ddt() at a certain timestep in Verilog-A

rhanna said:

Is it possible to find V(d[i]) at a certain timestep? How?

What do you mean by this? You can plot the signal if you wish - is that what you want?

Andrew

Hi Andrew,

V(d[i]) is ddt() of the signal V(x_node[i])

I know that the simulator chooses deltaT but I need to find the derivative at some specific timesteps to use it in the same code

for example something like:  ddt(V(x_node[i]) + 0.5*(1-alpha)*h*V(d[i])) , but this will be just ddt(V(x_node[i]))

Is there a way to implement this?

Thanks

I'm not sure you really made it much clearer - what isn't clear is what you mean by "some specific timesteps". Perhaps you are wanting to have some expression which uses the value from some fixed delay beforehand (in which case you could use the delay() function), or maybe you could use @(timer(...)) to regularly sample the value and derivative and then combine the current and previous samples in some way. However, I'm just guessing what your actual intention is here - that's what's not obvious from your question.

Andrew

Hi Andrew,

I have discrete state-space equations 
in the form:
Xi = A.Xi-1 + B.Ui
Yi = C.Xi

I need to convert these difference equations into differential to be solved by the simulator
for example by the aid of forward Euler method we can say:
Xi = X + (1-alpha).h.ddt(X)           at t = ti-1 + alpha.h
Xi-1 = X - alpha.h.ddt(X)

after plugging Xi and Xi-1, the simulator solves for X

As forward Euler is not so efficient, I'm trying to approx. the state-space equations using Runge-Kutta method RK4 https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods:
where Xi = X + (1-alpha).h.1/6.(K1 + 2K2 + 2K3 + K4)
K1: derivative at initial point, K2 and K3 are mid-point derivatives and K4 is the derivative at the endpoint.

In Verilog-A:
K1 <+ ddt(V(X_node));
but how can K2, K3, K4 be implemented?

I hope my idea is clear now.

Thanks for your care.

I'm a bit rusty on this, but I think the best solution would be to convert the discrete state-space representation into the z-domain and then you can use the zi_zp or zi_nd functions in Verilog-A to easily represent the system in the z-domain. See this http://mocha-java.uccs.edu/ECE5710/ECE5710-Notes05.pdf (I'm sure there are many other places that would describe this). This seems the simplest way - utilising a built-in feature of Verilog-A rather than doing all the work yourself.

Otherwise you could probably implement some kind of sampling using @(timer) as I suggested to do what you're asking, but I've never tried doing that to implement a differential equivalent of the state space equations, so I can't say for certain (and I don't have the time to think it through, do experiments and so on) whether it is the right approach or not. 

Andrew

Thanks, Andrew for your time.

Maybe with MATLAB also, there can be a move to the Z domain then by d2c() we can transform the TF to the S domain. The problem that the sampling time is fixed.

In order to have a variable timestep, the simulator should solve the differential equations which should come from the state-space difference equations.

Regarding the usage of @timer(start_time, period)

the period is the time taken from the start_time at which there's a value for X or ddt(X). How can I know this during the runtime?

Regards,

Ramy

Ramy,

I thought you said your state space equations were discrete state space equations. If so, transforming them to the Z-domain makes most sense, and then there is a discrete time step. The zi_zp or zi_nd functions (and the other combinations) are told the period for the z-domain system. The simulator itself can still have variable time steps.

Maybe I'm misunderstanding something in what you're asking. If they were continuous time state space equations, you could convert them to a Laplace transform (s-domain) representation and use the laplace_zp, laplace_nd functions in Verilog-A.

Andrew.

Dear Andrew,

The state-space equations are discrete, let's say the output is Y[n] = C.X[n]. The Z-transform filter output is Y(z). I'm looking for Y(t).

How can that be reached through Verilog-A? 

Ramy

Ramy,

That’s exactly what the zi_* functions do. You give them the zeros and poles or numerator and denominator of the z-domain transfer function, the discrete period and the time domain input signal and it will give you the time domain response. 

Andrew

Thanks for your explanation. 

This would help a lot if the system is linear.

In some cases, the system is non-linear for example: X[n+1] = tanh(A.X[n] + B.U[n] + C) and it will be so hard to find a TF.

Do you have any advice on how can I deal with this or in this case I have to work in the time domain?

Ramy