Control of Linear Systems

Table of Contents

1

a) Block Diagram Reduction

b) Routh stability

2

a) State Space Model: Simulation Diagram

b) Pole-Zero Cancellation

(i) Stability

(ii) Controllability

(iii) Observability

c) Rank Test

i) Controllability

ii) Observability

3) State Variable Feedback

a) State Space Model

b) State Variable Feedback Design

1) Carry Out Controllability test

2) Decide where to place the closed-loop poles.

3) Multiply out the poles to get the closed-loop characteristic equations

4) Evaluate $\mathit{Ac}=A–\mathit{BK}$

5) Evaluate the actual closed-loop characteristic equation as [ $\mathit{\lambda I}–\mathit{Ac}=0]$

6) Equate the desired and actual CLCE

c) Simulation Diagram

d) Steady State Error and Tracking Controller

4) PI Controller

a) Design Requirements

b) Steady State Error

b) Verification by Simulink

Works Cited

Appendix

Word count for main content:  1965

1

a) Block Diagram Reduction

Consider the following block diagram:

$\mathit{ N}=10$

Figure 1

Now considering the section highlighted in Figure 1. With the feedback loop rule let $G=\frac{15}{s^2+11s+10},\mathit{ and},\mathit{ H}=2$

$\frac{G}{1+\mathit{GH}}$

$=\frac{\left(\frac{15}{s^2+11s+10}\right)}{1+\left(\frac{15*2}{s^2+11s+10}\right)}$

$=\frac{15}{s^2+11s+40}$

Following the combinations rule, the cascade becomes;

$=\frac{1}{s}*K_B*\frac{15}{s^2+11s+40}*\frac{(s+2)}{(s+5)}=\frac{15K_B(s+2)}{s(s+5)(s^2+11s+40)}=\frac{\mathit{15}{\mathit{K}}_{\mathit{B}}\mathit{s}\mathit{+}\mathit{30}{\mathit{K}}_{\mathit{B}}}{{\mathit{s}}^{\mathit{4}}\mathit{+}\mathit{1}{\mathit{6}\mathit{s}}^{\mathit{3}}\mathit{+}\mathit{95}{\mathit{s}}^{\mathit{2}}\mathit{+}\mathit{200}\mathit{s}}$

Figure 2 – System of Q1 reduced

To reduce the feedback further let $G=\frac{15K_{B}s+30K_B}{s^4+1{6s}^3+95s^2+200s}$

and $H=1$

;

$\frac{15K_{B}s+30K_B}{s^4+1{6s}^3+95s^2+200s+15K_{B}s+30K_B}=\frac{15K_{B}s+30K_B}{s^4+1{6s}^3+95s^2+(200+15K_B)s+30K_B}$

Hence the Closed Loop Character Equation being:

${\mathit{s}}^{\mathit{4}}\mathit{+}\mathit{1}{\mathit{6}\mathit{s}}^{\mathit{3}}\mathit{+}\mathit{95}{\mathit{s}}^{\mathit{2}}\mathit{+}\mathit{(}\mathit{200}\mathit{+}\mathit{15}{\mathit{K}}_{\mathit{B}}\mathit{)}\mathit{s}\mathit{+}\mathit{30}{\mathit{K}}_{\mathit{B}}$

b) Routh stability

A system is stable when the output is able to return to its original value after a disturbance from an impulse causing a change in the input.

For a characteristic equation of $a_n{s}^n+a_{n–1}{s}^{n\mp 1}+\dots +a_{1}s+a_0=0,$

Stability is achieved if all the coefficients are positive. As a negative coefficient would result in a root with a positive real part, which causes the output to tend towards infinity (instability). If any are of zero value, the system is considered to be critically stable (Dutton, Thompson, & Barraclough, 1997). 

To find the value of K_B to where the system is operating in a stable manner, the Routh stability criterion will be used (Table 1).

Table 1 – Sample of the Routh array

${\mathit{s}}^{\mathit{n}}$	$a_1$	$a_{n–2}$	$a_{n–4}$
${\mathit{s}}^{\mathit{n}\mathit{–}\mathit{1}}$	$a_{n–1}$	$a_{n–3}$	$\dots$
$\mathit{⋮}$	$b_1$ $b_1=\frac{{(a}_{n–1}*a_{n–2})–(a_n*a_{n–3})}{a_{n–1}}$	$b_2$ $b_2=\frac{{(a}_{n–1}*a_{n–4})–(a_n*a_{n–5})}{a_{n–1}}$	$b_3$
${\mathit{s}}^{\mathit{1}}$	$y_1$	$y_2$	$\dots$
${\mathit{s}}^{\mathit{0}}$	$z_1$	$\dots$

Replacing the variables with the appropriate value gives;

