Filter Output Energy

On Oct 19, 6:48&#4294967295;am, Jerry Avins <j...@ieee.org> wrote:
> On 10/18/2011 11:37 AM, dvsarwate wrote:
>
>
>
> > On Oct 18, 5:53 am, "commsignal"<commsignal@n_o_s_p_a_m.yahoo.com>
> > wrote:
> >> If a signal with energy 'E' is filtered with a filter having unit energy,
> >> what is the energy of the output signal?
> >> Thanks.
>
> > Exercise: The signal is sqrt{2E} cos(2*pi*1000*t),
> > a signal of energy E at 1 kHz. it is passed through
> > a bandpass filter with center frequency 100 MHz
> > and bandwidth 1 MHz. &#4294967295;What do you think is the
> > energy of the output signal?
>
> > Assuming that "filter having unit energy" means that
> > the impulse response of the filter is a unit-energy signal,
> > all that can be said that the output signal will have
> > energy _at most_ E, and this extreme value will occur
> > if the filter is matched to the signal.
>
> 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?
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.

a resistor isn't lossless...I^2 R but two capacitors as a voltage
divider is I suppose.

On Oct 20, 6:01&#4294967295;am, dvsarwate <dvsarw...@yahoo.com> wrote:
> On Oct 19, 10:20&#4294967295;am, Clay <c...@claysturner.com> wrote:
>
>
>
> > Even if the resistance is zero, you create electromagnetic radiation
> > which takes away the extra energy.
>
> Absolutely true. &#4294967295;Indeed, the loss of energy due to
> electromagnetic radiation must be accounted for
> even when the resistance is not zero. &#4294967295;When the
> resistance is not zero, current flows for all t > 0,
> and steady state is reached only at t = oo. &#4294967295;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. &#4294967295;So, for *each* R > 0, no matter how small
> we choose R &#4294967295;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 &#4294967295;electromagnetic radiation? &#4294967295;I don't
> like this explanation, but I am aware that "different
> strokes for different folks"...
>
> Dilip Sarwate

If a magnet attracts a nail (mass m) from a distance h from the
ground, then it has done mgh work in doing so. When you remove the
nail that energy has to go back into the magnetic field.

HardySpicer  <gyansorova@gmail.com> wrote:

>On Oct 20, 6:01&#4294967295;am, dvsarwate <dvsarw...@yahoo.com> wrote:

>> Absolutely true. &#4294967295;Indeed, the loss of energy due to
>> electromagnetic radiation must be accounted for
>> even when the resistance is not zero. &#4294967295;When the
>> resistance is not zero, current flows for all t > 0,
>> and steady state is reached only at t = oo. &#4294967295;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. &#4294967295;So, for *each* R > 0, no matter how small
>> we choose R &#4294967295;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 &#4294967295;electromagnetic radiation? &#4294967295;I don't
>> like this explanation, but I am aware that "different
>> strokes for different folks"...

>If a magnet attracts a nail (mass m) from a distance h from the
>ground, then it has done mgh work in doing so. When you remove the
>nail that energy has to go back into the magnetic field.

Assuming it did not become magnetized.  Then some energy has
also gone into rearrangement of states within the nail.

Steve

On Wed, 19 Oct 2011 19:54:18 +0000, Steve Pope wrote:

>>If a magnet attracts a nail (mass m) from a distance h from the ground,
>>then it has done mgh work in doing so. When you remove the nail that
>>energy has to go back into the magnetic field.
> 
> Assuming it did not become magnetized.  Then some energy has also gone
> into rearrangement of states within the nail.
> 
> Steve

I'm pretty sure that if the nail has become magnetized then the force 
needed to remove the nail will be higher -- otherwise you could 
demagnetize a magnet by repeatedly magnetizing other objects with it, and 
I've never seen that phenomenon.

-- 
www.wescottdesign.com

On 10/19/2011 5:45 PM, Tim Wescott wrote:
> On Wed, 19 Oct 2011 19:54:18 +0000, Steve Pope wrote:
>
>>> If a magnet attracts a nail (mass m) from a distance h from the ground,
>>> then it has done mgh work in doing so. When you remove the nail that
>>> energy has to go back into the magnetic field.
>>
>> Assuming it did not become magnetized.  Then some energy has also gone
>> into rearrangement of states within the nail.
>>
>> Steve
>
> I'm pretty sure that if the nail has become magnetized then the force
> needed to remove the nail will be higher -- otherwise you could
> demagnetize a magnet by repeatedly magnetizing other objects with it, and
> I've never seen that phenomenon.

Nor have I.

A complete analysis accounts for the energy lost as heat and sound when 
the nail strikes the magnet. nevertheless, the magnet's external field 
is the same before the nail was picked up and after it is removed.

Do you like there conundrums? Here's another. Two identical springs are 
identically distorted and tied with acid-proof string into the new 
shape. One is set aside, the other, dissolved in acid. From the unetched 
spring, the energy (0.5kx^2) can be recovered. What happened to the 
energy stored in the spring that was etched?

