Re: Electrical circuits and charge - caution, lengthy

Since my reply is quick, and your post is long, forgive my top-posting. The mathematician's reply is completely correct. Going from the differential equation to the given integral equation simply is not mathematically correct in of itself. There must be another (probably unstated) physical assumption in play. The OP's suggestion "the charge that passed the point from beginning of time to current time (where negative infinity is beginning of time)" is exactly the reasoning I would have used. In addition there are questions of convergence, as the OP suggested, that the EE book has completely overlooked. But these issues of convergence will be trivially handled if one makes the physically very reasonable assumption that i(t) is zero if one goes sufficiently far back in time. I should also add that your reply also makes perfect sense as well.


Brian VanPelt wrote:
I am a reading of the math group sci.math where a person asked the
following question.


Op's original post:


I am pretty familiar with first year calculus (including FTOC and
convergence of improper integrals).

I am now reading an EE circuit theory book where there is a common
calculus step in several derivations that I can not follow 100%.

For example:

i(t) = dq(t)/dt (i=current is time rate of change of q=charge)

Then book would say, therefore....

q(t) = int[from -inf to t] i(t)dt


This happens a lot in the book, where the "reverse of a derivative
formula" results in an improper integral. Now, it makes sense when I
think of things like this as "the charge that passed the point from
beginning of time to current time (where negative infinity is
beginning of time)," but what exactly is the calculus step taken? I
mean, how do we get an improper integral from the original derivative
formula (what manipulation or theory or reasoning do we use to do that
step)? What about issues of convergence?


Basically, what allows me to do such a step?

z(t) = dy(t)/dt ==> y(t) = int[from -inf to t] z(t)dt

(assume only that y(t) is differentiable on the interval considered)


My reply


The integral undoes the derivative.

Start with

i(t) = dq(t)/dt

and integrate both sides with respect to t

Int( i(t) dt ) = Int( dq(t)/dt dt)

From here, insert any limits of integration that are needed.

Int[-inf to t] i(t) dt = Int[-inf to t] dq(t)/dt dt

Int[-inf to t] i(t) dt = q(t) - lim(t --> -inf) q(t)

I am guessing that it must be the case that lim(t --> -inf) q(t) = 0,
which makes sense and the charge must be decaying.

Hope this helps,

Brian


Math professor reply


I am pretty familiar with first year calculus (including FTOC and
convergence of improper integrals).

I am now reading an EE circuit theory book where there is a common
calculus step in several derivations that I can not follow 100%.

For example:

i(t) = dq(t)/dt (i=current is time rate of change of q=charge)

Then book would say, therefore....

q(t) = int[from -inf to t] i(t)dt
I hope it's assuming somewhere that the limit of q(t) as t ->
-infinity is 0.

This happens a lot in the book, where the "reverse of a derivative
formula" results in an improper integral. Now, it makes sense when I
think of things like this as "the charge that passed the point from
beginning of time to current time (where negative infinity is
beginning of time)," but what exactly is the calculus step taken? I
mean, how do we get an improper integral from the original derivative
formula (what manipulation or theory or reasoning do we use to do that
step)? What about issues of convergence?


Basically, what allows me to do such a step?

z(t) = dy(t)/dt ==> y(t) = int[from -inf to t] z(t)dt

(assume only that y(t) is differentiable on the interval considered)

It's not true then...
The Fundamental Theorem of Calculus tells you, if z(t) is continuous
then y(b) - y(a) = int_a^b z(t) dt. Assuming the limit of y(a) as a ->
-infinity
is L, that implies y(b) - L = int_{-infinity}^b z(t) dt (that improper
integral being defined as the limit of integrals from a to b as a -> -infinity). So your formula is right if L happens to be 0, but
not
otherwise. If the limit of y(a) as a -> -infinity exist, then the
improper
integral doesn't exist either.
.

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