Variation of Poles: Response Time & Sampling Rate

[Ioannis86]

Homework Statement

Hello, I would like your comments on my observations. I have a set of data from a step experiment, one input (volts D.C), one output (temperature). By doing system identification, the resultant model is second order and its poles lie at 0.99 and 0.8 (z-plane)

Now for the set of data, the sampling rate is 5 seconds. What I did is I took the same set of data and made a new one by appending one sample per 10 samples (50 seconds sampling rate). The parameters of the new model give me the poles at 0.92 and 0.65.

In a closed loop system I know that the closer a pole is to zero, it indicates a faster response. Does that also applies for my open loop transfer function?
Physically, from the data, I can understand that by using 50sec sampling rate there is a bigger temperature change at every sample as compared to 5 sec rate.

 

[lippyka]

So does that mean that, by increasing the sampling rate, the system is more output sensitive?Homework Equations Poles of a transfer function indicate the response time of the system. A pole closer to the origin indicates a fast response.The Attempt at a SolutionYes, the poles of an open loop transfer function indicate the response speed of the system. A pole closer to the origin indicates a faster response. Therefore, the poles of your second model at 0.92 and 0.65 are closer to the origin and would indicate a faster response than the first model at 0.99 and 0.8.Increasing the sampling rate does make the system more output sensitive. This is because the temperature change is greater at each sample, which means the system is able to adjust to changes in the environment more quickly.