# Galiphrey's little web site

*Waves*

I was building some speakers, and it started me wondering
about some things. And I thought that you, the web-site goer, may like to wonder as well!
I won’t claim that nobody’s ever
thought about it before, but I can claim that
Thanks are owed to my dad for noodling
through much of this with me, which is part of the fun as well!
From him you can see the value of actually understanding something,
rather than just being able to work fancy equations. That’s
not to say that understanding can’t come from math; definitely it
does! But it’s easy to fall into the doing of math for its own
sake. If long equations are the only way you have of explaining
something, then you probably don’t really understand it. On
the other hand as well, if the math doesn’t work out right, it’s
more likely your fault than its.
The power of an electrical signal into
a resistive load is unrelated to its frequency. But the power
of an acoustic signal is proportional to the square of its frequency
(as well as the square of its amplitude). Same with light.
Why is this discrepancy?
The tempting correlation between mechanical
amplitude and electrical amplitude, whether of V or I, is the wrong
analogy. But that doesn’t mean there’s not a correct analogy.
It is this: acoustic amplitude corresponds to electric CHARGE.
Please imagine 9 electrical RMS watts
at 10 kHz into an 8-ohm, 93 dB/W/m tweeter. RMS power is Vpeak^2
/ 2R, so it’s a voltage swing of +/- 12 volts. (Let’s assume
no harmonics, power-factor = X throughout the example, etc..)
If 100% efficiency for this particular tweeter corresponds to say 110
dB/W/m (due to its dispersion pattern), then 93 dB/W/m means the efficiency
is -17 dB, or about 2%. So we have
Frequency: 10 kHz Power-in: 9 W rms Amplitude-in: 12 V peak Power-out: 180 mW rms dB at one meter: 93 + 9.5 = 102.5 dB Now imagine 9 electrical RMS watts at
100 Hz into an 8-ohm, 93 dB/W/m woofer (and imagine its dispersion pattern
is the same as the tweeter, just for imagination’s sake).
Frequency: 100 Hz Power-in: 9 W rms Amplitude-in: 12 V peak Power-out: 180 mW rms dB at one meter: 93 + 9.5 = 102.5 dB I’ve tried to choose realistic numbers.
You really could find a normal woofer and a tweeter with these characteristics,
approximately. Now compare in your imagination or in your living
room the amplitude (excursion) of the woofer cone with that of the tweeter
cone. The same power in. The same power out. The same
electrical amplitude in. But the mechanical amplitude out is MUCH
DIFFERENT. Obviously, the tweeter doesn’t have to move nearly
as far to provide the same amount of acoustical (mechanical) power as
the woofer. Similarly, you can pluck a guitar string and see that
the amplitude is greater for lower notes, even though the energy imparted
by your plucker is the same. So, all I’m trying to get at is
that clearly the FREQUENCY of an acoustic wave is involved with its
power. In fact, the power of an acoustic wave is proportional
to the square of the frequency as well as the square of the amplitude.
We really don’t have to worry too much about the frequency response
of the human ear, and wonder whether that’s involved to a large extent;
it’s not. The same thing is true for light--a UV photon carries
more energy than an IR photon. SO! Why is it not true for
an electric signal (remember, we’re forgetting about reactive loads
for the sake of the illustration.) So that’s my question.
Why does frequency affect power for sound and light, but not electric
signals?
Mechanical amplitude doesn’t correspond
to amplitude of V or I as U and I are tempted to assume, but rather
to Q. There’s the correct analogy. Here’s what happens
if you keep going with this analogy:
Compressibility does delay change of
force, just as C delays change of V (and each by 90 degrees). Mass does
delay change of velocity, just as L delays change of I (and each by
90 degrees). Larger masses do have more kinetic energy capacity (due
to the speed of light limitation), just as larger L can store more energy.
Deeper pistons can store more potential energy, just as larger C can
store more energy. Power really is force times velocity, just
as V * I. Distance per second really is velocity, just as charge
per second is current. Also, F = M * A means Vl(t) = L*di/dt.
Which is true. Ohm’s law makes sense, too. If you require
greater force to maintain a certain velocity than your neighbor, it’s
because you’re met with linearly greater friction, which will produce
linearly greater heat.
The ideal L-C oscillator is analogous
to a swinging pendulum (lossless ball bearings, in a vacuum, etc.)
The swinging pendulum slides energy back and forth between kinetic (L)
and potential (C) at a resonance determined by the mass (L) and length/weight
(C). Potential energy plus kinetic energy is constant, just as
energy in L plus energy in C is constant. The same analogy is
the [mass-less] piston that compresses a [mass-less] gas in a [mass-less]
cylinder connected to, well, a mass. [And nothing involved may
conduct heat away from the compressed gas.] If you supply
energy in the form of force over a distance, the piston will pump forever
at a resonance, trading potential energy with kinetic, just like the
pendulum, and just like the L-C oscillator. If you change the
weight of the pendulum without changing its mass, by taking it to the
moon, then you’ve increased C (length/weight) which will reduce the resonant frequency.
