# DC gain from Nyquist Plot

In the solution they have given like this

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 $K_v=2$

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.

• Whats wrong in my method can you please tell? – Rohit Oct 26 '17 at 15:31
• also how they wrote its a unit step input without giving in the question. Please answer – Rohit Oct 26 '17 at 15:37
• 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. – Suba Thomas Oct 26 '17 at 15:45
• Look up the definition of dc gains (or steady-state gains). The input is assumed to be a unit step input. – Suba Thomas Oct 26 '17 at 15:46
• Sorry I just saw its was typo I know its a type 1 system.because one 180 semicircle is there. – Rohit Oct 26 '17 at 15:49