1.) What happens when i daisy chain inputs?
or channel jump the channels from different
amp designs?
a.) Does the impedance get mismatched?
because the two amps have different
circuits and are different designs?
b.) Does it change the Frequency Response?
c.) Does it load down the inputs?

example: guitar goes in fender normal input#1
jack.normal input#2 jack goes to a
Orange amp input jack#1 with a daisy
chain cord.
input jack#2 of orange amp another
daisy chain cord goes to a Matchless amp
input jack#1

2.) The Signal parallels off to the other amp but
is there any REFLECTED load current or voltage
when channel jumping?


3.) Does anyone know any cool Channel jumping
path ways?

1.) Y cord impedance going into 2 amps channels?

2.) Daisy Chain inputs going into 2 amp channels impedance?

3.) Im looking at it in a impedance way Y cord VS Daisy Chaining
input channels

• http://www.customaudioelectronics.com/frequently_asked_questions.htm

Daisy chain uses both jacks using both 68K resistors
example- guitar goes into 68k input resistor hi input jack
The Low input jack 68K input resistor is daisy chained
to the other amps channel input so there is another
68K

So Daisy chain uses 3 input 68K resistors so the impendances
is different because there is 3 68Ks resistors in parallel

The Y cord just uses 2 input 68K resistors i don't know if its in
parallel i know the signal from the guitar is parallel to the
amp channals
example: guitar output gets Y cord the Y cord splits to Two
different different amps channels one being a marshall
and other amp being a fender amp. So there is only
2 68K resistors and not in parallel i think

So the Y cord uses 2 68k resistors

The Daisy chain uses 3 68K resistors in parallel or more if you
have more and more amps

So the impedances are way diffeerent from using a Y cord than
Dasiy chain

what do you think im guessing here

a) 2 amps, both input 1 with Rin=1Meg with "Y" cord - guitar "see" the two 1Meg loads in parallel, or 1Meg/2=500k. Three amps - 1Meg/3=333k.

b) 2 amps, "daisy chaining" - guitar in input 1 (Rin=1Meg), then out through input 2 to the second amp and its input 1 again (Rin=1Meg). BUT, we now have two 68k resistors in series between inputs 1&2 on the first amp, or in another words, guitar "see" 1Meg in parallel, then 2*68k=136k in series, then 1Meg in parallel. The resultant resistance is now (1000k+136k)//1000k = 531,8 kOhms. Not much of the difference, but if we connect the third amp on the same way (another 2*68k in series, and 1Meg in parallel) then resultant resistance is (531,8+136)//1000k = 400k, larger then "triple" Y-cord result of 333k. But, first amp get full signal voltage, and other two little attenuated signal through 136k/1Meg combinations.
Normally, this is correct if the inputs are wired at the same way, you can try it with ohmmeter, amps switched off, cables installed.

-All in all, you can try both solutions, but personally I don`t like either of them, as I said, our tiny voltage/high impedance source (guitar PUs) probably needs better solution, see splitter, etc, theory.