A. BACKGROUND:

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.

5. 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)

Outputs:

[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.

B. CLARIFICATION:

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

C. QUESTION

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:

3. 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:

[dx/dt_dynamic;d2x/dt2_dynamic]=A•[x_dynamic;dx/dt_dynamic]+B•[u;du/dt]

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:

[y;dy/dt]=C•[x_dynamic;dx/dt_dynamic]+D•[u;du/dt]

[y;dy/dt]->[x_quasistatic;dx/dt_quasistatic]

[x_dynamic;dx/dt_dynamic]->[0;0]