On Oct 18, 2:00�pm, dvsarwate <dvsarw...@yahoo.com> wrote:> On Oct 18, 1:48�pm, Jerry Avins <j...@ieee.org> wrote: > > > > > A real stumper for many beginners is, "If the energy doesn't come out of > > the filter, where does it go?" There's no problem with digital filters. > > After all, it's merely numbers. But what about analog filters made of > > lossless components? > > So here is a question about analog lossless components. > A capacitor C is charged to V volts and thus is storing energy > 0.5CV^2. �It is connected in parallel to an identical capacitor > through lossless (zero-resistance) wires, and so charge flows > from one capacitor to the other till each capacitor is holding > half the charge and thus has voltage V/2 across it. �Now the > energy being stored is 0.5C(V/2)^2 + 0.5C(V/2)^2 = 0.25CV^2. > What happened to the rest of the energy? > > --Dilip Sarwate > > P.S. �Answers of E = mc^2 will receive zero credit!photons!
Filter Output Energy
Started by ●October 18, 2011
Reply by ●October 19, 20112011-10-19
Reply by ●October 19, 20112011-10-19
On Oct 19, 8:30�am, dvsarwate <dvsarw...@yahoo.com> wrote:> On Oct 19, 12:17�am, Tim <t...@seemywebsite.please> wrote: > > <<<<<material snipped>>>>> > > > If pressed, my answer to both of them would be to express the "perfect" > > components as lossy components taken to the limit as the loss goes to > > zero. �Then you'll find that infinite current squared times zero > > resistance equals infinite power, for just the right sort of zero time > > that adds up to your lost energy. > > Yes indeed. �But taking things to the limit is done all the time. > Who among the readers of this note has _not_ used 1 as the > value of (sin x)/x at x = 0, even though, if pressed, the person > might concede that at x = 0 the value of (sin x)/x is undefined > and the value 1 that is being used is the limiting value of > (sin x)/x as x tends to 0 ? > > Dilip Sarwate > > P.S. �For the record, here is my take on the solution to the riddle > > Suppose the connecting wires have resistance R > 0. > Then I(t), the current flowing decays away exponentially > and the capacitors reach steady state at t = oo. �If > you calculate the energy expended in the resistance > due to I(t), it is _exactly_ the energy that is missing. > Curiously, int_0^infty I^2(t) R dt �does _not_ depend > on the value of R at all (remember that I(t) is of the > form exp(-t/RC) etc). �So, if we let R -> 0, the initial > current is higher, and the decay is more rapid (time > constant of circuit is getting smaller) but the energy > expended is the same, and using the continuity type > arguments that we like, the same amount is expended > at R = 0. �An instantaneous infinite current delivers > finite energy into a zero resistor in zero time! �A'i'nt > math wonderful?Even if the resistance is zero, you create electromagnetic radiation which takes away the extra energy. Clay
Reply by ●October 19, 20112011-10-19
On Tue, 18 Oct 2011 14:52:57 -0400, Jerry Avins <jya@ieee.org> wrote:>I won't give it away, but instead pose a mechanical version....>What happened to the rest of the energy?Conservation of Momentum states that momentum is conserved, but not necessarily energy. This applies not only to linear momentum and velocity, but to their analogs in angular (L=Iw, E=�Iw�) and electrical (Q=CV, E=�CV�) spaces. Looking at simple linear Conservation of Momentum applied to a collision of two objects: (m1)(v1) + (m2)(v2) = (m1)(V1) + (m2)(V2) where v1, v2 are the velocities before collision, and V1, V2 are the velocities after collision. Let's try a test case. Let m2=m1, v1=K, v2=0: (m1)(K) + (m1)(0) = (m1)(V1) + (m1)(V2) K = V1 + V2 Iff the collision is elastic, then kinetic energy is preserved: �(m1)(v1)� + �(m2)(v2)� = �(m1)(V1)� + �(m2)(V2)� �(m1)(v1)� + �(m1)(0)� = �(m1)(V1)� + �(m1)(V2)� (K)� = (V1)� + (V2)� Combining the two: (K = V1 + V2)� = (V1)� + (V2)� (V1)� + 2(V1)(V2) + (V2)� = (V1)� + (V2)� 2(V1)(V2) = 0 so either V1 or V2, but not both, must be zero. Now let's consider the analog of the problem at hand, in which case m2=m1, v1=K, v2=0, and V2=V1 (meaning that the two objects stuck together after the collision): (m1)(v1) + (m2)(v2) = (m1)(V1) + (m2)(V2) (m1)(K) + (m1)(0) = (m1)(V1) + (m1)(V1) K = V1 + V1 V1 = �K The kinetic energy before collision is: e = �(m1)(v1)� + �(m2)(v2)� = �(m1)(K)� + �(m1)(0)� = �(m1)(K)� The kinetic energy after collision is: E = �(m1)(V1)� + �(m2)(V2)� = �(m1)(�K)� + �(m1)(�K)� = �(m1)(�K)� + �(m1)(�K)� = �(m1)(K)� (half the initial energy) Gotta love the consistency.
Reply by ●October 19, 20112011-10-19
On Oct 19, 11:44�am, Greg Berchin <gjberc...@chatter.net.invalid> wrote:> On Tue, 18 Oct 2011 14:52:57 -0400, Jerry Avins <j...@ieee.org> wrote: > >I won't give it away, but instead pose a mechanical version. > ... > >What happened to the rest of the energy? > > Conservation of Momentum states that momentum is conserved, but not necessarily > energy. This applies not only to linear momentum and velocity, but to their > analogs in angular (L=Iw, E=�Iw�) and electrical (Q=CV, E=�CV�) spaces. > > Looking at simple linear Conservation of Momentum applied to a collision of two > objects: > � � � � (m1)(v1) + (m2)(v2) = (m1)(V1) + (m2)(V2) > where v1, v2 are the velocities before collision, and V1, V2 are the velocities > after collision. > > Let's try a test case. Let m2=m1, v1=K, v2=0: > � � � � (m1)(K) + (m1)(0) = (m1)(V1) + (m1)(V2) > � � � � K = V1 + V2 > > Iff the collision is elastic, then kinetic energy is preserved: > � � � � �(m1)(v1)� + �(m2)(v2)� = �(m1)(V1)� + �(m2)(V2)� > � � � � �(m1)(v1)� + �(m1)(0)� = �(m1)(V1)� + �(m1)(V2)� > � � � � (K)� = (V1)� + (V2)� > > Combining the two: > � � � � (K = V1 + V2)� = (V1)� + (V2)� > � � � � (V1)� + 2(V1)(V2) + (V2)� = (V1)� + (V2)� > � � � � 2(V1)(V2) = 0 > so either V1 or V2, but not both, must be zero. > > Now let's consider the analog of the problem at hand, in which case m2=m1, v1=K, > v2=0, and V2=V1 (meaning that the two objects stuck together after the > collision): > � � � � (m1)(v1) + (m2)(v2) = (m1)(V1) + (m2)(V2) > � � � � (m1)(K) + (m1)(0) = (m1)(V1) + (m1)(V1) > � � � � K = V1 + V1 > � � � � V1 = �K > > The kinetic energy before collision is: > � � � � e = �(m1)(v1)� + �(m2)(v2)� > � � � � � = �(m1)(K)� + �(m1)(0)� > � � � � � = �(m1)(K)� > > The kinetic energy after collision is: > � � � � E = �(m1)(V1)� + �(m2)(V2)� > � � � � � = �(m1)(�K)� + �(m1)(�K)� > � � � � � = �(m1)(�K)� + �(m1)(�K)� > � � � � � = �(m1)(K)� (half the initial energy) > > Gotta love the consistency.You have a rotating version of Newton's cradle. Clay
Reply by ●October 19, 20112011-10-19
On Oct 19, 10:20�am, Clay <c...@claysturner.com> wrote:> > Even if the resistance is zero, you create electromagnetic radiation > which takes away the extra energy. >Absolutely true. Indeed, the loss of energy due to electromagnetic radiation must be accounted for even when the resistance is not zero. When the resistance is not zero, current flows for all t > 0, and steady state is reached only at t = oo. However, in this infinite time interval, this loss of energy due to electromagnetic radiation must be quite small, because on the macroscopic level, the energy lost as heat in the (nonzero) resistance plus the energy stored in the two capacitors at t = oo is **exactly** equal to the initial energy stored in one capacitor at t = 0, that is, the law of conservation of energy holds. So, for *each* R > 0, no matter how small we choose R to be, the loss of energy due to electromagnetic radiation is essentially 0, but at R = 0, all the missing energy (or maybe an amount significantly larger than "essentially 0") suddenly disappears via electromagnetic radiation? I don't like this explanation, but I am aware that "different strokes for different folks"... Dilip Sarwate
