# Steady State Error of controlled loop system

I am trying to calculate the steady state error of the following system but unable to do it. I have used MATLAB and calculated the steady state error to be 0.1128 but don't understand the steps that I need to do to calculate this.

Thanks

Here is how you proceed :

1. Reduce the system to either unity feedback system or a Single block representing the closed loop transfer function. In your case, reducing to a single block should be easy.

2. For a unity feedback system , as shown

Then from the diagram

E(s) = R(s) - Y(s) = R(s) - G(s)E(S)


Or

E(s) = R(s) / [ 1 + G(s)]


If you reduce it to a single block, Then

E(s) = R(s) - R(s)G(s) = R(s)[ 1 - G(s)]


If you have non unity feedback system, then

G'(s) = G(s) / [ 1 + G(s)H(s) - G(s)]


will reduce it to the form in figure.

1. For steady state error, you need to specify the input. 3 inputs are used :

• Step with Laplace transform 1/s
• Ramp with Laplace transform 1/s^2
• Parabolic with Laplace transform 2/s^3
2. Substitute the transform of input into the equation for your error obtained in step 2.

3. Use the final value theorem acc to which

E(infinity) = lim s-> 0 [ s.E(s)]

• Thanks for the help. I have followed your steps and did the following: 1/s(1-700s+700/S^4+19.5s^3+99.5s^2+117s+736), then i did E(infinity) = lim s-> 0 [ 1-700/736]= 0.04891. Which is not correct. Dec 13, 2014 at 12:16
• Oh sorry forget to mention K = 0.375. Dec 13, 2014 at 12:49
• I done it again with gain k=0.375 and I get 0.1206 but I should get 0.1128. Dec 13, 2014 at 12:57
• I've found my mistake (and trying to correct it, will update soon). Maybe I should stop answering unless I am 1000 % sure :/ Dec 13, 2014 at 12:59
• I used this source ctms.engin.umich.edu/CTMS/index.php?aux=Extras_Ess and my textbook. If that doesnt solve the problem either, please uncheck the answer so I can remove it. Either way, apologies for incorrect info. Dec 13, 2014 at 13:55