I'd like to pick \$k_d\$ using root locus method, but have problems deriving the necessary transfer function of the system presented below. Assume \$k_p\$ is fixed. The question originates from Randal Beard's paper: "Quadrotor dynamics and control", p.42. The answer is actually given there, but for a slightly different block diagram and with no derivation. So it's the derivation that matters to me most.
As far as I understand the method in question, I need obtain an equation:
\$ 1 + k_d P(s) = 0 \$,
but don't know how to derive \$ P(s) \$.
If you're able and willing to help, please don't only provide the solution – I need to know how the solution was obtained to be able to help myself in the future. Any hints appreciated.
EDIT: What I already tried out is simplifying the above block diagram into the form below:
Then we have:
\$ \Large L(s) = \frac{G(s)H(s)}{1 + k_d G(s)H(s)} \$ %Inner loop's transfer function
\$ \Large R(s) = \frac{L(s)/s}{1 + k_p L(s) / s} = \frac{L(s)}{s + k_p L(s)} \$, substituting for \$ L(s) \$ we have:
\$ \Large R(s) = \huge \frac{\frac{G(s)H(s)}{1 + k_d G(s)H(s)}}{s + k_p \frac{G(s)H(s)}{1 + k_d G(s)H(s)}} = \Large \frac{G(s)H(s)}{s(1+k_d G(s)H(s)) + k_p G(s)H(s)}\$.
So the point is, how to convert \$ R(s) \$ denominator into: \$ 1 + k_d P(s) = 0 \$.
Best Answer
OK, I got it. To obtain the necessary transfer function in Evan's form, one has to assume \$ \phi_c = 0 \$. Then, the block diagram in question can be converted into:
So \$ P(s) = \Large \frac{p}{\alpha} = \frac{G(s)H(s)}{1 + \frac{k_p}{s} G(s)H(s)}\$.