Steady state error and PID controller

See also problem

Problemè 1

In Lab 4, We used a PD compensator to control our ball and beam apparatus. The transfer function of our PD compensator was as follows:

$$G_C(s)=K_{D}s+K_P$$

However, we did not use the compensator in this form. The transfer function we used in lab was as follows:

$$G_C(s)=K_C(s+z)$$

Question

Solve for $K_C$ and $z$ in terms of $K_P$ and $K_D$.

Given

$$\begin{array}{rl}{G}_C(s)& =K_C(s+z)\\ G_C(s)& =K_{D}s+K_P\end{array}$$

Or it can be written as:

$$(K_C-K_D)s+K_{C}z-K_P=0$$

To solve for the characteristic equation, we can set the coefficients of $s$ and the constant term to zero:

$$\begin{array}{rl}{K}_C-K_D& =0\\ K_{C}z-K_P& =0\end{array}$$

Thus, we can solve for $K_C$ and $z$ as follows:

$$\begin{array}{rl}{K}_C& =K_D\\ z& =\frac{K_P}{K_C}=\frac{K_P}{K_D}\end{array}$$

Problemè 2

Given that the transfer function of our Ball and Beam plant used in the previous lab is as follows:

$$G(s)=\frac{0.419}{s^2}$$

And given that the controller is applied to the plant in cascade configuration, find:

2.a

Static error constant for position (position constant)

This is a Type-2 system, thus position constant $K_p=\infty$

2.b

Static error constant for velocity (velocity constant)

Velocity constant $K_v={\lim }_{s\to 0}sG(s)={\lim }_{s\to 0}\frac{0.419s}{s^2}=\infty$

2.c

Static error constant for acceleration (acceleration constant)

Acceleration constant $K_a={\lim }_{s\to 0}{s}^{2}G(s)={\lim }_{s\to 0}\frac{0.419s^2}{s^2}=0.419$

2.d

Steady-state error for a step input $u(t)$

For a step input $R(s)=\frac{1}{s}$, the steady-state error is given by:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{R(s)}{1+K_{p}G(s)}=\underset{s\to 0}{\lim }\frac{1/s}{1+\infty }=0$$

2.e

Steady-state error for a ramp input $tu(t)$

For a ramp input $R(s)=\frac{1}{s^2}$, the steady-state error is given by:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{sR(s)}{1+K_{v}G(s)}=\underset{s\to 0}{\lim }\frac{s/s^2}{1+0.419}=\frac{1}{0.419}\approx 2.39$$

2.f

Steady-state error for a parabolic input $t^{2}u(t)$

For a parabolic input $R(s)=\frac{1}{s^3}$, the steady-state error is given by:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{s^{2}R(s)}{1+K_{a}G(s)}=\underset{s\to 0}{\lim }\frac{s^2/s^3}{1+0.419}=\frac{1}{0.419}\approx 2.39$$

Problemè 3

We will be augmenting our controller to include an integrator. The transfer function of our new PID compensator will be as follows;

$$G_C(s)=K_{D}s+K_P+\frac{K_I}{s}$$

Given that the transfer function for our plant has not changed, and given that this controller is also applied to the plant in cascade configuration.

The closed-loop transfer function is

$$\frac{Y(s)}{R(s)}=\frac{G(s)G_C(s)}{1+G(s)G_C(s)}=\frac{\frac{0.419}{s^2}\ast (K_{D}s+K_P+\frac{K_I}{s})}{1+\frac{0.419}{s^2}\ast (K_{D}s+K_P+\frac{K_I}{s})}=\frac{0.419\ast (K_{D}s+K_P+\frac{K_I}{s})}{s^2+0.419\ast (K_{D}s+K_P+\frac{K_I}{s})}$$

3.a

Static error constant for position (position constant)

$$K_P=\underset{s\to 0}{\lim }{G}_C(s)G(s)=\underset{s\to 0}{\lim }\frac{0.419}{s^2}(K_{D}s+K_P+\frac{K_I}{s})=\infty$$

3.b

Static error constant for velocity (velocity constant)

$$K_V=\underset{s\to 0}{\lim }s{G}_C(s)G(s)=\underset{s\to 0}{\lim }s\frac{0.419}{s^2}(K_{D}s+K_P+\frac{K_I}{s})=0.419K_D+\frac{0.419K_P}{s}+\frac{0.419K_I}{s^2}=0.419K_I$$

