Recent content by Razvan

I just wanted to make sure it is correct. These are the plots for the z transforms. I don't know how to plot only at discrete values. Are these correct?

This is what the output looks like for an RC circuit with R=C=1, the output being the voltage on the capacitor and the input a unit step. The plot is made both using a ZOH and not using a ZOH. Are my computations correct? When considering a ZOH, the input must be considered as being sampled, so...

I still don't think I understood correctly, I apologize. If I want to simulate what will happen, then yes, I need to add the effect of the ZOH to the computation. But when writing the algorithm, why exactly do I need to "create" myself the effect of the ZOH inside the controller? Isn't this...

Let's take a concrete example of a continuous controller, like G(s) = (s+2)/(s+10). It has a G(z) equivalent. Having the transfer function, I can create the algorithm based on the difference equation. Now, if I consider a sampler + a ZOH before the controller, I get a new transfer function...

If I were to write the "algorithm" for such a system, wouldn't it be simply: read(Vin); Vout = Vin/2; // output = {previous outputs} + {previous/current inputs}, the difference equation the z transform represents output(Vout); ? And this sequence would be executed at the beginning of every...

The ultimate goal of such a design is to get to a difference equation which can be implemented using a "computer", correct? So why do I need to take into consideration the effect of the ZOH when converting to the z-domain? If from the s-domain I realize that the input signal needs to be...

I have some questions related to how a discrete control system is designed. One method is to design the controller in the continuous time domain, arriving at a transfer function (in the s-domain). After that, a transfer function for the ADC system must be taken into consideration. I will suppose...

Thank you very much for your help. I am looking forward to future collaborations.

In the irreversible case, ##\frac{T_f}{T_0}## is linear in ##\frac{m_1-m_2}{m_1}##. $$\frac{T_f}{T_0} = 1+ (\frac{m_1-m_2}{m_1})\frac{1-\gamma}{\gamma}$$ In the reversible case, if we use $$(1+x)^a = 1+\frac{a}{1!}x + \frac{a(a-1)}{2!}x^2 + \cdots$$ we get $$\frac{T_f}{T_0} =...

From ##C_p = C_v + R## we can express ##V_f## as $$V_f = V_0(1+\frac{1}{\gamma}\frac{m_1-m_2}{m_2})$$ And from $$\frac{T_f}{T_0} = \frac{P_fV_f}{P_0V_0} = \frac{m_2}{m_1}(1+\frac{1}{\gamma}\frac{m_1-m_2}{m_2})$$ results $$T_f = T_0\frac{m_2}{m_1}(1+\frac{1}{\gamma}\frac{m_1-m_2}{m_2})$$

From $$m_2\frac{d^2x}{dt^2}=F_I-m_2g \vert \cdot \frac{dx}{dt}$$ $$m_2\frac{dx}{dt}\frac{d^2x}{dt^2}=F_I\frac{dx}{dt}-m_2g\frac{dx}{dt}$$ On the left side there is $$m_2va = m_2\frac{d}{dt}(\frac{1}{2}v^2)$$ On the right side there is no major change. The work done on the piston and mass has...

Ok. Please do not let me take up your time. I wish you a wonderful day.

I will write the steps, so other readers can benefit from this as well. Because ##PV^{\gamma}## is constant, from the ideal gas law (##PV=nRT##), we get that ##TV^{\gamma-1}## is constant. Finally, we have $$T_0V_0^{\gamma-1} = T_fV_f^{\gamma-1}$$ $$T_f = T_0 (\frac{V_0}{V_f})^{\gamma-1} = T_0...

An adiabatic reversible expansion respects $$PV^{\gamma} - const.$$ So the final volume is $$V_f = (\frac{m_1}{m_2})^\frac{1}{\gamma}V_0.$$

Yes.