Table 2 – Routh array in question

${\mathit{s}}^{\mathit{4}}$	$\mathit{1}$	$\mathit{95}$	$\mathit{30}{\mathit{K}}_{\mathit{B}}$
${\mathit{s}}^{\mathit{3}}$	$16$	$200+15K_B$	$0$
${\mathit{s}}^{\mathit{2}}$	$b_1=\frac{{(a}_{n–1}*a_{n–2})–(a_n*a_{n–3})}{a_{n–1}}$ $b_1=\frac{(16*95)–(1*200+15K_B)}{16}$ ${\mathit{b}}_{\mathit{1}}\mathit{=}\frac{\mathit{1320}\mathit{–}\mathit{15}{\mathit{K}}_{\mathit{B}}}{\mathit{16}}$	$b_2=\frac{\left(16*30K_B\right)–(1*0)}{16}$ ${\mathit{b}}_{\mathit{2}}\mathit{=}\frac{\mathit{480}{\mathit{K}}_{\mathit{B}}}{\mathit{16}}$	$b_3=0$
${\mathit{s}}^{\mathit{1}}$	$c_1=\frac{\left(\frac{1320–15K_B}{16}*\left(200+15K_B\right)\right)–(16*\frac{480K_B}{16})}{\frac{1320–15K_B}{16}}$ $c_1=\frac{–225{K_B}^2+9120K_B+264000}{1320–15K_B}$	$c_2=0$	$\dots$
${\mathit{s}}^{\mathit{0}}$	$d_1=0$	$\dots$

Here we assume that all of the elements in the first column are positive to have a stable system.

$b_1>0$	$c_1>0$
$\frac{1320–15K_B}{16}>0$	$\frac{–225{K_B}^2+9120K_B+264000}{1320–15K_B}>0$
$\mathit{88}\mathit{>}{\mathit{K}}_{\mathit{B}}$	$–225{K_B}^2+9120K_B+264000>0$
	$K_{B1}=60.067{1,\mathit{ K}}_{B2}=–19.53$

For stability K_B must be in the range;

$\mathit{–}\mathit{19}\mathit{.}\mathit{53}\mathit{<}{\mathit{K}}_{\mathit{B}}\mathit{<}\mathit{60}\mathit{.}\mathit{07}$

2

a) State Space Model: Simulation Diagram

Consider the following state space model of a system;

$A=\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right],\mathit{ B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathit{ C}=\left[\begin{array}{ccc}0& 1& 0\end{array}\right],\mathit{ D}=\left[0\right]$

$\dot{x}=\mathit{Ax}+\mathit{Bu},\mathit{ y}\left(t\right)=\mathit{Cx}+\mathit{Du}$

Substituting the appropriate values:

$\left[\begin{array}{c}\dot{x_1}\\ \dot{x_2}\\ \dot{x_3}\end{array}\right]=\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u\left(t\right)$		$y=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[0\right]u$
$\dot{x_1}=a_{11}{x}_1–x_3$	$\dot{x_2}=x_1–3x_2+2x_3$	$\dot{x_3}=–4x_3+u\left(t\right)$	$y=x_2$

This gives the following simulation diagram:

Figure 3  – Simulation diagram of the system in Q2

b) Pole-Zero Cancellation

Reducing the diagram presents the system’s transfer function.

Figure 4 – Simulation diagram in state space domain

By block diagram reduction of the feedback loop; G = $\frac{1}{s}$

and H is respectively 4, -a₁₁ and 3:

Figure 5

Orange Section: $\frac{1}{s+4}$

Purple Section: $\frac{1}{s–a_{11}}$

Green Section: $\frac{1}{s+3}$

Figure 6

Figure 7 – Using the parallel rule

Figure 8 – Using the cascade rule

Resulting in;

_­ $\frac{Y\left(s\right)}{U\left(s\right)}=G\left(s\right)=\frac{2s–2a_{11}+1}{(s–a_{11})(s+4)(s+3)}$

(i) Stability

Consider the poles of the characteristic equation: $\left(s–a_{11}\right)\left(s+4\right)\left(s+3\right)$

.

For an unstable system, the pole ${‘a}_{11}‘$

must have a negative real part. As a positive pole would result in the output to tend to infinity and the system will be unstable. (Dutton, Thompson, & Barraclough, 1997)

(ii) Controllability

Considering Figure 7, if $a_{11}=–3$

cascading the 3 blocks would result in the cancellation of the pole of the 3^rd block ( $\frac{1}{s+3}$

). The pole-zero cancellation in this sense would consequently make the system uncontrollable as the pole of the 3^rd block cannot be driven from the plant input. Although, the system can still be observable as the variable $a_{11}$

will still contribute to the output.

