Part 9 - General controller structure

Introduction

In this part we’ll see how we define a general feedback loop.

General controller structure

Assume the general structure for a feedback loop.

However, we now have some noise that is added (in theory the sign can be whatever) to our out signal. Additionally, the noise, $V(s)$ will go through a process, $G_v(s)$.

Along with this noise we have some noise in the feedback loop as well, so-called measurement error. We’ll call the new feedback signal for $Y_m = Y(s) - W(s)$, where $W(s)$ is the extra noise.

Mathematically we will describe this as: $$ \begin{align*} E(s) & = R(s) - Y_m(s) \\ & = R(s) - (Y(s) - W(s)) \\ & = R(s) - Y(s) + W(s) \end{align*} $$

$$ Y(s) = G_v(s)V(s) + F(s)G(s) [R(s) - Y(s) + W(s)] \\ Y(s) = G_v(s)V(s) + F(s)G(s)R(s) - F(s)G(s)Y(s) + F(s)G(s)W(s) \\ Y(s) + F(s)G(s)Y(s) = G_v(s)V(s) + F(s)G(s)R(s) + F(s)G(s)W(s) \\ Y(s)(1 + F(s)G(s)) = G_v(s)V(s) + F(s)G(s)R(s) + F(s)G(s)W(s) \\ Y(s)(1 + L(s)) = G_v(s)V(s) + L(s)R(s) + L(s)W(s) \\ Y(s) = \dfrac{1}{1 + L(s)} G_v(s)V(s) + \dfrac{L(s)}{1 + L(s)} R(s) + \dfrac{L(s)}{1 + L(s)} W(s) \\ $$

Let’s denote $\dfrac{1}{1 + L(s)}$ as $S(s)$, weird notation, but let’s go with it. Let’s denote $\dfrac{L(s)}{1 + L(s)}$ as $T(s)$.

We call $S(s)$ for the sensitivity function. $T(s)$ is called for the complementary sensitivity function.

Meaning that: $$ S(s) + T(s) = 1 \ | \ \text{Not hard to see} $$

We also have a name for $\dfrac{G_v(s)}{1 + L(s)} = G_v(s) S(s)$, noise-sensitivity function.

What is the error in $E(s)$?

The impact from $R(s)$ of the overall error is found by setting $V(s) = 0, W(s) = 0$

If we wanted to know the impact from $V(s)$ we set the others to 0.

$$ \begin{align*} E(s) & = R(s) - Y(s) \\ & = R(s) - T(s)R(s) & = R(s)(1 - T(s)) & = R(s)S(s) \end{align*} $$

If we want $y(t)$ to follow $r(t)$, $S(s)$ needs to be small, which is equivalent to saying $T(s)$ should be “large”.

Let’s do an actual example now.

Example

We have a car system, where $y(t)$ is velocity of the car. $F_d(t)$ is a traction force. $u(t)$ is throttle.

In the $s$-domain we have: $$ Y(s) = \dfrac{5}{1 + 5s} F_d(s) $$

$$ F_d(s) = \dfrac{0.1}{1 + s}U(s) $$

$$ U(s) = \dfrac{(1 + s)F_d(s)}{0.1} $$

The car has a cruise control - this is a feedback loop with a PI-controller. $$ F(s) = K_p \cdot \dfrac{1 + T_i s}{T_i s} $$

We know that the transfer function from throttle $\rarr$ velocity is: $$ G(s) = \dfrac{Y(s)}{U(s)} = \dfrac{\dfrac{5}{1 + 5s} F_d(s)}{\dfrac{(1 + s)F_d(s)}{0.1}} = \dfrac{0.5}{(1 + 5s)(1 + s)} $$

The loop transfer function is therefore: $$ L(s) = F(s)G(s) = K_p \cdot \dfrac{1 + T_i s}{T_i s} \cdot \dfrac{0.5}{(1 + 5s)(1 + s)} $$

If we choose a time constant, such as $T_i = 5$, we get: $$ L(s) = F(s)G(s) = \dfrac{0.1K_p}{s(1 + s)} $$

Let’s study $S(s)$ and $T(s)$ now.

$$ S(s) = \dfrac{1}{1 + L(s)} = \dfrac{1}{1 + \dfrac{0.1K_p}{s(1 + s)}} = \dfrac{s(1 + s)}{s(1 + s) + 0.1K_p} = \dfrac{s^2 + s}{s^2 + s + 0.1K_p} $$

$$ T(s) = \dfrac{L(s)}{1 + L(s)} = \dfrac{\dfrac{0.1K_p}{s(1 + s)}}{1 + \dfrac{0.1K_p}{s(1 + s)}} = \dfrac{0.1K_p}{s(1 + s) + 0.1K_p} = \dfrac{0.1K_p}{s^2 + s + 0.1K_p} $$

We can view this as a second-order system. $$ \omega_n = \sqrt{0.1K_p} $$

$$ \zeta = \dfrac{1}{2 \omega_n} = \dfrac{1}{2 \sqrt{0.1K_p}} $$