)=K(K(x-y)-y)=K^2x-K^2y-Ky \\ y+K^2y+Ky&=K^2x \tag{4}\label{4} \end{align}$$

For the output to be the same as the input it means $x=y$ ($=1$ in this case) and, when you replace that in $\eqref{4}$ you'll end up with:

$$1+K^2+K=K^2\quad\Rightarrow\quad K=-1$$

Since the gain of the current source must not change -- it was with this gain that the transfer function worked with the capacitors -- it follows that the resistors must have a negative value, which is absurd.

So, the states of the system are built-in, it's what makes the system a 2nd order, or Nth order, they are fixed (for LTI). The state of the output is a completely different dish. Just because the derivative of the output tends towards zero it doesn't mean that the system's states are zero. As I said in the comments, too, it is precisely because of those states that the system's output got to be in a (quasi-)steady-state. In other words: the system's output state is not the same as a system's state -- the former deals with the rate of change of the output, the latter with the memory of the system.


It's possible that the language is a barrier, I'm not native English, but what I was trying to say in my last comment was that there is no such thing as zero derivatives. It's just an impression. I'll repost the picture from the comments:

no such thing as zero derivatives

Look at the time scale: 90 - 100 seconds. The transfer function is unchanged. If you calculate the maximum output at 90 s it will be:


At 100 s it's $1.93\cdot10^{-22}$. By all accounts the impulse response at that time is considered zero, well within "stable" limits, and oscillators or not, those would be considered to be in steady-state. And yet, the output and the state are different, still.

Here's yet another way of looking at it: if what you say is true, that is, if the states can be considered zero, relative to the change in output, then the ratio between the output and the states should converge towards 1 (considering 0/0=1, as a limit).

Let's test that, too, shall we? Here is the same impulse response, on the same 100 s timescale, plotted alongside the variation of the state (bottom), versus the arc tangent of their ratio (top). The arc tangent is chosen to keep the plotting within readable limits, otherwise there will be divisions by zero and those will appear as humongous spikes, obscuring the message. If what you say is true, that the rate of change is zero, then the angle should converge towards zero, in a similar way in which the output or the state do, exponentially. In fact, you know what?, why not stretch the timescale to 1500 s, where the maximum amplitude should be 1.9e-326:

derivatives everywhere

Whaddayaknow? The ratio keeps on going. If you'll look at the Y-axis at the bottom plot (the output and the state) it's in the limits of the double range (1e-308). It can't plot the last part, it's so small. I think you'd agree that it can be considered well within the steady-state at that point (~1.4 ks). And yet, the top plot keeps on going, steady as hell. It's only in the last 20 s or so that the plot fails (you can see that it can't go all the way until the end), and that's because it had reached the limits of double. The output is simply the value of the ratio of the residuals at that point -- which is, still, not zero. Even then.

I don't know what other proofs you need. You seem to be fixed on the idea that an equilibrium implies zero rate of change of states relative to the I/O, that the $\dot{x}$ or $\ddot{x}$ end up zero -- that is false. As long as there is even a single state the derivatives will never be zero. Just because practical measures consider 1e-10 to be more than below steady-state it doesn't mean that things stop moving. Because, ultimately, that's what a zero derivative implies: no rate of change. But that is never true with states (unless dead).

  • $\begingroup$ Thank you for all this and I agree! But are we talking past each other? (1) I do not consider the output’s states variables to always be equal to the system’s state variables (2) when switched from capacitor to resistor, I can’t help but see that as just turning off the dynamics of system’s state variables since those no longer vary (3) but when switched to the resistor it is then the output that is driven by input, but I get it the system's 2nd order characteristics are built into that driven relation equation, but doesn't mean that system's state variables are varying because they're not $\endgroup$
    – L92MD14
    Commented Jun 8, 2022 at 14:51
  • $\begingroup$ (4) So that 1/3 that appears in output = 1/3*input is just the structural aspects of that 2nd order system but the system's state variables do not vary (5) when the switch occurs I think that the system is in quasi-static equilibrium & if the input doesn’t vary then in static equilibrium, albeit the systems structural aspects of it being second order are built in (6) IDK that when the output is driven by the 1/3•input (when system is quasi-steady I call it quasi-static), that its cause is to do with the states solely, but for sure the systems state variables don't vary, but that isn’t my point $\endgroup$
    – L92MD14
    Commented Jun 8, 2022 at 15:07
  • $\begingroup$ @L92MD14 As hinted somewhere in there, the states are never zero, because as soon as there is one state, the system's response gets an exponential term, which means even if you run out of precision, the response doesn't. Which means what you perceive to be equilibrium is, in fact, a struggle, which involves dynamics. You might argue that there is noise, but that will only strengthen the point of view: zoom in close enough and you'll see that the x, x', and x" are, actually, different -- the Y axis is femto Volts, can be barely plotted, but different. $\endgroup$ Commented Jun 8, 2022 at 16:32
  • $\begingroup$ I think all of what you and I are saying is in agreement, I had said that the system state variable X is non-zero, but that other system state variables DX/DT and D2X/DT2 are both zero for quasi-static equilibrium. And as for a complete static case, where the system state variable X doesn’t vary but that shouldn’t mean it is zero. So then there’s no exponential issue right? Anyway cant it be so that although sys state variables are never zero that they still are just not varying? As in there is not time rate of change of sys state variables. So either quasi-static or static equilibrium exist. $\endgroup$
    – L92MD14
    Commented Jun 8, 2022 at 17:42
  • $\begingroup$ @L92MD14 I would've posted it as a comment but then I realized it goes on for too long, so I've updated the answer. I'm not really sure what more I can add at this point, so this will probably be my last edit. My conclusion is you're overthinking it and, in the process, you're making some assumptions that turn out not to be true. But, as you say, maybe there'a a language barrier, who knows? $\endgroup$ Commented Jun 8, 2022 at 20:09

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.