 (iii) Observability

For the system to become unobservable, $a_{11}=–4$

. Cascading would result in a pole-zero cancellation between the zero in the middle block and the pole of the first. Resulting in unobservability because the associated mode will not carry the dynamic response of the zero to the plant output as it will not be ‘visible’. This can still be controllable as the mode can still be driven from the input.

For the system to be both unobservable and uncontrollable, the pole and zero that cancel each other out must reside in the same block. (Dutton, Thompson, & Barraclough, 1997)

c) Rank Test

i) Controllability

Consider the controllability matrix $T_c$

;

$T_c=\left[\begin{array}{ccc}B& \mathit{AB}& \mathit{AAB}\end{array}\right]$

$A*B=\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]*\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ –4\end{array}\right]$

$A*\mathit{AB}=\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]*\left[\begin{array}{c}1\\ 2\\ –4\end{array}\right]=\left[\begin{array}{c}{a}_{11}–4\\ –13\\ 16\end{array}\right]$

Therefore,

$T_c=\left[\begin{array}{ccc}0& 1& a_{11}–4\\ 0& 2& –13\\ 1& –4& 16\end{array}\right]$

To check if the system is controllable, use the rank test as follows; Starting with the largest sub matrix (itself);

$\left|T_c\right|=\left| \left[\begin{array}{ccc}0& 1& a_{11}–4\\ 0& 2& –13\\ 1& –4& 16\end{array}\right] \right|\mathit{ }=+0\left|\begin{array}{cc}2& –13\\ –4& 16\end{array}\right|–0\left|\begin{array}{cc}1& a_{11}–4\\ –4& 16\end{array}\right|+1\left|\begin{array}{cc}1& a_{11}–4\\ 2& –13\end{array}\right|$

$\left|T_c\right|=0–0+(\left(1*–13\right)–\left(2*{(a}_{11}–4)\right))$

$\left|T_c\right|=–5–2a_{11},\mathit{ }{a}_{11}=–3$

$\left|T_c\right|=1$

For the system to be uncontrollable, it mustn’t be full rank i.e. the determinant of the controllability matrix must be zero at this point. The rank = 2; when looking into the submatrix of $\left[\begin{array}{cc}2& –13\\ –4& 16\end{array}\right]$

ii) Observability

Using the Observability matrix $T_o$

;

$T_o=\left[\begin{array}{c}C\\ \mathit{CA}\\ \mathit{CAA}\end{array}\right]$

As, $\mathit{CA}=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]*\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]=\left[\begin{array}{ccc}1& –3& 2\end{array}\right]$

 

$\mathit{CA}*A=\left[\begin{array}{ccc}1& –3& 2\end{array}\right]*\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]=\left[\begin{array}{ccc}{a}_{11}–3& 9& –13\end{array}\right]$

$T_o=\left[\begin{array}{ccc}0& 1& 0\\ 1& –3& 2\\ a_{11}–3& 9& –13\end{array}\right]$

To check if the system is observable, use the rank test as follows; starting with the largest sub matrix (itself);

$\left|T_o\right|=+0\left|\begin{array}{cc}–3& 2\\ 9& –13\end{array}\right|–1\left|\begin{array}{cc}1& 2\\ a_{11}–3& –13\end{array}\right|+0\left|\begin{array}{cc}1& –3\\ a_{11}–3& 9\end{array}\right|$

$\left|T_c\right|=0–(\left(1*–13\right)–\left({(a}_{11}–3)*2\right))+0$

$\left|T_o\right|=7+2a_{11},\mathit{ }{a}_{11}=–4$

$\mathit{ }\left|T_o\right|=–1$

For the system to be unobservable, it mustn’t be full rank i.e. the determinant of the observability matrix must be zero at this point.  But it is partially as the rank = 2; considering the submatrix of $\left[\begin{array}{cc}–3& 2\\ 9& 13\end{array}\right]$

. So the rank deficiency is 1.

3) State Variable Feedback

a) State Space Model

 To obtain a state space model of the system, consider the transfer function’s controllable canonical form;

Figure 9 – Laplace Transfer function version and State space representation in the controllable canonical form

$\mathit{As},\mathit{ G}\left(s\right)=\frac{s+2}{s^3+{4s}^2+s–6}$

$n=3,\mathit{ }{a}_3=–6,\mathit{ }{a}_2=1,\mathit{ }{a}_1=4,\mathit{ }{b}_o=0,\mathit{ }{b}_1=0,\mathit{ }{b}_2=1,\mathit{ }{b}_3=2$

G(s) in the controllable canonical form;

$A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 6& –1& –4\end{array}\right],\mathit{ B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathit{ C}=\left[\begin{array}{ccc}2& 1& 0\end{array}\right],\mathit{ d}=\left[0\right]$

b) State Variable Feedback Design

SVF takes the state variables of a system’s output and feeds them back into the input of said system for improving the performance of the closed-loop system. (Dutton, Thompson, & Barraclough, 1997)

