First Order Systems

General Form

$$\dot{x}(t)+\frac{1}{\tau }x(t)=Ku(t)$$

$x(t)\to$ state || output, $u(t)\to$ input $K\to$ gain, $\tau \to$ time constant

Varying of $\tau$ changes the speed of the system, and varying the gain of the system changes the magnitude of the output with respect to the input

Laplace:

$$\frac{1}{\tau s+1}$$

Proportional Control:

We have here our controller ($G_1(s)=K_p$), and our plant ($G_2(s)=\frac{1}{\tau s+1}$) This means that:

Our Steady State Response

$$\begin{array}{rl}\frac{C(s)}{R(s)}& =\frac{\frac{K_p}{\tau s+1}}{1+\frac{K_p}{\tau s+1}}=\frac{K_p}{1+\tau s+K_p}\\ \text{Given: }R(s)& =\frac{1}{s}\text{ for step input}\\ C(s)& =\frac{K_p}{s(1+\tau s+K_p)}\\ c_{ss}& =\underset{s\to 0}{\lim }\{sC(s)\}=\underset{s\to 0}{\lim }\{\frac{K_p}{\tau s+1+K_p}\}=\frac{K_p}{1+K_p}\\ c_{ss}& =\frac{K_p}{1+K_p}\end{array}$$

Our Steady State Error

$$\begin{array}{rl}\frac{E(s)}{R(s)}& =\frac{1}{1+\frac{K_p}{\tau s+1}}=\frac{\tau s+1}{1+\tau s+K_p}\\ \text{Given: }R(s)& =\frac{1}{s}\text{ for step input}\\ E(s)& =\frac{\tau s+1}{s(1+\tau s+K_p)}\\ e_{ss}& =\underset{s\to 0}{\lim }\{sE(s)\}=\underset{s\to 0}{\lim }\{\frac{\tau s+1}{\tau s+1+K_p}\}=\frac{1}{1+K_p}\\ e_{ss}& =\frac{1}{1+K_p}\end{array}$$

PD Control:

We have here our controller ($G_1(s)=K_p+K_{d}s$), and our plant ($G_2(s)=\frac{1}{\tau s+1}$)

Once again, if we assume a unity step input, or $R(s)=\frac{1}{s}$:

Our Steady State Response

$$\begin{array}{rl}\frac{C(s)}{R(s)}& =\frac{\frac{K_p+K_{d}s}{\tau s+1}}{1+\frac{K_p+K_{d}s}{\tau s+1}}=\frac{K_p+K_{d}s}{1+(\tau +K_d)s+K_p}\\ \text{Given: }R(s)& =\frac{1}{s}\text{ for step input}\\ C(s)& =\frac{K_p+K_{d}s}{s(1+(\tau +K_d)s+K_p)}\\ c_{ss}& =\underset{s\to 0}{\lim }\{sC(s)\}=\underset{s\to 0}{\lim }\{\frac{K_p+K_{d}s}{1+(\tau +K_d)s+K_p}\}=\frac{K_p}{1+K_p}\\ c_{ss}& =\frac{K_p}{1+K_p}\end{array}$$

Our Steady State Error

$$\begin{array}{rl}\frac{E(s)}{R(s)}& =\frac{1}{1+\frac{K_p+K_{d}s}{\tau s+1}}=\frac{\tau s+1}{1+(\tau +K_d)s+K_p}\\ \text{Given: }R(s)& =\frac{1}{s}\text{ for step input}\\ E(s)& =\frac{\tau s+1}{s(1+(\tau +K_d)s+K_p)}\\ e_{ss}& =\underset{s\to 0}{\lim }\{sE(s)\}=\underset{s\to 0}{\lim }\{\frac{\tau s+1}{1+(\tau +K_d)s+K_p}\}=\frac{1}{1+K_p}\\ e_{ss}& =\frac{1}{1+K_p}\end{array}$$

