System Dynamics and Control: Module 10 - First-Order Systems

Introduction of the canonical first-order system as well as a characterization of its response to a step input.

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[@stanleyohore8024]

Hello, I have a system where the step input is 2v, and the output is 5v and a controller constant k, can we say the dc gain is 2.5?

[@hillrickc]

If the steady-state output of the system is 5 V for a 2-V step, then the DC gain of the system is indeed 2.5. That, however, doesn't necessarily mean the controller constant K is 2.5.

[@michealscofield4991]

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[@kristofeh]

good video. one question about canonical form though, at the end you say it would be incorrect to take the gain from the transfer function of 1/(0.5s + 0.5), but isn't the canonical form 1/s+a rather than 1/s+1 suggesting that you would need to look at another term to find the canonical form

[@hillrickc]

In the video the canonical form is given as K/(tau*s+1) ... so for the transfer function 1/(0.5s+0.5) some students think the DC gain is K = 1 and the time constant is tau = 0.5. This is incorrect. In reality, you need to divide the numerator and denominator by 0.5 to get it into the canonical form, 2/(s+1). Therefore, the true DC gain is K = 2 and the true time constant is tau = 1.

[@kirillolkhovsky9160]

how can i get a TF if my input is a ramp input and i only have given th e graph?

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[@nikks135]

Hello sir, i have watched maximum all of your videos, i was wondering that can you upload a video regarding "how to convert non linear equation to linear equation."

[@ahmedmessoud5410]

can you help me? how do you determine the value of tau

[@MyQwertyuioplk]

Is it possible to access the presentation sir

[@hillrickc]

Try this link: www.dropbox.com/s/u7k7cf24md4zx1r/Slides.zip?dl=0

[@mirzatalha5076]

what is significance of initial slope

[@hillrickc]

I am not sure what you mean with your question, but you can use the initial slope of a first-order step response to estimate the time constant. If you look at the figure shown around 23 minutes, it shows that the initial slope equals 1/time constant (if you extend the line, it crosses the steady-state value at time t = time constant). Alternatively, I tend to look at the time it takes to reach 63.2% of the total change since I think it is less sensitive to error.

[@cogitateandabet]

20:46