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相机帧率与曝光时间之间的关系(待完善)

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Hamamatsu Learning Center: Digital Camera Readout and Frame Rates (fsu.edu)
Frame Rate (fps) = 1 / (Frame Acquisition Time + Frame Read Time)

Frame acquisition time components:

Frame read time components:

Frame Acquisition Time = (TPR × Nclear) + Topen + Texp + Tclose

where T(PR) and N(clear) are the time required to clear the parallel register and number of clear cycles performed, and T(open)T(exp), and T(close) represent the shutter opening delay time, the exposure time, and the shutter closing delay time, respectively.

Frame Read Time = TSR + (Trow × Nrow) + (TSD × Ndiscard) + (tread × Nread)

where T(SR) represents the time required to clear the serial register, T(row) and N(row) are the time required for a parallel row shift and the number of rows in the array, T(SD) and N(discard) are the serial discard time and the number of pixels not being read, t(read) is the serial conversion time per pixel and N(read) represents the number of pixels to be read. This latter number is, at maximum, the total pixel array size, and is reduced in accordance with subarray readout and/or pixel binning operations.

 

由于这些 CCD 设计包括一个感光传感器区域和一个遮光存储区域,用于将帧传输到串行寄存器,因此曝光和读出两个过程可以在时间上重叠。相机系统可以提供可选择的曝光和读出模式,根据这两种操作是按顺序独立执行还是同时执行分为非重叠和重叠模式。无论曝光时间如何,高灵敏度操作通常都需要非重叠操作。

 

非重叠模式:

 

当 CCD 在非重叠模式下运行时,可以指定任何曝光时间并完全完成,并在曝光结束时依次读取。图 4 显示了说明这种曝光读出模式的时序图。在非重叠模式下,对序列中的每一帧重复相同的循环; CCD清除残留电荷,在规定的曝光时间内对电荷进行积分,电荷从感光阵列转移到掩蔽存储阵列(分离存储部分或行间阵列),最后读出电荷。根据具体情况,曝光时间可能比读取时间更短或更长,每帧的总时间为两个间隔的总和(在没有机械快门延迟的情况下),因为清除时间和移位所需的时间到存储阵列的数据非常快,并且在确定帧速率方面并不重要。这种操作模式提供了与传统全帧设备架构相似的性能。时序图(图 4)说明了一系列三个图像帧的曝光和读出时间序列。使用示例中显示的任意时间,10 毫秒 (ms) 曝光和 50 毫秒帧读取时间,获取三帧的总时间为 180 毫秒(3 x 10 ms + 3 x 50 ms)。因此相应的帧速率为每秒 16.7 帧(3 帧/0.180 秒)。

 

 

相机重叠模式(overlap)下的两种曝光情况:

重叠模式用于需要记录动态过程的应用,其中需要连续成像以提供足够的时间分辨率。为了最大限度地增加用于数据收集的时间比例,CCD 会连续运行(100% 占空比)。初始曝光后,数据被转移到帧传输阵列,下一次曝光间隔立即开始,同时读取前一帧。

该序列继续时,帧到帧的时序由曝光时间或帧读取时间决定,取决于哪个更长。因此,最小曝光时间等于帧读取时间。在编程曝光时间小于读出时间的情况下,序列中的第一帧按照编程的确切时间曝光,随后的帧按照读出时间曝光。序列时序实际上由较长持续时间的读出周期控制。图 5 显示了重叠模式操作的时序图,其中编程曝光时间为 10 毫秒,CCD 读出时间为 50 毫秒,如前一个(非重叠模式;图 4)示例中所使用。在此模式下获取 3 帧所需的总时间为 160 毫秒,根据一次 10 毫秒曝光和两次 50 毫秒曝光计算,这三个读取周期重叠,每次 50 毫秒(10 毫秒 + 3 x 50小姐)。在重叠模式下运行会导致三帧序列的时间减少 20 毫秒,相应的帧速率为每秒 18.8 帧(3 帧/0.160 秒)。在这种类型的序列中,第一帧的曝光时间比后续帧短,并且图像强度通常不匹配。

When the programmed exposure time is less than the frame read time, the following general expression can be used to calculate the total time required to capture a specified number of frames (N):

TN = (Tread × N) + Texp

where T(N) is the total time required to capture a sequence of N frames, T(read) is the single-frame read time, and T(exp) represents the programmed exposure time.

A second possible timing variation occurs during operation in overlap mode when exposure time is greater than frame read time, and therefore controls the timing of sequential frames. As a result, each frame in a sequence is exposed for the exact time specified in the system control software, and all images are of equal intensity. As illustrated by the timing diagram for this operation mode (Figure 6), after the initial frame of a sequence is exposed, readout takes place during exposure of the each subsequent frame, and the sequence ends with a final readout-only cycle. The timing diagram illustrated is for three frames exposed for 75 milliseconds each, and a CCD frame read time of 50 milliseconds. The sequence of three frames requires 275 milliseconds, calculated as three exposure intervals and one additional readout cycle (3 x 75 ms + 50 ms), and results in a frame rate of 10.9 frames per second (3 frames/0.275 second).

The following equation is utilized to calculate the sequence capture time for N frames in overlap mode when the programmed exposure time is greater than the frame read time:

TN = (Texp × N) + Tread

where T(N) is the total time required to capture a sequence of N frames, T(exp) represents the programmed exposure time, and T(read) is the single-frame read time.

 

标签:完善,read,frame,相机,毫秒,时间,time,曝光
来源: https://www.cnblogs.com/KobeBLN/p/15023268.html