Jerry
-- 
Engineering is the art of making what you want from things you can get.

On 10/19/2011 8:44 AM, Greg Berchin wrote:
> 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.

So, in this electrical system of 2 capacitors, there is conservation of 
charge when they are connected.  Not conservation of energy.  That's the 
gross answer to the question (which I thought we were avoiding).

If we look a bit broader, there is conservation of energy in a closed 
system.  Was this supposed to be a closed system?

If one neglects EM radiation, then the energy goes to heat that remains 
in the "system" if it's closed.

I suppose the dog clutch example, the clutch also makes noise in 
addition to making hot dogs.

Fred

Tim <tim@seemywebsite.please> wrote:

(snip)

>>>> 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?

(snip)
>> I won't give it away, but instead pose a mechanical version. Two
>> identical flywheels, moment of inertia I, can be coupled by a dog
>> clutch. Initially, one rotates at angular velocity w radians/second and
>> the other is stationary. Without exerting any forces outside of the
>> system, the flywheels are coupled by the dog clutch. Momentum being
>> conserved, they then rotate together at w/2. Before they were coupled,
>> the total energy was 0.5Iw^2, all in the rotating part. After the
>> coupling, the energy of the flywheels together is 0.5(2I)(w/2)^2, or
>> 0.25Iw^2. What happened to the rest of the energy?

> IMHO the appropriate answer to both Jerry's and Dillip's questions are 
> "you are asking for a reasonable physical explanation of the behavior of 
> a physically unreasonable system".

For the capacitor, you can't get away from inductance, even if
you make the lead length zero.  

There is a chapter in "Feynman Lectures on Physics" where he tries
to make a highest resonant frequency LC circuit.  First, decrease C
by increasing the plate separation.  Then decrease L until you have
only one wire with no turns.  Then put more and more L wires in
parallel, and eventually you end up with a cylindrical can, or
what is called a resonant cavity in microwave electronics.
The sides, and even the end plates themselves will contribute to 
the inductance.

And even for superconductors.

> Invoking physics (energy balance) in the same context as physically 
> impossible devices (infinitely strong dog clutches, superconducting 
> arcless switches, etc.) is a good brain teaser, but it's not necessarily 
> fair or in any way realistic.

> 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.  Or use Jerry's dodge and invoke 
> inductance in the wires (and flex in the shafts attached to the dog 
> clutch) and find that the energy remains stored in the system after all.

-- glen

Tim Wescott <tim@seemywebsite.com> wrote:
> On Wed, 19 Oct 2011 19:54:18 +0000, Steve Pope wrote:

>>>If a magnet attracts a nail (mass m) from a distance h from the ground,
>>>then it has done mgh work in doing so. When you remove the nail that
>>>energy has to go back into the magnetic field.
 
>> Assuming it did not become magnetized.  Then some energy has also gone
>> into rearrangement of states within the nail.

> I'm pretty sure that if the nail has become magnetized then the force 
> needed to remove the nail will be higher -- otherwise you could 
> demagnetize a magnet by repeatedly magnetizing other objects with it, and 
> I've never seen that phenomenon.

There is a demagnetizing that occurs to permanent magnets when
they don't have a closed magnetic loop.  You put a keeper across
a horseshoe magnet when it isn't being used.  I first remember
the story back to toy train engines, where if you take the motor
apart, the magnet will lose much of its magnetization.

There does have to be some frictional loss, or hysteresis loss,
though, in magnetizing a hard (permanent) magnetic material.  
My though, is that the attractive force is reduced, but I believe
also that the removal force is higher.  

Consider the power needed to power an electromagnet while it is
picking up the nail, and while the nail is being removed.

-- glen

Jerry Avins <jya@ieee.org> wrote:

(snip)
> A complete analysis accounts for the energy lost as heat and sound when 
> the nail strikes the magnet. nevertheless, the magnet's external field 
> is the same before the nail was picked up and after it is removed.

> Do you like there conundrums? Here's another. Two identical springs are 
> identically distorted and tied with acid-proof string into the new 
> shape. One is set aside, the other, dissolved in acid. From the unetched 
> spring, the energy (0.5kx^2) can be recovered. What happened to the 
> energy stored in the spring that was etched?

The stored spring energy is stored in the electromagnetic effects
between electrons and atoms as the crystal distorts.  It is, then,
presumably released as the atoms are released from their strained state.

To do the comparison, you should disolve an undistorted spring also.

Anyway, the whole system will be warmer, though there is usually
a lot of heat from disoving metals in acids, so you might not
notice.

-- glen

Tim Wescott <tim@seemywebsite.com> wrote:

(snip)
>> 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?

OK, who first came up with L'Hopital's rule, and, if someone
else, why does it have L'Hopital's name on it?

-- glen