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1. ### Zero order hold -- Discrete control systems

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?
2. ### Zero order hold -- Discrete control systems

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...
3. ### Zero order hold -- Discrete control systems

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...
4. ### Zero order hold -- Discrete control systems

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...
5. ### Zero order hold -- Discrete control systems

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...
6. ### Zero order hold -- Discrete control systems

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...
7. ### Zero order hold -- Discrete control systems

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 $$\frac{T_f}{T_0} =... 10. ### Adiabatic irreversible expansion From ##C_p = C_v + R## we can express ##V_f## as$$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})$$11. ### Adiabatic irreversible expansion From$$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)$$On the right side there is no major change. The work done on the piston and mass has... 12. ### Adiabatic irreversible expansion Ok. Please do not let me take up your time. I wish you a wonderful day. 13. ### Adiabatic irreversible expansion I will write the steps, so other readers can benefit from this as well. Because ##PV^{\gamma}## is constant, from the ideal gas law (##PV=nRT##), we get that ##TV^{\gamma-1}## is constant. Finally, we have$$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.

I would like to see how it could be modeled, but I am afraid I do not have enough knowledge in this area. Despite that, if you are willing to explain it, I would be more than happy to follow it. All that I can think about that situation is that the parameters of the gas would no longer be...

The only reason I decided to assume that the piston is not frictionless is because I know exactly how much of the work done turns into heat that way. So, the differential equation that models the process is $$dU = -PdV + \frac{F}{A} dV$$ $$\frac{C_v}{R}(PdV+VdP) = -PdV + \frac{F}{A} dV$$ Which...

Thank you for your answer. But the overall effect of letting the gas "expand freely" (finite difference between the outside and the inside pressure) or assuming that the piston is not frictionless and the process is done quasi-statically would be the same: adding heat to the gas when it could be...

This is the page: http://en.wikipedia.org/wiki/Irreversible_process.

This example is taken from the wikipedia page describing irreversible processes. I just want to make sure I understand correctly why the initial state can't be reached anymore. I assume the transitions to be quasi-static, but there is friction between the piston and the cylinder. If so...
21. ### Body effect for NMOS transistor

"However the holes could also come from a p-doped region that is in close proximity to e MOS capacitor such as the source/drain region of a p-MOSFET." Maybe this could happen inside a mosfet as well. I thought about this only because if the electrons do come from the source, then the electric...
22. ### Body effect for NMOS transistor

I found my post (http://electronics.stackexchange.com/questions/145582/why-gate-source-and-not-gate-body-voltage) and the first answer contains this quote from a book (I hope I am allowed to add links to other sites): "In the above discussion you may have wondered, “Where do the electrons come...
23. ### Body effect for NMOS transistor

I was told (and I think the person who told me quoted from a book, so it should be true) that the inversion layer is created using electrons from the source terminal because free electrons in the bulk appear only because of the thermal agitation (so their number is limited) and it would take too...
24. ### Body effect for NMOS transistor

So I was right, the negative ions repel the electrons that would normally form the channel. I have another question, somehow related. Is is correct that the channel is mainly formed of electrons that come from the source terminal? Thank you very much for your time!
25. ### Body effect for NMOS transistor

But if the holes are attracted by the bulk terminal, leaving under the gate negative ions, why wouldn't just a small positive voltage on the gate attract more electrons to form the channel? Is it because the negative ions somehow "repel" the electrons that would normally start forming the channel?
26. ### Charge near grounded plane

This was one of my misunderstandings: I thought that if the plane wasn't grounded, it wouldn't generate any electric field because it would be electrically neutral (which is wrong). But making the plane grounded, and considering the outside charge q, we can say that the plane isn't neutral...
27. ### Charge near grounded plane

I was thinking if the plane had a very small thickness (but not flat), what would be the correct way of finding the electric field caused by the induced charges without using the image method. So I thought that a small cylinder (or rectangular parallelepiped) with one base inside the conductor...
28. ### Charge near grounded plane

So in order to get the correct electric field, the gaussian surface should be for example a very thin cylinder with one base inside the plane and the other outside?
29. ### Charge near grounded plane

But if Gauss' method is applied on a surface that surrounds the plane, wouldn't the charges inside cancel out?
30. ### Charge near grounded plane

That is where my problem comes from. So the charges that have accumulated on the top surface of the plane "come from the ground" and make the plane (if we consider the plane and the "ground" to be separate objects) electrically negative, generating an electric field outside. If we considered...