I'm having some trouble with the following question:
Consider a system with transfer function $G(s)$, placed in a negative unity feedback loop. Knowing that the closed-loop system is stable and has a steady-state error of $1$ when we input a step of amplitude 2 (this is $u(t)=2,t\geq0$ and $u(t)=0,t<0$), what is the value of $G(0)$?
I draw the following block diagram for the system:
I got the following expression for $E(s)$: $$E(s)=\frac{1}{1+H(s)}U(s)$$
We know that $e_{ss}=\lim_{t\to\infty}e(t)=\lim_{s\to 0}sE(s) = 1$. Assuming the continuity of $H$, we get: $$\frac{2}{1+H(0)}=1 \implies H(0)=1$$
Now, We get the following expression for the transfer function $G$: $$Y(s)=H(s)E(s)=\underbrace{\frac{H(s)}{1+H(s)}}_{G(s)}U(s)$$
So, assuming the continuity of $G$: $$G(0)=\frac{H(0)}{1+H(0)} = 1/2$$
The problem is, according to my teacher, the correct answer is $G(0)=1$. I don't see what I did wrong. Where is the mistake?