free works, not Pro

Mobile apps
#1

Hi
I’ve spent the last couple of hours trying to get my IC pro to connect. I have cubase 6.5 and win 7 and I tried most of the suggestions in this forum and on other places.

The thing that makes me frustrated is that it connects without any trouble when I use the free version Cubase IC but not with the Pro. The free one works both with WPA and WEP security

Am I correct when I say that if the free version works, there shouldn’t be any network problems? And why can’t Pro connect?

Best regards
Mikael

#2

Hello Mikael,

I can only imagine that you have installed the older version of the remote SKI that only supported Cubase iC and not iC Pro.
Did you download and install the latest version of the remote SKI?
http://www.steinberg.net/en/support/content_and_accessories/steinberg_ski_remote.html

#3

Hi!
I have the new version of SKI. I both tried updating from 1.0 and then I tried to remove it from the computer and re-installed it. Still the Cubase IC connects immediately and in Pro it can’t find my computer. Any more suggestions? I’ve tried to disable all firewalls, changed to WEP security, checked bonjour in msconfig, used the same wifi, authorized my computer in Itunes, held my thumbs and prayed to God , without any result :wink:

/Mikael

#4

Hi,


It seems pretty odd indeed.
What do you mean by not working though?
Do you see your computer on the connection page of Cubase iC Pro?
Or it somehow hangs when attempting to connect afterward?


Cheers!

#5

When I open IC Pro the connection page says “no computer found”. In the free version it just connect without any questions.

/Mikael

#6

Hi Mikael,


Don’t hold it against us but we just want to check that twice.
Could you confirm me the Steinberg Remote component’s version number mentioned in your Device manager?


Cheers

#7

Hi
I understand :slight_smile:
Under Device setup/steinberg SKI Remote it says Version 1.0.3.1791

/Mikael

#8

That’s interesting. If I get time, I’ll try the free version. The paid version is of no use at all. It will not work as documented a number of times already.