Finding Transfer Functions from Response Graphs

[@trenteg11]

How would you deal with a graph that is a 2nd order response?

[@olayomateoreynaud9956]

great video! How would you do it for a second-order system?

[@NitBeanTheMachine]

Wait - if you're taking the inverse Laplace of Y(s), this would make the output y(t) (output varying with time) which was already given in the initial graph. How are you getting that y(t) is actually a deviation from the steady state value?

[@catherinegodbout-lavoie2533]

K=2 because K = y2-y1/u2-u1 y=process response and u=excitation (impulse here) and y2=2 y1=0 u2 =1 u1-0 so: K= 2-0/1-0 =2

[@ananthurajagopal9854]

What is theta? Is it time delay?

[@VincentStevenson]

Yes, theta is the time delay.

[@ubg4618]

is exp(-t/3) = e^(-t/3) ?

[@VincentStevenson]

Yes

[@ubg4618]

@@VincentStevenson If i am given the input signal as a step function 1/s and the graph of the output signal where it shows that gain is 1. Can I then calculate the tau more precise? And how do i write the equivalent transfer function (k /ts + 1) of the system that produced the input/ouput pair?

[@henrikt1051]

finaly I understand it!

[@soneng9078]

How do you know it is 5?

[@VincentStevenson]

I arbitrarily gave it a value for the sake of having numbers in this example.

[@joedorseyjr]

@allen thompson 1- (1/e^(1)) = .632120558 that's where the value comes from. Which is derived from the equation Y(tau) = Yf - Yf*e^(-t/tau) where t and tau are equal.