click once did not end the trial


click once did not end the trial

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nakayama
nakayama
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Hi 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.

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1s three tones(second).zip (16 views, 604.00 KB)
Dave
Dave
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nakayama - Monday, June 25, 2018
Hi 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>

?

nakayama
nakayama
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Dave - Monday, June 25, 2018
nakayama - Monday, June 25, 2018
Hi 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>

?

Thank Dave for your advice. The reason I put <trial tap2>  because I want the response doesn't affect the tone presentation.
the participants hear the interval T1 demarcated by two tones(t1, t2), and click somewhere at (t3) to reproduce the interval by equaling T1 with t3-t2.
because of timing noise, t3-t2 might be smaller or larger than T1, but the response is still recorded.
the third tone will serve as a helping signal to keep tack on T1 and its presentation isn't affected by t3.
which has three outcomes:
if t3 is before the 3rd tone presentation, t3 is recorded and the trial ends.
if t3 is in the 3rd tone presentation, t3 is recorded and the trial ends.
if t3 is after the 3rd tone presentation, t3 is recorded and the trial ends.
i tried to lengthen the tone presentation to 80ms, but issues still occurs. is there a way to put <sound5> and <tap2> together?
Thank youWink

 


Dave
Dave
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nakayama - Monday, June 25, 2018
Dave - Monday, June 25, 2018
nakayama - Monday, June 25, 2018
Hi 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>

?

Thank Dave for your advice. The reason I put <trial tap2>  because I want the response doesn't affect the tone presentation.
the participants hear the interval T1 demarcated by two tones(t1, t2), and click somewhere at (t3) to reproduce the interval by equaling T1 with t3-t2.
because of timing noise, t3-t2 might be smaller or larger than T1, but the response is still recorded.
the third tone will serve as a helping signal to keep tack on T1 and its presentation isn't affected by t3.
which has three outcomes:
if t3 is before the 3rd tone presentation, t3 is recorded and the trial ends.
if t3 is in the 3rd tone presentation, t3 is recorded and the trial ends.
if t3 is after the 3rd tone presentation, t3 is recorded and the trial ends.
i tried to lengthen the tone presentation to 80ms, but issues still occurs. is there a way to put <sound5> and <tap2> together?
Thank youWink

 


> is there a way to put <sound5> and <tap2> together?

That is sort of what I proposed in the previous reply, I thought, i.e. have <trial sound5> last indefinitely and register the response

<trial sound5>
/ stimulustimes = [0=sound.tone]
/ ontrialend = [trial.sound5.resetstimulusframes()]
/ validresponse = (lbuttondown,screen)
/ responsetrial = (lbuttondown, start2)
/ inputdevice = mouse
</trial>

instead of farming that out to a separate <trial> (i.e. <trial tap2>Wink. And if you want to make sure that the tone is played to completion even if the response occurs *while* the tone is still playing, then setting /erase = false in <sound tone> should do the trick.

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