
2*pi*f*t
Enthusiast
Bi-wiring revisited....
here to see all thess big doin's. But I'm having a little trouble here, John. I might could see
your arguement if it were, say:
But, we don't have two sources as your graph indicates, we have the lower (revised)
one with only a single voltage source (amp.) As in
Now, I learned that cos(a) + cos(b) = 2*cos(a+b)*cos(a-b)
So, for say:
v1(t) = 1*cos(2*pi*100*t) (Woofer Signal = cos(a))
and
v2(t) = 1*cos(2*pi*5000*t) (Tweeter Signal = cos(b))
where f1 = 100Hz and f2 = 5,000Hz, respectively. And V1=V2. Our output signal is
vS= cos(5100*pi*t)&cos(4900*pi*t) and looks like this:
But, I'm sure I'm in error, so I wait for your correction.
You have been after me about not understanding the bi-wire thingy, so I moseyed on overCool...
I will not hold you to your hat offer.
Give me a few minutes to draw up a supporting diagram..for now, here's a graph to consider, it shows the dissipation differences between a mono and a biwire configuration..
![]()
Ok..now with that to see, a little splainin...
The currents and the resistance are normalized, so the peak woof and tweet currents are 1, and the resistance of the wire is also normalized to 1.
The woof current is the blue line, the diss is (Asquared) I<sup>2</sup>R, so as you can see, it peaks at 1, and is always positive (real dissipation cannot be negative). It is your basic sine squared waveform, a direct result of the woof current flowing through a wire by itself. Note that for an ideal load, this is also exactly the same dissipation time profile scaled differently.
The tweet current is the magenta line, called Bsquared. again, I squared R...
With biwiring, the total wire dissipation is of course, the sum of the two, A squared plus B squared, and the load dissipation at the speakers is an exact scale of the wire dissipation.
here to see all thess big doin's. But I'm having a little trouble here, John. I might could see
your arguement if it were, say:

But, we don't have two sources as your graph indicates, we have the lower (revised)
one with only a single voltage source (amp.) As in

Now, I learned that cos(a) + cos(b) = 2*cos(a+b)*cos(a-b)
So, for say:
v1(t) = 1*cos(2*pi*100*t) (Woofer Signal = cos(a))
and
v2(t) = 1*cos(2*pi*5000*t) (Tweeter Signal = cos(b))
where f1 = 100Hz and f2 = 5,000Hz, respectively. And V1=V2. Our output signal is
vS= cos(5100*pi*t)&cos(4900*pi*t) and looks like this:

But, I'm sure I'm in error, so I wait for your correction.
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