I wouldn’t call an envelope signal AC or DC, because it only triggers when you give it a gate signal. No gate signal = 0V envelope CV! Maybe if you were to trigger one regularily (with a clock etc.), then the wave might be considered AC.
Different instruments use different CV, the most common being 0 to +5V. One exception is the Voyager’s Pan CV, which is -2.5 to +2.5V
I was told by Moogmusic that a voltage over +5V will be current limited, so will not damage the circuits. Be careful though! You can do interesting things by going outside the 0 to +5V window though (see later).
Negative voltages give interesting results! Patch the ENV. OUT of the MF-101 to the 251 to get an inverted envelope signal. You may wish to attenuate the signal, so you can control the envelope amount. Patch the inverted envelope back into CUTOFF on the 101. Now when you trigger the envelope, the cutoff frequency will be lowered according to the amount you set on the attenuator. This is in contrast to the normal operation of the pedal. Remember, if you are lowering the voltage, you must first start with a higher frequency, so turn the CUTOFF knob up so that its voltage may be lowered by the inverted envelope.
You could also try patching the same inverted envelope to MF-104(X) TIME input. Then the repeat rate would increase as the envelope is triggered.
If the normal range (0 to +5V) is exceeded, this will sometimes do nothing, but on LFOs the range of the LFO’s rate is increased. I guess you could call it under or over-clocking. A negative (e.g. -1V) applied to a rate input will decrease the rate accesible (i.e. provide lower frequencies) than the knob allows normally. Similarly, but opposite, when +5V is exceeded.
These are dependent on the knob position and are additive. The inverted envelope lowers the CUTOFF voltage, but not below zero; e.g. CUTOFF normally at +4V, recieves a -1.5V signal, drops to 2.5V. Underclocking an LFO means going below zero Volts. It just depends how the CV destination works.
0 to +5 is not the same as -5 to +5, as the latter represents a change in voltage of 10V, the former 5V. However, a change in voltage from 0 to either +5 or -5V represents a change of 5V, so if the functions with respect to time are the same, they are opposite signals (you would get the other by placing one through an inverter circuit).
These signals actually have the same phase but opposite amplitute. This is because, if they start at the same time and have the same frequency, they peak at the same time. If we added these together, we would get 0V (any number plus its opposite = zero):
---------------------------------- +5 Volts
//////////////
-----------------------------------0 Volts
//////////////
---------------------------------- -5 Volts
Say both signals had the same amplitude (0 to +5V):
///////
---------------0 Volts and at the same time:
///////
---------------0 Volts
These are in phase, but have the same amplitude, so when we add them together, the signal doubles, going from 0 to +10 V (at the peak we have +5 + +5 = 10V)
Now say we have two signals out of phase (by 180 degrees), with the same amplitude:
////////
------------------0 Volts and
////////
------------------0 Volts
Now, the resultant voltage will always be +5V. Look at the signals, when one peaks at +5, the other is 0V and vice versa. Somewhere in the middle, the two will add to give +5V. (This is the principle behind a bridge diode, it converts AC to DC).
And now +5 and -5V DC:
------------------------------- +5 Volts
------------------------------- 0 Volts
------------------------------- -5 Volts
All the way along we have +5 and -5V, so add them together to get zero.
(Two times +5V = +10V, if you had two +5V running together.)
So +5 and -5 are opposite in magnitude, only wave functions (voltage/vibrations etc. changing with time) can be out of phase.
I hope that explains it.
I’m going to go and play the Voyager…