Help: unit-step response of a first order system

  • Thread starter Jonny2011
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Homework Statement



A first-order system

G(s)=[tex]\frac{k}{s+a}[/tex]

and its response to a unit step input are shown in the figure below.

173799238.jpg


Determine the system parameters a and K


2. The attempt at a solution

I’ve tried to find K & a using matlab and read a lot but didn’t find the right topic about this problem. I need hints guys to solve .
 
Last edited:

Answers and Replies

  • #2
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[tex]u(t) \Longleftrightarrow \frac{1}{s}[/tex]
[tex] y(t)=x(t)*h(t) \Longleftrightarrow X(s)H(s)[/tex]
[tex] Y(s) = \frac{k}{s(s+a)}=\frac{k}{as}-\frac{k}{a(s+a)}[/tex]
[tex]\frac{k}{as}-\frac{k}{a(s+a)} \Longleftrightarrow \frac{k}{a}(1-e^{-at})[/tex]

Now, it seems you have a tangential line and knowledge for what change in time it takes for that tangent to arrive to the maximal height of your output. So let derive an answer from this given:

We shall call the function for the tangent z(t) to avoid confusion with my previously used y(t). Thus:

[tex] z - z_1 = m(t - t_1) [/tex]
[tex] z_1 = y(t_1) = \frac{k}{a}(1-e^{-at_1})[/tex]
[tex] m = y'(t_1) = ke^{-at_1}[/tex]

Plugging in the last two into the first:

[tex] z - \frac{k}{a}(1-e^{-at_1}) = ke^{-at_1}(t-t_1)[/tex]

We then solve for what new time, t_2, will give the maximal height, which is z = k/a:

[tex] \frac{k}{a} - \frac{k}{a}(1-e^{-at_1}) = ke^{-at_1}(t_2-t_1)[/tex]
[tex] t_2 = \frac{1}{a} + t_1[/tex]

or

[tex] t_2 - t_1 = \Delta t = \frac{1}{a} \rightarrow a = \frac{1}{\Delta t}[/tex]

We then need simply to apply the known maximal height of one to solve for k:

[tex] \frac{k}{a} = 1 \rightarrow k = a = \frac{1}{\Delta t}[/tex]
 
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  • #3
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thank you that was really helpful.

so we have t1=0 & t2=0.1 by applying the a=k=1/t2-t1

we get a=k=10

G(s)=[tex]\frac{10}{s+10}[/tex]

I checked it in matlab and it was correct.
 

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