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why the eproach here is so different..

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dknguyen

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When you do KVL or KCL with capacitors and inductors you get integrals or derivatives because the voltage and current formulas of those components have integrals/derivatives. It just looks like good ol' KVL to me.

And where do you see Norton and Thevnin? I don't see that anywhere.
 
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wizard

Member
I do not know what you really mean by your question?!

You may know, Kvl And KCL are universal formulas in Electronics branch. Some times Northon/Thevenin Theorems or Voltage/Current divder formulas can help to simply calculte the parameters of a circuit (the voltage of the branches and the current of the nodes) without (just some times) using KVL/KCL which maybe take more time or can reducing a complex circuit to an equal simpler one. But anyhow KVL/KCL are universal for any RLC circuit.

The voltage across a Capacitor is calculated by the integral formula which is in the pic:
Capacitor - Wikipedia, the free encyclopedia
 
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wizard

Member

MrAl

Well-Known Member
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Hi,


Since most of the work is already done for us here (all the integrals written out)
the only thing left to do is to find the new limits of integration. This is very
obvious if you look at the integrals for V2 and V3 and then try to figure
out the new limits of integration for V23.
In words:
If we integrate a function from -inf to t and then subtract from that the integral
of that same function integrated from -inf to t-1, that means we are left with
a much 'smaller' section to integrate over.
Geometrically, if we subtract a large area from a somewhat larger area, we are left
with a smaller area.

See it now?
 
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Hi,


Since most of the work is already done for us here (all the integrals written out)
the only thing left to do is to find the new limits of integration. This is very
obvious if you look at the integrals for V2 and V3 and then try to figure
out the new limits of integration for V23.
In words:
If we integrate a function from -inf to t and then subtract from that the integral
of that same function integrated from -inf to t-1, that means we are left with
a much 'smaller' section to integrate over.
Geometrically, if we subtract a large area from a somewhat larger area, we are left
with a smaller area.

See it now?
i understand all of it.
my main problem is with the actual meaning
the question says

"find V23 as a function of f(x)"


they are using an integral
this integral represent a "memory" of a capacitor all the voltage which is being passed threw from minus infinity till "t".

finding V23 as a function of f(x)
means we are looking for a voltage in only one point of time.(not all the point from minus infinty till t).
i would do kvl kcl and solve diff equation
why i cant do it here???

why are they using memory integral
?
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,


Are you looking at a different picture than the one you posted in the
beginning of this thread? Perhaps you should go look at that and make
sure it is the right one, because i do not see that question you are saying
that they are asking. I dont see any "f(x)" anywhere in that pic.
 
sorry its blurred

on the left current course if written
f(t)


on the right current course if written
f(t-1)

and i should have rephrase the question
find V23 as a function of f(x)

now knowing all that

can you solve the misunderstanding that i present in post 6
(the one before your last one)
 

MrAl

Well-Known Member
Most Helpful Member
Hello again,


Actually you still havent told me anything new. I knew about f(t) and f(t-1)
but there is nothing there that says "f(x)" anywhere.

Do you really mean f(lambda) perhaps?

Or, do you want to find the voltage gradient between 2 and 3 where x would
be the distance?

I can only guess as to what you are asking until you answer these questions.
 
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MrAl

Well-Known Member
Most Helpful Member
Hi,

So you are looking for a solution like:

y=x(f(t))

where x is some function that will work no matter what f(t) is, or just
for some particular f(t) ?
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,


Sorry, i can not remember the trick that makes this work.
It's been quite a while since i worked on the more theoretical aspects
of some of these circuits.


That reminds be though, have you had Z transforms yet?
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,


What did you end up with?
 

MrAl

Well-Known Member
Most Helpful Member
Hello again,


Oh well i thought we knew that solution already?
 
yes
:) we knew

so i got to the conclusion that whenever i have
an undefined source
then i will be doing that integral subtraction way
 
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MrAl

Well-Known Member
Most Helpful Member
Hi again,


I think i remember the solution now, but you'll need to use Z transforms for this
to make any sense to you. I hope you have had at least an introduction, but
if not it will have to wait for a later time.

Here is the rationale:

In a sampled system, z^-1 can be interpreted as a delay in the signal
such as f(t)*z^-1. Since in the problem we have to subtract f(t) from
it's own delayed version f(t-1) delayed by exactly 1, we can write the
solution to the sampled system like this:
f(t)*1-f(t)*z^-1

or simply:
f(t)*(1-z^-1)

(we also need to integrate BTW).

Now if we find the continuous time expression of this we get a new time expression,
and i bet you already guessed what it is (he he):

Intg f(t)*(1-z^1) dt=Intg[t-1 to t] f(t) dt

In other words, in the sampled time domain when we subtract a function delayed by 1
from the original function, we get f(t)*(1-z^-1), which is equivalent to the
continuous time domain Intg[t-1 to t] f(t) dt (after an integration).
So this should prove that that integral shown in the problem is correct.

It's been a while since i looked at these theoretical topics, so there may be some
limits as to the range or type of functions that this works for, but i would bet that
the linear functions work with this idea.
 
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