Quadrature Encoder speed measurement

取消
显示结果 
显示  仅  | 搜索替代 
您的意思是: 
已解决

Quadrature Encoder speed measurement

跳至解决方案
1,372 次查看
dh1
Contributor II

Hello

In the reference manual of the IMXRT1060X, there is a chapter 56.3.5 Speed Measurement Method, which explains how to measure the speed of a quadrature encoder signal.

Is there a code example available, which implements this algorithm?

Thanks and regards,
Daniel

0 项奖励
回复
1 解答
1,344 次查看
_Leo_
NXP TechSupport
NXP TechSupport

Hi,

Thank you so much for your interest in our products and for using our community.

For the Quadrature Decoder (QDC) module, only the SDK examples enc_basic and enc_index_interrupt are available, which only compute position. No examples are available for the Speed Measurement Method.

Sorry for the inconvenience.

Have a nice day!

在原帖中查看解决方案

0 项奖励
回复
4 回复数
1,345 次查看
_Leo_
NXP TechSupport
NXP TechSupport

Hi,

Thank you so much for your interest in our products and for using our community.

For the Quadrature Decoder (QDC) module, only the SDK examples enc_basic and enc_index_interrupt are available, which only compute position. No examples are available for the Speed Measurement Method.

Sorry for the inconvenience.

Have a nice day!

0 项奖励
回复
1,296 次查看
dh1
Contributor II

Hi,

Looking at it again, it seems the "Speed Measurement Method" topic is misleading. It refers to the register POSDPER. But this controller (imx RT106x) does not even have this register?

Or am I overseeing something?

Regards,
Daniel

0 项奖励
回复
1,274 次查看
_Leo_
NXP TechSupport
NXP TechSupport

Apologies. That's correct, the IMXRT1060 does not support this register. It is an typo that I will report to the corresponding department.

In such a case, you will need to create an algorithm using a general purpose timer as the time base connected to the TRIGGER and take a snapshot of the POSD register in order to calculate speed.

On the other hand, you can use IMXRT1160/70 where it is supported.

Wish it helps you.

0 项奖励
回复
1,316 次查看
dh1
Contributor II

Hi,

Thank you for your answer. That is strange, because the description seems so detailed that somebody must have tested this somehow.

Best regards,
Daniel

0 项奖励
回复