analysis of PD controller vs static-gain

Question

I have a plant $G(s)$ which gives position and velocity as output $$ G = [G_{ru}(s)\quad G_{vu}(s)]^T $$ where $G_{ru}$ is the integral of $G_{vu}$, and $u$ is the input to the system. Now if I want to control the system with a proportional-derivative law I can write

$$ u = C e_r $$, with $$C = (k_p +s k_d)$$ where $e_r$ is the error with respect to the reference position (e.g., a step command),

I can build the sensitivity and complementary transfer functions as

$$ S = (1+G_{ru}C)^{-1} $$

$$ T = G_{ru}C(1+G_{ru}C)^{-1} $$

When I instead consider the system as having 2 outputs,

and I define a new controller as

$$C_{2} = [k_p\quad k_d]$$,

and the corresponding sensitivities as $$S_2 = (I_2 + GC_2)^{-1}$$ $$T_2 = GC_2(I_2 + GC_2)^{-1}$$

we can see that $S_2$ and $T_2$ are now $2 \times 2$ transfer matrices.

The question is: why $T_2(1,1)$ differs from $T$ since they have the same controller and represent the same thing (in this case the how the position behaves given a step command)?

The corresponding bode plots are depicted here below. Only the dc gain is the same, but the transient is quite different. Why does this happen?

Answer 1

Both the (negated) output and reference are needed to calculate the error. So in the scalar case the reference also gets multiplied by $s\,k_d$, while $T_2(1,1)$ only considers the contribution of $r$ to $y_1$. In order to get the correct results you should also add the contribution of $s\,r$, which could be expressed as $s\,T_2(1,2)$ or $T_2(2,2)$.