3.c

Static error constant for acceleration (acceleration constant)

$$K_A=\underset{s\to 0}{\lim }{s}^2{G}_C(s)G(s)=\underset{s\to 0}{\lim }{s}^2\frac{0.419}{s^2}(K_{D}s+K_P+\frac{K_I}{s})=0.419K_P$$

3.d

Steady-state error for a step input $u(t)$

For a step input $R(s)=\frac{1}{s}$, the steady-state error is given by:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{sR(s)}{1-\frac{C(s)}{R(s)}}=\underset{s\to 0}{\lim }s\frac{1/s}{1-\frac{C(s)}{R(s)}}=\frac{1}{1+K_P}=0$$

3.e

Steady-state error for a ramp input $tu(t)$

For a ramp input $R(s)=\frac{1}{s^2}$, the steady-state error is given by:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{s^{2}R(s)}{1-\frac{C(s)}{R(s)}}=\underset{s\to 0}{\lim }\frac{s^2/s^2}{1-\frac{C(s)}{R(s)}}=\frac{1}{K_V}=\frac{1}{0.419K_I}$$

3.f

Steady-state error for a parabolic input $t^{2}u(t)$

For a parabolic input $R(s)=\frac{1}{s^3}$, the steady-state error is given by:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{s^{3}R(s)}{1-\frac{C(s)}{R(s)}}=\underset{s\to 0}{\lim }\frac{s^3/s^3}{1-\frac{C(s)}{R(s)}}=\frac{1}{K_A}=\frac{1}{0.419K_P}$$

Problemè 4

Ideally you want your controller design to reject a step disturbance input at $D(s)$. This means that in the steady state for $D(s)=\frac{1}{s}$, the output $Y(s)$ is unchanged.

4.a

Ignoring the input $R(s)$, what is the transfer function $\frac{E(s)}{D(s)}$ in terms of $G_1(s)$ and $G_2(s)$?

To find the transfer function $\frac{E(s)}{D(s)}$, then the transfer function $\frac{E(s)}{D(s)}$ is given by:

$$\frac{E(s)}{D(s)}=\frac{G_1(s)G_2(s)}{1+G_1(s)G_2(s)}$$

4.b

For $G_1(s)=K_C(s+z)$ and $G_2(s)=\frac{0.419}{s^2}$ what is the steady state error resulting from step inputs $R(s)=\frac{A}{s}$ and $D(s)=\frac{B}{s}$

The steady-state error to step input $R(s)=\frac{A}{s}$ is given by:

$$e_{ss}(R)=\underset{s\to 0}{\lim }\frac{A}{s}(\frac{1}{1+L(s)})$$

with $L(s)=G_1(s)G_2(s)=\frac{0.419K_C(s+z)}{s^2}$

$$e_{ss}(R)=\underset{s\to 0}{\lim }\frac{A}{s}(\frac{1}{1+\frac{0.419K_C(s+z)}{s^2}})=\frac{A}{0.419K_C}$$

The steady-state error to step input $D(s)=\frac{B}{s}$ is given by: $e_{ss}(D)=\frac{B}{0.419K_C}$

Thus, the total steady-state error is $e_{ss}=e_{ss}(R)+e_{ss}(D)=\frac{A+B}{0.419K_C}$

4.c

For $G_1(s)=K_{D}s+K_P+\frac{K_I}{s}$ and $G_2(s)=\frac{0.419}{s^2}$ what is the steady state error resulting from step inputs $R(s)=\frac{A}{s}$ and $D(s)=\frac{B}{s}$

$$L(s)=G_1(s)G_2(s)=\frac{0.419(K_{D}s+K_P+\frac{K_I}{s})}{s^2}=\frac{0.419K_D{s}^2+0.419K_{P}s+0.419K_I}{s^3}$$

The steady-state error to step input $R(s)=\frac{A}{s}$ is zero:

$$e_{ss}(R)=\underset{s\to 0}{\lim }\frac{A}{s}(\frac{1}{1+L(s)})=\underset{s\to 0}{\lim }\frac{A}{s}(\frac{1}{1+\frac{0.419K_D{s}^2+0.419K_{P}s+0.419K_I}{s^3}})=0$$