We have an open loop system with an input u , a transfer function G(s) and an output y. We apply the following inputs $$u_i(t)=\sin(ω_it), i=1,2,...6 $$ and get the following responses
These are of the form $$y(t)=Y_1\sin(ω_it+ψ_i)$$ Seeing some of the graphs I though G(s) is a differentiator but this doesn't hold true for every graph. The solution manual gives $$G(s)=\frac{s}{(s+10)^2}$$ Any ideas ?
We have $$G(s)U(s)=Y(s)$$ Taking the Laplace transforms : $$G(s)\frac{ω_i}{s^2+ω_i^2}=Y_1\frac{s\sinψ_i+ω_i\cosψ_i}{s^2+ω_i^2} \\ G(s)ω_i=Y_1(s\sinψ_i+ω_i\cosψ_i)$$ Doesn't seem like the (s+10)^2 denominator will show up.