Let the system design yield a 20% overshoot and a 10-90% rise time of about 1s. These requirements are given for the assumption that they will be suitable for this stability design for the gain matrix K. The Design procedure to stabilise a system with a state feedback controller goes as follows;

1) Carry Out Controllability test

In controllable canonical form the system is considered completely state controllable, so a controllability test is unnecessary (Dutton, Thompson, & Barraclough, 1997). Being controllable, it’s possible to move the open-loop poles of the system to any arbitrary closed-loop location. As the system is already considered to be controllable, there is no need to perform a stabilizable test, which involves using a sensitivity matrix such as the following $S\left({\lambda }_i\right)=[A–I{\lambda }_i\mathit{ B}]$

. An observable system is able to restructure the state vector completely by the system’s output (Dutton, Thompson, & Barraclough, 1997). This is possible for the system in question, therefore an observability test is unnecessary.

2) Decide where to place the closed-loop poles.

Figure 10 – The curve of a standard second order system with the addition of an imaginary curve (red) which has a 20% overshoot

Respecting the requirements, observe a curve that would have a 20% maximum overshoot, such as the red curve in Figure 10. From observation, this red curve shows to be between a damping ratio $\left(\zeta \right)$

of 0.4 (yellow) and 0.5 (blue). By interpolation, let the damping ratio to be of 0.46 in value.

As the desired rise time of 10% to 90%; approximately $\mathrm{ }{\mathrm{w}}_{\mathrm{n}}\mathrm{t}=0.5\mathrm{ and }2.1\mathrm{ }$

respectively. The rough rule states: ${\mathrm{W}}_{\mathrm{n}}=\frac{1.8}{{\mathrm{t}}_{\mathrm{r}}}$

. When the desired time of ${\mathrm{t}}_{\mathrm{r}}=1$

;

${\mathrm{W}}_{\mathrm{n}}=\frac{1.8}{1}=1.8\mathrm{ rad}/\mathrm{s}$

Consider the standard second order system;

$\frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}$

The characteristic equation [ $s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2$

] becomes;

$s^2+\left(2*0.46*1.8\right)s+{1.8}^{2\mathit{ }}\mathit{ }={\mathit{ s}}^2+1.656s+3.24$

Which represents poles of $s= \mathit{–}\mathit{0}\mathit{.}\mathit{828}\mathit{ }\mathit{\pm }\mathit{1}\mathit{.}\mathit{598}\mathit{j}$

. As the system in question is actually in third order a third pole is required. To not interfere with the behaviour of the 2^nd order poles, this pole must be at least x10 than that of the other poles’ real part. Therefore, the third pole will be $s_3= –8$

.

3) Multiply out the poles to get the closed-loop characteristic equations

Obtaining the desired closed loop characteristic equation (DCLCE) as

$\left(s+0.828+1.598j\right)\left(s+0.828–1.598j\right)\left(s+8\right)=0$

$\mathit{DCLCE}\mathit{=}{\mathit{s}}^{\mathit{3}}\mathit{+}{\mathit{9}\mathit{.}\mathit{6565}\mathit{s}}^{\mathit{2}}\mathit{+}\mathit{16}\mathit{.}\mathit{4872}\mathit{s}\mathit{+}\mathit{25}\mathit{.}\mathit{9136}\mathit{=}\mathit{0}$

4) Evaluate $A_c=\left[A–\mathit{BK}\right]$

See Appendix for proof and reasoning for matrix K.

To obtain the closed loop plant matrix ( $A_c$

) substitute in the appropriate values;

$A_c=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 6& –1& –4\end{array}\right]–\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\left[\begin{array}{ccc}{K}_{11}& K_{12}& K_{13}\end{array}\right]\mathit{ }=\mathit{ }\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 6–K_{11}& –1–K_{12}& –4–K_{13}\end{array}\right]$

5) Evaluate the actual closed-loop characteristic equation as [ $\mathit{\lambda I}–A_c=0]$

The actual closed loop pole locations are the result of the eigenvalues of this matrix i.e. the roots of the CLCE given as;

$\left|\mathit{\lambda I}–A_C\right|=0$

$\left|\lambda \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]–\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 6–K_{11}& –1–K_{12}& –4–K_{13}\end{array}\right]\right|=\left| \left[\begin{array}{ccc}\lambda & –1& 0\\ 0& \lambda & –1\\ –6+K_{11}& 1+K_{12}& \lambda +4+K_{13}\end{array}\right] \right| =0$

Expanding by column 1 gives;

$=+\lambda \left|\begin{array}{cc}\lambda & –1\\ 1+K_{12}& \lambda +4+K_{13}\end{array}\right|–0\left|\begin{array}{cc}–1& 0\\ 1+K_{12}& \lambda +4+K_{13}\end{array}\right|+(–6+K_{11})\left|\begin{array}{cc}–1& 0\\ \lambda & –1\end{array}\right|$

