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PID controller

proportional control

definition

$K_p e(t) = K_p [u(t) - y(t)]$

Example: Given the closed-loop transfer function is $T(s) = \frac{G_p(s)}{1+G_p(s)} = \frac{1}{s+2}$

adding proportional

closed-loop transfer function is:

T(s) = \frac{K_p G_p}{1 + K_p G_p}

integral control

"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=Latex, block/.style={draw, minimum width=1.5cm, minimum height=1cm}]\n\n% Nodes\n\\node[draw, circle, minimum size=0.5cm] (sum) {}; % Summing junction\n\\node[block, right=2cm of sum] (compensator) {$\\frac{K_I}{s}$};\n\\node[block, right=2.5cm of compensator] (Gp) {$\\frac{1}{s+1}$};\n\\node[below=1.5cm of compensator] (feedback) {feedback};\n\n% Labels\n\\node[above=0.1cm of compensator] {compensator};\n\\node[above=0.1cm of Gp] {$G_p(s)$};\n\n% Input and Output\n\\node[left=1cm of sum] (input) {R(s)};\n\\node[right=1cm of Gp] (output) {C(s)};\n\n% Arrows (Forward path)\n\\draw[->] (input) -- (sum.west);\n\\draw[->] (sum.east) -- (compensator.west);\n\\draw[->] (compensator.east) -- (Gp.west);\n\\draw[->] (Gp.east) -- (output);\n\n% Feedback path\n\\draw[->] (output.east) -- ++(1,0) |- (feedback) -| (sum.south);\n\n% Plus and Minus signs\n\\node at (0.2, 0.5) {$+$};\n\\node at (0.2, -0.5) {$-$};\n\n\\end{tikzpicture}\n\\end{document}"

source code

Closed loop here yields

T(s) = \frac{K_I}{s^2 + s + K_I}

steady state error is 0, while steady-state output is 1

PI control

proportional-integral

"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=Latex, block/.style={draw, minimum width=1.5cm, minimum height=1cm}]\n\n% Nodes\n\\node[draw, circle, minimum size=0.5cm] (sum) {}; % Summing junction\n\\node[block, right=2cm of sum] (compensator) {$G_c = \\frac{K_I}{s} + K_p$};\n\\node[block, right=2.5cm of compensator] (Gp) {$\\frac{1}{s+1}$};\n\\node[below=1.5cm of compensator] (feedback) {feedback};\n\n% Labels\n\\node[above=0.1cm of compensator] {compensator};\n\\node[above=0.1cm of Gp] {$G_p(s)$};\n\n% Input and Output\n\\node[left=1cm of sum] (input) {R(s)};\n\\node[right=1cm of Gp] (output) {C(s)};\n\n% Arrows (Forward path)\n\\draw[->] (input) -- (sum.west);\n\\draw[->] (sum.east) -- (compensator.west);\n\\draw[->] (compensator.east) -- (Gp.west);\n\\draw[->] (Gp.east) -- (output);\n\n% Feedback path\n\\draw[->] (output.east) -- ++(1,0) |- (feedback) -| (sum.south);\n\n% Plus and Minus signs\n\\node at (0.2, 0.5) {$+$};\n\\node at (0.2, -0.5) {$-$};\n\n\\end{tikzpicture}\n\\end{document}"

