Effect on delay tail only

[IronMan]

Do I understand it correctly that it is possible to ONLY use an effect on a certain effect (so for example if you only want reverb on the delay but not on the normal signal).
I've read the tutorial on axefxwiki (http://axefxwiki.guitarlogic.org/index.php?title=Tutorials#Effect_on_delay_tail_only) but I don't understand the mechanism.

As you can see in the picture above the normal signal is feeded into the rotary and the rotary has a delay. Wouldn't it be the same as if you routed it without the send / return (just with a separate signal)? Can someone please explain how this works? It would be even better if someone can make a screenshot with the new editor since I don't understand the mechanism at all.

Thanks!

 

[VegaBaby]

The above picture is a little different. What that pic descibes is a feedback loop, where the delays are being fed back through the rotary. In away it's the same what you're asking, just a bit more extreme if you want.

If you want only the delay pass through a reverb, easiest way to do is: you have two rows. One is the amp row, the second for the delay. After the amp/cab you route down into the delay (set to 100% mix) from there (same row) into the reverb (mixed in to taste). Then you connect your reverb out back up into the amp row. And as you'll hear...only the delay has reverb .

Basically you have two ways of routing, series and parallel. What I described is parallel in which you're running the reverb in series to the delay.

 

[chris]

i've never used the send return....

in that setup above, wouldn't the rotary signal be put back into the delay? and then would THAT delayed signal get sent through the loop over and over?

 

[VegaBaby]

i've never used the send return....

in that setup above, wouldn't the rotary signal be put back into the delay? and then would THAT delayed signal get sent through the loop over and over?

Yes. Each repeat would get more of the Rotary. It's a cool way to get tape like delay lines etc...

 

[chris]

i've never used the send return....

in that setup above, wouldn't the rotary signal be put back into the delay? and then would THAT delayed signal get sent through the loop over and over?

Yes. Each repeat would get more of the Rotary. It's a cool way to get tape like delay lines etc...

OH SNAP! I'm trying this when i can! i need that kinda sound...

 

[VegaBaby]

i've never used the send return....

in that setup above, wouldn't the rotary signal be put back into the delay? and then would THAT delayed signal get sent through the loop over and over?

Yes. Each repeat would get more of the Rotary. It's a cool way to get tape like delay lines etc...

OH SNAP! I'm trying this when i can! i need that kinda sound...

Do a search in the Wiki. Back then, Scarr posted quite a few very cool examples for the feedback loop. Be careful with levels though (you'll see...)

 

[IronMan]

Yes. Each repeat would get more of the Rotary. It's a cool way to get tape like delay lines etc...

I'm not sure whether I understand it correctly now. Let me try to explain why the feedback loop will add more and more Rotary to the signal.

If you forget the send / return you will just have a Rotary which has Delay added. So if the delay is send back with the send / return you will get the already processed signal with more Rotary, right? And since it's a loop it will continue to loop and add more and more Rotary? But wouldn't you also get more and more delay this way? Wouldn't this get out of hand quickly? How can you keep it under control?

Just one more thing though. If you connect the Graph Eq to the Delay instead of the Rotary. This way you will not have Rotary on the original signal and only on feedback loops of the delay? Is this correct?

 

[VegaBaby]

Yes. Each repeat would get more of the Rotary. It's a cool way to get tape like delay lines etc...

I'm not sure whether I understand it correctly now. Let me try to explain why the feedback loop will add more and more Rotary to the signal.

If you forget the send / return you will just have a Rotary which has Delay added. So if the delay is send back with the send / return you will get the already processed signal with more Rotary, right? And since it's a loop it will continue to loop and add more and more Rotary? But wouldn't you also get more and more delay this way? Wouldn't this get out of hand quickly? How can you keep it under control?

Just one more thing though. If you connect the Graph Eq to the Delay instead of the Rotary. This way you will not have Rotary on the original signal and only on feedback loops of the delay? Is this correct?

That's a 'yes' on evrything you said / asked.

To control the feedback of the delay, make sure to set feedback in the actual delay block to a minimum and then you control the amount of feedback inside the loop block.

If you go from the GEQ directly to the delay you're only using the rotary as a toneshaper for the delay block. You can add as many as you like and build your own little custom delay that way.

Of course, if you don't use the feedback loop, you could also connect the delay first and then go into the rotary, that way only the delay would get rotary. This would of course have to be in a parallel routing like I descibed above, where you should set delay mix to 100%.