+xHi Dave,
I designed a timing task in which the tones beeps three times in sequence. Participants are to click the screen at any time pint after the offset of the 2nd tone. The trial ends after one click.
However, when testing the syntax, the trial did not end right after one click, especially when the click time coincides with (somewhere at) the offset of the 3rd tone.
I wish to know how to fix this problem.
I drew a graph to explain the trial sequence.
Thank you in advance.
Best Wishes.
If I'm seeing this correctly, the trial playing the 3rd sound is extremely short, almost certainly too short to register a response by itself
<trial sound5>
/ stimulustimes = [0=sound.tone]
/ ontrialend = [trial.sound5.resetstimulusframes()]
/ trialduration = 50/ responsetrial = (noresponse,tap2)
/ validresponse = (mousemove,noresponse,lbuttondown,screen)
/ responsetrial = (lbuttondown, start2)
/ inputdevice = mouse
</trial>
and if a participant tries to respond at around the time that trial ends, there is a chance neither <trial sound5> nor <trial tap2> will register that -- the latter, because it's not ready to accept responses yet. What I don't understand is why there is a <trial tap2> at all. Why don't you have <trial sound5> play back the sound and listen for a response indefinitely, i.e. something like
<trial sound5>
/ stimulustimes = [0=sound.tone]
/ ontrialend = [trial.sound5.resetstimulusframes()]
/ validresponse = (lbuttondown,screen)
/ responsetrial = (lbuttondown, start2)
/ inputdevice = mouse
</trial>
?