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评估频率位移探测器在听到复音分音中的部分音的能力中可能起的作用。

Assessing the possible role of frequency-shift detectors in the ability to hear out partials in complex tones.

机构信息

Department of Experimental Psychology, University of Cambridge, Cambridge, UK.

出版信息

Adv Exp Med Biol. 2013;787:127-35. doi: 10.1007/978-1-4614-1590-9_15.

Abstract

The possible role of frequency-shift detectors (FSDs) was assessed for a task measuring the ability to hear out individual "inner" partials in a chord with seven partials uniformly spaced on the ERBN-number (Cam) scale. In each of the two intervals in a trial, a pure-tone probe was followed by a chord. In one randomly selected interval, the frequency of the probe was the same as that of a partial in the chord. In the other interval, the probe was mistuned upwards or downwards from the "target" partial. The task was to indicate the interval in which the probe coincided with the target. In the "symmetric" condition, the frequency of the mistuned probe was midway in Cams between that of two partials in the chord. This should have led to approximately symmetric activation of the up-FSDs and down-FSDs, such that differential activation provided a minimal cue. In the "asymmetric" condition, the mistuned probe was much closer in frequency to one partial in the chord than to the next closest partial. This should have led to differential activation of the up-FSDs and down-FSDs, providing a strong discrimination cue. Performance was predicted to be better in the asymmetric than in the symmetric condition. The results were consistent with this prediction except when the probe was mistuned above the sixth (second highest) partial in the chord. To explain this, it is argued that activation of FSDs depends both on the size of the frequency shift between successive components and on the pitch strength of each component.

摘要

频率位移探测器(FSD)的可能作用在一项任务中得到了评估,该任务旨在测量在具有七个等距 ERBN 数(Cam)刻度上的内部分音的和弦中听到单个“内”分音的能力。在试验的两个间隔中的每一个中,都有一个纯音探针,然后是一个和弦。在一个随机选择的间隔中,探针的频率与和弦中的一个分音相同。在另一个间隔中,探针从“目标”分音向上或向下失谐。任务是指示探针与目标相吻合的间隔。在“对称”条件下,失谐探针的频率在 Cam 之间位于和弦中的两个分音的频率之间的中间位置。这应该导致向上-FSD 和向下-FSD 的激活大致对称,使得差分激活提供最小的线索。在“不对称”条件下,失谐探针的频率与和弦中的一个分音比与下一个最近的分音更接近。这应该导致向上-FSD 和向下-FSD 的激活差异,提供强烈的辨别线索。预测在不对称条件下的表现要好于对称条件。结果与这一预测一致,除了当探针在和弦中的第六个(第二高)分音以上失谐时。为了解释这一点,有人认为 FSD 的激活不仅取决于连续分量之间的频率偏移大小,还取决于每个分量的音强。

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