# analysis of PD controller vs static-gain

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?

## 1 Answer

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 $$T_2(2,2)$$.

• thanks. This was indeed causing the difference. Jan 17, 2020 at 5:32