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enter image description here


In the solution they have given like this

enter image description here

Here I did't get how they directly wrote DC gain 1?


I solved like this

\$GM=\frac{1}{\left |G(s)\right||H(s)|}\$ Here H(s)=1

From the plot GM=0.5

So G(s) will be 2 also System is type 2 so DC gain will be \$K_v=2\$

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If it is a type 1 system, it follows that the steady-state error is \$0\$, from which it follows that the DC gain will be \$1\$.

As an example consider \$\frac{3}{s (s+1) (s+2)}\$, whose Nyquist plot is similar to the one you have. The closed-loop system is \$\frac{3}{s (s+1) (s+2)+3}\$. In the limit as \$s\$ approaches 0, it will be \$1\$.

More generally, consider a type 1 system \$\frac{k \ n(s)}{s \ d(s)}\$. The closed loop system is \$\frac{k \ n(s)}{s \ d(s)+k \ n(s)}\$. In the limit as \$s\$ approaches 0, it will simplify to \$\frac{k \ n(s)}{k \ n(s)}\$, which is \$1\$.

The system is type 1, because as \$\omega\$ decreases towards 0 the plot has goes to infinity almost along the imaginary axis. This is characteristic of a type 1 system.

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  • \$\begingroup\$ Whats wrong in my method can you please tell? \$\endgroup\$ – Rohit Oct 26 '17 at 15:31
  • \$\begingroup\$ also how they wrote its a unit step input without giving in the question. Please answer \$\endgroup\$ – Rohit Oct 26 '17 at 15:37
  • \$\begingroup\$ First, it's not a type 2 system. If it was, the Nyquist plot will be along the real axis as \$\omega\$ decreases to zero. Second, the question has to do with dc gains, and you are computing gain margin and the static velocity error constant for some reason. \$\endgroup\$ – Suba Thomas Oct 26 '17 at 15:45
  • \$\begingroup\$ Look up the definition of dc gains (or steady-state gains). The input is assumed to be a unit step input. \$\endgroup\$ – Suba Thomas Oct 26 '17 at 15:46
  • \$\begingroup\$ Sorry I just saw its was typo I know its a type 1 system.because one 180 semicircle is there. \$\endgroup\$ – Rohit Oct 26 '17 at 15:49

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