Control Systems

Modeling a system

Series RC circuit

Let \(I(t)\) = output (current), \(v(t)\) = input (applied voltage).

Time-domain equation (series resistor \(R\) and capacitor \(C\)):

\[v(t)=R\,I(t)+\dfrac{1}{C}\displaystyle\int_{t_0}^{t} I(\tau)\,d\tau\]

Assuming zero initial conditions, take Laplace transform (differentiate both sides of the time-domain equation for simplification):

\[V(s)=R\,I(s)+\dfrac{1}{sC}\,I(s)=\Big(R+\dfrac{1}{sC}\Big)I(s)\]

Transfer function from input voltage to current:

\[ \frac{I(s)}{V(s)}=\frac{1}{R+\dfrac{1}{sC}} =\frac{sC}{sRC+1}. \]

Two-node/two-branch network

KVL in Laplace transform domain

\[ \begin{pmatrix} V(s) \\ 0 \end{pmatrix} = \begin{pmatrix} R_1+\dfrac{1}{sC_1} & -\dfrac{1}{sC_1} \\[8pt] -\dfrac{1}{sC_1} & R_2+\dfrac{1}{sC_1}+\dfrac{1}{sC_2} \end{pmatrix} \begin{pmatrix} I_1(s) \\ I_2(s) \end{pmatrix}. \]

Solve using Cramer's rule. Let the coefficient matrix be \(A(s)\) and \(\Delta(s)=\det A(s)\).

\[ \Delta(s) =\Big(R_1+\frac{1}{sC_1}\Big)\Big(R_2+\frac{1}{sC_1}+\frac{1}{sC_2}\Big) -\Big(\frac{1}{sC_1}\Big)^2. \]

Transfer function is given by

\[ I_1(s)=\frac{\det A_1(s)}{\Delta(s)} \quad\text{and}\quad I_2(s)=\frac{\det A_2(s)}{\Delta(s)}, \]

Here, \(\det A_1(s)\) is determinant of matrix formed by replacing first column of \(A\) with \(\begin{pmatrix}V(s)\\0\end{pmatrix}\), \(\det A_2(s)\) is determinant of matrix formed by replacing second column of \(A\) with \(\begin{pmatrix}V(s)\\0\end{pmatrix}\).

Control Systems — Lecture 2 (MathLaTeX + SVG)

Mechanical damper system

Damper equation:

\[ F = B\,\frac{dx}{dt} \]

Newton’s law for \(m_2\) gives:

\[ F - k(x_2 - x_1) - B\left(\frac{dx_2}{dt} - \frac{dx_1}{dt}\right) = m_2\,\frac{d^2x_2}{dt^2} \]

Newton's law for \(m_1\) gives:

\[ k(x_2 - x_1) + B\left(\frac{dx_2}{dt} - \frac{dx_1}{dt}\right) = m_1\,\frac{d^2x_1}{dt^2} \]

The equations can be written in a matrix form:

\[ \begin{pmatrix} F \\ 0 \end{pmatrix} = \begin{pmatrix} k + B\frac{d}{dt} + m_2\frac{d^2}{dt^2} & -k - B\frac{d}{dt} \\[6pt] -k - B\frac{d}{dt} & k + B\frac{d}{dt} + m_1\frac{d^2}{dt^2} \end{pmatrix} \begin{pmatrix} x_2 \\ x_1 \end{pmatrix} \]

Laplace transform form of the equation is:

\[ \begin{pmatrix} F(s) \\ 0 \end{pmatrix} = \begin{pmatrix} k + sB + s^2 m_2 & -k - sB \\[6pt] -k - sB & k + sB + s^2 m_1 \end{pmatrix} \begin{pmatrix} X_2(s) \\ X_1(s) \end{pmatrix} \]

Using Cramer’s rule, we can find:

\[ \frac{X_2(s)}{F(s)} \quad\text{and}\quad \frac{X_1(s)}{F(s)}. \]

Equivalent electric circuit for mechanical damper system

In the damper system, we have these components:

\[F = m \frac{d^2 x}{dt^2} \quad\quad F = B \frac{dx}{dt} \quad\quad F = kx\]

To convert it into an equivalent electrical circuit, we use the following relations for electrical component

Force–current analogy

Resistor

\[ I = \frac{V}{R} \]

Inductor

\[ I = \frac{1}{L} \int V \, dt \]

Capacitor

\[ I = C \frac{dV}{dt}\]

Let \(\frac{dx}{dt} = v\). Two equations \(\Rightarrow\) Two nodes.

Force–voltage analogy

Resistor

\[V = IR\]

Inductor

\[ V = L \frac{dI}{dt} \]

Capacitor

\[ V = \frac{1}{C} \int I \, dt \]

Let \(\frac{dx}{dt} = i\). Two equations \(\Rightarrow\) Two loops.

Scanned notes

These scanned lecture notes are from the course Control Systems at Indian Institute of Technology, Bombay. Use them at your own discretion. If you would like to help digitize these lecture notes, contact the editor.

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Author

Anurag Gupta is an M.S. graduate in Electrical and Computer Engineering from Cornell University. He also holds an M.Tech degree in Systems and Control Engineering and a B.Tech degree in Electrical Engineering from the Indian Institute of Technology, Bombay.


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