[sdiy] Re: crystal clear

[Czech Martin]

don't forget that the average power spectra of 
classical music look different from
electric instruments music, and certainly different
for "academic" electronic music, where perhaps
sine waves bounce arround, causing unusually high
power at 1kHz and above.


Some PA systems can not really deal with sine
wave compositions, because of power and
distortion, well , at least those that I had
for my stuff.

m.c.


-----Original Message-----
From: Magnus Danielson [mailto:cfmd at swipnet.se]
Sent: Montag, 12. Mai 2003 17:30
To: debus at cityweb.de
Cc: synth-diy at dropmix.xs4all.nl
Subject: Re: [sdiy] Re: crystal clear


From: Ingo Debus <debus at cityweb.de>
Subject: Re: [sdiy] Re: crystal clear
Date: Mon, 12 May 2003 16:07:05 +0200

> Don Tillman wrote:
>  > Imagine a simple biamping system:
>  >   Low amp: 100 watts
>  >   High amp: 50 watts
>  >
>  > The obvious single amp version of this would suggest that a full range
>  > amp equivalent would be 150 Watts.
>  >
>  > But because the maximum amplifier voltage is proportional to the
>  > square root of the maximum amplifier power, the full range amp's power
>  > rating will need to be able to handle the sum of those voltages:
>  >
>  >   Full range amp: (sqrt(100)) + sqrt(50))^2 => 291 watts
>  >
>  > This is somewhat bogus reasoning because you will only use those
>  > 291 watts during occasional peaks.  But it does mean that biamping
>  > allows you to handle more peak power than you'd expect.
> 
> I don't quite understand why you can add the voltages here, while when
> stacking two full range systems you add the powers.

OK. Consider that we run a 2-way system and we toss two signals, one for each
of the drivers and both at maximum power. Let's say we toss a 100 Hz sine
signal for the bas at 100 W and a 1 kHz sine signal for the top driver at 50 W
(50 W at 1 kHz is a bit high, but right now it's just a picked frequency and
it doesn't really affect the line of argumentation that it's a few octaves of).

Now, if we assume that both speakers have a 8 Ohm pure resistive impedance at
their frequencies (i.e. 100 Hz for the bas, 1 kHz for the top) and that the
passive cross-over is ideal such that it present 8 Ohm pure resistive impedance
at both these frequencnies, have no leakage of 1 kHz on the bas output and no
leakage of 100 Hz on the top output and in addition is totally free of loss,
then we have very simple formulas.

The power of the bas and top element becomes:

                2
               U
                b
P  = 100 W = -----
 b           Re(I)

               2
              U
               t
P  = 50 W = -----
 t          Re(I)

Giving the RMS power of:
       _________           _
U  = \/ 100 * 8 | = 20 * \/2 V
 b
       ________
U  = \/ 50 * 8 | = 20 V
 t

Giving the peak voltages of

^          _
u = U  * \/2 = 40 V
 b   b

^          _          _
u = U  * \/2 = 20 * \/2 V
 t   t

Right?

Now, since we have a linear system we know that the superposition rule holds,
so both signals just sum up linearly (since we also do not experience any
interference which would cause a more complex vector summation). So, for the
amp we get

^   ^    ^
u = u  + u
 a   b    t

For an amp able to handle the impedance, we clip in voltage just before we clip
in current, so it suffice to say we clip in voltage, so the voltage limit would
have us at the peak voltage of both signals. However, the amp rating is for a
single sine at about the clip, so the power-rating of the amp (which is a
different thing from what the amp actually delivers) then becomes:

     ^     ^    ^
     u     u  + u
      a     b    t
U  = --- = ------- = U  + U
 a     _       _      b    t
     \/2     \/2

        2     2            2
       U     U  + 2U U  + U
        a     b     b t    t
P  = ----- = ---------------
 a   Re(I)        Re(I)

Now, if we compare this with putting amplifiers for each of the bas and top
speaker elements, we have a need for an excess rating of

               2U U
                 b t
P  - P  - P  = -----
 a    b    t   Re(I)

This is what keeps messing with your mind. A single sine and a dual sine isn't
the same signal and some assumptions comes to shame.

I don't think I need to mention the situation for a three-way or four-way
system, since they are just extrapolition on the same line of arguments. So
even when assuming an ideal situation in terms of losses and impedances, you've
lost dynamics in your system just for running it passively.

OK, now somebody got to be satisfied since I did bring the formulas alongside
my argumentations! ;O)

Especially in a live system is 2-3 dB of extra dynamics worth alot. Ever looked
at professional amp-prizes?

Cheers,
Magnus