Reply by ●October 19, 20112011-10-19
On Wed, 19 Oct 2011 05:30:15 -0700, dvsarwate wrote:> On Oct 19, 12:17 am, Tim <t...@seemywebsite.please> wrote: > > <<<<<material snipped>>>>> > >> If pressed, my answer to both of them would be to express the "perfect" >> components as lossy components taken to the limit as the loss goes to >> zero. Then you'll find that infinite current squared times zero >> resistance equals infinite power, for just the right sort of zero time >> that adds up to your lost energy. > > Yes indeed. But taking things to the limit is done all the time. Who > among the readers of this note has _not_ used 1 as the value of (sin > x)/x at x = 0, even though, if pressed, the person might concede that at > x = 0 the value of (sin x)/x is undefined and the value 1 that is being > used is the limiting value of (sin x)/x as x tends to 0 ?I use sinc(x), or I explicitly put the limit in there. Doesn't everyone? -- www.wescottdesign.com
Reply by ●October 19, 20112011-10-19
On Wed, 19 Oct 2011 12:19:47 -0500, Tim Wescott wrote:> On Wed, 19 Oct 2011 05:30:15 -0700, dvsarwate wrote: > >> On Oct 19, 12:17 am, Tim <t...@seemywebsite.please> wrote: >> >> <<<<<material snipped>>>>> >> >>> If pressed, my answer to both of them would be to express the >>> "perfect" components as lossy components taken to the limit as the >>> loss goes to zero. Then you'll find that infinite current squared >>> times zero resistance equals infinite power, for just the right sort >>> of zero time that adds up to your lost energy. >> >> Yes indeed. But taking things to the limit is done all the time. Who >> among the readers of this note has _not_ used 1 as the value of (sin >> x)/x at x = 0, even though, if pressed, the person might concede that >> at x = 0 the value of (sin x)/x is undefined and the value 1 that is >> being used is the limiting value of (sin x)/x as x tends to 0 ? > > I use sinc(x), or I explicitly put the limit in there. Doesn't > everyone?Interesting side note: if you solve the integral that coughs up (sin x)/ x, but replace the cos(w*x) with its Taylor series you get an answer that never requires you to divide by zero. It requires you to do a Taylor's series out to some absurd number of terms, but you never have to divide by zero. That's just a "gee whiz" thing. But if you want to solve B(t) = int_0^T_s e^{At} dt, then doing it "the normal way" gets you B(t) = A^{-1}(e^{AT_s} - I) This is fine, unless A is singular. Then you're just totally and completely hosed. But if you do the Taylor's expansion on e^{At} in the original integral, you get B(t) = sum_n=0^\infty (At^n)/(n!) And that is not only mathematically possible, it's also numerically tractable for quite a few realistic matrices A. -- www.wescottdesign.com
Reply by ●October 19, 20112011-10-19
dvsarwate <dvsarwate@yahoo.com> wrote:>When the >resistance is not zero, current flows for all t > 0, >and steady state is reached only at t = oo. However, >in this infinite time interval, this loss of energy due >to electromagnetic radiation must be quite small, >because on the macroscopic level, the energy lost >as heat in the (nonzero) resistance plus the energy >stored in the two capacitors at t = oo is **exactly** >equal to the initial energy stored in one capacitor >at t = 0, that is, the law of conservation of energy >holds. So, for *each* R > 0, no matter how small >we choose R to be, the loss of energy due