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Mostly, whenever it is asked to find the DC gain of A system; in the solutions provided, they find the steady state error coefficient (corresponding to the Type of the system). But I figured out that finding the DC gain this way is equivalent to simply finding the open loop DC gain, and not the closed loop DC gain for the system. So is it something understood , that whenever we are asked find the DC gain we must find the DC gain of open loop transfer function?

For example for this question:

Question: A ramp input applied to a unity feedback system results in 5% steady state error. The Type number and zero frequency gain of the system are?

The solution describes the zero frequency gain of the system equal to the Velocity Error Constant, which is equal to 1/0.05=20. Isn't this the DC gain of open loop instead of the complete 'SYSTEM'?

Where am I going wrong?

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    \$\begingroup\$ I must admit that I do not fully understand the question.....because: Whenever I am asked to find the gain (or the DC gain) of a system which has feedback, it is important to add the information if the open-loop or closed-loop gain is relevant for the problem. For a steady-state error at the output, it is of course the closed-loop gain and for all questions related to stability properties it is the loop gain (open-loop) which is important. \$\endgroup\$
    – LvW
    Commented Mar 22, 2020 at 11:50

1 Answer 1

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The question dictates that a Ramp input has a steady state error, therefore it must be a Type 1 (1st order) System. enter image description here I remember this from Uni. in '70's but found on web here.
p.5 on http://www.cs.mun.ca/av/old/teaching/cs/notes/steady_quad.pdf

From Steady state error, \$e_{ss}\$ using the final value theorem as follows;
E(s) is the Laplace transform of the error signal, e(t).

\$e_{ss}=e(t)_{t→∞}=sE(s)_{s→0}\$

Control System errors for unity negative feedback

Step response = \$\dfrac{1}{1+k_p}\$

  • this is a zero frequency response normalized to 1 unit input

Ramp response = \$\dfrac{1}{k_v}\$

  • this is a velocity dependent input rate

The overall steady state error is the sum from each input.

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  • \$\begingroup\$ Thanks for the answer but I am looking for some information on the DC gain... \$\endgroup\$
    – Bhuvnesh
    Commented Nov 24, 2019 at 16:45
  • \$\begingroup\$ No information is given for DC input or step gain error, so no conclusion for Av(0Hz) is possible \$\endgroup\$ Commented Nov 24, 2019 at 16:47
  • \$\begingroup\$ But the answer to DC gain is given as 20. How did they do it then? This question is from the Gate exam! \$\endgroup\$
    – Bhuvnesh
    Commented Nov 24, 2019 at 17:01
  • \$\begingroup\$ I found the proof and assumptions I was making. So the DC gain is the Kv \$\endgroup\$ Commented Nov 24, 2019 at 19:46
  • \$\begingroup\$ Yeah... But doesn't it turn out that Kv is actually open loop gain, and not the closed loop system gain? We must find out the closed loop gain when it is asked about the 'system gain'! \$\endgroup\$
    – Bhuvnesh
    Commented Nov 25, 2019 at 16:35

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