Steady State Error Derivation, formulas, and its jaw-dropping examples

Hello everyone, I hope that you all are doing good in your lives. As we all know that control system is in wide use in our day-to-day lives; for example in air conditioners, tanks, and automatic irons, etc. However, it is also applicable in industries as a transportation system, quality checker, power system, and much more. So one of the important factors of the control system is a steady-state error which we are going to focus on today. Here we are going to include topics like derivation, formulas, and examples on steady-state error. Besides if you want to have a review of the control system what it is basically then go through the link below.

Control System (Features, Types, block diagram, advantages and disadvantages)

What is a Steady State Error?

Steady-state error (ess) is defined as the difference or say the variance between the input and the output response of the system within the time interval it goes to infinity i.e the steady-state level. However, the error depends on the type of signal at the input. Besides the calculation of steady-state error is only possible for a system that is at a stable state level.

Steady-State Error of a Unity feedback system

So consider a system with negative unity feedback i.e. the value of feedback in the path is 1. The signal flow diagram of the system is;

Steady state error of unit feedback

    \[\frac {C(s)} {R(s)} = \frac {G(s)} {1 + G(s)} \end.\]

    \[C(s) = \frac {R(s)G(s)} {1 + G(s)} \end.\]

Thus the o/p of the summing point is;

    \[E(s) = R(s) - C(s)\end.\]

A steady-state error of system with feedback

The system below is an example of a system with negative feedback.ess with fedback system

So we have the value of the output summing point which we are going to use for the derivation of steady-state error of system with a feedback element.

    \[E(s) = R(s) - B(s)\end.\]

However, the value of B(s) = C(s).H(s) and C(s) = E(s).G(s)

    \[\therefore E(s) = R(s) - E(s). G(s). H(s)\end .\]

    \[\therefore E(s) + E(s) .G(s). H(s)  = R(s) \end .\]

    \[\therefore E(s) [1 + G(s). H(s) ]  = R(s) \end.\]

    \[\therefore E(s) = \frac {R(s)} {1 + G(s). H(s)}  \end .\]

Whereas for unity feedback system the error is;

    \[\therefore E(s) = \frac {R(s)} {1 + G(s)}  \end.\]

    \[\therefore steady~ state~ error~ e_s_s = \lim_{t \rightarrow \infty}e(t) \end.\]

Static Error Coefficient

However it is defined as the ability of the system to reduce the static error coefficient of a particular system. There are three main types of static error coefficients that we are about to discuss in the further part of the blog.

Position error coefficient

However, consider a system with a open loop transfer system and step signal as an input response.

    \[E_s_s = \lim_{s \rightarrow 0} s. E(s) \end .\]

    \[E_s_s = \lim_{s \rightarrow 0} s \frac {R(s)} {1 + G(s) H(s)}\end .\]

Since here the value of R(s) = A/s;

    \[E_s_s = \lim_{s \rightarrow 0} s \frac {\frac {A} {s} } {1 + G(s) H(s)}\end .\]

    \[\therefore E_s_s = \lim_{s \rightarrow 0} \frac {A } {1 + G(s) H(s)}\end.\]

    \[\therefore E_s_s = \frac {A } { \lim_{s \rightarrow 0} 1 + G(s) H(s)}\end .\]

However for such a system within a limit G(s)H(s) is constant and called as positional error coefficient (Kp) of the system.

    \[\since K_p = \lim_{s \rightarrow 0} G(s)H(s)\end.\]

    \[\therefore E_s_s = \frac {A } { 1 + K_p}\end.\]

Velocity Error Coefficient

However, consider a system with a open loop transfer system and ramp signal as an input response.

    \[E_s_s = \lim_{s \rightarrow 0} s.E(s) \end.\]

    \[E_s_s = \lim_{s \rightarrow 0} s \frac {R(s)} {1 + G(s) H(s)}\end.\]

Since here the value of R(s) = A/s2;

    \[E_s_s = \lim_{s \rightarrow 0} s \frac {\frac {A} {s^2} } {1 + G(s) H(s)} \end.\]

    \[\therefore E_s_s = \lim_{s \rightarrow 0} \frac {A } {s(1 + G(s) H(s) )}\end.\]

    \[\therefore E_s_s = \lim_{s \rightarrow 0} \frac {A } {s + s (G(s) H(s) ) }\end.\]

    \[\therefore E_s_s =\frac {A } { \lim_{s \rightarrow 0} s + s ( G(s) H(s) ) }\end .\]

    \[\therefore E_s_s =\frac {A } { \lim_{s \rightarrow 0} s (G(s) H(s)) }\end.\]

However for such a system within a limit sG(s)H(s) is constant and called as velocity error coefficient (Kv) of the system.

    \[\since K_v = \lim_{x \rightarrow 0} sG(s) H(s) \end.\]

    \[\therefore E_s_s = \frac {A } { K_v}\end.\]

Acceleration Error Coefficient

However, consider a system with a open loop transfer system and parabolic signal as an input response.

    \[E_s_s = \lim_{s \rightarrow 0} s.E(s) \end.\]

    \[E_s_s = \lim_{s \rightarrow 0} s \frac {R(s)} {1 + G(s) H(s)}\end.\]

Since here the value of R(s) = A/s3;

    \[E_s_s = \lim_{s \rightarrow 0} s \frac {\frac {A} {s^3} } {1 + G(s) H(s)} \end .\]

    \[\therefore E_s_s = \lim_{s \rightarrow 0} \frac {A } {s^2(1 + G(s) H(s) )}\end.\]

    \[\therefore E_s_s = \lim_{s \rightarrow 0} \frac {A } {s^2 + s^2 (G(s) H(s) ) }\end.\]

    \[\therefore E_s_s =\frac {A } { \lim_{s \rightarrow 0} s^2 + s^2 ( G(s) H(s) ) }\end .\]

    \[\therefore E_s_s =\frac {A } { \lim_{s \rightarrow 0} s^2 (G(s) H(s)) }\end.\]

However for such a system within a limit s^2G(s)H(s) is constant and called as acceleration error coefficient (Ka) of the system.

    \[\since K_a = \lim_{x \rightarrow 0} s^2G(s) H(s) \end.\]

    \[\therefore E_s_s = \frac {A } { K_a}\end.\]

Solved Example

Q.1 A unity feedback system has a transfer function G(s) = 40(s+2) / s(s+1)(s+4) and thus determines the type of the system, all error coefficients, and also the error for a ramp input signal of magnitude 4.

Solution:-

    \[G(s) = \frac {40(s+2)} {s(s+1)(s+4)}\end.\]

However from the above equation we can conclude that there is only one pole at the origin and thus it is named as Type 1 system.

Secondly calculating the error coefficients;

    \[K_p = \lim_{x \rightarrow 0} G(s) H(s) \end.\]

    \[= \lim_{x \rightarrow 0} \frac {40 (s + 2)} {s (s + 1) (s + 4)} = \infty \end.\]

    \[K_v = \lim_{x \rightarrow 0} s G(s) H(s) \end.\]

    \[= \lim_{x \rightarrow 0} s \frac {40 (s + 2)} {s (s + 1) (s + 4)} = 20 \end.\]

    \[K_a = \lim_{x \rightarrow 0} s^2 G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} s^2 \frac {40 (s + 2)} {s (s + 1) (s + 4)} = \infty \end.\]

And lastly the steady state error for ramp input of magnitude 4 is;

    \[e_s_s = \frac {A} {k_v} = \frac {4} {k_v} = \frac {4} {20} =0.2 \end.\]

Q.2 Find the value of error coefficients and steady-state error for the system with input 3 + t + t2 and open-loop transfer function i.e.

    \[G(s)H(s) = \frac { 10( s + 2 )( s + 3  )}{ s ( s + 1 )( s + 4 )( s + 5)}\end.\]

Solution:- Firstly calculating the values of error coefficients

    \[K_p = \lim_{x \rightarrow 0} G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} \frac {10 (s + 2) ( s + 3 ) }{ s ( s + 1 )( s + 4 )( s + 5)} = \infty \end.\]

    \[K_v = \lim_{x \rightarrow 0} s G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} s \frac {10 (s + 2) ( s + 3 ) }{ s ( s + 1 )( s + 4 )( s + 5)}  = 3 \end.\]

    \[K_a = \lim_{x \rightarrow 0} s^2 G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} s^2 \frac {10 (s + 2) ( s + 3 ) }{ s ( s + 1 )( s + 4 )( s + 5)} = 0 \end.\]

However the given input equation is 3 + t + t2 from which we can get the magnitude of the of the individual type input signals i.e step (A1), ramp (A2), and parabolic (A3).

Thus A1 = 3, A2 = 1 and A3 = 1

And lastly calculating the steady state error from the found magnitude;

    \[For ~ step ~input \end.\]

    \[e_s_s = \frac {A1} {1 + k_p} = \frac {3} {1 + k_p} = \frac {3} {1 + \infty} = 0 \end.\]

    \[For ~ ramp ~input \end.\]

    \[e_s_s = \frac {A2} {k_v} = \frac {1} {k_v} = \frac {1} {3} = 0.34 \end.\]

    \[For ~ parabolic ~input \end.\]

    \[e_s_s = \frac {A3} {k_a} = \frac {1} {k_a} = \frac {1} {0} = \infty \end.\]

Q.3 For the transfer function G(s) = K / s(s + 1)(1 + 0.4s) and r(t) as 4t and k =2 find the value of ess, then assume the value of steady state error to be 0.2 and fin its corresponding alue of k

Solution:- Firstly calculating value of error coefficients with the available transfer function;

    \[K_p = \lim_{x \rightarrow 0} G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} \frac {k }{ s ( s + 1 )( s + 4 )( 1 + 0.4s)}\end.\]

    \[= \lim_{x \rightarrow 0} \frac {2}{s  (1 )( 4 )( 1 )} = \infty \end.\]

    \[K_v = \lim_{x \rightarrow 0} s G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} s \frac {k }{ s ( s + 1 )( s + 4 )( 1 + 0.4s)}\end.\]

    \[ = \lim_{x \rightarrow 0} \frac {2}{ (1 )( 4 )( 1 )} = 2\end.\]

    \[K_a = \lim_{x \rightarrow 0} s^2 G(s) H(s)\end.\]

    \[= \lim_{x \rightarrow 0} s^2 \frac {k }{ s ( s + 1 )( s + 4 )( 1 + 0.4s)}\end.\]

    \[=\lim_{x \rightarrow 0} s\frac {2}{ (1 )( 4 )( 1 )} = 0 \end.\]

from the input signal r(t) = 4t we will get the value of magnitude as 4 thus we can get the steady state error as;

    \[e_s_s = \frac {A} {k_v} = \frac {4} {k_v} = \frac {4} {2} = 2 \end.\]

Secondly we have to calculate the value of k for the ess 0.2

    \[for ~ steady ~ state ~ error ~ K_v = k  \end.\]

    \[e_s_s = \frac {A} {k_v} = \frac {A} {k} = \frac {4} {k}  \end.\]

    \[\therefore 0.2 = \frac {4} {k}\end.\]

    \[\therefore k = 0.2 \end.\]

Conclusion

Here we are at the end of the blog. I hope that you got answers to all your questions and are satisfied with it. If you did like the blog then please do share it with other and also mention below the part which you like the most. Besides if you do have any doubts regarding the topic then please feel free to ask in the comments or contact us through the pages available. And also if any there any suggestions regarding the topic then please mention

Regards;

Have a nice day 🙂

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