Closed Loop Time Constant

$${\tau }_{cl}=\frac{\tau +K_d}{K_p+1}$$

Integral Control

We have here our controller ($G_1(s)=\frac{K_i}{s}$), and our plant ($G_2(s)=\frac{1}{\tau s+1}$) So it follows:

$$\begin{array}{rlrl}\frac{G(s)}{R(s)}& =\frac{\frac{K_i}{\tau s^2+s}}{1+\frac{K_i}{\tau s^2+s}}=\frac{K_i}{\tau s^2+s+K_i},& \frac{E(s)}{R(s)}& =\frac{1}{1+\frac{K_i}{\tau s^2+s}}=\frac{s(\tau s+s)}{\tau s^2+s+K_i}\\ & & & \\ \text{Given }& R(s)=\frac{1}{s}\text{:}& & \\ C(s)& =\frac{K_i}{s(\tau s^2+s+K_i)},& E(s)& =\frac{\tau s+s}{\tau s^2+s+K_i}\\ c_{ss}& =\underset{s\to 0}{\lim }\{sC(s)\}=\underset{s\to 0}{\lim }\{\frac{K_i}{\tau s^2+s+K_i}\}& e_{ss}& =\underset{s\to 0}{\lim }\{sE(s)\}=\underset{s\to 0}{\lim }\{\frac{s(\tau s+s)}{\tau s^2+s+K_i}\}\\ c_{ss}& =1& e_{ss}& =0\end{array}$$

Finally expanding this:

$$\begin{array}{rlrl}{\omega }_{n,cl}& =\sqrt{\frac{K_i}{\tau }},& {\xi }_{cl}& =\frac{1}{2\sqrt{\tau K_i}}\end{array}$$

This is interesting as we know to avoid an oscillatory response we must have:

$$\begin{array}{rl}\frac{1}{2\sqrt{\tau K_i}}\ge 1,& \text{ as }\xi \ge 1\text{ for a stable system}\\ \Rightarrow & \sqrt{\tau K_i}\le 0.5\\ \Rightarrow & K_i\le \frac{0.25}{\tau }\end{array}$$

PI Control

We have here our controller ($G_1(s)=K_p+\frac{K_i}{s}$), and our plant ($G_2(s)=\frac{1}{\tau s+1}$)

$$\begin{array}{rlrl}\frac{G(s)}{R(s)}& =\frac{K_{p}s+K_i}{\tau s^2+(K_p+1)s+K_i},& \frac{E(s)}{R(s)}& =\frac{s(\tau s+1)}{\tau s^2+(K_p+1)s+K_i}\\ & & & \\ \text{Given }& R(s)=\frac{1}{s}\text{:}& & \\ C(s)& =\frac{K_{p}s+K_i}{s(\tau s^2+(K_p+1)s+K_i)},& E(s)& =\frac{s(\tau s+1)}{\tau s^2+(K_p+1)s+K_i)}\\ c_{ss}& =\underset{s\to 0}{\lim }\{sC(s)\}=\underset{s\to 0}{\lim }\{\frac{K_{p}s+K_i}{\tau s^2+(K_p+1)s+K_i}\}& e_{ss}& =\underset{s\to 0}{\lim }\{\frac{s(\tau s+1)}{\tau s^2+(K_p+1)s+K_i}\}\\ c_{ss}& =1& e_{ss}& =0\end{array}$$ $$\begin{array}{rlrl}{\omega }_{n,cl}& =\sqrt{\frac{K_i}{\tau }},& {\xi }_{cl}& =\frac{K_p+1}{2\sqrt{\tau K_i}}\end{array}$$

Which means to avoid an oscillatory response:

$$\frac{K_p+1}{2\sqrt{\tau K_i}}\ge 1⟹K_p\ge 2\sqrt{K_i\tau }-1$$

PID Control

$${\omega }_{n,cl}=\sqrt{\frac{K_i}{\tau +K_d}},{\zeta }_{cl}=\frac{K_p+1}{2\sqrt{K_i\left(\tau +K_d\right)}}$$