technical discussion of highpass filtering.

[thch 0]

just something i was looking at today.

The basic highpass filter will allow for higher peak voltages then the input has.

basically -- if you put in a 1V peak square wave, you can get peaks of 2V from a highpass filter. this is easy to show with RC circuits such as a simple cap inline with the speaker.

Square wave goes to 1V. cap charges to 1V. square wave goes to -1V, cap is still charged to 1V. voltage across speaker is now -2V. cap will charge to -1V, square wave chages to +1V, speaker sees +2V. note that RMS power is lower, but peak power is higher.

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This brings up two issues:

1.) the peak values can be high, thus higher power amplifiers may be needed to allow for this. basically the peak-power output becomes the metric of choice, but peak power is misrepresented... tragically, some amps may allow for a higher peak output power then 2*RMS for low power signals such as these.

2.) highpass filters focus on differences. this means that the output may need to be able to make very large voltage swings very quickly. possibly the amplifier's slew rate may not be high enough.

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now first let me cite some flaws in my argument and tell you that this is for discussion, and as such should not be accepted as "important". in engineering there are often things that do happen, but are not important. some effect might create distortion at 300khz which would be inaudible and isn't too important.

flaw 1 -- square waves. amps aren't made for square waves. audio bandwidth is only 20-20khz, square waves have harmonics up to daylight frequencies... so a true square wave input shouldn't be assumed.

flaw 2 -- no statistical data. i have not compiled any data as to common songs' characteristics. its possible that natural musical songs do not have properties that emphasize this.

[flakko 1]

:blink:

i really need to read up on this stuff...

[Acidburn 10]

gEEk

most of that went over my head

[///M5 2,833]

just something i was looking at today.

The basic highpass filter will allow for higher peak voltages then the input has.

basically -- if you put in a 1V peak square wave, you can get peaks of 2V from a highpass filter.  this is easy to show with RC circuits such as a simple cap inline with the speaker.

uh, if you want people to comment on your "theory" perhaps stating it mathematically since it is "simple" here would help. show us the circuit and the math that shows 2V....

[thch 0]

(Vin)--------| |-----+----/\/\/\-------------(GND)

this is the best i can do for a circuit. Vout is seen from the middle node to ground and is the voltage across the resistor.

Vin -> 1V, capacitor charges to +1V.

(+1V)-------|+1 0|----+------/\/\/\------(0V)

Vout = 0V.

Vin -> -1V, and at that instant:

(-1V)-------|+1 0|------+-----/\/\/\-------(0V)

KVL: -(-1V) +1V +Vout = 0

Vout = -2V

(+1V)-------|-1 0|------+-----/\/\/\-------(0V)

if the cap charges to -1V, then the input switches to +1V, the output will become:

-(+1V) -1V +Vout=0, thus Vout = +2V

----------------------------------------------------

obviously this example is applicable for square waves of freqeuncy, f < 1/(8 pi R C)

4 R C for 4 time constants to "fully" charge the capacitor, and 2 pi for radian to Hz conversion.

of course there will be overshoot for square waves of higher frequency, but not the doubling shown here.

these high peaks will be short lived, and the output will quikly move to 0V, given by (after the switching) |Vout| = 2 e^(-t/RC)

peak power is thus 4x what is normally possible. (lets assume R=1). RMS power is lowered though:

P = 1/(4RC) Integral(0 to 4RC)(4e^(-2t/RC) dt)

P ~ 1/2

(middle steps for anyone interested)

u = -2t/RC

du = -2/RC dt

dt = -RC/2 du

P = 1/RC INT ( -RC/2 e^u du)

the intgral of e^u du is merely e^u, and e^u is only signifigant at u = 0. at u = -8, the value of e^u is pretty insignifigant, so i'll ignore it.

P is then 1/RC * RC/2

of course for lower frequency square waves you'd have a lower RMS power. because the 1/4RC term gets lower and lower, while the integral pretty much stays as RC/2.

a typical square wave would have an RMS power, and peak power, of 1 for a 1ohm load.

[Tirefryr 1,742]

How does this pertain? Music is AC voltage. .

[ANeonRider 3]

hmm, square waves. Shouldn't be any in any decently tuned system.

[Tirefryr 1,742]

hmm, square waves. Shouldn't be any in any decently tuned system.

Shouldn't be any in any system.

[thch 0]

1.) square waves are AC. you can define a frequency even.

2,) square waves merely show the point the easiest -- they are the worst case scenario. further analysis with square waves is fairly easy.

[Supa_c 1]

I try to stay away from square waves.

They dont sound nice

[thch 0]

you know, its funny. earlier yesterday i said "any atom with more or fewer electrons then protons would be an ion". a friend said "no, if it had more it would not be an ion -- it would be a cation".

"why is it everyday i have to explain basic set theory to someone"

i guess a day has passed.

analogy time:

cations are a subset of ions

square waves are a subset of signals requiring high peak powers.

[Supa_c 1]

I dont understand geek talk.

[Acidburn 10]

you know, its funny.  earlier yesterday i said "any atom with more or fewer electrons then protons would be an ion".  a friend said "no, if it had more it would not be an ion -- it would be a cation".

"why is it everyday i have to explain basic set theory to someone"

i guess a day has passed.

analogy time:

cations are a subset of ions

square waves are a subset of signals requiring high peak powers.

anions and cations are funny

but you know what's funnier?

cat-ions

personally, i cant wait to see ions made of cat