TI中文支持网在28335的eQEP模块的reference Guide一书中，有个计算转速的公式：
v(k)=x(k)-x(k-1)/T
其说明如下：
Estimation based on Equation 1 has an inherent accuracy limit directly related to the resolution of the
position sensor and the unit time period T. For example, consider a 500-line per revolution quadrature
encoder with a velocity calculation rate of 400 Hz. When used for position the quadrature encoder gives a
four-fold increase in resolution, in this case, 2000 counts per revolution. The minimum rotation that can be
detected is therefore 0.0005 revolutions, which gives a velocity resolution of 12 rpm when sampled at 400
Hz. While this resolution may be satisfactory at moderate or high speeds, e.g. 1% error at 1200 rpm, it
would clearly prove inadequate at low speeds. In fact, at speeds below 12 rpm, the speed estimate would
erroneously be zero much of the time
在这里我没看明白的有两点：
1、最小可检测圈速为0.0005是如何计算的？
2、速度精度12rpm是如何计算的？
求教了
Jay：
1. 说明假设编码器的线数为500，使用正交编码，编码器每转一圈，QEO能捕获的脉冲数就是2000，所以分辨率为1/2000=0.0005
2. 说明假设QEP采样频率为400Hz，那么可以计算到速度精度为0.0005*400*60=12rpm. 这也就是说编码器的旋转速度不能低于12rpm，如果那样速度估算将产生较大误差。
在28335的eQEP模块的reference Guide一书中，有个计算转速的公式：
v(k)=x(k)-x(k-1)/T
其说明如下：
Estimation based on Equation 1 has an inherent accuracy limit directly related to the resolution of the
position sensor and the unit time period T. For example, consider a 500-line per revolution quadrature
encoder with a velocity calculation rate of 400 Hz. When used for position the quadrature encoder gives a
four-fold increase in resolution, in this case, 2000 counts per revolution. The minimum rotation that can be
detected is therefore 0.0005 revolutions, which gives a velocity resolution of 12 rpm when sampled at 400
Hz. While this resolution may be satisfactory at moderate or high speeds, e.g. 1% error at 1200 rpm, it
would clearly prove inadequate at low speeds. In fact, at speeds below 12 rpm, the speed estimate would
erroneously be zero much of the time
在这里我没看明白的有两点：
1、最小可检测圈速为0.0005是如何计算的？
2、速度精度12rpm是如何计算的？
求教了
Bruce：
回复 Jay:
非常感谢 jay shen 的答案，谢谢了，受益匪浅！
在28335的eQEP模块的reference Guide一书中，有个计算转速的公式：
v(k)=x(k)-x(k-1)/T
其说明如下：
Estimation based on Equation 1 has an inherent accuracy limit directly related to the resolution of the
position sensor and the unit time period T. For example, consider a 500-line per revolution quadrature
encoder with a velocity calculation rate of 400 Hz. When used for position the quadrature encoder gives a
four-fold increase in resolution, in this case, 2000 counts per revolution. The minimum rotation that can be
detected is therefore 0.0005 revolutions, which gives a velocity resolution of 12 rpm when sampled at 400
Hz. While this resolution may be satisfactory at moderate or high speeds, e.g. 1% error at 1200 rpm, it
would clearly prove inadequate at low speeds. In fact, at speeds below 12 rpm, the speed estimate would
erroneously be zero much of the time
在这里我没看明白的有两点：
1、最小可检测圈速为0.0005是如何计算的？
2、速度精度12rpm是如何计算的？
求教了
Bruce：
请问一下，速度计算频率（velocity caculation rate）=400HZ是什么意思？
跟500线/转的编码器输出有什么关系？
在28335的eQEP模块的reference Guide一书中，有个计算转速的公式：
v(k)=x(k)-x(k-1)/T
其说明如下：
Estimation based on Equation 1 has an inherent accuracy limit directly related to the resolution of the
position sensor and the unit time period T. For example, consider a 500-line per revolution quadrature
encoder with a velocity calculation rate of 400 Hz. When used for position the quadrature encoder gives a
four-fold increase in resolution, in this case, 2000 counts per revolution. The minimum rotation that can be
detected is therefore 0.0005 revolutions, which gives a velocity resolution of 12 rpm when sampled at 400
Hz. While this resolution may be satisfactory at moderate or high speeds, e.g. 1% error at 1200 rpm, it
would clearly prove inadequate at low speeds. In fact, at speeds below 12 rpm, the speed estimate would
erroneously be zero much of the time
在这里我没看明白的有两点：
1、最小可检测圈速为0.0005是如何计算的？
2、速度精度12rpm是如何计算的？
求教了
Jay：
回复 Bruce:
400Hz是假定的QEP采样频率。
采样频率与编码器线数共同决定了转速精度。
在28335的eQEP模块的reference Guide一书中，有个计算转速的公式：
v(k)=x(k)-x(k-1)/T
其说明如下：
Estimation based on Equation 1 has an inherent accuracy limit directly related to the resolution of the
position sensor and the unit time period T. For example, consider a 500-line per revolution quadrature
encoder with a velocity calculation rate of 400 Hz. When used for position the quadrature encoder gives a
four-fold increase in resolution, in this case, 2000 counts per revolution. The minimum rotation that can be
detected is therefore 0.0005 revolutions, which gives a velocity resolution of 12 rpm when sampled at 400
Hz. While this resolution may be satisfactory at moderate or high speeds, e.g. 1% error at 1200 rpm, it
would clearly prove inadequate at low speeds. In fact, at speeds below 12 rpm, the speed estimate would
erroneously be zero much of the time
在这里我没看明白的有两点：
1、最小可检测圈速为0.0005是如何计算的？
2、速度精度12rpm是如何计算的？
求教了
Bruce：
最小可检测转速= 采样频率x分辨率x60
那采样频率越高，最小可检测转速越大，不合逻辑啊？
求解释……
在28335的eQEP模块的reference Guide一书中，有个计算转速的公式：
v(k)=x(k)-x(k-1)/T
其说明如下：
Estimation based on Equation 1 has an inherent accuracy limit directly related to the resolution of the
position sensor and the unit time period T. For example, consider a 500-line per revolution quadrature
encoder with a velocity calculation rate of 400 Hz. When used for position the quadrature encoder gives a
four-fold increase in resolution, in this case, 2000 counts per revolution. The minimum rotation that can be
detected is therefore 0.0005 revolutions, which gives a velocity resolution of 12 rpm when sampled at 400
Hz. While this resolution may be satisfactory at moderate or high speeds, e.g. 1% error at 1200 rpm, it
would clearly prove inadequate at low speeds. In fact, at speeds below 12 rpm, the speed estimate would
erroneously be zero much of the time
在这里我没看明白的有两点：
1、最小可检测圈速为0.0005是如何计算的？
2、速度精度12rpm是如何计算的？
求教了
Martin Yu：
编码器的线数定了之后，分辨率固定（如1/2000=0.0005），意味着电机转速太低时，一个采样周期内电机没有转过一个编码器刻度距离，即没有发出一个脉冲；如果提高采样频率，即缩短采样周期，则需要提高电机转速才能让编码器发出速度脉冲，因此“那采样频率越高，最小可检测转速越大”。