to >electromagnetic radiation is essentially 0, but at >R = 0, all the missing energy (or maybe an amount >significantly larger than "essentially 0") suddenly >disappears via electromagnetic radiation? I don't >like this explanation, but I am aware that "different >strokes for different folks"...I'm not sure I agree the EM radiation is always really close to zero. Do the following: assuming no resistance, compute the velocity and acceleration of the electrons moving from one capacitor to the other. Figure out how much EM radiation is associated with this acceleration (Maxwell's equations). Then figure out how much less acceleration there is with a reasonably low value for the resistance. I bet it's not that much less and you still get most of the EM. Indeed I bet I can produce a visible spark with such a setup. Steve
Reply by ●October 19, 20112011-10-19
On 10/19/2011 1:01 PM, dvsarwate wrote:> On Oct 19, 10:20 am, Clay<c...@claysturner.com> wrote: > >> >> Even if the resistance is zero, you create electromagnetic radiation >> which takes away the extra energy. >> > > > Absolutely true. Indeed, the loss of energy due to > electromagnetic radiation must be accounted for > even when the resistance is not zero. When the > resistance is not zero, current flows for all t> 0, > and steady state is reached only at t = oo. However, > in this infinite time interval, this loss of energy due > to electromagnetic radiation must be quite small, > because on the macroscopic level, the energy lost > as heat in the (nonzero) resistance plus the energy > stored in the two capacitors at t = oo is **exactly** > equal to the initial energy stored in one capacitor > at t = 0, that is, the law of conservation of energy > holds. So, for *each* R> 0, no matter how small > we choose R to be, the loss of energy due to > electromagnetic radiation is essentially 0, but at > R = 0, all the missing energy (or maybe an amount > significantly larger than "essentially 0") suddenly > disappears via electromagnetic radiation? I don't > like this explanation, but I am aware that "different > strokes for different folks"...Being old enough to have actually seen a working spark-gap transmitter, I rather like the radiation solution. Indeed, there is no sudden transition to radiation loss at R=0, but the portion of loss attributable to radiation increases as R-->0. Dilip's **exactly** isn't quite true. The energy dissipated in the resistor is the difference between half the initial capacitor energy and what it radiated. The merit of these absurdities lies in getting newbies to think rather than just to solve equations. They also can serve to drive home that lumped constants are useful but unrealizable abstractions. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●October 19, 20112011-10-19
On Oct 19, 7:00�am, dvsarwate <dvsarw...@yahoo.com> wrote:> On Oct 18, 1:48�pm, Jerry Avins <j...@ieee.org> wrote: > > > > > A real stumper for many beginners is, "If the energy doesn't come out of > > the filter, where does it go?" There's no problem with digital filters. > > After all, it's merely numbers. But what about analog filters made of > > lossless components? > > So here is a question about analog lossless components. > A capacitor C is charged to V volts and thus is storing energy > 0.5CV^2. �It is connected in parallel to an identical capacitor > through lossless (zero-resistance) wires, and so charge flows > from one capacitor to the other till each capacitor is holding > half the charge and thus has voltage V/2 across it. �Now the > energy being stored is 0.5C(V/2)^2 + 0.5C(V/2)^2 = 0.25CV^2. > What happened to the rest of the energy? > > --Dilip Sarwate > > P.S. �Answers of E = mc^2 will receive zero credit!That's an old one.