$=\mathit{ \lambda }\left(\left(\mathit{ \lambda }*\left(\lambda +4+K_{13}\right)\right)–\left(–1*\left(1+K_{12}\right)\right)\right)+(–6+K_{11})\left(\right(–1*–1)–\left(\lambda *0\right))$

$\mathit{Actual\; CLCE}={\lambda }^3+\left(4+K_{13}\right){\lambda }^2+\left(1+K_{12}\right)\lambda +\left(–6+K_{11}\right)=0$

6) Equate the desired and actual CLCE

By equating the desired and actual CLCE the values of the K matrix can be gained;

${\lambda }^3+\left(4+K_{13}\right){\lambda }^2+\left(1+K_{12}\right)\lambda +\left(–6+K_{11}\right)=s^3+{9.6565s}^2+16.4872s+25.9136$

Poles (s) and eigenvalues ( $\lambda$

) are the same: $s=\lambda$

. The following will equate the coefficients of both actual and desired;

$\left(4+K_{13}\right)=9.6565$ $K_{13}=5.6565$

$\left(1+K_{12}\right)=16.4872$ $K_{12}=15.4872$

$\left(K_{11}–6\right)=25.9136$ $K_{11}=31.9136$

$\mathit{K}\mathit{=}\left[\begin{array}{ccc}\mathit{31}\mathit{.}\mathit{91}& \mathit{15}\mathit{.}\mathit{49}& \mathit{5}\mathit{.}\mathit{66}\end{array}\right]$

c) Simulation Diagram

$A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 6& –1& –4\end{array}\right],\mathit{ B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathit{ C}=\left[\begin{array}{ccc}2& 1& 0\end{array}\right],\mathit{ d}=\left[0\right]$

$\dot{x}=\mathit{Ax}+\mathit{Bu},\mathit{ y}\left(t\right)=\mathit{Cx}+\mathit{Du}$

Substituting in the appropriate matrices;

$\left[\begin{array}{c}\dot{x_1}\\ \dot{x_2}\\ \dot{x_3}\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 6& –1& –4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u\left(t\right)\mathit{ }y=\left[\begin{array}{ccc}2& 1& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[0\right]u$

$\dot{x_1}=x_2$

$\dot{x_2}=x_3$

$\dot{x_3}=6x_1–x_2–4x_3+u\left(t\right)$

$y=2x_1+x_2$

This gives the following simulation diagram:

Figure 11 – Plant with full State variable feedback in MATLAB

d) Steady State Error and Tracking Controller

Using the Closed loop system from part c), the following shows the system’s response when subjected to a unit step input.

 

Figure 12 – Closed-loop step response with amplitude against time (seconds) (left) and magnified to show the intersection at 10% and 90% rise time (right)

Figure 12 shows the system exhibits a response where the settling point is approximately 0.077, with a 30% maximum overshoot as it peaks at 0.1. The rise time from 10% to 90% shows to be 585ms with a settling time of 7s. The system seems to have a steady state error of 0.92, from a unit step input. Using a tracking controller which introduces integrators can reduce the steady state error but doing so will increase the order of the system (Dutton, Thompson, & Barraclough, 1997).

Figure 13 – Block diagram form of a general tracking system

 The design procedure is similar to that of SVF but with the following changes; for step 2) extra poles can be considered. A new unknown element $K_i$

is introduced (see Figure 13), therefore in step 5) for evaluating $A_c$

  becomes $A_c=\left(\begin{array}{cc}A–\mathit{BK}& B{K}_i\\ –C& 0\end{array}\right)$

, where the size of K = [number of plant i/p x number of states] and Ki = [number of plant i/p x number of plant o/p]. Additionally, step 6) will be equating the coefficients to solve for K and K_i.

4) PI Controller

a) Design Requirements

$N=10$

Figure 14 – System in question for Q4

With the feedback rule;

$\frac{G}{1+G\left(s\right)G_c\left(s\right)H\left(s\right)}$

$=\frac{\frac{{23(K}_{p}s+K_i)}{s^2+10s}}{1+\frac{{23(K}_{p}s+K_i)}{s^2+10s}}$

$=\frac{{23(K}_{p}s+K_i)}{s^2+(10+{23K}_p)s+23K_i}$

Equating to a second order system;

$\frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}=\frac{{23(K}_{p}s+K_i)}{s^2+(10+{23K}_p)s+23K_i}$

The settling time ( $t_s$

) is the time taken for the response to reach and stay within a certain range around the final value. The design specifications require a settling time with a 2% criterion ( $x=2$

)

$t_S=–\frac{\ln \left(\frac{x}{100}\right)}{\zeta {\omega }_n}=\frac{4}{\zeta {\omega }_n}$

As the settling time is desired to be less than 1s, the design will have a target that follows a settling time of 0.1s, to ensure that the system is well within the specifications.

