inverse signal of a phasor?
Hi,
Im trying to do something that seems quite simple, but its got me stumped. Im trying to invert the phasor signal, so I can dial how much it is inverted. A simple patch beneath explains the problem:
Would be grateful for any help
Cheers
N
Hi
I'm afraid I'm not 100% clear on what you wish to achieve, even from looking at your patch, but here is a simpler way to reverse (I think you mean reverse and not invert) a phasor ramp:
HTH
Brendan
Thanks. Interesting way of doing it. Im trying to adjust the phase of these, so hoping to find a solution where I can bring side B in-line with side A as well
- for messages i would use [line] and not [phasor~] ith [snapshot]
- you generally invert by [* -1.], [+ 1.] or [expr -$f1+1.]
- for the transition/modulation you´re looking for, you could mix the 0. 1. (normal ramp) against 0.5 0.5, and 0.5 0.5 against 1. 0. (inverted ramp) using simple math.
actually, i would do that in a bipolar range (-1 1).
then you can just mix against 0. 0. by multiplying, or just mix against the inverted ramp by adding it to the normal ramp.
normal ramp:
[line 25]
[* 2.]
[- 1.]
inverted ramp:
[line 25]
[* -2.]
[- 1.]
modulation:
just add the inverted ramp times two to the normal ramp
using expression, that should fit all into one object:
normal ramp:
[line 25]
[expr ($f1*2)-1]
inverted ramp:
[line 25]
[expr -(($f1*2)-1)]
with modulation input:
[line 25]
[expr (($f1*2)-1) + (-(($f1*2)-1)*2.)*($f3)]
...where $f3 is the modulation input (=your knob, but with a range of 0.-1.)
everything below is my name and is not required for the patch.
-110
I'm not sure what you mean by "how much it's inverted"; what is the in-between state?
Are you trying to offset the phase of these or just invert them?
If you're trying to offset: