Question

(19) For the unity feedback system whose transfer funcetion is shown below, deter- 19.) For the unity feedback system whose transfer function is shown below, deter- mine the steady-state error for a: . Step input, rt) u(t) Ramp input, r(t) tul) 250 s(s 2Xs +5)

Answer 1

Given the open loop transfer function is

Given the open loop transfer function is G (s) = 250/s (s + 2) (s + 5) The block diagram of the unity feedback will as shown below from the above block diagram, the error will be given by E (s) = R (s) - Y (S) rightarrow E (s) = R (s) - G (s) E (s) rightarrow E (s) + G 9s) E (s) = R (s) rightarrow e (s) = R (s)/1 + g S() rightarrow = 1/ 1 + 250/s(s + 2)(s + 5) R (s) rightarrow E (s) = s(s + 2) (s + 5)/s(s + 2) (s + 5) + 250 R (s) According to final value theorem, the final value of any function can be given by lim _ t rightarrow infinity e(t)

The block diagram of the unity feedback system will be as shown below.

from the above block diagram, the error will be given by

$E(s)=R(s)-Y(s)$

According to final value theorem, the final value of any function can be given by

Using the above theorem, the steady-state error is given by

So, the steady state error is given by

(a) Step Input

Now the Laplace transform of the input is

The steady state error is given by

(b) Ramp input

The Laplace transform of the input is

$R(s)=rac{1}{s^2}$

The steady state error is