$t_S=0.1=\frac{4}{\zeta {\omega }_n}$

$\mathit{Therefore},\mathit{ \zeta }{\omega }_n=40$

The following equation of coefficients applies; Note: According to the graph of Figure 10, take the most appropriate curve where $\zeta =0.8$

.

$0.1=\frac{4}{0.8{\omega }_n},\mathit{ }{\omega }_n=50$	$2\zeta {\omega }_{n}s=\left(10+{23K}_p\right)s,\mathit{ As}, 2\zeta {\omega }_n=80$
${{\omega }_n}^2=23K_i{=50}^2$	$80=\left(10+{23K}_p\right)$
${\mathit{K}}_{\mathit{i}}\mathit{=}\mathit{108}\mathit{.}\mathit{6957}$	${\mathit{K}}_{\mathit{p}}\mathit{=}\frac{\mathit{70}}{\mathit{23}}\mathit{=}\mathit{ }\mathit{3}\mathit{.}\mathit{0435}$

To solve for K,

$K{{\omega }_n}^2={23(K}_{p}s+K_i),\mathit{ and},\mathit{ }{{\mathit{ \omega }}_n}^2=23K_i$

$\mathit{K }23K_i={23(K}_{p}s+K_i)$	$=\frac{K_{p}s}{K_i}+\frac{K_i}{K_i}$	$=\frac{K_{p}s}{K_i}+1$

$\mathit{K}=\frac{3.0435\mathit{ s}}{108.6957}+1= \mathit{0}\mathit{.}\mathit{028}\mathit{ s}\mathit{+}\mathit{1}$

Therefore,

$\frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}$

$=\frac{{23(K}_{p}s+K_i)}{s^2+(10+{23K}_p)s+23K_i}$

$= \frac{70\mathit{ s}+2500}{s^2+80s+ 2500}$

b) Steady State Error

To check the steady state error, consider the final value theorem;

$e_{\mathit{ss}}=e\left(\infty \right)=\underset{\mathit{t }\to \infty }{\lim }e\left(t\right)=\underset{s\to 0}{\lim }s*E\left(s\right)$

_­As $E\left(s\right)=\frac{1}{1+G\left(s\right)G_c\left(s\right)H\left(s\right)}\mathit{ R}\left(s\right)$

.  With a ramp input of $R\left(s\right)=\frac{1}{s^2}$

;

$\underset{s\to 0}{\lim }s*E\left(s\right)=s* \frac{1}{1+\left(K_p+\frac{K_i}{s}\right)*\left(\frac{23}{s+10}\right)} \frac{1}{s^2}$

$=\frac{1}{1+\frac{{23(K}_{p}s+K_i)}{s^2+10s}}*\frac{1}{s}$

$=\frac{1}{s+\frac{{s\left(23\right(K}_{p}s+K_i\left)\right)}{s(s+10)}}$

$=\frac{1}{s+\frac{{23(K}_{p}s+K_i)}{s+10}}$

$\mathit{As\; s}=0;\mathit{ }=\frac{10}{23K_i}$

$e_{\mathit{ss}}=0.004$

As the steady state error experienced is within the specification of being less than 0.1 when subjected to a ramp input, the designed control system meets the specifications.

b) Verification by Simulink

The following figures have verified the design of part (a) and shows to satisfies the specification of settling less than 1 second for a unit step input (Figure 15). It is shown that the steady state error exhibited with a ramp input being approximately 0.004, similar to that calculated analytically (Figure 17).

With the addition of the PI controller, it also shows the difference on how the addition of the proportional controller it decreases the stead state error. But adding in an integral making the PI controller, further reduces the stead state error but does introduce slight oscillation with a unit step input.

Figure 15 – Block diagram (left) and response (right) of the guidance system with PI controller when subjected with a unit step input

Figure 16 – Block diagram (left) and response (right) of the guidance system with PI controller when subjected with a unit ramp input

 

Figure 17 – To show that the steady state error experienced with a ramp input is of 0.004227.

Works Cited

Dutton, K., Thompson, S., & Barraclough, B. (1997). The Art of Control Engineering. Essex: Addison Wesley-Longman.

Appendix

Regarding question 1)

For simplification, the model can be reduced to an open loop transfer function. This is defined as follows;

Figure 18– a basic closed loop system diagram

Where,

$C=\mathit{EG\; and\; E}=R–\mathit{CH}$

$C=\left(R–\mathit{Ch}\right)G$

$C+\mathit{CHG}=\mathit{RG}$

$\frac{\mathit{output}}{\mathit{input}}=\frac{C}{R}=\frac{G}{1+\mathit{GH}}$

This is also referred to as the block diagram reduction rule for feedback loops, as shown in Figure 19.

