What i am requested to do is to Find

for

.
I realized that the diode is initialy open because voltage is negative, and in time

it closes (it doesnt close at t = 0 because the diode has the same voltage as the capacitor which is -1 and it takes time to become positive),

my problem is that i don't know how to prove that the voltage on the capacitor is continuous between

and

?
usually i would try to write the diode as a current source which would behave like a step function,
but i dont know how to write it here.
I prefer a mathematical proof for the continuity because it helps me understand it better.

thanks very much to everyone who can help

 

The voltage across ANY capacitance is continuous. The energy stored in capacitor is a function of the capacitance and the voltage. Since energy cannot be created nor destroyed, but only converted to another form or moved to someplace else, the energy stored in the capacitor must be continuous.

Another way of looking at it is that the time rate at which the energy changes in capacitor is, by definition, power. In order for energy to be discontinuous, the power at the discontinuity must be infinite.

Now, having said that, it IS possible for the voltage to be discontinuous, but it requires the capacitance to be discontinuous. You would then find the new capacitance and use the energy before the change to determine the voltage after the change. But since most capacitances are related to the physical geometry of conductors, claiming a discontinuous change in capacitance is also a hard sell.

 

Why would the diode behave like a current source?

 

The voltage across ANY capacitance is continuous. The energy stored in capacitor is a function of the capacitance and the voltage. Since energy cannot be created nor destroyed, but only converted to another form or moved to someplace else, the energy stored in the capacitor must be continuous.

Another way of looking at it is that the time rate at which the energy changes in capacitor is, by definition, power. In order for energy to be discontinuous, the power at the discontinuity must be infinite.

Now, having said that, it IS possible for the voltage to be discontinuous, but it requires the capacitance to be discontinuous. You would then find the new capacitance and use the energy before the change to determine the voltage after the change. But since most capacitances are related to the physical geometry of conductors, claiming a discontinuous change in capacitance is also a hard sell.

in the homework we get the voltage on capacitor and current on inductor are sometimes not continuous when there is a dirac selta entrance,
but this is as you said - not a finite amount of voltage or current,
so I guess what you said is true, but I personally like to prove it mathemaically and not just physically and you do that by the differential equation that you get:
you know that the biggest derivative (and in first order circuits like what I posted here it's the first derivative) acts like the right side of the equation,
for example, in an equation like this:

you know that:

and

, which means that x is continuous.

so I want to get in this problem to an equation like that that will prove that

is continuous

Why would the diode behave like a current source?

usually when the problem i get is more simple i can convert my diode to a current source,
for example if for t < 0 you know that the diode is open and there is a steady current

and for 0 < t you know that the diode is closed you can convert the diode to a current source with value of

, in this form you can get a differential equation and prove continuity in the way i written above.

 

The use of delta functions with regard to circuits are essentially to instantly inject charge into a capacitor or current into an inductor in order to satisfy the boundary conditions. Essentially, you have an equation that models the circuit after some configuration has changed (a switch is thrown or, in this case, a diode begins conducting). But your equation is ONLY valid AFTER that time. So you set up the equation as if there is no energy stored in the components at that time and then use delta functions to kick the initial energy conditions into the circuit to account for the effect of all time prior to the event taking place.


I'm still not following your current source model for the diode. I think perhaps I see what you are doing and why, but if so it is a real kludge. Which is not to say that it might not be reasonable -- but it becomes very case-specific to justify that it is a reasonable model for that specific analysis.

 

No differential equation, by itself, is going to prove that the voltage across a capacitor has to be continuous. Think about it. The current in a capacitor does not have to be continuous and, likewise, the voltage across and an inductor does not have to be continuous.

The requirement for the continuity stems from physics, so to turn it into a mathematical proof, you have to impose the physical constraints. I've given you the necessary constraints -- power, which is the time rate of change of energy, must be finite. So you tie the voltage across a capacitor to the energy stored in the capacitor to the power applied to or by the capacitor and then impose the constraint that power must be finite.

One way to do this is to assume that a step change in voltage on a capacitor is possible. Then find the power associated with that step change and show that it is not finite. Since power must be finite, you have shown, via contradiction, that step changes in capacitor voltage are impossible. That will provide your mathematically rigorous prove.

 

The use of delta functions with regard to circuits are essentially to instantly inject charge into a capacitor or current into an inductor in order to satisfy the boundary conditions. Essentially, you have an equation that models the circuit after some configuration has changed (a switch is thrown or, in this case, a diode begins conducting). But your equation is ONLY valid AFTER that time. So you set up the equation as if there is no energy stored in the components at that time and then use delta functions to kick the initial energy conditions into the circuit to account for the effect of all time prior to the event taking place.


I'm still not following your current source model for the diode. I think perhaps I see what you are doing and why, but if so it is a real kludge. Which is not to say that it might not be reasonable -- but it becomes very case-specific to justify that it is a reasonable model for that specific analysis.

all what you written here makes a lot of sense, thank you.
but im still looking for a mathematical way to prove the continuity, I find that when I prove things mathematically it helps me understand the material better, even physically.
so do you know any way to do this, maybe something similar to the biggest derivative method, that will fit this problem?

edit: I didnt see your second comment, i read it now.

 

No differential equation, by itself, is going to prove that the voltage across a capacitor has to be continuous. Think about it. The current in a capacitor does not have to be continuous and, likewise, the voltage across and an inductor does not have to be continuous.

The requirement for the continuity stems from physics, so to turn it into a mathematical proof, you have to impose the physical constraints. I've given you the necessary constraints -- power, which is the time rate of change of energy, must be finite. So you tie the voltage across a capacitor to the energy stored in the capacitor to the power applied to or by the capacitor and then impose the constraint that power must be finite.

One way to do this is to assume that a step change in voltage on a capacitor is possible. Then find the power associated with that step change and show that it is not finite. Since power must be finite, you have shown, via contradiction, that step changes in capacitor voltage are impossible. That will provide your mathematically rigorous prove.

I really like your last idea of prooving by contradiction,
thank you very much