Okay, two instrument signals combine, in a mixing resistor. At various times there will be cancellation of the signals. I heard this in the past when plugging two guitars into the same amp; there would be instances of no sound at all and instances of garbled yuk!
Question: How do mixing resistors work? Do mixing resistors really work at all?
silverfox.
Mixing resistors question
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martin manning
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Re: Mixing resistors question
For a full cancellation to happen, it requires the two signals to be of equal amplitude and 180 deg out of phase, as described, -but- if the same situation should occur with two inputs in a large mixing desk with smilar control settings, exactly the same thing would happen.....
In theory, this situation can also occur acoustically in a room, but reflections will usually be so plentiful that phase and amplitude is smeared enough to limit the effect. Weird things can and does happen, though, also acoustically.....
In theory, this situation can also occur acoustically in a room, but reflections will usually be so plentiful that phase and amplitude is smeared enough to limit the effect. Weird things can and does happen, though, also acoustically.....
Last edited by Aurora on Thu Jun 11, 2015 7:54 pm, edited 1 time in total.
Re: Mixing resistors question
In the technical world, you can model any signal source as an ideal voltage source in series with some resistor/capacitor/inductor network. The ideal voltage source has an impedance of zero, meaning it can supply infinite current if asked. The R/C/L network supplies the real-world parts.
A guitar pickup, for instance, looks like an ideal voltage source in series with about 4K to 18K of wire resistance in the coils of the pickup, about 2H to 4H (yes, Henries!) of inductance, and some capacitance which shunts the inductor at higher frequencies. The henries of inductance is the obvious elephant in the room, and the reason we need 1M or more input impedances.
When you stick this source into your amp, you add the 68K mixing resistor network and the 1M or more resistance of the input grid leak resistor. What matters to the input tube is only the voltage across the input resistor, so the pickup resistance/inductance/capacitance is in series with the 68K network as the upper resistor of a voltage divider.
At low frequencies, where the inductive impedance is low, you get mostly the resistors cutting signal: maybe 6K plus the 68K for 74K as an upper resistor and 1M for a lower resistor means you get 1M/(74K +1M) = 0.93, or 93% of your signal getting through. At high frequencies - let's say 5kHz - a 2H inductor has an impedance of Xl = 2*pi*F*L =6.28*5000*2
or 62.8K.
So at 5kHz, the pickup resistance plus inductance looks like 66.8K, the series 68K is added, and the tube grid sees 1M/(66.8K+68K+1M) = 0.88, so you have lost some more signal; you have 88%, not 93%. Still, not a huge loss.
If we add another 68K to ground by plugging into the "low gain" input, now you are seeing a low frequency signal of 68K/(74K+68K) = 0.4788, about half, so your signal at the input grid, where it counts, is half of what it was at low frequencies. At 5kHz, it's 0.335, or 33% of the original signal.
That's how the high and low gain inputs work.
If you plug in two guitars, each guitar has its own pickup impedance in series with its signal, and each one sees its own 68K to the 1M at the input grid.
And now we're finally getting down to your question: each guitar has that "ideal voltage source" back inside the pickup where we can't get at it, in series with the pickup internal impedance. So each guitar loads the other guitar's signal down in this setup. One guitar sees a series impedance of its own R+L for its pickups, a 68K resistor, then as a load, the other guitar's 68K input resistor and the other guitar's R+L pickup impedance.
If the pickups and such are identical, you'd expect the signals to both be dropped by 50%, about, when plugged in, because they load each other.
In practice, you'll notice that I left out worrying about that the guitar's controls were doing to its own internal impedance. I did that because this is already too long and most readers' eyes will be crossed by now.
But in the two-guitar case, each guitar's controls also load the other guitar, and all that keeps them from interacting painfully is the series 68K resistors.
So - what they do is let the signals add, and keep two input signals from loading/interacting as much as they would otherwise.
A guitar pickup, for instance, looks like an ideal voltage source in series with about 4K to 18K of wire resistance in the coils of the pickup, about 2H to 4H (yes, Henries!) of inductance, and some capacitance which shunts the inductor at higher frequencies. The henries of inductance is the obvious elephant in the room, and the reason we need 1M or more input impedances.
When you stick this source into your amp, you add the 68K mixing resistor network and the 1M or more resistance of the input grid leak resistor. What matters to the input tube is only the voltage across the input resistor, so the pickup resistance/inductance/capacitance is in series with the 68K network as the upper resistor of a voltage divider.
At low frequencies, where the inductive impedance is low, you get mostly the resistors cutting signal: maybe 6K plus the 68K for 74K as an upper resistor and 1M for a lower resistor means you get 1M/(74K +1M) = 0.93, or 93% of your signal getting through. At high frequencies - let's say 5kHz - a 2H inductor has an impedance of Xl = 2*pi*F*L =6.28*5000*2
or 62.8K.
So at 5kHz, the pickup resistance plus inductance looks like 66.8K, the series 68K is added, and the tube grid sees 1M/(66.8K+68K+1M) = 0.88, so you have lost some more signal; you have 88%, not 93%. Still, not a huge loss.
If we add another 68K to ground by plugging into the "low gain" input, now you are seeing a low frequency signal of 68K/(74K+68K) = 0.4788, about half, so your signal at the input grid, where it counts, is half of what it was at low frequencies. At 5kHz, it's 0.335, or 33% of the original signal.
That's how the high and low gain inputs work.
If you plug in two guitars, each guitar has its own pickup impedance in series with its signal, and each one sees its own 68K to the 1M at the input grid.
And now we're finally getting down to your question: each guitar has that "ideal voltage source" back inside the pickup where we can't get at it, in series with the pickup internal impedance. So each guitar loads the other guitar's signal down in this setup. One guitar sees a series impedance of its own R+L for its pickups, a 68K resistor, then as a load, the other guitar's 68K input resistor and the other guitar's R+L pickup impedance.
If the pickups and such are identical, you'd expect the signals to both be dropped by 50%, about, when plugged in, because they load each other.
In practice, you'll notice that I left out worrying about that the guitar's controls were doing to its own internal impedance. I did that because this is already too long and most readers' eyes will be crossed by now.
But in the two-guitar case, each guitar's controls also load the other guitar, and all that keeps them from interacting painfully is the series 68K resistors.
So - what they do is let the signals add, and keep two input signals from loading/interacting as much as they would otherwise.
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schaublin65
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Re: Mixing resistors question
Hi,
that seems simple enough!
Thanks R.G.
take care
John
that seems simple enough!
Thanks R.G.
take care
John
I never considered this part of the equation
RG "So each guitar loads the other guitar's signal down in this setup"
This seem most likely what was occurring with the amp. It was a low end tube amp from the mid 60's and more likely, but unknown, the inputs were not properly isolated and the two guitars were interacting more so then at the input stage... At times there were dead spots in the output and you could just tell the guitars were clashing somewhere in the signal chain.
your explanation also clears up the 68k or whatever value, input resistors scheme.
Thanks,
silverfox.
This seem most likely what was occurring with the amp. It was a low end tube amp from the mid 60's and more likely, but unknown, the inputs were not properly isolated and the two guitars were interacting more so then at the input stage... At times there were dead spots in the output and you could just tell the guitars were clashing somewhere in the signal chain.
your explanation also clears up the 68k or whatever value, input resistors scheme.
Thanks,
silverfox.