source code

Closed-loop transfer function

T(s) = \frac{K_I + sK_p}{s^{2} + (1+K_p)s + K_I}

PC: impact on speed of response
IC: force steady-state error to 0

derivative control

"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=Latex, block/.style={draw, minimum width=1.5cm, minimum height=1cm}]\n\n% Nodes\n\\node[draw, circle, minimum size=0.5cm] (sum) {}; % Summing junction\n\\node[block, right=2cm of sum] (compensator) {$G_c = K_D s$};\n\\node[block, right=2.5cm of compensator] (Gp) {$\\frac{1}{s+1}$};\n\\node[below=1.5cm of compensator] (feedback) {feedback};\n\n% Labels\n\\node[above=0.1cm of compensator] {compensator};\n\\node[above=0.1cm of Gp] {$G_p(s)$};\n\n% Input and Output\n\\node[left=1cm of sum] (input) {R(s)};\n\\node[right=1cm of Gp] (output) {C(s)};\n\n% Arrows (Forward path)\n\\draw[->] (input) -- (sum.west);\n\\draw[->] (sum.east) -- (compensator.west);\n\\draw[->] (compensator.east) -- (Gp.west);\n\\draw[->] (Gp.east) -- (output);\n\n% Feedback path\n\\draw[->] (output.east) -- ++(1,0) |- (feedback) -| (sum.south);\n\n% Plus and Minus signs\n\\node at (0.2, 0.5) {$+$};\n\\node at (0.2, -0.5) {$-$};\n\n\\end{tikzpicture}\n\\end{document}"

source code

T(s) = \frac{K_D s}{(1+K_D)s + 1}

introduces an open-loop zero
$K_D$ increases system might not be stable

in second-order system

$T(s) = \frac{s K_D P \omega_n^2}{s^{2}+ 2(\zeta + \frac{K_D}{2} P\omega_n)\omega_n s + \omega_n^2}$
damping effect $\zeta^{'} = \zeta + \frac{K_D}{2}P\omega_n$

PID control

G_C(s) = K_p + \frac{K_I}{s} + K_D s

in time domain:

u(t) = K_P e(t) + K_I \int_{0}^{t} e(\eta) d\eta + K_D \frac{d(e(t))}{dt}

Component	Discrete-Time Equation
Proportional	$u(k) = K_P e(k)$
Integral	$u(k) = K_I T \sum_{i=1}^{k} e(i)$
Derivative	$u(k) = \frac{K_D}{T}[e(k) - e(k-1)]$

approximate of PID controller: $u(k) = K_P e(k) + K_I T \sum_{i=1}^{n} e(i) + \frac{K_D}{T}[e(k) - e(k-1)]$

proportional control

definition

$K_p e(t) = K_p [u(t) - y(t)]$

Example: Given the closed-loop transfer function is $T(s) = \frac{G_p(s)}{1+G_p(s)} = \frac{1}{s+2}$

adding proportional

closed-loop transfer function is:

T(s) = \frac{K_p G_p}{1 + K_p G_p}

integral control

"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=Latex, block/.style={draw, minimum width=1.5cm, minimum height=1cm}]\n\n% Nodes\n\\node[draw, circle, minimum size=0.5cm] (sum) {}; % Summing junction\n\\node[block, right=2cm of sum] (compensator) {$\\frac{K_I}{s}$};\n\\node[block, right=2.5cm of compensator] (Gp) {$\\frac{1}{s+1}$};\n\\node[below=1.5cm of compensator] (feedback) {feedback};\n\n% Labels\n\\node[above=0.1cm of compensator] {compensator};\n\\node[above=0.1cm of Gp] {$G_p(s)$};\n\n% Input and Output\n\\node[left=1cm of sum] (input) {R(s)};\n\\node[right=1cm of Gp] (output) {C(s)};\n\n% Arrows (Forward path)\n\\draw[->] (input) -- (sum.west);\n\\draw[->] (sum.east) -- (compensator.west);\n\\draw[->] (compensator.east) -- (Gp.west);\n\\draw[->] (Gp.east) -- (output);\n\n% Feedback path\n\\draw[->] (output.east) -- ++(1,0) |- (feedback) -| (sum.south);\n\n% Plus and Minus signs\n\\node at (0.2, 0.5) {$+$};\n\\node at (0.2, -0.5) {$-$};\n\n\\end{tikzpicture}\n\\end{document}"

source code

Closed loop here yields

T(s) = \frac{K_I}{s^2 + s + K_I}

steady state error is 0, while steady-state output is 1

PI control

proportional-integral

"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=Latex, block/.style={draw, minimum width=1.5cm, minimum height=1cm}]\n\n% Nodes\n\\node[draw, circle, minimum size=0.5cm] (sum) {}; % Summing junction\n\\node[block, right=2cm of sum] (compensator) {$G_c = \\frac{K_I}{s} + K_p$};\n\\node[block, right=2.5cm of compensator] (Gp) {$\\frac{1}{s+1}$};\n\\node[below=1.5cm of compensator] (feedback) {feedback};\n\n% Labels\n\\node[above=0.1cm of compensator] {compensator};\n\\node[above=0.1cm of Gp] {$G_p(s)$};\n\n% Input and Output\n\\node[left=1cm of sum] (input) {R(s)};\n\\node[right=1cm of Gp] (output) {C(s)};\n\n% Arrows (Forward path)\n\\draw[->] (input) -- (sum.west);\n\\draw[->] (sum.east) -- (compensator.west);\n\\draw[->] (compensator.east) -- (Gp.west);\n\\draw[->] (Gp.east) -- (output);\n\n% Feedback path\n\\draw[->] (output.east) -- ++(1,0) |- (feedback) -| (sum.south);\n\n% Plus and Minus signs\n\\node at (0.2, 0.5) {$+$};\n\\node at (0.2, -0.5) {$-$};\n\n\\end{tikzpicture}\n\\end{document}"

source code

Closed-loop transfer function

T(s) = \frac{K_I + sK_p}{s^{2} + (1+K_p)s + K_I}

PC: impact on speed of response
IC: force steady-state error to 0

derivative control

"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\n\\begin{document}\n\\begin{tikzpicture}[auto, node distance=2cm, >=Latex, block/.style={draw, minimum width=1.5cm, minimum height=1cm}]\n\n% Nodes\n\\node[draw, circle, minimum size=0.5cm] (sum) {}; % Summing junction\n\\node[block, right=2cm of sum] (compensator) {$G_c = K_D s$};\n\\node[block, right=2.5cm of compensator] (Gp) {$\\frac{1}{s+1}$};\n\\node[below=1.5cm of compensator] (feedback) {feedback};\n\n% Labels\n\\node[above=0.1cm of compensator] {compensator};\n\\node[above=0.1cm of Gp] {$G_p(s)$};\n\n% Input and Output\n\\node[left=1cm of sum] (input) {R(s)};\n\\node[right=1cm of Gp] (output) {C(s)};\n\n% Arrows (Forward path)\n\\draw[->] (input) -- (sum.west);\n\\draw[->] (sum.east) -- (compensator.west);\n\\draw[->] (compensator.east) -- (Gp.west);\n\\draw[->] (Gp.east) -- (output);\n\n% Feedback path\n\\draw[->] (output.east) -- ++(1,0) |- (feedback) -| (sum.south);\n\n% Plus and Minus signs\n\\node at (0.2, 0.5) {$+$};\n\\node at (0.2, -0.5) {$-$};\n\n\\end{tikzpicture}\n\\end{document}"

source code

T(s) = \frac{K_D s}{(1+K_D)s + 1}

introduces an open-loop zero
$K_D$ increases system might not be stable

in second-order system

$T(s) = \frac{s K_D P \omega_n^2}{s^{2}+ 2(\zeta + \frac{K_D}{2} P\omega_n)\omega_n s + \omega_n^2}$
damping effect $\zeta^{'} = \zeta + \frac{K_D}{2}P\omega_n$

PID control

G_C(s) = K_p + \frac{K_I}{s} + K_D s

in time domain:

u(t) = K_P e(t) + K_I \int_{0}^{t} e(\eta) d\eta + K_D \frac{d(e(t))}{dt}

Component	Discrete-Time Equation
Proportional	$u(k) = K_P e(k)$
Integral	$u(k) = K_I T \sum_{i=1}^{k} e(i)$
Derivative	$u(k) = \frac{K_D}{T}[e(k) - e(k-1)]$

approximate of PID controller: $u(k) = K_P e(k) + K_I T \sum_{i=1}^{n} e(i) + \frac{K_D}{T}[e(k) - e(k-1)]$

PID controller

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proportional control

adding proportional

integral control

PI control

derivative control

PID control

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