If you change the mass of the pendulum without changing its weight,
by figuring out exactly how much mass to add such that it weighs the
same on the moon as it originally did on earth, and then you move it
to the moon... well then you’ve increased L which will reduce
the resonant frequency.
Please take a mass in your hand and move
it to and fro, pretending to be a linear motor. Make some funny
sound-effects. Notice that you have to start pulling back with
your arm BEFORE the mass reaches its peak position. Unless there
is loss, you have to start pulling back 90 degrees before peak.
This is a braking phase during which the mass gives back to you
100% of the energy you gave it (which you are free to loose on your
own.) If there is loss (a load which does not return energy),
the angle is somewhere between 0 and 90 degrees, and this angle (cosine
of) determines how much energy goes into the load vs. how much is reflected
back at you. If you are bad at reclaiming reflected energy, as
likely you are, you can add something spongy, or springy, such as a
gas-compressing piston. You can tune either the frequency or
the piston size (or gas density?) to balance with the mass, in order
to reach resonance, where all the energy bounced from L is reclaimed
by C, which bounces it back at L, and leaves you to worry about more
important things like what’s for dinner. At resonance, there
is no reflected power, so it doesn’t matter if you’re prone to loosing
it. Resonance eliminates the braking phase.
All the delays by 90 degrees we’ve
talked about so far are for pure sine waves. What if it’s not
a pure sine wave? Anything that’s not a sine wave is several
sine waves of various frequencies and phases. So for example,
mass on its own delays change of velocity by exactly 90 degrees for
Now, please exert a sinusoidally oscillating
force on a mass, holding the amplitude of the force-wave constant as
we change frequency and mass. Low frequencies will result in larger
velocities than high frequencies because the mass has more time to accelerate
before the force reverses direction. Small masses will also result
in larger velocities than larger masses. This is identical to
the low-pass filtering effect of an inductor. The longer a voltage
is applied before reversing direction, the greater current is allowed
to flow through the inductor. And because large inductors take
longer to charge, smaller inductors will sooner conduct more current.
So, small inductors pass higher frequencies just as small masses translate
force to velocity faster. And a small mass cannot be charged with
as much kinetic energy (either thought of in absolute terms due to the
speed of light, or thought of at the same square velocity) as a larger
one.
This is a bigger challenge for your imagination
I’m afraid, but give it a try. Please imagine a mass-less cylinder
of mass-less gas, compressible by a mass-less piston. (Or a mass-less
spring if that’s easier.) No mass, and no friction, and nothing
involved may conduct heat away from the compressed gas, etc.
This is a capacitor. When starting at a zero-energy state, any
force at all on the piston immediately results in an enormous velocity
(because there’s no mass.) And the counter-force is zero.
But only for as long as you haven’t moved yet (which at the speed
of light isn’t very long, as it turns out.) Half of a split
instant later, a counter-force begins to grow, and the velocity decreases.
The longer you exert your constant force, the closer to unity the counter-force
becomes, and the closer to zero the velocity becomes. And neither
ever reaches its target--you have an eternity to exert this constant
force before reaching full-charge. Therefore, high frequencies
of the same sinusoidal force-wave are met with less counter-force and
yield greater velocities than lower frequencies. This is exactly
the behavior of a capacitor.
Light’s produced when an electron gets
all excited and hops up and down through energy shells. The movement
of the charge of the electron could be thought of as current, and it
induces a magnetic field which spends itself into an electric field
which then spends itself into a magnetic field, and so on. That’s
the wave perspective; the particle perspective (which to me is more
mystical) is that the energy gained from the fall is emitted as a photon. The L-C resonator slides energy back
and forth between an electric field in C and a magnetic field in L.
An electromagnetic wave also slides energy back and forth between an
electric field and an [orthogonal] magnetic field. When an electron hops through shells,
the reason it takes and gives energy is because it's changing [orbit]
altitudes within an electric field set up by the protons of the atom.
An electron gains energy when moving in, and spends energy to move out.
Like everything else, an electron seeks the lowest energy state to which
it is welcome. If you pull an electron out by a few shells, it
takes more and more energy the further you pull, and when you let go,
it bounces back in with a vigor that’s greater from higher altitudes,
and sounds just like this:
So the amount of energy lost or gained during the shell-hop depends
on the strength of the electric field (number of protons), and on the
change of potential (distance jumped). It takes more energy to jump
up by 6 inches on earth than it does on the moon. If an electron
moves through one volt of potential, the energy spent (or gained, depending
on direction) is called an electron-volt. This is a specific [and very
small] amount of energy. |