System Identification
An Oven
Guided Problem
This is a plot of the time response of a small warming oven.  Room temperature is apparently 25oC.  When the oven is turned full ON you get the response shown. We'll assume that full ON is an input step of 1.0.


We will assume that the output of the system is the temperature rise above room
Here's the data with the steady-state subtracted out.  What conclusions can we draw from the data?
Is the system first order?

Problem

Q1  Is the system first order?

The system is probably not first order.  If the system were first order, then the response would change slope immediately after the application of the step.   Instead, there is a slight delay in the response as it starts to grow.  That's pretty good evidence for a second time
        The system is indeed second order or higher.  Now here's another question.

        Now, we need to investigate the system dynamics.  We have concluded that the system is at least second order, but the later part of this transient response looks like a single time-constant response.  Here are some observations that can be made.

Problem

P1  How much more time does it take for the system to go 63% of the way from 20 to 40?

Enter your answer (in sec.) in the box below, then click the button to submit your answer.

Your grade - on a scale from zero to ten is: 


Continuing on:

 Before we leave, we should note that the actual time constant is really 5 seconds,
 so we'll use that value as necessary.

        Here's a video of the expression: (40 - (40 - Intercept)e-t/t).

  • You can control the intercept.

Problem

P2  What intercept works best?

Enter your answer (in sec.) in the box below, then click the button to submit your answer.

Your grade - on a scale from zero to ten is: 


Enter your answer by clicking the button above.  Then scroll down.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

The correct intercept is -10.  At least that seems best to the author.

        The expression we get for the time response for larger times is:

 g(t) = 40 - 50e-t/t

We can now subtract (40 - 50e-t/t) from the total response.

  • In the graph at the right, the difference between the long term and the actual response looks like a decaying exponential that starts at 10 and is over in a few seconds.
  • In the movie below, we've changed the scales so you can see this better.  The graph shows a fast time constant - the dotted red curve.
  • You can control the time constant.

Problem

P3  What time constant works best?  Here, "works best" means that it best fits the difference between the actual data and the single time constant expression for larger times.

Enter your answer (in sec.) in the box below, then click the button to submit your answer.
Your grade - on a scale from zero to ten is: 

Enter your answer by clicking the button above.  Then scroll down.

































        OK, you've gotten all the data you need to describe the step response.  In the response pictured here, the expression for the response is:

        The Laplace transform of this response is:

 F(s) = 10/(s + 1) - 250/(5s + 1) + 40/s

 and this simplifies to:

 F(s) = 40/[(s + 1)(5s + 1)s]

        Now, if the step response - which is the response to 1/s - is as given above, then the transfer function, G(s), must be:

 G(s) = 40/[(s + 1)(5s + 1)]

and we have solved the problem we set out to solve!"
 

           We can comment on what we have found here.

        We can gain some understanding of this if we examine the response of a
 system with two real poles.