Figure 19 – Illustration of the Feedback loop rule

Calculation for equation 1a)

$\frac{G}{1+\mathit{GH}}=\frac{\left(\frac{15}{s^2+11s+10}\right)}{1+\left(\frac{15*2}{s^2+11s+10}\right)}$

Multiply the numerator and denominator by $(s^2+11s+10)$

, gives;

$=\frac{\left(\frac{15(s^2+11s+10)}{s^2+11s+10}\right)}{{(s}^2+11s+10)+\left(\frac{30(s^2+11s+10)}{s^2+11s+10}\right)}$

$=\frac{15}{s^2+11s+10+30}$

$=\frac{15}{s^2+11s+40}$

Calculation for equation 1b) when solving for K_B from c₁

$–225{K_B}^2+9120K_B+264000>0$

Here we solve for $K_B$

using the quadrating formula,

$\frac{–b\pm \sqrt{b^2–4\mathit{ac}}}{2a},\mathit{ where\; a}=–225,\mathit{ b}=9120\mathit{ and\; c}=264000$

$=\frac{–\left(9120\right)\pm \sqrt{{\left(9120\right)}^2–\left(4*–225*264000\right)}}{2(–225)}=\frac{9120\pm 12426.80973\mathit{ j}}{450}$

$=\frac{9120\pm 12426.80973\mathit{ j}}{450}$

$x_1=60.067{1,\mathit{ x}}_2=–19.53$

Regarding Question 2.

2b i)

The eigenvalues of the matrix A are also referred to as the poles of the transfer function, which can be obtained using the following equation for stability;

$\left|\mathit{\lambda I}–A\right|=0$

$\left|\lambda \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]–\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]\right|=0$

$\left|\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]–\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]\right|=0$

$\left| \left[\begin{array}{ccc}\lambda –a_{11}& 0& –1\\ –1& \lambda +3& –2\\ 0& 0& \lambda +4\end{array}\right] \right|=0$

Expanding by column 2 gives;

$=+0\left|\begin{array}{cc}–1& –2\\ 0& \lambda –4\end{array}\right|–(\lambda +3)\left|\begin{array}{cc}\lambda –a_{11}& –1\\ 0& \lambda –4\end{array}\right|+0\left|\begin{array}{cc}\lambda –a_{11}& –1\\ –1& –2\end{array}\right|$

$=–(\mathit{ \lambda }+3)\left(\right(\left(\lambda –a_{11}\right)*\left(\lambda –4\right))–\left(0*–1\right))$

$=–(\mathit{ \lambda }+3)\left(\lambda –a_{11}\right)\left(\lambda –4\right)$

As expected, the eigenvalues present the same poles as the LTF model.

2b ii)

Here we consider the state vector differential equation;

$\dot{x}=\mathit{Ax}+\mathit{Bu}$

$y\left(t\right)=\mathit{Cx}+\mathit{Du}$

 Taking Laplace transforms of the state equations [ $\dot{x}=\mathit{Ax}+\mathit{Bu}$

] gives;

$\mathit{sX}\left(s\right)–X\left(0\right)=\mathit{AX}\left(s\right)+\mathit{BU}\left(s\right)$

$\mathit{sX}\left(s\right)–\mathit{AX}\left(s\right)=X\left(0\right)+\mathit{BU}\left(s\right)$

$(\mathit{sI}–A)X\left(s\right)=X\left(0\right)+\mathit{BU}\left(s\right)$

For the X(s) to become the subject; $X\left(s\right)=\frac{X\left(0\right)+\mathit{BU}\left(s\right)}{\mathit{sI}–A}$

$X\left(s\right)={(\mathit{sI}–A)}^{–1}(X\left(0\right)+\mathit{BU}\left(s\right))$

With assuming zero initial conditions, where X(0)=0, the previous takes the form of;

$X\left(s\right)={(\mathit{sI}–A)}^{–1}\mathit{BU}\left(s\right)$

Substituting in the appropriate values of matrix A and B gives;

$={\left(s\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]–\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right]\right)}^{–1}*\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

$={\left(\left[\begin{array}{ccc}{s–a}_{11}& 0& 1\\ 1& s+3& 2\\ 0& 0& s+4\end{array}\right]\right)}^{–1}*\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

See appendix for the steps of calculation for the inverse of matrix (sI-A).

$=*\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

$=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

2b iii)

A completely observable system is able to restructure the state vector completely from measurements by the system’s output. It  can also be referred to as ‘dual controllability’ as it uses similar functions to that of controllability, but focuses on the C matrix rather than the B. (Dutton, Thompson, & Barraclough, 1997)

Instead of concentration on the state vector, the output vector is in interest;

$y\left(t\right)=\mathit{Cx}+\mathit{Du}$

Taking Laplace transforms;

$Y\left(s\right)=\mathit{CX}\left(s\right)+\mathit{DU}\left(s\right)$

As $X\left(s\right)={(\mathit{sI}–A)}^{–1}\mathit{BU}\left(s\right)$

, Y(S) becomes;

$Y\left(s\right)=C{(\mathit{sI}–A)}^{–1}\mathit{BU}\left(s\right)+\mathit{DU}\left(s\right)$

$Y\left(s\right)=\{C{\left(\mathit{sI}–A\right)}^{–1}B+D\}U\left(s\right)$

$\frac{Y\left(s\right)}{U\left(s\right)}=\left\{C{\left(\mathit{sI}–A\right)}^{–1}B+D\right\}=G\left(s\right)$

Therefore, the transfer function is obtained as:

$G\left(s\right)=C{\left(\mathit{sI}–A\right)}^{–1}B+D$

Here the focus is on the matrix of ${C\left(\mathit{sI}–A\right)}^{–1}$

$A=\left[\begin{array}{ccc}{a}_{11}& 0& 1\\ 1& –3& 2\\ 0& 0& –4\end{array}\right],\mathit{ B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathit{ C}=\left[\begin{array}{ccc}0& 1& 0\end{array}\right],\mathit{ D}=\left[0\right]$

Substituting the matrices gives;

${C\left(\mathit{sI}–A\right)}^{–1}=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]*{\left(\left[\begin{array}{ccc}{s–a}_{11}& 0& 1\\ 1& s+3& 2\\ 0& 0& s+4\end{array}\right]\right)}^{–1}*\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

$=\frac{\left[\begin{array}{ccc}0& 1& 0\end{array}\right]}{{–s}^3+\left(a_{11}–7\right)s^2+\left(12+7a_{11}\right)s+12a_{11}}*{\left(\left[\begin{array}{ccc}{s–a}_{11}& 0& 1\\ 1& s+3& 2\\ 0& 0& s+4\end{array}\right]\right)}^{–1}*\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Regarding question 3

3b)

Consider a basic closed-loop system as a state space model as follows;

Figure 20 – Closed loop state space model

By returning the output ‘state vectors’ of the system back into the output via a matrix K, the following model is generated;

Figure 21 – State space model with a state variable feedback

With the addition of K;

$u=r–\mathit{Kx}$

the state vector equation becomes;

$\dot{x}=\mathit{Ax}+\mathit{Bu}$

$\dot{x}=\mathit{Ax}+B(r–\mathit{Kx})$

$\dot{x}=\left(A–\mathit{BK}\right)x+\mathit{Br}$

Allow, $A_C=(A–\mathit{BK})$

. Hence,

$\dot{x}=A_{C}x+\mathit{Br}$

Here ‘r’ represents the closed-loop input vector, whilst $A_c$

represents the closed-loop A matrix. Therefore, the eigenvalues of $A_c$

, now respectively represent the poles of the closed-loop system.

Consider the matrix of K being the size of [the number of inputs x number of states]. For this situation, there is 1 input and 3 states to consider i.e. K = 1x 3 matrix;

$\left[\begin{array}{ccc}{K}_{11}& K_{12}& K_{13}\end{array}\right]$

Regarding Question 4)

Consider the following transfer functions

$G\left(s\right)=\frac{23}{s+10}\mathit{ and }{G}_c\left(s\right)=K_p+\frac{K_i}{s}$

With the Cascade Rule:

$G_c\left(s\right).G\left(s\right)=\left(K_p+\frac{K_i}{s}\right)*\left(\frac{23}{s+10}\right)$

$=\left(\frac{K_p}{1}+\frac{K_i}{s}\right)*\left(\frac{23}{s+10}\right)$

$=\left(\frac{K_{p}s+K_l}{s}\right)*\left(\frac{23}{s+10}\right)$

$=\frac{{23(K}_{p}s+K_i)}{s^2+10s}$

With the feedback rule: $\frac{G}{1+\mathit{GH}}$

, where $G=\frac{{23(K}_{p}s+K_l)}{s^2+10s}$

, H=1

$=\frac{\frac{{23(K}_{p}s+K_i)}{s^2+10s}}{1+\frac{{23(K}_{p}s+K_i)}{s^2+10s}}$

$=\frac{{23(K}_{p}s+K_i)}{s^2+10s+{23(K}_{p}s+K_i)}$

$=\frac{{23(K}_{p}s+K_i)}{s^2+(10+{23K}_p)s+23K_i}$

Extras of 4b)

Figure 22 – Block diagram (left) and response (right) of the guidance system with PI controller when subjected with a unit ramp input